Beating the Big Numbers on High Rollers

Trebek and Lee. They're cops!

Trebek and Lee. They’re cops!

When you think about dice and game shows, the show you probably think of first is the Heatter-Quigley show High Rollers, hosted by Alex Trebek and his awesome 70’s fro for two separate runs on NBC, and then again by Wink Martindale in 1987. Through every incarnation of the show, the bonus round remained the same: the Big Numbers.  Based on the old gambling game Shut the Box, it heavily relied on luck, but also had some strategy in how you played it.  It also seemed extremely difficult, as players rarely walked away winners.  We wondered if the difficulty of the game was due to poor strategy, or if the game was simply stacked against the contestant.  To answer that question, we first had to figure out just what the optimal strategy should be.

The rules are simple enough. The contestant is faced with the numbers 1 through 9.  In order to win, they must eliminate each number, which they do by rolling two dice.  After each die roll, the contestant chooses from the numbers they have remaining either a number or combination of numbers that equal to the number rolled, which are then eliminated from further consideration.  If the contestant manages to eliminate all nine numbers before rolling a number that cannot be matched, they win the grand prize.

During the course of the game, the contestant can also earn “insurance markers”. Every time they roll doubles, they earn an insurance marker, which essentially give the contestant an extra life – if they roll a total which they cannot match, they may instead roll again.  Contestants can use insurance immediately, even on the same roll that earned them the insurance.

The Big Numbers board in the 1987 revival, in all its glorious 80s-ness.

The Big Numbers board in the 1987 revival, in all its glorious 80s-ness.

Sometimes you can only match a roll one way.  But most of the time, especially in the first few rolls, you have several ways to match a total with the numbers remaining.  For example, there are 12 possible ways of matching a roll of 12 on the first roll!  Which of these 12 would give you the best chance of winning the game?

We decided to take a brute force approach to evaluating this game.  Since there are nine different numbers, and each number has two states (either still on the board or removed), that means that there are 512 (2 to the 9th power) possible combinations of numbers that you could be left with at some point during the game.  While that would be a large number to work out by hand, luckily we can make the computer do most of the heaving lifting for us.

First, we can trivially work out the probabilities of every situation where there are no choices to be made.  For example, consider the situation where only the number 7 is lit up and we have no insurance markers.  The only way to win in this case is to roll a 7. Seasoned gamblers would know that the chances of rolling 7 on a pair of dice is 16.66% chance, or ⅙.  However we have to remember that not all non-seven rolls will lose the game – a roll of doubles grants an insurance marker that we could immediately cash in.  That means that of the 36 possible rolls, 6 are winners, 6 let us roll again, and 24 are losers, which means the chances of winning are 6 / 30, or 20%.

Once we have these easy cases calculated, we can start working on the cases where we have a choice.  We used an iterative process for this.  We made up a list of all number combinations that we could not evaluate as above, and looked at each of them.  If we came up with a case where every option’s chances of winning had not yet been determined, we skipped it for the moment.  If we could figure out every choice’s chances of victory, we could then figure out which option would provide the greatest chance of victory, and then determine what the overall chances of winning were from that configuration of numbers.

For example, say we have the numbers 3, 4, and 7 left on the board.  We could not have figured out the chances of victory from this position earlier, since if we roll a 7, we can do one of two things: take off the 3 and 4 leaving the 7, or remove the 7 and leave the 3 and 4.  But now that we’ve determined the chances of winning from all of the simple cases, we can use those to figure out this case as well.  We’ve already figured out that the chances of victory if we have a bare 7 left on the board is 20%.  It turns out that the chances of victory with 3 and 4 on the board are slightly better: 20.83%.  That means in this instance, we should eliminate the 7 if we roll a seven with the dice, since a board with a 3 and a 4 left on it would be easier to deal with than a board with only a 7 on it.  Combine that with the chances of winning when we roll the other good numbers, none of which require making a choice (3, 4, 10, 11), we can determine that the overall chances of victory from this position are 7.64%.

After iterating through the list of unsolved configurations several times, we eventually discovered both the best strategy to use and the chances of winning from every possible combination of numbers.  From the starting position, where all nine numbers are lit and you have no insurance markers, you have a 17.1% chance of knocking all the numbers off.

Gene has a 9.32% chance of winning right now. How'd he do?

Gene has a 9.32% chance of winning right now. How’d he do?

Can we make any characterizations about the best strategy?  In general, it seems that the best strategy is to remove the largest numbers from the board that you can from your roll.  This means that on your first roll, if your roll is less than nine, you should remove just the number you rolled.  If you roll 10 or more, you should remove the 9 and either the 1, 2, or 3, depending on if you rolled 10, 11, or 12.

I do say this is the best strategy in general, but there appear to be a large number of exceptions.  Say you rolled a five to start with, and took off the 5.  If on your second roll you rolled another five, you might be inclined, following the rule of thumb above, to knock off the 4 and the 1.  Doing this leaves you with a 7.01% chance of winning.  If instead you removed the 3 and 2, you’d have a 7.47% chance of winning.  This is just one of many cases where following the general strategy is not optimal.  We tried to figure out if there was a common thread to these exceptions, but nothing jumped out at us.  Even so, We would expect that a person following the basic strategy and ignoring these exceptions would probably only cost themselves a few tenths of a percent on their overall winning percentage.

If you’d like to play around with these results, I’ve included a little widget at the end of this post for you to play with.  Highlight the numbers remaining on the board and the number of insurance markers you have, and it’ll outline the best strategy to follow at that point, as well as your chances of victory.  Have fun!

Chasing Down The Best Chaser

Chasers

Pictured Left to Right: Velma, Scooby, Daphne, Shaggy, Fred.

Earlier this year, Jenny Ryan joined the cast of the ITV hit show The Chase as the fifth resident chaser. The Vixen will take her place among trivia’s rogues gallery alongside Mark “The Beast” Labbett, Shaun “The Dark Destroyer” Wallace, Anne “The Governess” Hegerty, and Paul “The Sinnerman” Sinha. Given Jenny’s trivia bona fides (QI elf, Only Connect series champ, University Challenge, Mastermind, and Fifteen-to-One alumna), it’s no surprise that she fit right in alongside the others, who between them have over 700 episodes of experience with crushing the hopes and dreams of unlucky contestants. But, after all those episodes, who among them is the best at their job? Who’s the one you would least want to meet at night in a dark alley of trivia?  (Note: Given that Ryan has not had much time to accumulate data, we will ignore her for the purposes of this question.)

The chasers have two chances to catch contestants and eliminate them during the show. First, they can eliminate contestants individually during each contestant’s head-to-head round. Finally, they can eliminate the team as a whole by catching them during the final chase. Let’s look at each round individually and see what data we can get.

Head-to-Head Round

chase_tableDuring the head-to-head round, the contestant is tasked with getting multiple choice questions right, each correct answer allowing the contestant to take one step towards victory and earning money for the communal team bank. The chaser starts eight steps away from the finish. The contestant can choose to start either four, five, or six steps away from the finish (and thus starting with either a four, three, or two step head start on the Chaser), with the further starting locations worth more money. The Chaser and the contestant are asked the same questions, with each right answer bringing them closer to the finish line.  If the contestant manages to stay ahead of the Chaser and reach the finish line, they put their earned cash into the communal bank, and earn a spot in the Final Chase at the end of the show. If the Chaser catches up to them before that happens, they don’t earn the money and are eliminated.

It is tough to get an accurate read on the Chaser’s abilities from the data that we have in this round. We are relying on the results provided by the Chase Wikia, which only gives a one-line summary of how each episode finished. It would be great if we could watch all 707 episodes from the first eight seasons of the show and keep detailed records of the Chaser’s correct answer rate, but that may be a project for when somebody finally invents the 28-hour day. Still, we do have a record of how often each Chaser catches a contestant in this round. Can we do anything with that?

chase_h2h_table

Before we put too much stock in these numbers, I do want to share my reservations about it. First, we do not have the data of how many contestants opted to start with a two, three, or four step head start. We can assume that each chaser gets about the same amount of contestants to step closer or further away, but it introduces a small element of imprecision we can’t address.

Another issue that muddies the water is that some contestants are uncatchable. If a contestant answers every question correctly (or answers incorrectly a fewer amount of times than their head start), then the performance of the Chaser is moot. It will always be chalked up as a loss. It doesn’t seem right that there are times that a Chaser could answer every question correctly or every question incorrectly and have it look the same either way in the data.

Finally, the bigger issue is that these numbers are so close as not to be statistically significant. Even though each Chaser has faced down between 600 and 800 contestants each, that’s still too small of a sample size to say that these numbers are definitive. The margin of error of each of these numbers at a 95% confidence interval is around 3.5% percent, as illustrated in this chart.

chase_h2h_2

In this graph, we’ve highlighted the margin of error in red, representing where we think each Chaser’s true value could fall within a 95% certainty.  You can see that even Paul’s low mark could theoretically still be the highest among the four.

So it’s clear that this analysis isn’t the best determination of Chaser performance. Can we do better in the Final Chase?

Final Chase

Chasers

Pictured Left to Right: Ginger, Baby, Scary, Sporty, Posh

After every contestant has had a chance to play head-to-head against the Chaser, those who survive are brought back to try and win the communal bank. The team is given 2 minutes to answer as many questions as they can. The Chaser then gets another two minutes to try to match the score set by the contestants. If they do that, then it’s game over for the team. If the Chaser falls short, then the surviving team members split the bank. To even the playing field, the contestants are given two major advantages. Firstly, they earn a head start equal to the number of surviving contestants. Secondly, anytime that the Chaser misses a question during their turn, the clock is stopped and the contestants get a chance to answer it themselves. If they get the question correct, they push the Chaser back one step.

The data we have for this round is of a much higher quality. We know how many contestants the team has left and how well they scored during their two minutes. We also have the Chaser’s final score, and how long the Chaser took to catch the team if the Chaser won. What we’d like to do is use the Chaser’s score as the performance metric in this round, but we’ll need to do a few things first to take care of the variable conditions of this round.

chase_final_picThe most obvious place to start is by normalizing the amount of time each Chaser has to answer questions. Since the game is over as soon as the Chaser meets the score set by the contestants, we need to figure out what the Chaser would have scored had they had their full two minutes. That’s simple enough: We will give them credit for the missing time by assuming that they will continue to answer questions at the same rate. Thus, if a Chaser catches a team with a score of 15 with 30 seconds left, we will treat that as a score of 20. If the Chaser fails to catch the contestants, their score will not change, as they used the entire two minutes.

Now, there are a couple of issues with treating the scores this way. If a Chaser has to chase down a small score, it’s possible that they might take a couple of seconds extra to think about each question before answering. On the flip side, if they have to chase down a large score, they may rush and become prone to more mistakes. Also, (and I have no proof of this except my own anecdotal experience of watching the show) I feel that the host, Bradley Walsh, will speed up his reading of the questions if time is winding down and the Chaser is close to the target. While these are things we need to be aware of, I still feel comfortable about normalizing the scores in this way.

Chasers

Pictured Left to Right: Niall, Liam, Harry, Louis, Zayn

The other thing we need to control for is the number of opponents that the Chaser is facing. Since the contestants get offered any questions that the Chaser misses, and their right answers are deducted from the Chaser’s score, the number of contestants left on the team has a direct impact on the Chaser’s final score. Analyzing the Chaser’s scores as a function of team size tells us that a two or three player team will earn around one more pushback than a one player team, while a full four player team earns around 1.5 pushbacks more than the single player. If the Chaser faces a multi-person team, we will give them credit for these extra pushbacks so the data is normalized to a Chaser facing a single player.

Now that we’ve eliminated all variables outside the control of the Chaser, here’s each Chaser’s average performance.

chase_final_table

Mark, Anne, and Paul are all very close, but Shaun ends up averaging almost 2 questions less. This is borne out by using each chaser’s raw winning %: Mark, Anne, and Paul win about three-quarters of the time, while Shaun’s victory rate is only two-thirds.

This data does not suffer from the issues of the data from the head-to-head round. This data is a metric of raw performance on the part of the Chasers; we have eliminated any effects the contestants have on this score. It is also significant to a 95% confidence level. The margin of error on these numbers is between 0.6 and 0.7 of a question for each Chaser, which means that while we can’t say that Mark, Anne or Paul are better than one another, they all have performed better than Shaun.

Extrapolation

There’s something else that we can do with this data that’s pretty cool. To illustrate this, let’s take a look at a graph of the frequency of Mark’s normalized scores, rounded to the nearest whole number.

chase_labbett_bellcurve

Say, that kinda looks like a bell curve, doesn’t it? Doing some normality testing on the data bears this hypothesis out; this data likely conforms to a normal distribution. The other Chasers’ data has the same feature. Therefore, since we know each Chasers’ average performance and standard deviation during the Final Chase, we can extrapolate upon this data and determine the odds of a Chaser beating any given score by fitting a normal distribution to each Chaser’s average score and standard deviation.

For example, let’s assume that a full team of 4 sets a score of 17 during their final chase. Not too shabby, right?  What’s the likelihood that each Chaser will chase down that score?

Since our averages are normalized for a 1 person team, and this example uses a 4 person team, we will add 1.5 to their final score to represent the greater number of pushbacks that the team will score. So, the Chaser will have to score at least 18.5 points in order to catch the team. What is the team’s chance of victory against each Chaser?

chase_final_example

Here you can see just how much an effect that two question difference between Shaun and the other three has. Facing Paul, Anne, or Mark, the team has less than a 1 in 4 chance of victory. Up against Shaun, the team will run out winners 42% of the time.

Here’s the full graph that shows the chance that a team will beat each Chaser based on their final score (before pushbacks).

chase_answers_graph

Pictured Left to Right: Wasp, Hulk, Iron Man, Thor, Ant-Man

Pictured Left to Right: Wasp, Hulk, Iron Man, Thor, Ant-Man

So who is the best Chaser? With the data we have right now, it’s hard to tell. I’d be inclined to call it a dead heat between Mark and Anne, with Paul just a nose behind them, and Shaun a bit further back. Despite this gap, I want to stress that Shaun is still an formidable opponent, and if the contestants facing him are expecting an easy game, they’re going to be disappointed.  Time will tell how Jenny will fit into this group, but given her quizzing pedigree I expect her to do just as well as the other four regulars.

2015 Jeopardy Tournament of Champions: Semifinal Update

Couple of random thoughts before revealing my predictions for the ToC Semifinals:

– My system did pretty darn well this year, getting 4 of the 5 winners of the semifinals correct. Granted, it wasn’t much of a radical prediction to say that Matt Jackson and Alex Jacob would win their games. However, I’d argue that Kerry Greene, despite nominally being the top seed in her game, was not an obvious favorite, nor would Catherine Hardee be easy to pick out as a favorite to win from the third lectern. The system’s one miss was favoring Greg Seroka and Kristin Sausville over the eventual winner from Tuesday’s game, Brennan Bushee, though to be fair Bushee won the game from last place by being the only player to get Final Jeopardy correct.
– The Wild Card cutoff point was higher than average this year at $14,000. The average over the history of the tournament (after doubling the scores of the pre-double dollars era) stood at $10,464. Anecdotally, I’d think that may be the effect of almost all players going into their games with the goal of not necessarily winning, but playing to hit a self-determined goal score that would earn them a wild card. With much more data out there about things like historical wild card totals, I wonder if this is going to lead to a situation where the wild card cutoff will always be higher than historically expected. Then again, last year’s cutoff was $9,100, and most of the data was available then too, so it’s just as likely there’s no great reason for this year’s cutoff being so much higher.
– I would love to know how the Jeopardy team selects the semifinal matchups. I’ve tried to come up with some set of seeding rules, but nothing I can find explains the matchups perfectly. The only rules that I know for sure are that players will not face their opponents from their quarterfinal match, and two people with the same first name will not play each other. This is different from the quarterfinals, where the games are fairly obviously seeded so that in each match one of the top 5 players (ranked by games and money won) plays one of the second five and one of the bottom five.
– I need to thank Andy Saunders of The Jeopardy Fan for his guesses as to what the semifinal matchups would be, which turned out to be correct and give me a little more time to run the numbers. Jeopardy didn’t officially release the matchups until Monday morning (as far as I saw through the official channels), which is slightly annoying for those of us in the game-show-data-analysis business.

toc-sf1
Our prediction of Jackson vs. Jacob vs. [Seroka/Sausville/Hardee] isn’t going to be happening, since neither Greg Seroka nor Kristin Sausville made the second week, and Catherine Hardee is playing Matt Jackson in Wednesday’s game. Instead, the favorite for that third slot becomes Dan Feitel, winner of the Semifinal Matchup sweepstakes. He had the biggest movement in our prediction engine thanks to staying out of the path of the two juggernauts, increasing his chances of winning the tournament from 4% to 14%.

toc-sf2

Alex dominated his quarterfinal game, becoming the only player to have a lock game last week. We see no reason to expect a different result from his semifinal matchup against Brennan Bushee and Vaughn Winchell.

toc-sf3

If anybody is going to keep us from a final of M Jackson v. A Jacob v. AN Other, Catherine Hardee has the best chance of doing it. She could actually outbuzz Jackson, possibly the first time he’s ever had to face somebody who could do that.  If she can keep her number of wrong answers down and take a few Daily Doubles, she could still certainly crash the finals.  However, the smart money still has to be on Jackson winning this matchup.

toc-sf-odds

Thanks to a slightly easier Semifinal matchup, Alex Jacob takes the tag of favorite by a clear margin over Jackson. Both men are close to 1-in-4 odds of taking the title. As prevously stated, Dan Feitel moves from his quarterfinal position of “best of the rest” into a solid third place thanks to avoiding the two favorites. Good luck to all participants, and here’s hoping that the games to come are just as fun, interesting, and exciting as last week’s games.

Predicting the 2015 Jeopardy Tournament of Champions

Trivia Christmas is coming a bit early this year. Instead of the usual 18 month wait between events, we’re getting another installment of the Jeopardy Tournament of Champions a mere twelve months since the last incarnation. It’s always fun to try to predict how these things will go, so we’ll see if we can better our efforts from last year, where we had eventual winner Ben Ingram rated as the fifth most likely winner in a fairly wide-open field. This year, however, the field has broken down into three distinct groups.  The field is led by two co-favorites, each of whom have a one-in-five chance of victory.  There are three dark horse candidates for the title, with chances of victory ranging from 11% to 14%. Finally, we have the remaining ten contestants, none of whom have more than a 1-in-25 chance of taking home the title.

Before we reveal our predictions, let’s do a quick recap of our methodology. We analyse players’ ability using two metrics. The first is Buzzer Percentage, which attempts to measure how often a contestant will buzz in. Since we can’t measure this directly, we use a Monte Carlo simulation to estimate it based on how often a player has successfully rung in during their previous games. The second metric is Precision, which is the percentage of times that they are correct on clues where they buzzed in (ignoring Final Jeopardy & Daily Doubles). For reference’s sake, the average Jeopardy contestant has a buzzer percentage around 60% and a precision around 87%.  This system is based on the system that the designers of Watson used to simulate human opponents for Watson to practice against.

Once we have those numbers, we run a bunch of simulated games using each player’s Buzzer Percentage and Precision, and determine how likely they are to win their quarterfinal games. Using the results of these simulations, we track how likely a contestant is to win their quarterfinal or qualify for a wild card, and use that to weight a player’s potential matchups in the semifinals and finals.  Here’s what our simulation gives us for each of this week’s quarterfinal games:

Monday's Odds

Monday’s predictions.

If you want a dark horse candidate for your those office Jeopardy pools I know you’re all in, Catherine Hardee is a decent candidate. Despite winning one game fewer than Dan Feitel and Vaughn Winchell, her strong Buzzer Percentage and respectable precision gives her a slight edge in Monday’s matchup.

Tuesday's Game

Tuesday’s predictions.

Two of the three dark horse candidates face off on Tuesday’s show. Greg Seroka and Kristin Sausville both have a decent chance of taking down the whole tournament. Brennan Bushee is a bit unfortunate in his draw; despite his strong precision he could get squeezed out by two of the tournament’s strongest buzzers.

Wednesday's Odds

Wednseday’s Predictions.

96.8% precision. Ninety-six point eight percent precision. That is a crazy, insane, whacked-out number, and helps make Matt Jackson, unsurprisingly, one of the favorites for the tournament. Jackson draws a fairly easy group for his first game. John Schultz’s metrics are among the weakest in the field, and are actually comparable to an average contestant. Jennifer Giles could take an upset win, and in doing so become only the second Teacher’s Tournament winner to make the second week.  The first, of course, being Colby Burnett, who won the whole thing in 2013.

Thursday's Odds

Thursday’s predictions.

There’s no nice way to put this: the lineup for Thursday’s match is by far the weakest. All three contestants are very conservative buzzers; expect a higher than average number of clues going untried. Kerry Greene is at least precise when she buzzes in, which gives her quite an edge over the error-prone Andrew Haringer and Elliot Yates.  Any way you slice it, this game will give an automatic semifinal berth to a player who may have had a hard time qualifying in other games.

Friday's Odds

Friday’s predictions.

Alex Jacob has the busiest buzzer finger in the entire tournament. When paired with his excellent precision, it makes him the co-favorite to win the entire tournament. Scott Lord and Michael Bilow will have a hard time matching up.

Complete Picture
Alex Jacob and Matt Jackson are so close that it’s fair to call them co-favorites to win the tournament.  Assuming that they manage to avoid each other in the semifinals (which is what caused most of their eliminations in our simulations), they should get to lock horns in a great two-day final.  Chances are they’ll meet one of Greg Seroka, Kristin Sausville, or Catherine Hardee there.  At that point, the only winners I’d be willing to predict for certain would be the audience, who would get to watch an exhibition in excellent Jeopardy gameplay.

Lightning Round: Matt Jackson, New Jeopardy Record Holder?

So things have been quiet here for a few months.  I’ve been very busy at work lately, but I do have a couple of articles that are very close to being finished that will be up in the next few weeks. One is a treatise on Daily Double wagering that I’ve been working on for the better part of a year, and another article will be evaluating the performances of the Chasers on ITV’s The Chase.  However, current events have prompted me to write a Lightning Round article about this man, who has polarized fans of Jeopardy over the past two weeks:

The owner of this smile is Matt Jackson, a paralegal from DC who yesterday became the 5th person ever to reach 10 wins, putting him 5th on the all-time win list behind Arthur Chu (11), David Madden (19), Julia Collins (20), and, of course, Ken Jennings (74).  Given Jackson’s performances so far, how many wins is he likely to finish his run with?  Could we be looking at a new record holder?

I’ve taken a look at this sort of thing before, back in June of 2014 after Julia Collins had finished her 20 game run.  I’m going to use the same methodology here: look at Jackson’s game situations heading into Final Jeopardy, and determine how often Jackson should be expected to win if he continues in that fashion.

In Jackson’s 10 games so far, he has achieved 8 lock games and 2 crush games heading into Final Jeopardy.  The lock games are easy to deal with – Jackson wins those 100% of the time.  That leaves the 20% of the time when Jackson is leading by more than 2/3s of his nearest opponent’s score.  In order to lose a game that you are crushing heading into Final Jeopardy, two things need to happen: you need to respond incorrectly to Final, while your nearest opponent needs to respond correctly.  So far, Jackson has a 60% correct response rate in Final Jeopardy.  I’ll use the historical correct answer percentage for an average contestant in Final Jeopardy to determine the chance that his trailing opponent answers correctly, which is 48.8%.  Since both events have to happen in order for Jackson to lose, we multiply the chances that Jackson misses (40%) by the chances that his opponent answers correctly (48.8%).  This means that the chance that Jackson loses in a crush situation is 19.6%.  Or, in other words, Jackson wins a crush 80.4% of the time.

So, 80% of the time, he locks up the game before Final and wins.  20% of the time, he has a crush heading into Final and wins 80.4% of the time.  Combine those two probabilities, and you come up with an impressive 96.1% win rate.  That is very impressive, close to Ken Jennings’ 97.0% win rate and well ahead of the third place win rate, David Madden’s 85.6%.

Does that mean he’s a threat to Jennings’ record?  It’s not very likely.  Jennings was very good but also very lucky, and outperformed his expectation (a mere 31 wins) by a large margin.  A person with a 96.1% chance of winning would be expected to “only” win 24.56 games before losing.  In Jackson’s case, we can add his 10 wins to that total to get our current estimate: an astounding but far from record-setting 34 games won.  I predict his current chances of the setting the record at 7.2%: possible but unlikely.

Of course, this analysis is predicated on him keeping up his pace of dominating the first two rounds before heading into Final Jeopardy.  Should he start to leave more openings for his opponents to catch him in Final, or (gasp) actually come into Final behind at some point, his expected win total would plummet.  Still, as long as he keeps up this level of performance, I’d expect to see Matt Jackson on our screens for some time to come.

Lightning Round: 500 Questions

500QABC’s latest Big Event Game Show Thing, 500 Questions, is currently in the middle of its nine night run.  And while pundits are keeping tabs on whether or not the show will actually manage to ask 500 questions during its entire run (spoiler alert: no), a question asked on the LearnedLeague forums got my attention.  A user asked what the chances were of a contestant actually completing the titular 500 questions.  That sounds like something we can look into.  So, we’re starting a new series here at Game Show Theory, The Lightning Round, devoted to questions about game shows that are interesting, but don’t really qualify for a full strategic breakdown.

Let’s have a quick refresher of 500 Questions’ rules.  A contestant is asked trivia questions one at a time, up to a theoretical total value of 500 questions.  Answering them correctly can earn money, which they secure after every 50 questions.  However, if they ever miss three questions in a row, they’re off the show.  There are some different types of question, and the presence of another player who may occasionally make life difficult for the contestant, but for our purposes we are going to ignore their effect on the game.

Question 5: How many fingers am I holding up?

Question 5: How many fingers am I holding up?

So, how likely is it that a contestant sees all of their 500 questions? I know of no simple probability distribution to address this question, so we’re going to take a slightly more manual and iterative approach to the problem.  We’re going to break the problem down by calculating the chance of a contestant surviving 1 question, 2 questions, 3 questions, and so on, up to the goal of 500 questions.

Let’s work through an example.  Let’s assume a prospective contestant will give a correct answer to a question a respectable 60% of the time.  Figuring out the chances of surviving the first two questions is trivial – it’s 100%, since there is no way to get three wrong answers in a row yet.  The first chance of losing comes at 3 questions.  The player would have to get the first three questions wrong in a row, which translates to 3 straight 40% shots:

CodeCogsEqn

The chances that the player would go three-and-out is 6.4%, meaning that 93.6% of the time they’re still in the game after three questions.

With that done, let’s work out the chances that the player bombs out after 4 questions.  You might initially think at first that it’s the same as above, 6.4%, but actually there’s a couple of wrinkles we have to consider.  First of all, we need to factor in the chances that they’ve already been defeated, since you can’t lose after 4 questions if you’ve already lost after 3 questions.  Secondly, losing at 4 questions not only requires the contestant to have gotten questions 2, 3, and 4 incorrect, but also must have gotten question 1 correct.  If Question 1 was answered incorrectly, there is no way the player can get three in a row wrong on question 4.  Either they answer questions 2 and 3 wrong, in which case they’ve already been eliminated, or they answer one of those questions correctly, in which case Question 4 can’t be the third wrong answer in a row.

This gives us the following formula:

CodeCogsEqn (1)

Factoring all that in, we now have a 3.6% chance of losing after question 4.  Combined with the chances of losing after question 3, the total chances of survival are now a hair above 90%.  If 10% of the time, the player will be out after only 4 questions, their chances of surviving 500 questions is not looking too strong.

Calculating the odds of losing on questions 5 and beyond can be calculated in the same way as question 4.  We multiply the chance that we are still in the game at that point by the chance of answering one question correct and three questions wrong.

Question 500: Mark Burnett's Beard. Seriously, WTF?

Question 500: Mark Burnett’s Beard. Seriously, WTF?

As you might imagine given the results so far, the results do not make pleasant reading for our hypothetical player.  They only have a 50% of getting to question 19, well before they have the chance to make any money.  They’re only going to be able to bank the money earned in their first 50 questions 14.8% of the time.  And the chances of getting through all 500 questions?  0.0000003%.  That’s about a 1 in 300 million chance,  meaning they have as much chance of surviving 500 questions as they have of dying  as a result of a shark attack.  Put another way, our 60% contestant should stick to playing Powerball – they’ll have about twice the chance of winning the jackpot there.

What if we increase the contestant’s average question get rate?  Here’s a chart that breaks down the chances of players of different strengths hitting the 25, 50, 100, 250, and 500 question milestones:

500q-chart

If a player wanted to have a 50/50 shot of getting through the 500 questions, they need to be very, very good.  A player would have to get 88.37% of their questions right to stand a break-even chance of finishing the game, a number I would expect only trivia elite could get close to achieving.

One of the decisions that the producers of 500 Questions have made is to be outspoken in calling their contestants geniuses.  They better be – only geniuses stand a chance of performing well.

How to become Richer than Uncle Pennybags on Monopoly Millionaire’s Club

There have been many attempts to translate Monopoly, often considered one the worst board games ever, to television. A series of pilots in the late 80s failed to get off the ground.  ABC ran a pretty good attempt of bringing the game to the small screen as a 13 week summer series in 1990, paired with the epic Super Jeopardy! tournament. In more recent times, elements of the board game have been found on the Hub’s Family Game Night.  And now, with backing by the Multi-State Lottery Association, a new show based on Monopoly has hit syndication, Monopoly Millionaire’s Club.

logo

In an alternate universe, we’d be watching “The Landlord’s Game Millionaires’ Club”.

Monopoly Millionaire’s Club is a special subgenre of game show called a lottery show.  Depending on where you live, that term either has you nodding your head in recognition (if you live in a state with a proud history of lottery shows like Illinois or California), furrowing your brow in confusion (if you live in most of the rest of the US), or nodding your head in recognition even though you’re wrong (if you live in the UK, where “lottery show” has a different connotation).  In the US, a lottery show is a game show that is produced by a lottery commission.  It looks, sounds, and smells like a game show to the untrained eye, but there are three key differences that separate a lottery show from a standard game show.

  1. Contestants are chosen through the lottery.  Often times lottery commissions will run a series of special tickets where either the main prize or a “second chance” prize is an appearance on one of their shows as a contestant.  This is the only method of contestant recruitment, which provides a different type of contestant than the carefully controlled contestant selection process of most game shows.
  2. By law, all competition must be luck based, not skill based.  In order for the lottery commission  to get away with giving away prizes on these shows, they have to ensure that it truly remains a lottery.  You’ll never see any trivia questions or obstacle courses be played on a lottery show.
  3. The prizes on a lottery show are much, much bigger than on regular shows.  It makes sense, since the lottery commission is bankrolling the show, that they can afford to give away large amounts of money in prizes.  On most game shows nowadays, winning even a 6-figure sum is quite rare.  On lottery shows, you’ll see 6 and 7 figure prizes won with regularity.
Your contestant selection process.

Your contestant selection process.

Lottery shows are a perfect subject for us here at Game Show Theory.  To prevent the games from becoming a boring procession of “pick a number, win some money”, the show will often include an element of risk that the contestant must face, choosing to put their current winnings at jeopardy in order to win more. It’s these decisions that allow us to analyze these games and develop strategies for them.

On MMC, contestants are chosen from the audience of lottery winners to come and play a game for themselves and their section of the audience.  All money won is split half by the player, and half with their audience section.  So far, there are 8 games in the rotation, each one inspired by a different aspect of Monopoly.  Each game can see the player win a maximum of $100,000, but each game could also see a too-greedy player leave with $0.

I took a look at each game and determined the optimal strategy for playing it, usually by brute-forcing my way through all possible decision paths possible in the game.  Then, I determined the average amount of money a player would win by following that strategy, as well as their chances of winning the top prize or nothing.  It turns out that there is a large difference in what you can expect to win in each game.  The 8 games, placed in order from least profitable to most, are:

8. Community Chest

Average: $16,339.77

Chance of Winning $100,000: 4.5%

Chance of Winning $0: 31.3%

Any resemblance this game has to Deal or No Deal is purely coincidental.

cc3

Purely. Coincidental.

cc1

10 identical sealed boxes, $100,000, just one question. Have I said purely coincidental yet?

The contestant is faced with 10 boxes, each with a money amount in them, from $500 up to $5,000.  Pick a box, win that amount of money.  Simple enough.  However, there’s a temptation.  The remaining boxes all have their money amounts double.  If the contestant wants to, they can give back the money they have and select a different.  If they choose to do that, they box they select must contain a prize equal or greater in value to the one they give back.  If that’s the case, then the remaining boxes double again and the same offer is made.  This process can be repeated until the contestant wins a box worth $100,000 (if doubling a box’s value would take it past $100,000, the value instead gets set at $100,000.)  If they choose a box with a smaller value than the one they give back, they lose and win nothing.

cc2

Yeah, you’re not buying the “purely coincidental” line anymore, are you?

The strategy formula for this game is rather simple.

 

cc_equation

M is the value you currently own, n is the number of boxes left in the game, and the sum of Mp is the total money left in the remaining boxes (ignoring values that would end the game).  As long as this formula is true, you should play on.  If it’s false, you should stop with your current winnings.

This game is by far the worst game to play, partly because it’s so easy to lose before accumulating a decent sum.  For example, if you select the $5,000 box with your first selection, you’re facing a choice of either walking with that paltry sum or playing on with a 4/9 chance of being knocked out.  You have to cheat death, so to speak, many times before being able to even have a chance to earn $100,000 in this game, much less actually win it.  An average value of over $16,000 may not sound bad, but when the best game on the list has an average value three times that, you see just how bad of a game this is.


 

7. Advance to Boardwalk

Average: $28,708.46

Chance of Winning $100,000: 16.4%

Chance of Winning $0: 33.8%

Unfortunately, nothing to do with the spin-off board game from the 80’s.  The game is played on a fourteen step board.  Each step has a money amount on it starting at $1,000 and increasing by $1,000 up to $13,000 on Step 13.  Step 14, however is Boardwalk, and worth $100,000 if landed on by exact count (if a contestant rolls a number that would take them beyond Boardwalk, they do not move).  The contestant rolls a die, and moves the number of spaces rolled.  Each money amount that is landed on (not passed over) is added to their bank.  The catch is, they cannot roll the same number twice.  They are given one freebie which allows them to ignore their first duplicated roll, but any more duplicate rolls after that bankrupts them and ends the game.

atb2

The Money … Ladder? How about Money Midway?

Despite the simplicity of the game, it took a lot of effort to work out the correct strategy to this game.  We can lay out a couple of obvious ground rules.   Firstly, you should keep rolling as long as you still have your Free Roll token, since you are in no danger.  Just as obviously, you should stop whenever all of the safe rolls would take you beyond Boardwalk, as there’s no gain to be had.

What to do on the first 4 rolls is simple enough: you roll, no matter what.  There’s no combination of rolls that would give you a high enough bank to not make rolling again worthwhile.

Turn 5 is the big question.  I’ve tried to suss out some simple rules for playing this turn, but as far as I can tell there are no hard and fast rules.  I can say that you should generally play on as long as you can still land on Boardwalk, but even a rule as simple as that has exceptions.  The best that I can do is give you this (ahem) easy-to-read chart.

 

atb_chart

(Click to embiggen)

atb3

Turning the sides of the die red when they become bad rolls is a very swish touch.

That’s not pixel art or a quilt pattern, that’s the chart that tells you what to do on Turn 5. The big numbers correspond to what you rolled in the first (going up and down), and second (going left to right) turns.  Doing that will lead you to another 36-cell chart, where you can find your action by cross-referencing your third (going up and down) and fourth (going left to right) rolls.  If the box is gold, then you’ve already landed on Boardwalk, congratulations!  Similarly, if the box is black, you’ve already lost the game.  Otherwise, if the box is green, you should play on, and if it’s red you should quit.

Turn 6 is a little simpler to understand.  You can create some rules based on the space you’re currently on:

– $13,000: Always stop.

– $12,000: Roll on if 2 is still a good number and you have less than $20,000 OR exactly $22,000.

– $11,000: Roll on if 3 is still a good number and you have less than $22,000.

– $10,000: Roll on if 4 is still a good number and you have less than $23,000.

– $9,000: Roll on if 5 is still a good number.

Turn 7, if we get that far, is very easy to play.  Always stop.  It’s never worth it to play on.


 

6. Electric Company

Average: $33,074.05

Chance of Winning $100,000: 11.8%

Chance of Winning $0: 26.5%

ec1

I told you our electric bill would skyrocket if you turned on every light in the house!

The player is presented with 25 light bulbs, and 10 switches.  Each switch will light up a number of light bulbs from 1 to 10.  Each number is associated with only one switch, so if you find a switch that lights up 4 light bulbs, you know none of the other switches will also light 4 bulbs.

The goal of the game is to light as many bulbs as you can.  Each bulb lit is worth an increasing amount of money, up to $100,000 with bulb 24.  Light up bulb 25, however, and you cause a blackout, losing all your money.  You can stop at any time, leaving with the money accumulated up to that point.

We can’t give you a strict dividing line as to when you should stop and when you should continue, since that depends on the configuration of switches left in play.  What we can give you is a formula for figuring out if you should continue or not.  It’s exactly the same as the formula we used in Community Chest:

cc_equation

M is the amount of money you currently have, n is the number of switches left to pull, and Sum(Mp) is the total of the possible outcomes if you were to pull the next switch.  Again, as long as the left side of the equation is smaller than the right side, you should play on.

ec2

You want switches? We got switches!

Let’s walk through the game play in the first episode as an example.  At the first time the contestant faces a decision, she’s lit up 19 light bulbs, worth $20,000.  She has 7 switches left, worth 1, 2, 3, 5, 7, 8, and 10.  Thus, the left side of our equation is $20,000 * 7 = $140,000.  The possible outcomes if she went on are $25,000, $30,000, $40,000, $100,000, and $0 three times over.  The sum of those outcomes is $215,000.  Thus it is the right decision to move onward.

She moved onward, and flipped the switch worth 3 bulbs, increasing her total to $40,000.  Now there are 6 switches left, worth 1, 2, 5, 7, 8, and 10 bulbs.  The left side of the equation is $240,000, while the right side is only worth $150,000.  She should stop, which is what she chose to do.


 

5. Park It!

Average: $34,217.59

Chances of Winning $100,000: 32.3%

Chance of Winning $0: 56.9%

pi1

Two words that strike fear in every driving test taker.

The Monopoly Parking Garage (must have been in an edition I didn’t own) has five floors, each of which can take one car (just like the real Atlantic City, right?).  10 cars are waiting to be parked, each carrying a value between $1,000 and $10,000.  The contestant chooses a car, and once the value of the car is revealed, chooses which level of the garage to park it in.  The cars must be parked in order of value, with the highest value car on the top floor, and cannot be changed once parked.  If a car cannot be parked, it’s game over.  However, successfully parking five cars will win $100,000.  At any time, the contestant can walk away with the value of the parked cars.

This is a difficult game to win, as the majority of the time you will win nothing.  On the other hand, it’s an easy game to play, as the best strategy when it comes to placing the cars is almost always the intuitive, common sense strategy, trying to keep as many cars in play with each placement.  About the only surprise that will happen commonly comes on the first placement.  If you find the $2,000 or $9,000 car, they should be placed on the 2nd and 4th floors, respectively, to prevent a potential game over on turn 2.

pi2

The best solution for games where everybody wants to play as the race car.

The first time you should consider stopping is after placing the third car, and that’s only if you have three of the seven remaining cars able to be placed.  As long as you can still place a majority of cars, it’s right to continue.  After placing the fourth car, compare your bank to the following formula:

pi_equation

C is the number cars that will fit in the final open slot.  If your bank is bigger than that number, then stop.


 

4. Block Party

Average: $36,434.83

Chance of Winning $100,000: 26.1%

Chance of Winning $0: 41.6%

bp1

Yes, they animate a game board in the middle of the bigger game board that makes up the set. It’s Boardception.

The contestant is presented with an array of 12 face down cards. 8 of the cards are associated with a color group on the Monopoly board, and have a cash value starting with $1,000 for Mediterranean/Baltic, and going up to $20,000 for Park Place/Boardwalk. Picking those cards lights up the respective monopoly on the board, and banks the associated cash.  3 of the cards are strikes.  Get 2 strikes, and your bank is cut in half, though you may continue to play.  Find all three strikes, and you leave with nothing.  The remaining card is the “Block Party” card.  Find that card and you can light up one whole side of the board, up to 2 monopolies.  If you light up all eight monopolies before finding the third strike, you win $100,000.  And, of course, you can leave at any time and win the value of your bank.

The Thrill of Victory.

The Thrill of Victory.

The first strategic question is how to use the Block Party card – to maximize money won or to maximize blocks won? Testing both scenarios showed that there was a clear winning strategy: light up as much as you can, regardless of their value.  It is much more profitable in the long run to light up the Dark Purples and Light Blues and only put $3,000 in your bank, than to light up only the Dark Blues for $20,000.

The second and much harder question to answer is when you you should stop.  After playing around with a bunch of different strategies, the best formula I have discovered for determining whether or not to quit is as follows:

If Block Party card is on the board:

bp_eq1

After Block Party card is found:

bp_eq2

bp2

The Agony of Defeat.

S is the number of strike cards left on the board, and P is the number of property cards left.  For example, at the start of the game, the formula would be as such:

bp_eq3

Not coincidentally, that’s pretty close to the average value of the game.  As long as that value is greater than the value of your bank, keep picking cards.


 

3. No Vacancy

Average: $38,729.46

Chance of Winning $100,000: 4.6%

Chance of Winning $0: 2.3%

nv1

From Yelp: “Very expensive for the area. Front desk forced us all to stay on the same floor. 2 Stars.”

The contestant is tasked with filling up all three floors of a Monopoly hotel.  Each floor has seven rooms.  On each turn, five cars come out, of which the contestant must select one.  Each car has a unique number of people in it, from 1 to 5.  The contestant has to choose which floor to put those people on; there must be enough room left on the chosen floor, and all people in a car must be put on the same floor.  If none of the floors have enough open rooms to fit the number of people in the chosen car, the contestant wins nothing.  Each room filled on the bottom floor is worth $1,000, the middle floor $2,000, and the top floor $3,000.  Fill up all the rooms, and you win $100,000.

This game is one of the hardest to win, but also one of the hardest to win nothing.  The best strategy is to stop playing once your top floor has 4 or more rooms filled and the other two floors have 3 or more rooms filled (thus making the car with five people in it unplayable).  There are a few exceptions to that rule where you should continue:

– 4 Rooms filled in the top and 3 Rooms filled in the middle.

– 3 Rooms filled in the middle and the bottom filled completely.

– 4 Rooms filled top, the middle filled completely, and 3 Rooms filled bottom.

– 4 Rooms filled top, and both middle and bottom filled.

– 3 Rooms filled bottom, and both the top and middle filled.

nv2

Can I ask the drivers if they drove in the the carpool lane on their way here?

Following this strategy maximizes your winnings, but will see you ending the game prematurely about 93% of the time.

What was really surprising to me was the optimal strategy to use when filling rooms. My assumption going in was that it would be very straightforward: fill the top row until you come to a number you can’t fill there, then move on to the middle row and then the bottom.  It turns out that the strategy to maximize your winnings is much, much more complex.  For example, on the first turn you should fill the rooms on the top floor, unless you pick the car which has 4 people in it, in which case they should go in the middle row!  I tried sussing out some placement rules to use, but I couldn’t generate an easy-to-follow list of rules.  I placed the optimal strategy in a spreadsheet if you’re interested in the gory results.


 

2. Bank Buster

Average: $42,592.69

Chance of Winning $100,000: 24.6%

Chance of Winning $0: 19.0%

bb1

Say we get into the cage, and through the security doors there and down the elevator we can’t move, and past the guards with the guns, and into the vault we can’t open…

The contestant is presented with a bank vault, locked with six locks.  Twelve keys are given to the contestant to select one at a time.  Each lock has two different keys that will unlock it.  Unlock a lock, and an amount of money is added to your bank depending on the lock, from $6,000 up to $20,000.  However, if you select a key that fits an already opened lock, that lock gets re-locked for good, and the money taken away from you.  You need to unlock five of the six locks to earn the top prize of $100,000.  If you manage to re-lock two locks, thus making the game unable to be won, you lose everything.

bb2

Five keys too many for Jack Narz, seven keys too few for Richard Bacon.

Not only is this game very likely to end well for the contestant, but the optimal strategy is also very simple.  The first time you want to consider stopping is after your 4th key.  If you have re-locked a lock and are in danger of losing, stop if your bank is $26,000 or more (except in the very specific case that you’ve unlocked the $6K and $20K locks, and the re-locked lock is the $7K lock), otherwise go on.  After you’ve picked 5 keys, stop the game with $21,000 or more in your bank (unless you have exactly $21,000 and the $20K lock is still available to open).  Finally, stop with 6 keys if your bank has reached $34,000.  The game is guaranteed to end one way or another after 7 keys have been chosen.


 

1. Ride the Rails

Average: $49,942.86

Chance of Winning $100,000: 39.5%

Chance of Winning $0: 7.1%

rtr1

I’ve sold monorails to Chesapeake, Gulf Coast, and Washington, and, by gum, it put them on the map!

This game is pretty similar to ITV’s 2009 show The Colour of Money.  That should have sent a shiver down your spine if you’re unlucky enough to remember The Colour of Money.

The contestant is presented with a list of ten railroad lines.  Each line has a railroad engine carrying between one and ten boxcars, called “cash cars”, and a caboose.  Each train has a unique number of cash cars between 1 and 10, like the switches in Electric Company.  Once a contestant chooses a line, that line’s train will slowly make it’s way across the floor, revealing its cash cars.  Each cash car revealed is worth money: $1,000 for the first line chosen, $2,000 for the second, $3,000 for the third, and $5,000 for the fourth.  At any time, the contestant can stop the train and bank the value of the cash cars revealed.  This is important, because if the caboose comes out, all of the money earned on that train is lost (money won on previous trains is kept, though).  If the contestant manages to earn $50,000 after four trains, then their total is doubled to $100,000.

Thanks to combinatorics, we know there are only 5,040 different combinations of trains that could be selected.  We can also create an exhaustive list of strategies, where a strategy is the number of cars we will let pass before hitting the brakes in each round.  Since the possible number of cars in a train changes in later rounds based on what was picked in earlier rounds, we’ll define strategies in terms of the length of possible trains in each round.  For example, if a strategy told us to stop after the 3rd train in the second round, that would mean stopping after 3 cars if we picked a train from 4-10 in the first round, or stopping after 4 cars if we pick a train from 1-3 instead.  To reflect the incentive of trying to reach $50,000 to win the maximum prize, we’ll also add a special strategy that can be chosen in the fourth round: “End”, where we will let the train continue until we’ve hit $50,000 or we see the caboose.

rtr3

Stop on a Whammy!

After testing every possible strategy against every possible configuration of trains, the winning strategy is fairly simple: 6, 5, 4, End.  In other words, we let 6 cars pass in the first round before stopping.  In the second round, let the number of cars pass equal to the 5th longest train in the second round.  In the third, we want to see a number of cars equal to the 4th longest train left on the board.  Finally, we go for $50,000 or bust in the last round.  This strategy will win us, on average, almost $50,000, making this game the most profitable game to play.


 

BONUS: Go for a Million

Average: $176,140+

Chance of Winning $1,000,000: 7.8%

Chance of Winning $0: 15.2%

gftm1

The big board in action, with the (sigh) “Monopoly Rock ‘n Roller” die roller in the middle.

The end game is almost identical to the end game to the 1990 version of Monopoly, for those of us who remember it.  The contestant starts on GO, and has 5 rolls to traverse the 40 squares of the board and return back to GO.  Along the way, the contestant earns money for each square he lands on, which he can stop with at any time.  If the contestant rolls doubles, he gets another roll, but like in the board game, three straight doubles and you’re off to jail with no money.  Likewise, landing on Space #30, “Go To Jail”, is also an instant lose.  You can also lose by landing on Community Chest and Chance spaces; contestants have to draw a card when they land on that space, which like the game could be a “Go To Jail” card.  If you make it around the board and pass GO, you win $200,000.  However, if you land right on GO, you win $1,000,000 for yourself, and your audience section gets a rolling Audience jackpot.

gftm5

“Can’t I just pay $50 and try again?”

In order to play this, you first have to give back whatever you’ve won in the first game.  Since the expected value of the game is over $176,000, you should always give back anything you won previously, even a maximum prize of $100,000.  I’m not able to calculate the exact value of the game, since I don’t have a list of how much money you win by landing on every property.  The figure listed above is only counting the $200,000 for passing GO, and $1,300,000 for landing on GO, assuming the value of the Audience jackpot to be $300,000.

gftm_win

What happens when you win $1,000,000. Sliding into GO optional.

Also, I don’t really have a strategy for whether you should stop or not.  I doubt that you should ever stop, since the chances of failure are so small compared to the potential payout.  I suppose it might be possible if, as a worst case scenario, you are on Indiana Avenue, seven squares away from “Go To Jail”, after having rolled two doubles with only one roll left.  Since you have a 33% chance of failure in that scenario (⅙ of the time you’ll roll a 7 to land in jail, ⅙ of the time you’ll roll your 3rd double) with no chance of hitting the jackpot, you’re probably better off quitting then.  But situations like that are so rare that it’s not worth the effort to try to quantify them.  Just keep rolling, and hopefully you’ll be rolling in the dough before long.

1000 Heartbeats: When to Cashout?

Logo

You know what they say about the Deputy Undersecretary of the Interior, they’re only 1,000 heartbeats away from the Presidency.

ITV debuted a new game show on February 23rd, 1000 Heartbeats.  The main gimmick of the show is that the contestant’s own heartbeat determines how long they have to play.  It’s a stylish show, complete with a live string quartet providing the music at the same tempo as the contestant’s heart rate, and has been getting favorable reviews.

The contestant is given a “clock” of 1000 of their own heartbeats (measured by a well-hidden heart monitor) to play a series of minigames testing the contestant’s skills in anagrams, mental arithmetic, and general knowledge. Each successfully completed minigame increases the potential winnings of the contestant, up to a maximum of £25,000.  However, if they run out of heartbeats, their game is over and they leave empty-handed.  After each game is completed, the contestant is given a preview of the next game and, taking into consideration the number of heartbeats they have remaining, may either play on for more money or stop.  However, before the contestant can walk away with their money, they must play one final minigame with the remainder of their heartbeats, named Cashout.

gameplay

My cardiologist’s new stress test has yet to be endorsed by the American Heart Association.

Cashout is a fairly simple game.  As their heartbeats tick down, the contestant is given a series of True or False statements, and must correctly answer 5 in a row in order to win.  Giving an incorrect answer not only forces them to start their chain of 5 answers over again, but also deducts 25 heartbeats from their clock.  It’s an effective denouement, and has provided us with tense finishes already in the show’s short history.  But watching it got me thinking – how many heartbeats would you want to bring into Cashout to maximize your chance of success? And when during the course of the game is it ideal to play Cashout instead of pressing onward for a potentially higher payday?

strings

People were shocked to see the bold new direction being taken by Brian Eno.

Before we can measure how successful a contestant will be in Cashout, we first need to determine what metrics we can measure that will allow us to estimate a contestant’s success.  I’ve identified three metrics: what their heart rate is, how often they give the correct answer, and how long each question takes to read and answer.  Using those three variables, we can figure out the odds of completing a sequence of 5 correct answers, and how many heartbeats would elapse during that time.

These metrics will be different for each player, but for the purposes of this article, we will create an average contestant, using the data from the contestants who played during the first five episodes.  In Cashout, the average contestant would answer 70.14% of the True/False questions correctly, while taking 6.45 seconds per question with a heart rate of 128 BPM.  Using these values, we ran a Monte Carlo simulation to determine the chance that the average contestant will successfully complete Cashout given the number of heartbeats they started with.  The results are displayed in this graph.

Cashout Chart

It’s a curve that starts declining fairly gently, but increases in slope before crashing to 0% around 70 heartbeats   That’s the minimum number of heartbeats needed to see and answer 5 questions correctly with no wrong answers  It may not have have happened during the first week of shows, but it should happen 17.5% of the time.  Starting Cashout with anything above 250 heartbeats leads to a better than 50% chance of succeeding.

We can use these values to help decide whether or not to proceed to the next round during gameplay.  At any given point in a player’s game, we can calculate the Expected Value (EV) of the game, which is the amount of money the player would win on average if they played Cashout.  That value is determined by the amount of money banked so far multiplied by their chances of completing Cashout with their remaining heartbeats.  For example, if a contestant has £500 banked, and has a 90% chance of winning Cashout with their remaining heartbeats, the EV of their game at that point would be £450.

If it’s a good idea for the contestant to play onward, the EV of their game after the next round must be higher. We can represent this in the following formula:

Equation1

M is the amount of money currently banked, H is the number of Heartbeats remaining, and C is the function that tells us the chances of winning Cashout given a number of heartbeats, as defined by the chart above.  M’ and H’ are the money and heartbeats left after the next round is played.

This function will be easier to work with if we divide both sides by M’, as so:

Equation2

Money LadderWe now see that the contestant should want to move on if their future chances are greater than their current chances multiplied by the ratio at which the banked money will increase.  Looking at the money ladder, we can see that in rounds 2, 3, 4, and 6 the money doubles, and M divided by M’ would be .5.  Thus, the contestant should move on if their chances of winning Cashout are greater than half of their current chances.  In rounds 3 and 7, the jump is greater, as the money is increased 150%.  M divided by M’ in this case would be .4, so the contestant’s future chances can drop by as much as 60% of their current level before moving on becomes a bad idea.

So, we know the ratio at which our money rises, and we can determine our current chances of winning Cashout.  But how can you determine your future chances?  After all, you don’t know exactly how many heartbeats you’ll have left.  This is the position in the game where an element of estimation comes into play.  Taking a look at your past performances as well as the difficulty level of the next game, you’ll have to estimate how many heartbeats you’ll think you’ll need to successfully complete the next game.

Here’s that data broken down in graph form.  The two lines represent the break-even points of how many heartbeats you can spend given your current number of heartbeats, depending on what round you are playing.

HB Chart

Click to see the full-size chart.

Let’s use this data to take a closer look at the two contestants who completed games on the show that aired on March 2nd.

contestant1

Luanne had a relatively poor first round of Contrast, and continued the theme with an even worse attempt at Unravel in round 2.  By the time Round 3 came along, she only had 366 heartbeats left to play Assemble. According to the chart above, Luanne should have continued to play if she thought that she could complete Assemble in fewer than 200 heartbeats.  Since she hadn’t done that yet in her game, it was probably a wise move that she opted to play an early Cashout.   Taking 366 heartbeats into Cashout should be enough for an average contestant to win about two-thirds of the time.  In this particular example, she managed to play Cashout successfully, ending with a scant 6 heartbeats remaining and taking home a hard-earned £500.

contestant2

Andy blew threw Contrast and Unravel before hitting a stumbling block in round 3 with Assemble.  He righted the ship in Round 4 with Link, and faced a decision on whether to play Keep Up, a mathematical game, in round 5.  It seems like most players are loathe to play these games requiring mathematical computation, but Andy opted to continue playing.  This could be seen as an aggressive move, but is a move that should lead to more money won as long as he spends fewer than 231 heartbeats on the game.  He only spent 190, which increased the expected value of his game by £414.25.  With only 204 heartbeats remaining, he had an easy choice to opt for Cashout at this point.  Unfortunately, he quickly exhausted his heartbeats with a series of wrong answers, and wound up not converting his banked £5,000.  Andy made a risky, yet mathematically sound choice to play Keep Up, but was not rewarded in the end.

Estimating how many heartbeats each minigame would take to complete would be very useful, We could could look at the past playings of each minigame and determine the time taken to complete it.  However, after only five episodes, the data we would get would not be very reliable due to the small sample size.  Perhaps if ITV gives this show the run it deserves, we will revisit this topic in a future article.

Jeopardy ToC Update: Semifinal Game 2

The Stats

You gotta hand it to Arthur Chu.  Despite being handed the toughest draw in the field, he’s making this look easy.

toc_sf2_table

Mark Japinga’s negging and Rider being crowded out on the buzzer could have made this a runaway, but Rider doubled up on a good Daily Double, and Japinga made a late run to remain relevant.

Daily Doubles

I’ll admit – as soon as Chu wagered $1,000 on his first Daily Double, my bad-bet-o-meter went crazy.  He found it early in the Jeopardy round, having called for the $600 and $800 clues in random categories looking for it as is his M.O.  He was trailing Japinga slightly at the time, $2,600 to $2,800, with Rider yet to open her mouth. I loaded up my still-in-development Daily Double evaluator, and fed it the parameters.  Here’s what it spat out at me:

dd_sf2

The five lines represent how comfortable you are with the category.  They have nothing to do with the expected difficulty of the clue, which has already been accounted for.  The table may be hard to read, but it suggests 3 wagers, depending on your level of confidence:

– $300-400 if you are not confident. Enough to take over the lead, but still enough to stay in touching distance if you miss it.

– $2,000 if you’re confidence is average. Beyond a certain point when you’re leading, every dollar you have is gives you a smaller chance of winning than the previous dollar. At this point, the diminishing returns of increasing your score catch up with the chances that you’ll actually get the clue right.

–  $2,400 if you’re supremely confident.  The system suggests the diminishing returns beyond that point are not worth an all-in wager, but I certainly wouldn’t begrudge one if that was your choice.

Chu’s selection of $1,000 is an odd one. It’s not conservative enough if you don’t like the category (which I imagine was Chu’s reasoning), as it leaves you at least two clues behind to catch up if you’re wrong. It’s also not aggressive enough to take advantage of the opportunity that a Daily Doubles represents.

Chu hit the first Daily Double in Double Jeopardy, and again wagered $1,000.  The situation was much different this time, with Chu on a commanding $13,200 over Japinga’s $3,600 and Rider’s $1,800.  Without looking too deep at the details this time, I think the two choices in wagers would be the minimum of $5 if you want to protect your lead, and a more aggressive wager of $6,000, looking to close out the game here but still keep twice Japinga’s score if you’re wrong.  I don’t begrudge Chu his wager that much, since I doubt there was any great difference between wagers of $5 and $1,000. And despite the disclaimer that this could be the stupidest thing she ever did, Rider’s true Daily Double later with her score of $4,600 a long way back of Chu’s $16,600 was the only real move she could have made, especially in a category that she seemed to like.  She answered correctly, and Japinga made a late run to set up a Final Jeopardy where all three players could still take the victory.

Final Jeopardy

The scores were Chu with $17,800, Rider with $12,800, and Japinga with $8,000.  I’ll leave the exact analysis of the situation to Keith at The Final Wager, who does a better job at it than I could ever hope to. But I do want to talk about something that I feel very strongly about which comes into play here: Stratton’s Dilemma.

Coined by Andy Saunders, Stratton’s Dilemma is the term for the situation Rider finds herself in, where she has to choose between two possible wagers. Let’s analyze her situation. We can assume that Chu will wager $7,801 or thereabouts, the amount needed to guarantee his victory as long as he gets Final Jeopardy correct. We must assume that Chu responds incorrectly, otherwise we have no chance of winning.  If Chu is incorrect, he will be left with $9,999. Thus, it is in our best interest to wager no more than $2,800, staying ahead of Chu no matter what our outcome is. However, the presence of Japinga complicates matters.  If Japinga wagers everything and doubles up, he’ll have $16,000.  To ensure that we remain ahead of Japinga in that situation, we have to bet at least $3,201.  Sometimes we can find a wager to satisfy both scenarios, but this is not one of those times. We have to choose one or the other, and know that some of the time we will choose incorrectly and lose when we could have won.  It’s an infuriating position. Which bet should we choose?

Well, let’s break down the scenarios.  Since each player can either be correct or incorrect in Final Jeopardy, there are 8 possibilities to consider.  Assuming that Chu wagers $7,801, and Japinga wagers his entire $8,000 (our worst case scenario), what happens when we bet $0?stratton1Compare that to what happens when we bet everything:

stratton2

We’ll always win in either case if we’re the only person to respond correctly. When we bet small, we will win if everybody answers incorrectly.  If we bet big, we will win if the leader misses and both we and the player in third respond correctly.  Which is more likely to happen?  If only we had a large repository of previously played Jeopardy games to look at…

I took a look at 2,184 games played over the last 10 years where all three players made it to Final Jeopardy.  Specifically, I counted the times when each of the above scenarios happened.  I separated the players by their ranking going into Final Jeopardy, so I know how often the leader missed Final Jeopardy while the other two got it right, or only the second-ranked player responded correctly, for example. And what were the results?stratton3

Unsurprisingly, the scenarios when all three players responded correctly and all three players responded incorrectly are more likely than any other case.  Players possess a shared knowledge base, and a question that one person knows is likely to also be known by the others, and vice versa.  The situation we’re hopping happens when we bet big, where 2nd and 3rd place answers correctly but the leader does not, is the least likely to happen. If we quantify our chances of winning using the above table, we have a 30.9% chance of victory with a small bet, while we only have a 19.1% chance of success if we wager big.  I don’t know about you, but if one choice of a dilemma increases my chances by over 60%, I wouldn’t call it much of a dilemma.

The Odds

Well, we’re going to get the matchup we wanted, Collins vs. Chu. And frankly, this tournament is now Chu’s to lose. Regardless of who wins the next match between Ben Ingram, Joshua Brakhage, and Sandie Baker, I can’t help but see Chu as a strong favorite, especially over a two-day affair where the normally high level of variance inherent in a Jeopardy match is lowered.

toc_sf2_odds

 

Jeopardy ToC Update: Semifinals Game 1

The Stats

The “curse” of second place continues.  This makes it the sixth straight time that the player with the second ranked chance by my system won the game.

toc_sf1_table

Terry O’Shea may have been extra choosy in her clue selection, but her 100% precision and her tendency to score rebounds off of the others still gave her a strong showing.  Jared Hall and Julia Collins played very similar games, but Hall found all three Daily Doubles.  Had he converted his all-in wager early in Double Jeopardy, the result may have been a lot different.  Instead, Julia played up to the expectations placed upon her, and became our first Finalist.

Strategy

Hall, as lampshaded by Trebek at one point, was the only player playing tonight who picked clues from the middle of the board at first, hunting for the Daily Double.  He was rewarded handsomely for his efforts, as all three Daily Doubles landed in his lap.  The first and third were rather straightforward efforts, with Hall having less than $1,000 in the Jeopardy round and $2,000 in Double Jeopardy, which almost always necessitates the maximum wager allowable wager of $1,000 or $2,000, depending on the round.  His second Daily Double, however, I think deserves some extra attention.

Hall had $4,200 early in the Double Jeopardy round, poised between Collins’ $6,800 and O’Shea’s $3,000.  The Daily Double was hiding in the $1,600 clue in the category Great American Novels.  The average player would probably pick a middling value without too much thought, probably about half their score or less.  Hall did what I imagine most good Jeopardy players would do, and bet it all.  There’s still enough money left on the board ($27,600 out of the $36,000 that started the round, as well as the other Daily Double) that even if you go bust, there’s still enough time to make a comeback.  And if you respond correctly and double up, you’ll have a lead and be well poised for the rest of the game.

I’ve mentioned that I’m working on a system that evaluates Daily Doubles and determines the expected win % of every possible wager, trying to find the “best” bet, if such a thing exists.  It’s still very rough around the edges, but I thought I’d give it this situation to mull over.  Here’s what it said:

dd_sf1

The five lines represent how comfortable you are with the category (not the clue value – that’s already been baked in to the system.), with red being the least comfortable to green being the most comfortable.  In the end, your comfort level with the category didn’t matter, as bets could be clustered into three categories:

– $0 – $1,200: Pretty bad.  You’re still going to be in second place no matter what.

– $1,200 – $2,600: Even worse.  Best case scenario is that you’re still trailing the leader, but now if you’re wrong you’re dumped down into 3rd place.

– $2,600+: Best. The benefit of being in first place if you are correct greatly outweighs the chance that you’ll be in third.  What’s interesting about this section is that your chances decrease as your wager increases beyond what it would take to get into first, indicating that perhaps what you gain by increasing your score above $6,801 is not worth what you lose if you by going further and further behind.  Like I said before, I don’t consider this system to be ready for prime-time, but it gives us something to think about: that a knee-jerk all-in bet in this scenario may be good, but not optimal.

Going into Final Jeopardy, it was still anybody’s game.  Collins led with $12,000, O’Shea had $8,200, and Hall was right behind them with $7,600.  This situation may look a little boring at first, but digging a little deeper into the math reveals a truly fascinating situation – and one that’s very scary for Collins, despite being in the lead.  Keith Williams of premier Jeopardy wagering blog The Final Wager has an excellent write-up and video detailing why, one that I highly recommend watching.

The producers of Jeopardy stop tape to give the players as much time as they want to calculate their wagers; Collins took 15 minutes to finally settle on hers.  In a cruel twist of fate, all of the strategizing proved moot, as only Collins was able to get the correct response in Final Jeopardy, and is our first confirmed finalist.  Well done to her.  She rode her luck a little, earning a wild card with a historically below-average score, and not getting any of the Daily Doubles in this match, but I don’t think anybody could begrudge her her place in the finals.

The Odds

Collins moves up to the top by virtue of having secured her finals place.  The remaining players’ chances only moved a fraction of a percentage point – Collins’s stats were darn near close to the average expected stats of the player coming out of the first semifinal.

toc_sf1_odds

We’ll see if tomorrow will get us the final match that everybody wants to see: Arthur Chu vs. Julia Collins, or if Mark Japinga or Rebecca Rider will upset the storyline.