Ask the Wizard: TV Game Shows - FAQ

• Player picks door 1 --> player wins
• Player picks door 2 --> player loses
• Player picks door 3 --> player loses

Following are what would happen if the player had a strategy of switching.

• Player picks door 1 --> Host reveals goat behind door 2 or 3 --> player switches to other door --> player loses
• Player picks door 2 --> Host reveals goat behind door 3 --> player switches to door 1 --> player wins
• Player picks door 3 --> Host reveals goat behind door 2 --> player switches to door 1 --> player wins

So by not switching the player has 1/3 chance of winning. By switching the player has a 2/3 chance of winning. So the player should definitely switch.

For further reading on the Monty Hall paradox, I recommend the article at Wikipedia .

1. 0.01
2. 1
3. 5
4. 10
5. 25
6. 50
7. 75
8. 100
9. 200
10. 300
11. 400
12. 500
13. 750
14. 1000
15. 5000
16. 10000
17. 25000
18. 50000
19. 75000
20. 100000
21. 200000
22. 300000
23. 400000
24. 500000
25. 750000
26. 1000000

Here is the flow of the game:

1. Player picks one case for himself
2. Player opens up six of the remaining 25 cases.
3. Banker makes an offer.
4. If player declines he opens five more of the 19 remaining cases.
5. Banker makes an offer.
6. If player declines he opens four more of the 14 remaining cases.
7. Banker makes an offer.
8. If player declines he opens three more of the 10 remaining cases.
9. Banker makes an offer.
10. If player declines he opens two more of the 7 remaining cases.
11. Banker makes an offer.
12. If player declines he opens one more of the remaining cases.
13. Keep repeating steps 11 and 12 until player accepts an offer or player has the last unopened case.

The following chart plots the player’s expected value and the banker’s offer.

The most obvious thing to be learned from these three charts are that the first four to six bank offers are terrible deals. The average suitcase has $131,477.54 before any are opened. To only offer $9000 to $13000 the first stage is a deal only a fool would make. However gradually the offers get better. Game 2 shows us the expected values were almost the same as the banker offers towards the end of the game when the player’s expected value was fairly low. However in games 1 and 3 when the expected values were higher the banker apparently was trying to take advantage of the risk averse nature of most people when large amounts are involved. I don’t know if it mattered but the contestant in game 2 appeared to be a gambler who wanted to win big. Based on comments by the host, who communicates to the banker by phone, the banker does appear to take the contestants words and actions into consideration. If I were in the banker’s shoes I would act much the same.

If the player is neither risk averse nor risk prone, and also ignoring tax implications, the player should keep refusing banker offers until one exceeds the average of the remaining suitcases. For most people the progressive nature of the income tax code favors taking a deal. As I have said before I would roughly say the value of money is proportional to the log of the amount. So the more wealth you have going into the game the more inclined you should be to gamble and refuse the banker offers. With such large amounts involved, no strategy will fit everybody. However I can fairly confidently say that the player should refuse the first four to six offers and then take the offers on a case by case basis (pun intended).

Links:
You can play Deal or No Deal at NBC.com.
Archive of past shows.

Let me explain what happened on Let’s Make a Deal. The contestant would choose one of three curtains. One would contain a very valuable prize and the other two smaller prizes. For the sake of argument let’s say behind one curtain was a car and behind the other two a goat. Then Monty would always, I repeat ALWAYS, open up one of the two unchosen curtains to reveal a goat. After hundreds of shows this would imply that Monty Hall (the host) knew where the car was and deliberately opened a curtain that revealed a goat. Obviously when the player chose his curtain the probability it held the car was 1/3 and the probability one of the two unchosen curtains held the car was 2/3. Monty is then predestined to open an unchosen curtain containing a goal. Predestined is the key word here. Because Monty can not open the player’s curtain at this stage the probability of the player’s curtain reveals the car stays at 1/3. The probability an unchosen curtain reveales the car remains at 2/3, however it is now all on one curtain. So after a goat is revelead the probability the player’s curtain has the car is 1/3 and the probability the other unopened curtain has the car is 2/3, making switching a wise choise.

The following table shows all the possible outcomes. In the case where the player chose the curtain with the car I had Monty opening a curtain arbitrarily. You can see that not switching results in a 1/3 probability of winning, and switching results in a 2/3 probability of winnning.

Meanwhile in Deal or No Deal nothing is predestined. Let’s assume on Deal or No Deal the amounts remaining were $0.01, $1, and $1,000,000. With three cases left it IS possible that the opened case will contain the million dollars. The following table shows the possible outcomes with three cases left. Remember, the player can not open his own case.

What the Deal or No Deal table shows is that with three cases left the probability the player opens the million dollar case is 1/3 (hopeless to win), the probability a switching player will win is 1/3, and the probability a switching player will lose is 1/3. Thus the odds are the same to switch cases. Once there are only two cases left the probability each case contains the larger prize is 50/50.

log(b) = 5.520696
b = 10^5.520696
b = $331,662.50.

So this hypothetical player should be indifferent at a bank offer of $331,662.50. The lesser your wealth going into the game the more conservatively you should play. Usually in the late stages of the game the bank offers are close to expected value, sometimes a little more bit more. The only rational case where a player could win the million is if he had a lot of wealth going into the game and/or the bank offers were unusually stingy. The producers seem to like hard-working middle class people, so we’re unlikely to see somebody who can afford to be cavalier when large amounts are involved. I have also never seen the bank make offers under 90% of expected value late in the game. The time when we will see somebody win the million is when a degenerate gambler gets on the show who can’t stop. When that happens I will be rooting for the banker.

I know that the equal probability of answering correctly among players is an unrealistic assumption. For one thing the leader probably has a higher probability of answering correctly. For another, each player knows the category before wagering. Give me a category of gambling and I’m close to a lock, but if is poetry, I’m doomed. So don’t write to me about how unrealistic my assumptions are, because I acknowledge that already. This is meant to be more of an exercise in game theory than practical advice for future contestants.

Next, we will need some data to look at. From j-archive.com we find that in season 22, in 2005, Final Jeopardy was answered correctly 43.80% of players. We also learn that all three contestants got it right 14.92% of the time, and all three got it wrong 24.86% of the time. Let p_n be the probability n players get the question right. The first equation to work with is:

p₀ + p₁ + p₂ + p₃ = 1

Substituting the known values for p₀ and p₃:

0.2486 + p₁ + p₂ + 0.1492 = 1.

p₁ + p₂ = 0.6022.

(1) p₁ = 0.6022 - p₂.

We can also construct the probability of any given player getting the question right as follows.

p₀×0 + p₁×(1/3) + p₂×(2/3) + p₃×1 = 0.4380.

p₁×(1/3) + p₂×(2/3) + 0.1492 = 0.4380.

p₁×(1/3) + p₂×(2/3) = 0.2888.

Multiplying both sides by 3:

(2) p₁ + p₂×2 = 0.8664.

Substituting the value for p₁ in equation (1) we get

0.6022 - p₂ + p₂×2 = 0.8664.

p₂ = 0.2642

Solving for p₁

p₁ = 0.6022 - p₂.

p₁ = 0.6022 − 0.2642 = 0.3380.

Based on the three-player probabilities above, in a two-player game the probability both get the question right is:

p₃ + (1/3)×0.2642 = 0.1492 + 0.2642×(1/3) = 0.2373.

The probability one gets it right is:

(2/3)×p₂ + (1/3)×p₁ = (2/3)×0.2642× + (1/3)×0.3380 = 0.4015.

The probability neither gets it right is:

(1/3)×p₁ + p₀ = 0.3380×(1/3) + 0.2486 = 0.3613.

It is obvious that if the leader has more than double the amount of the follower, he should bet less than the difference, to ensure victory. However, what should he do if his balance is less than double that of the follower? Let’s draw up a specific example where player A has $10,000 and player B has $8,000. To make matters easy, let’s restrict A’s options to making a big bet of $6001 (enough to guarantee victory with a correct answer) or a small bet of $1999 (leaving enough to guarantte victory if B answers incorrectly). Let’s restrict B’s options to a betting all or nothing.

At first glance, it would seem the right thing to do for player A to bet small, forcing player B to get the question right to win. That would give player A a 1 − 43.8% = 56.2% chance of winning. Assuming that strategy, player B would have to bet big to win. However, because of the correlation between players, if player B gets the question right, player A probably would too. The exact probability that player A gets the question right, given that player B got it right, by Bayes’ Theorem , is:

Probability(A and B correct)/Probability(B correct) = 0.2373/0.438 = 0.5417.

So, if player A knew player B would bet big, he should too. However, if player B knew player A would bet big, then he should bet small, and hope player A gets it wrong, ensuring a 56.2% chance of winning. Going further, if player A knew player B would be small, he would too, and have a 100% chance of winning. Assuming player A bet small, player B would of course bet big. And so we go around and around.

The optimal strategy for both players is to randomize their bet.

The following table shows the probability player A will win according to all four combinations of bets.

The next stop is hard to explain why, but for either player the optimal probability of either option is proportional to the absolute difference in values of the other option. Player A should bet high with probability proportional to abs(0.5620 − 1) = 0.4380. He should bet low with probability proportional to abs(0.799267- 0.4380) = 0.3613. So, the actual probability of A betting high should be 0.4380/(0.4380 + 0.3613) = 0.548002. The probability of going low is obviously 1 − 0.548002 = 0.451998.

By the same logic, player B should go high with probability proportional to abs(0.438000 - 1) = 0.562000, and low with probability proportional to abs(0.799267 − 0.562000) = 0.237267. The actual probability of going high should be 0.562000/(0.562000 + 0.237267) = 0.703145. Thus, the probability B should go low is 1 - 0.703145 = 0.296855.

The next table shows the probability of all four outcomes in strategy.

21

I can offer three theories why the banker sometimes offers more than the average of the remaining cases.

1. The show tries to portray the banker as sweating the money in his office. Howie Mandel is often commenting on the banker’s mood and tone of voice. Maybe it makes the show more dramatic to think of the banker as a risk-averse bean counter, preferring to cut his losses, than risk giving out a big prize.
2. The real banker truly is risk-averse. This is getting out of my area of expertise, but from my understanding, game and reality shows are usually produced by a company independent from the television network. These smaller companies will seek out an insurance company to mitigate the risk of contestants winning the larger prizes. In such a case, the insurance company would be the real banker, and may be influencing the behavior of the banker on the show. The insurance companies that insure odd-ball stuff like this are not gigantic, and may prefer playing it safe when large amounts are involved.

   In your example, the banker offer was 9.92% above expected value. If the banker were following the Kelly Criterion, such an offer would have been made with a total bankroll of only $782,008, which is less than the maximum prize. No self-respecting insurance company would be that conservative. Clearly, this reason alone cannot justify the offer in your example.

3. The show is trying to make the contestants look stupid and greedy. Shows like Are You Smarter than a Fifth Grader and the Tonight Show’s “Jaywalking” would not be successful if we didn’t find some satisfaction in laughing at the trivia-challenged. The shows Friend or Foe and The Weakest Link were outstanding at exposing greed in human nature. I must confess a sense of schadenfreude when a contestant refuses an above expected value offer, and walks with the lower amount in his case.

I tend to think the reason is a combination of these three reasons, but mainly the third.

If I ended this answer here, I’m sure I would get comments, questioning whether the hypothetical banker offers would have really been made. The implication being that they are puffed up for dramatic effect. I have recorded the specifics of 13 games. In one of them, with three cases left ($1,000; $5,000; and $50,000), the average was $18,667, and the offer was $21,000. That is 12.5% over the expected value. In another show, with two cases left ($400 and $750,000), the average was $375,200, and the offer was $400,000. That is 6.6% above expected value. So, I see no reason to question the integrity of the hypothetical offers.

Links:

Beating the Banker: Game Theoretic Applications in Deal or No Deal (PDF): This is a scholarly paper on both banker and player behavior in the U.S. version of Deal or No Deal. The paper suggests a formula for the banker offer, based on expected value, the round, standard deviation. However, the paper says the duel goals of the banker are to send the player home with as little as possible, and to keep him playing as long as possible. An above expected value offer in the last round goes against both goals.

Deal or no deal formula: This page shows old, and new, formulas for calculating the banker offer, based on the free game at the Deal or No Deal web site .

Let’s start with player C. He should anticipate that A might bet $6001, to stay above B if B is right. However, if A is wrong, that would lower him to $3999. C would need to bet at least $500, and be right, to beat A in such a scenario. However, in my opinion, if you must be right to win, you may as well bet big. So if I were C I would bet everything.

B is torn between betting big or small. A small bet should be $999 or less, to stay above C if C is correct. The benefit of a small bet is staying above C no matter what, hoping that A will go big, and be wrong. A big bet does not necessarily have to go the whole way, but it may as well. The benefit of a big bet is hoping that either A goes small, or goes big and is wrong, but both require B to be right.

A basically wants to go the same way as B. A small bet for A can be anything from $0 to $1000, which will stay above B if B bets $999. A big bet should be $6001, to guarantee a win if A is right, and still retain hope if B goes big, and all three players are wrong.

To help with the probabilities of the eight possible outcomes of right and wrong answers, I looked at the Final Jeopardy results for seasons 20 to 24, from j-archive.com . Here is what the results look like, where player A is the leader, followed by player B, and C in last.

Using the kind of game theory logic I explain in problem 192 at my site mathproblems.info , I find that A and B should randomize their strategy as follows.

Player A should bet big with probability 73.6% and small with probability 26.4%.
Player B should bet big with probability 67.3% and small with probability 32.7%.
Player C should bet big with probability 100.0%.

If this strategy is followed, the probability of each player’s winning will be as follows:

Player A: 66.48%
Player B: 27.27%
Player C: 6.25%

As an aside, based on the table above, the probability of the leader getting Final Jeopardy correct is 54.4%, for the second-place player, 49.8%, and 48.7% for the third-place player. The overall probability is 51.0%.

As a practical note, players do have knowledge of betting behavior. In my judgment, players tend to bet big more often than mathematically justified. Interestingly, I find wagering in Daily Double to be too conservative than mathematically justified. One of the reasons I believe Ken Jennings did so well was aggressive wagering on the Double Doubles. Anyway, in reality if I were actually on the show, I would assume the other two players would bet aggressively. So my actual wagers would be $6000 as A (being nice to B), $0 as B, and $3495 as C (leaving a little un-bet, in case A foolishly bets everything or all but $1, and is wrong).

Before somebody challenges me about how one could draw a random number in the actual venue, let me suggest the Stanford Wong strategy of using the second hand of your watch to draw a random number from 1 to 60.

Wikipedia page

The hostess conveniently tells you the average amount you need per remaining machine to reach your game. After hours of scribbling, I can’t come up with anything better than setting a stopping goal of about 25% higher than the required average. That is just an educated guess, so please don’t ask me to prove it is optimal. As you noted, also ride the gaps, never stopping just before an amount that was already picked.

When there are only two machines left, if the total amount needed is £13,000 or less, I would try to get it all in the second-to-last machine. If £14,000 or more, I would try to get half of it at the next machine.

If they should bring back this show, I hope my UK readers will let me know. This is the kind of puzzle that I could become obsessed with, like the Eternity puzzle, which was coincidentally (or not) also out of the UK.

P.S. Why do you spell "colour" with a u in the UK? It makes no sense to me.

Price Is Right web site

an old one

First, let’s calculate the expected value of a prize that is not paired with a second chance. The following table shows that average is $1371.74.

Second, calculate the average prize that does have a second chance. The following table shows that average is $225.

Third, create an expected value table based on the number of second chances the player finds. This can be found using simple math. For example, the probability of 2 second chances is (4/50)×(3/49)×(46/48). The expected win given s second chances is $1371.74 + s×$225. The following table shows the probability and average win for 0 to 4 second chances.

So the average win per punch (including additional money from second chances) is $1390.89.

The following table shows my strategy of the minimum win to accept, according to the number of punches remaining. Note the player can get to $1,400 with prizes of $1,000 + $250, + $100 + $50 via three second chances.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

For the benefit of other readers, let me review the rules first.

1. A team of players starts with $1,000,000.
2. The team is given a multiple choice question.
3. The team is to divide his money among the possible answers. Whatever money is put on the correct answer will move onto the next question.
4. The team must completely rule out at least one possible answer by not putting any money on it.
5. This process repeats for several rounds. The player is also given one chance to change his mind.

Obviously, if the team is sure of the answer then he should put all his money on the correct answer. If the team can narrow down the answer to two, but assigns each a 50% chance of being correct, then they should divide his money equally between the two choices.

Where it gets more difficult is if the team leans towards one answer but doesn’t completely rule out one or more of the others. Let’s look at an example. Suppose the team determines the probability of each correct answer as follows: A 10%, B 20%, C 30%, D 40%. How should they divide up his money?

I claim the answer is to follow the Kelly Criterion. Briefly, the team should maximize the log of his wealth with every question. To do this, you have to consider how much wealth you already have.

Let’s say your existing wealth, which you have accumulated independently of the show, is $100,000. It is your first question, so you have $1,000,000 of game show money to split up. First eliminate the option with the lowest probability, to conform with the show rules. Then you want to maximize 0.2×log(100,000+b*1,000,000) + 0.3×log(100,000+c*1,000,000) + 0.4×log(100,000+d*1,000,000), where lower-case a, b, and c refer to the portion placed on each answer.

This could be solved with calculus and solving a trinomial equation, trial and error, or my preference, the "goal seek" feature in Excel. Whatever you use, the right answer is to put 18.9% on B, 33.3% on C, and 47.8% on D.

Of course, nobody on the show is going to be able to do all this math in the time allowed, not to mention that you also have to move a lot of bundles of cash as well in that time. My more practical advice is to just divide up the money in proportion to your assessment of the probability of the answer’s being correct, assuming the least likely choice is not a possibility. In the example, that would cause a split of 22.2% on B, 33.3% on C, and 44.4% on D.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .