# TV Game Shows - FAQ

Tony C.

The first player should spin again if his first spin is 65 cents or less.

If any of the following conditions are true the second player should spin again.

- His score is less than the first player’s score.
- His score is 50 cents or less.
- His score is 65 cents or less
**and**he has tied the first player.

__Plinko__game on the Price is Right?

"Anonymous" .

From left to right the prizes are $100, $500, $1000, $0, $10000, $0, $1000, $500, $100. I would need to know the exact configuration of pegs on the board to do a perfect analysis but just eyeballing the board (see link above) I strongly feel the player should drop the puck directly over the $10,000 prize. Although it is bordered by two zeros all other prizes pale in comparison to the top prize. So the player’s strategy should be to maximize the probability of the top prize by dropping it directly above. To confirm to deny my hypothesis I did a search and there are lots of links devoted to the study of this game. This (www.amstat.org/publications/jse/v9n3/biesterfeld.html) is one of the better ones, which agrees with my conclusion. It states in part that the expected value of dropping the puck in the middle is $2557.91, on either side of the middle is $2265.92, and tapering off as you move away further from the center.

*Let’s Make a Deal*, there are three doors. For the sake of example, let’s say that two doors reveal a goat, and one reveals a new car. The host, Monty Hall, picks two contestants to pick a door. Every time Monty opens a door first that reveals a goat. Let’s say this time it belonged to the first contestant. Although Monty never actually did this, what if Monty offered the other contestant a chance to switch doors at this point, to the other unopened door. Should he switch?

"Anonymous" .

Yes! The key to this problem is that the host is predestined to open a door with a goat. He knows which door has the car, so regardless of which doors the players pick, he always can reveal a goat first. The question is known as the "Monty Hall Paradox." Much of the confussion about it is because often when the question is framed, it is incorrectly not made clear the host knows where the car is, and always reveals a goat first. I think put some of the blame on Marilyn Vos Savant, who framed the question badly in her column. Let’s assume that the prize is behind door 1. Following are what would happen if the player (the second contestant) had a strategy of not switching.

- 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.

Derek from Boston

I'm very familiar with this problem. I address it on my web site of math problems, problem number 6. There I address the general case, including not looking in the first envelope at all. However to answer your question we can not ignore the venue of where the game is taking place. You said it was a "game show." On most game shows $50,000 is a nice win. Few contestants on the Price is Right ever make it that high. I would guess that fewer than 50% of players on Who Wants to be a Millionaire get that high. Meanwhile wins of $25,000 are not unusual on game shows. Cars are won routinely on the Price is Right, which have values of about $25,000. The $32,000 level is a common win on Who Wants to be a Millionaire. The average win on Jeopardy per show is roughly $25,000. The great Ken Jennings averaged only $34,091 over his 74 wins. So, my point is that $50,000 is a nice win for a game show, and $100,000 wins are seen much less often that $25,000. Thus as a game show connoisseur it is my opinion that the other envelope is more likely to have $25,000 than $100,000. So I say in your example it is better to keep the $50,000. It also goes to show you can never assume the chances that the other envelope has half as much or twice as much are exactly 50/50. Once you see the amount and put it in the context of the venue it is being played you can make an intelligent decision on switching, which throws the 1.25x argument out the window.

- $0.50
- $1
- $2
- $5
- $10
- $25
- $50
- $75
- $100
- $150
- $250
- $500
- $750
- $1,000
- $1,500
- $2,000
- $3,000
- $5,000
- $7,500
- $10,000
- $15,000
- $30,000
- $50,000
- $75,000
- $100,000
- $200,000

The contestant selects one of the briefcases to be THEIR suitcase. Through a process of elimination, by opening the other suitcases, they try and work out how much money is their case, or whether it would be wiser to take a "Bank offer". The Bank Offers are based on, but not equivalent to, the arithmetic mean of the remaining briefcases. So, if there are mainly large valued briefcases remaining, there is a high chance that the contestant’s briefcase is valuable, and so the Bank Offer will be generous. Conversely, if the player has been less fortunate and opened the more valuable briefcases, then the Bank Offer will be low. What would be the best strategy to employ if you were a contestant on this game? A non-mathematical gut instinct strategy would be ignore the Bank offers and carry on opening cases until either the $200,000 was opened and eliminated, or both the $100,000 and $75,000 were opened and eliminated. What’s the math behind this game, Wizard?

Jacqui from Birmingham, England

Deal or No Deal just started here in the U.S.. The rules sound the same except our prizes go up to a million dollars as follows.

- 0.01
- 1
- 5
- 10
- 25
- 50
- 75
- 100
- 200
- 300
- 400
- 500
- 750
- 1000
- 5000
- 10000
- 25000
- 50000
- 75000
- 100000
- 200000
- 300000
- 400000
- 500000
- 750000
- 1000000

Here is the flow of the game:

- Player picks one case for himself
- Player opens up six of the remaining 25 cases.
- Banker makes an offer.
- If player declines he opens five more of the 19 remaining cases.
- Banker makes an offer.
- If player declines he opens four more of the 14 remaining cases.
- Banker makes an offer.
- If player declines he opens three more of the 10 remaining cases.
- Banker makes an offer.
- If player declines he opens two more of the 7 remaining cases.
- Banker makes an offer.
- If player declines he opens one more of the remaining cases.
- 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.

Next are the same table and charts for games 2 and 3.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 watch Deal or No Deal at NBC.com.

Archive of past shows.

Darren from Elk Grove, CA

As my December 26, 2005 column shows the banker offer is usually much less than the average of the remaining cases. However, hypothetically, if it always were, then every strategy would have the same expected value. The player would be indifferent at every offer.

Ken from Chester, NY

50-50

Jason from Pasadena, CA

No. I’m getting lots of people arguing with me about this one. Many writers claim that probabilities can not change if additional information is introduced. So if the probability starts at 1 in 26 then it must stay there. Contrary to what betting system salesmen say, probabilities indeed can change as additional information is introduced. I don’t want to try to teach basic probability here but any college level math book on conditional probability or Bayes’ Theorem should cover this topic nicely.

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.

### Let’s Make a Deal

Player Chooses |
Car | Curtain Opened |
Probability | Win by Switching |

1 | 1 | 1 | 0% | n/a |

1 | 1 | 2 | 5.56% | N |

1 | 1 | 3 | 5.56% | N |

1 | 2 | 1 | 0% | n/a |

1 | 2 | 2 | 0% | n/a |

1 | 2 | 3 | 11.11% | Y |

1 | 3 | 1 | 0% | n/a |

1 | 3 | 2 | 11.11% | Y |

1 | 3 | 3 | 0% | n/a |

2 | 1 | 1 | 0% | n/a |

2 | 1 | 2 | 0% | n/a |

2 | 1 | 3 | 11.11% | Y |

2 | 2 | 1 | 5.56% | N |

2 | 2 | 2 | 0% | n/a |

2 | 2 | 3 | 5.56% | N |

2 | 3 | 1 | 11.11% | Y |

2 | 3 | 2 | 0% | n/a |

2 | 3 | 3 | 0% | n/a |

3 | 1 | 1 | 0% | n/a |

3 | 1 | 2 | 11.11% | Y |

3 | 1 | 3 | 0% | n/a |

3 | 2 | 1 | 11.11% | Y |

3 | 2 | 2 | 0% | n/a |

3 | 2 | 3 | 0% | n/a |

3 | 3 | 1 | 5.56% | N |

3 | 3 | 2 | 5.56% | N |

3 | 3 | 3 | 0% | n/a |

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.

### Deal or No Deal

Player Chooses |
Million $ | Case Opened |
Probability | Win by Switching |

1 | 1 | 1 | 0% | n/a |

1 | 1 | 2 | 5.56% | N |

1 | 1 | 3 | 5.56% | N |

1 | 2 | 1 | 0% | n/a |

1 | 2 | 2 | 5.56% | Hopeless |

1 | 2 | 3 | 5.56% | Y |

1 | 3 | 1 | 0% | n/a |

1 | 3 | 2 | 5.56% | Y |

1 | 3 | 3 | 5.56% | Hopeless |

2 | 1 | 1 | 5.56% | Hopeless |

2 | 1 | 2 | 0% | n/a |

2 | 1 | 3 | 5.56% | Y |

2 | 2 | 1 | 5.56% | N |

2 | 2 | 2 | 0% | n/a |

2 | 2 | 3 | 5.56% | N |

2 | 3 | 1 | 5.56% | Y |

2 | 3 | 2 | 0% | n/a |

2 | 3 | 3 | 5.56% | Hopeless |

3 | 1 | 1 | 5.56% | Hopeless |

3 | 1 | 2 | 5.56% | Y |

3 | 1 | 3 | 0% | n/a |

3 | 2 | 1 | 5.56% | Y |

3 | 2 | 2 | 5.56% | Hopeless |

3 | 2 | 3 | 0% | n/a |

3 | 3 | 1 | 5.56% | N |

3 | 3 | 2 | 5.56% | N |

3 | 3 | 3 | 0% | n/a |

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.

Jason from Vancouver

When the prizes become life-changing amounts, the wise player should play conservatively at the expense of maximizing expected value. A good strategy should be to maximize expected happiness. A good function to measure happiness I think is the log of your total wealth. Let’s take a person with existing wealth of $100,000 who is presented with two cases of $0.01 and $1,000,000. By taking “no deal” the expected happiness is 0.5*log($100,000.01) + 0.5*log($1,100,000) = 5.520696. Let b be the bank offer where the player is indifferent to taking it.

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.

Eliot from Santa Barbara

First, for the sake of simplicity, I’m going to assume that for any given question, all players have the same probability of producing the correct answer. Also, questions range in difficulty, and this probability itself is random. Thus, the results among players will be correlated.

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(no longer available) 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:

p_{0} + p_{1} + p_{2} + p_{3} = 1

Substituting the known values for p_{0} and p_{3}:

0.2486 + p_{1} + p_{2} + 0.1492 = 1.

p_{1} + p_{2} = 0.6022.

(1) p_{1} = 0.6022 - p_{2}.

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

p_{0}×0 + p_{1}×(1/3) + p_{2}×(2/3) + p_{3}×1 = 0.4380.

p_{1}×(1/3) + p_{2}×(2/3) + 0.1492 = 0.4380.

p_{1}×(1/3) + p_{2}×(2/3) = 0.2888.

Multiplying both sides by 3:

(2) p_{1} + p_{2}×2 = 0.8664.

Substituting the value for p_{1} in equation (1) we get

0.6022 - p_{2} + p_{2}×2 = 0.8664.

p_{2} = 0.2642

Solving for p_{1}

p_{1} = 0.6022 - p_{2}.

p_{1} = 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:

p_{3} + (1/3)×0.2642 = 0.1492 + 0.2642×(1/3) = 0.2373.

The probability one gets it right is:

(2/3)×p_{2} + (1/3)×p_{1} = (2/3)×0.2642× + (1/3)×0.3380 = 0.4015.

The probability neither gets it right is:

(1/3)×p_{1} + p_{0} = 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.

### Probability Player A Wins

Player A | Player B | |

High | Low | |

High | 0.7993 | 0.438 |

Low | 0.562 | 1 |

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.

### Strategy Probabilities

Player A | Player B | ||

High | Low | Total | |

High | 0.385325 | 0.162677 | 0.548002 |

Low | 0.31782 | 0.134178 | 0.451998 |

Total | 0.703145 | 0.296855 | 1 |

Just my two cents, and no reply is necessary.

J.N.S. from Bellevue, WA

Thanks for not expecting a reply, but I usually do reply to game show questions. They claim in every episode that the amounts in the cases are randomly placed, and that neither Howie, nor the banker, know the results. This was never claimed on Let’s Make a Deal, where Monty Hall obviously did know. I too have seen the banker offer more than expected value as the last offer, especially when large amounts are involved. In my strong opinion, this is not because the banker knows what is inside the player’s case. In the 1950s there was a huge scandal when it became known that the show 21, as well as others, were fixed. There is no compelling reason to ruin a successful show, and the integrity of all game shows, to skim some prize money via the bank offers.

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

- 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.
- 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.

- 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**:

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.

Player A: $10,000

Player B: $8,000

Player C: $3,500

Eliot from Santa Barbara, CA

Let me start by making some assumptions. First, I’m going to assume that the three players have no prior knowledge of betting behavior in Final Jeopardy, except the probabilities of being correct in the table presented. Second, I’m going to assume that knowing the category is of no help. Third, I’m also going to assume that all three contestants want to go for the win, not wishing to take another player along in a tie.

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(no longer available). Here is what the results look like, where player A is the leader, followed by player B, and C in last.

### Possible Outcomes in Final Jeopardy

Player A | Player B | Player C | Probability |

Right | Right | Right | 21.09% |

Right | Right | Wrong | 9.73% |

Right | Wrong | Right | 10.27% |

Wrong | Right | Right | 8.74% |

Right | Wrong | Wrong | 13.33% |

Wrong | Right | Wrong | 10.27% |

Wrong | Wrong | Right | 8.63% |

Wrong | Wrong | Wrong | 17.92% |

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.

The player may yell "Stop!" at any time, and he will bank the amount showing on screen. If the player does not stop in time, and the machine runs out of money, then he banks nothing. A hostess provides statistics, such as number of machines left to pick, amount left to earn, average amount needed per machine to win, and what amounts remain in the machines.

A player can "play the gaps," in that if a run of machines have been picked, say, £4k, £5k, and £6k, a machine would be guaranteed to make it to £7,000 once it passes the £3,000 mark. My question is, what kind of strategy should a player use?

James Key from Louisville, KY

This is the kind of thing I could spend weeks analyzing. Unfortunately, I read your message almost three months after you wrote, due to a large backlog of “ask the Wizard” questions. The Wikipedia page seems to indicate that that show was a flop, and was canceled. However, it still makes for an interesting problem.

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.

Ibeatyouraces

For those not familiar with the rules, they are explained at the Price Is Right web site. Please take a moment to go there if you’re not familiar with the game, because I’m going to assume you know the rules. There are several YouTube videos of the game as well. Here is an old one, which shows a second chance, but the maximum prize at the time was $10,000 only. It is now $25,000.

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.

### Punch a Bunch Prize Distribution with no Second Chance

Prize | Number | Probability | Expected Win |

25000 | 1 | 0.021739 | 543.478261 |

10000 | 1 | 0.021739 | 217.391304 |

5000 | 3 | 0.065217 | 326.086957 |

1000 | 5 | 0.108696 | 108.695652 |

500 | 9 | 0.195652 | 97.826087 |

250 | 9 | 0.195652 | 48.913043 |

100 | 9 | 0.195652 | 19.565217 |

50 | 9 | 0.195652 | 9.782609 |

Total | 46 | 1.000000 | 1371.739130 |

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

### Punch a Bunch Prize Distribution with Second Chance

Prize | Number | Probability | Expected Win |

500 | 1 | 0.250000 | 125.000000 |

250 | 1 | 0.250000 | 62.500000 |

100 | 1 | 0.250000 | 25.000000 |

50 | 1 | 0.250000 | 12.500000 |

Total | 4 | 1.000000 | 225.000000 |

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.

### Punch a Bunch Prize Return Table

Second Chances | Probability | Average Win | Expected Win |

4 | 0.000004 | 2271.739130 | 0.009864 |

3 | 0.000200 | 2046.739130 | 0.408815 |

2 | 0.004694 | 1821.739130 | 8.551020 |

1 | 0.075102 | 1596.739130 | 119.918367 |

0 | 0.920000 | 1371.739130 | 1262.000000 |

Total | 1.000000 | 1390.888067 |

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.

### Punch a Bunch Strategy

Punches Remaining | Minimum to Stand |

3 | $5,000 |

2 | $5,000 |

1 | $1,400 |

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

Anon E. Mouse

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

- A team of players starts with $1,000,000.
- The team is given a multiple choice question.
- 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.
- The team must completely rule out at least one possible answer by not putting any money on it.
- 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.

- Players play one at a time while players yet to play are kept off stage and are given no information about how any previous player did.
- There are 100 cards, numbered 1 to 100.
- A player starts by choosing any card.
- After looking at it, the player may keep it or switch for a new card.
- All cards are dealt with replacement. In other words, old cards are put back in the deck, including after discarding.
- The player who draws the highest card wins.

Is there any positional advantage to going last in this game? What is the optimal strategy for each player?

Here is a YouTube video showing the game.

Dween

First, there is no positional advantage to acting last. Since the players are kept in a sound-proof booth while any previous players play, order doesn't matter.

Second, there must be a Nash Equilibrium to the game where a strategy to stand with a score of at least x points will be superior to any other strategy. The question is finding x.

What I did was ask myself what would be the strategy if instead of a card numbered 1 to 100, each player got a random number uniformly distributed between 0 and 1 and look for the point x where a perfect logician would be indifferent between standing and switching. With that answer, it is easy to apply the answer to a discrete distribution from 1 to 100.

I'll stop talking at this point and let my readers enjoy the problem. See the links below for the answer and solution.

Answer for a continuous distribution from 0 to 1.

Answer for a discrete distribution from 1 to 100.

For my solution, please click here (PDF).

This question was raised and discussed in my forum at Wizard of Vegas.

"Anonymous" .

For those unfamiliar with the rules, the player is given four price tags and must place them on four items. When he is done he pulls a lever which gives the number of correct matches. If the player has less than four correct, then he may rearrange the tags and try again. The player may try as many times as he can within 45 seconds.

My advice is to always submit a selection that has a chance of winning, given the previous history of selections and scores. If the first score is 0, then don't reverse two sets of two tags, but instead move everything by one position in either direction.

If you're not able to do the logic on the spot, then I spell it out for you below. To use this strategy, assign the different tags the letters A, B, C, and D. Then place them in the order shown, from left to right on the stage. Always start with ABCD. Then look up the score history below and choose the sequence of tags indicated for that score sequence.

If 0, then BCDA

If 0-0, then CDAB

If 0-0-0, then DABC (must win)

If 0-1, then BDAC

If 0-1-0, then CADB (must win)

If 0-1-1, then CDBA

If 0-1-1-0, then DCAB (must win)

If 0-2, then BADC

If 0-2-0, then DCBA (must win)

If 1, then ACDB

If 1-0, then BDCA

If 1-0-0, then CABD

If 1-0-0-1, then CBAC (must win)

If 1-1, then BDCA

If 1-1-0, then CABD

If 1-1-0-1, then CBAC (must win)

If 1-1-1, then BCAD (must win)

If 2, then ABDC

If 2-0, then BACD (must win)

If 2-1, then ACBD

If 2-1-0, then DBCA

If 2-1-1, then ADCB

If 2-1-1-0, then CBAD (must win)

The following table shows the probability of each number of total turns. The bottom right cell shows an expected number of turns of 10/3.

### Race Game

Turns | Combinations | Probability | Return |
---|---|---|---|

1 | 1 | 0.041667 | 0.041667 |

2 | 4 | 0.166667 | 0.333333 |

3 | 8 | 0.333333 | 1.000000 |

4 | 8 | 0.333333 | 1.333333 |

5 | 3 | 0.125000 | 0.625000 |

Total | 24 | 1.000000 | 3.333333 |

This question is discussed on my forum at Wizard of Vegas.

Naranjas1

As a word of explanation to other readers, let me explain what you're talking about. The Showcase Showdown is a game played on the game show The Price is Right. In the Showcase Showdown, each player takes his turn spinning a wheel which has an equal probability of stopping on every amount evenly divisible by .05 from .05 to 1.00 . If the player does not like their first spin they may spin again, adding the second spin to their first, however if they go over 1.00 they are immediately disqualified. In the event of a tie, each player will get one spin in a tie-breaker round, the highest spin wins. In the event of another tie, this process will repeat until the tie is broken.

The main purpose of the Showcase Showdown is to advance to the Showcase. However, there are also immediate cash prizes too, as follows:

- In the first round, if any player get a total of $1.00, whether in one sum or the sum of two spins, he shall win $1,000.
- In the first, and only first, tie-breaker round, if the wheel lands on $0.05 or $0.15, then the player shall win $10,000.
- In the first, and only first, tie-breaker round, if the wheel lands on $1.00, then the player shall win $25,000.

I explain the optimal strategy to the Showcase Showdown in column #101. Assuming that strategy is followed, the following table answers your questions and various other.

### Showcase Showdown Statistics

Question | Answer |
---|---|

Expected $1000 winners first round | 0.253790 |

Probability 2-player tie | 0.113854 |

Probability 3-player tie | 0.004787 |

Expected $10000 winners second round | 0.024207 |

Expected $25000 winners second round | 0.012104 |

Expected total prize money | $798.45 |

Probability any given player wins $1000 | 0.084597 |

Probability any given player wins $10000 | 0.008069 |

Probability any given player wins $25000 | 0.004035 |

The bottom row of the table shows that if you make the Showcase Showdown, without considering your order to spin, your chances of winning $25,000 is 0.004035, or 1 in 248.

This question is asked and discussed in my forum at Wizard of Vegas.

Jufo81

Let's call the former team of nine player team 1 and the one of six player team 2. The number of ways you could pick three players from team 1 and two from team 2 is combin(9,3)Ã—combin(6,2) = 1,260. The total number of ways to pick five out of 15 players is combin(15,5) = 3,003. So, the probability the first team is split 3/2 in favor of team 1 is 1,260/3,003 = 41.96%.

If that happened, then team 1 will have six players left and team 2 four players. The number of ways you could pick three players from team 1 and two from team 2 is combin(6,3)×combin(4,2) = 120. The total number of ways to pick five out of 10 players left is combin(10,5) = 252. So, the probability the second team is split 3/2 in of favor team 1, given that the first team is already split 3/2 that way, is 120/252 = 47.62%.

If the first two new teams are split 3/2, in favor of the former team 1, then the final team will be split 3/2 among the leftovers.

Thus, the answer to your question is 41.96% × 47.62% × 100% = 19.98%.

Formulas:

combin(x,y)=x!/((y!*(x-y)!)

x! = 1*2*3*...*x

This question is raised and discussed in my forum at Wizard of Vegas.

Tanko

Let me set the stage. On the June 3, 2019 James was within easy reach of breaking the record for total money won in regular games, which still stands at $2,520,700. James average win per game was much more than what he needed to break the record. So all eyes were watching on June 3 to see the record broken.

Instead, what happens is not only does James not break the record, but he loses. The winner, Emma, played a very strong strategic game as well as being good with the buzzer and simply answering correctly. She played just as James normally did. Going into Final Jeopardy the scores were:

- Emma — $26,600
- James — $23,400
- Jay — $11,000

In these situations, where second place has more than half of first place, and third place does not, it typically comes down to first and second place choosing to go high or low with their final wager. A high wager for first place is enough to lock in a win if correct. To be specific, two times the second place score less the first place score plus one dollar. That is exactly what Emma did with a wager of 2×$23,400 - $26,600 + $1 = $20,201. Most of the time, the first place player does this.

However, James didn't know what Emma would do when deciding his wager. The following table shows who would win according to what combination of wagers.

Click on image for larger version.

If Emma wagers at least $20,201 then she locks in a win if correct.

If Emma wagers low then she will win if either (a) James wagers low or (b) James wagers high and is wrong.

If James wagers high then he wins if (a) Emma goes high, Emma is wrong, and James is right, or (b) Emma goes low and James is right.

If James wagers low then he wins if Emma goes high and is wrong.

If perfect logicians were playing, both would randomize their decisions. However, rarely does the leader go low in these situations where he/she can be caught. If James anticipates Emma to go high, then he absolutely should go low. This way he doesn't have to get Final Jeopardy right to win, he just has to hope Emma blows it.

James actual bid was the correct amount to cover Jay if Jay bet everything and was right: $23,400 - 2×$11,000 - $1 = $1,399, which satisfied as a low wager for purposes of beating Emma.

If correct, James would get an extra $1,000 for coming in second, compared to third.

In conclusion, I completely reject the conspiracy theory that James threw the game. He played the right way and lost due to a combination of playing a strong competitor and what most people would call "bad luck."

**External Links**

- Jeopardy hall of fame
- James Holzhauer on Jeopardy — Discussion in my forum at Wizard of Vegas.

"Anonymous" .

Let me get a disclaimer out of the way first. The following analysis is based on statistical averages. An actual player should make mental adjustments for how well he knows the Final Jeopardy category as well estimating the opponent's chances of getting it correct.

To answer your question, I first looked at four seasons of data from the Jeopardy Archive to see the four possible combinations of the first (leading) and second place (chasing) player getting Final Jeopardy right and wrong.

### Final Jeopardy Scorecard

Leading Player | Chasing Player Correct | Chasing Player Incorrect | Total |
---|---|---|---|

Correct | 29.0% | 25.5% | 54.5% |

Incorrect | 17.7% | 27.8% | 45.5% |

Total | 46.8% | 53.2% | 100.0% |

Before going on, let's define some variables:

x = Probability leading player goes high.

y = Probability chasing player goes high.

f(x,y) = Probability of the high player winning.

Let's express f(x,y) in terms of x and y from the table above:

f(x,y) = 0.823xy + 0.545x(1-y) + 0.468(1-x)y + (1-x)(1-y)

f(x,y) = 0.810 xy - 0.455x - 0.532y + 1

To find the optimal values for x and y, let's take the derivative of f(x,y) with respect to both x and y.

f(x,y) d/dx = -0.455 + 0.810y = 0

Thus y = 0.455/0.810 = 0.562

f(x,y) d/dy = -0.532 + 0.810x = 0

Thus x = 0.523/0.810 = 0.657

So, the high player should wager high with probability 65.7% and the low player should wager high with probability 56.2%.

Based on watching, I think the high player wagers high greater than 65.7% of the time, thus if I were in second place, I would go low.

If both players following this randomizing strategy, the probability the leading player will win is 70.1%.

Putting all theory aside, if you're leading, predict what the chasing player will do and do the same. If you're chasing, predict the leading player's action and do the opposite. This strategy goes for all such tournaments.This question is raised and discussed in my forum at Wizard of Vegas.

"Anonymous" .

The reason is the Daily Doubles are placed in the bottom three rows 91.5% of the time. The following table shows their locations on the board over 13,660 Daily Doubles found.

### Daily Double Location

Row | Column 1 | Column 2 | Column 3 | Column 4 | Column 5 | Column 6 | |
---|---|---|---|---|---|---|---|

1 | 5 | - | 3 | 3 | 2 | 3 | 16 |

2 | 280 | 137 | 216 | 167 | 207 | 140 | 1,147 |

3 | 820 | 442 | 677 | 658 | 643 | 472 | 3,712 |

4 | 1,095 | 659 | 982 | 907 | 895 | 627 | 5,165 |

5 | 787 | 403 | 670 | 671 | 613 | 476 | 3,620 |

Total | 2,987 | 1,641 | 2,548 | 2,406 | 2,360 | 1,718 | 13,660 |

Source: J! Archive.

Here is the same data in the form of how often a Daily Double is found in each cell of the board.

### Daily Double Probability

Row | Column 1 | Column 2 | Column 3 | Column 4 | Column 5 | Column 6 | |
---|---|---|---|---|---|---|---|

1 | 0.0% | 0.0% | 0.0% | 0.0% | 0.0% | 0.0% | 0.1% |

2 | 2.0% | 1.0% | 1.6% | 1.2% | 1.5% | 1.0% | 8.4% |

3 | 6.0% | 3.2% | 5.0% | 4.8% | 4.7% | 3.5% | 27.2% |

4 | 8.0% | 4.8% | 7.2% | 6.6% | 6.6% | 4.6% | 37.8% |

5 | 5.8% | 3.0% | 4.9% | 4.9% | 4.5% | 3.5% | 26.5% |

Total | 21.9% | 12.0% | 18.7% | 17.6% | 17.3% | 12.6% | 100.0% |

The reason for searching for Daily Doubles is they are a good way to double your score. Most contestants will have a probability of about 80% to 90% of getting any given clue right. It is a great value to get even money on a wager you have an 80% to 90% chance of winning. A major reason James Holtzhauer won as much as he did was aggressive Daily Double searching and then going "all in" most of the time when he found them. It is also how he lost against Emma, when she employed the same strategy against him.

"Anonymous" .

For the benefit of readers who aren't familiar with the game, here is a video of it.

I submit that the following strategy results in a minimum average number of turns. There are many strategies that would tie it, but I don't think any can beat it.

To use the strategy, label the four price tags as 1, 2, 3, and 4. Place them on the four prizes according to the history of how many you got right in the past, starting with the first turn on the left.

### Race Game Strategy

History | Prize 1 | Prize 2 | Prize 3 | Prize 4 |
---|---|---|---|---|

None | 1 | 2 | 3 | 4 |

0 | 2 | 1 | 4 | 3 |

0,0 | 3 | 4 | 2 | 1 |

0,0,0 | 4 | 3 | 1 | 2 |

0,0,2 | 3 | 4 | 1 | 2 |

0,0,2,0 | 4 | 3 | 2 | 1 |

0,2 | 2 | 3 | 4 | 1 |

0,2,0 | 4 | 1 | 2 | 3 |

0,2,1 | 2 | 4 | 1 | 3 |

0,2,1,0 | 3 | 1 | 4 | 2 |

1 | 1 | 3 | 4 | 2 |

1,0 | 2 | 4 | 3 | 1 |

1,0,0 | 3 | 1 | 2 | 4 |

1,0,0,0 | 4 | 2 | 1 | 3 |

1,1 | 1 | 4 | 2 | 3 |

1,1,0 | 2 | 3 | 1 | 4 |

1,1,0,0 | 3 | 2 | 4 | 1 |

1,1,0,0,0 | 4 | 1 | 3 | 2 |

2 | 2 | 1 | 3 | 4 |

2,0 | 1 | 2 | 4 | 3 |

2,1 | 1 | 3 | 2 | 4 |

2,1,0 | 4 | 2 | 3 | 1 |

2,1,1 | 1 | 4 | 3 | 2 |

2,1,1,0 | 3 | 2 | 1 | 4 |

The next table shows the probability it will take 1 to 5 turns of the 24 possible ways to arrange the four price tags.

### Turns Required

Turns | Number | Probability |
---|---|---|

1 | 1 | 4.17% |

2 | 4 | 16.67% |

3 | 8 | 33.33% |

4 | 9 | 37.50% |

5 | 2 | 8.33% |

Total | 24 | 100.00% |

Taking the dot product, the average number of turns needed, under this strategy, is 3.29167.

This question is asked and discussed in my forum at Wizard of Vegas.