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. 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. December 23, 2008

This is a follow up on Deal or No Deal, which I watched for the first time recently. Your analysis assumes that the house doesn't know the value of the money in the suitcase. However, in the show I watched, in the endgame both contestants had selected a valuable case, and both were offered (or would have been offered, as one had already quit) above expected value (EV) deals. In the most extreme case, a player "would have been" offered $687K when the two dollar amounts left were $500K and $750K. The only rational explanation for this is that the banker knows the value of the player's suitcase and the deals offered are based on that.

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:

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. August 29, 2008

In a two-player case, what is optimal strategy in Final Jeoporday? — 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 it 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 by 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 guarantee victory if B answers incorrectly). Let's restrict B's options to betting all or nothing.

At first glance, it would seem the right thing to do for player A is 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 bet 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
Player A	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
Player A	High	Low	Total
High	0.385325	0.162677	0.548002
Low	0.31782	0.134178	0.451998
Total	0.703145	0.296855	1

So, if both players follow optimal strategy, the probability of player A winning is 0.385325×0.799267 + 0.162677×0.438000 + 0.31782×0.562000 + 0.134178×1 = 0.692023. Any deviation from this from either player will cause his probability of winning to decrease. December 14, 2007

Time for another Deal or No Deal question. Let's say after all the deals from the banker and guest appearances by Celine Dion, you're left with two suitcases, the $500,000 and the $1,000,000. The banker's offer will be slightly less than $750,000 I assume. Which would you choose? What if the two briefcase left were the $.01 and $1,000,000 one? I guess it's all a matter if you're a gambler or not, and nothing really to do with odds. The reason why I'm asking is I wonder if ANYBODY will ever win $1,000,000 (even if they've picked the magic briefcase). – 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. August 9, 2006

In your April 5, 2006 column you state that if there are only two cases left in Deal or no Deal and the million dollars is still in play then the probability my case has the million is 50-50. I disagree. Isn't this just a variation of the Monty Hall problem? That is, the million is more likely to be on the stage than in his case? -- 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 goat. 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. May 18, 2006

at the start of "deal or no deal" the odds of picking the 1,000,000 dollar case is 1 in 26. after eliminating all the cases exept 1, what are the odds that my case contains the million dollars. is it 50-50 or still 1 in 26 ? - Ken from chester,ny

50-50 April 5, 2006

Watching "Deal or no Deal". I realize the "offer" from the banker is just the remaining values of the cases divided by the number of cases [give or take rounding]. Is there ANY strategy to this game at all, or is "the deal" always just an OK thing to take? Does it depend on how many cases you have to open or anything? - 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. Jan. 14, 2006

My question is regarding the game show, Deal or No Deal, very popular in Australia and about to come to England. The contestant is shown twenty-six numbered briefcases, each containing a hidden amount of money, ranging from 50 cents to $200,000 as below.

$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? Thanks, - 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
1,000
5,000
10,000
25,000
50,000
75,000
100,000
200,000
300,000
400,000
500,000
750,000
1,000,000

Here is the flow of the game:

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

I have watched three games in their entirety. The following table shows the remaining prizes, average amount, and banker offer of what I will call "game 1".

Deal or No Deal Game 1

Initial Stage 1 Stage 2 Stage 3 Stage 4 Stage 5 Stage 6 Stage 7

$0.01 $0.01 $0.01

$1 $1 $1

$5 $5 $5 $5 $5

$10 $10 $10 $10 $10 $10 $10 $10

$25 $25 $25 $25

$50 $50

$75 $75 $75 $75 $75 $75 $75 $75

$100 $100 $100

$200 $200 $200 $200 $200

$300

$400 $400 $400 $400

$500 $500 $500 $500 $500 $500 $500

$750

$1000 $1000

$5000 $5000 $5000 $5000 $5000 $5000 $5000 $5000

$10000

$25000 $25000

$50000

$75000 $75000

$100000

$200000

$300000 $300000 $300000 $300000

$400000 $400000 $400000

$500000 $500000 $500000 $500000 $500000 $500000

$750000 $750000

$1000000 $1000000 $1000000 $1000000 $1000000 $1000000 $1000000 $1000000

Expected Value $152868 $147088 $164201 $188224 $250931 $201117 $251271

Offer $13000 $27000 $48000 $76000 $124000 $121000 $201000

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.

Deal or No Deal Game 2

Initial Stage 1 Stage 2 Stage 3 Stage 4 Stage 5 Stage 6 Stage 7 Stage 8 Stage 9

$0.01

$1 $1

$5 $5 $5 $5 $5 $5 $5 $5 $5

$10 $10

$25

$50 $50 $50 $50 $50 $50 $50 $50 $50 $50

$75 $75 $75

$100 $100 $100 $100 $100

$200 $200 $200 $200 $200 $200 $200 $200

$300 $300 $300 $300

$400

$500

$750 $750

$1000 $1000 $1000

$5000 $5000 $5000 $5000 $5000 $5000 $5000

$10000 $10000 $10000 $10000 $10000 $10000

$25000 $25000 $25000

$50000 $50000 $50000 $50000

$75000 $75000 $75000

$100000 $100000

$200000 $200000 $200000 $200000 $200000 $200000 $200000 $200000 $200000 $200000

$300000

$400000 $400000 $400000 $400000

$500000 $500000 $500000 $500000 $500000

$750000

$1000000 $1000000

Expected Value $118375 $84449 $105969 $89419 $35876 $41051 $50064 $66685 $100025

Offer $11000 $19000 $37000 $32000 $25000 $37000 $50000 $67000 $99000

Deal or No Deal Game 3

Initial Stage 1 Stage 2 Stage 3 Stage 4 Stage 5 Stage 6 Stage 7 Stage 8

$0.01

$1 $1

$5 $5 $5

$10 $10 $10

$25 $25 $25 $25 $25 $25

$50 $50 $50 $50 $50 $50 $50 $50 $50

$75 $75 $75 $75

$100 $100 $100 $100 $100

$200 $200 $200 $200 $200 $200 $200 $200 $200

$300 $300 $300 $300 $300 $300 $300

$400 $400

$500

$750

$1000 $1000 $1000 $1000

$5000 $5000 $5000

$10000

$25000 $25000 $25000 $25000 $25000

$50000 $50000

$75000

$100000 $100000

$200000 $200000

$300000

$400000 $400000 $400000

$500000 $500000 $500000 $500000 $500000 $500000 $500000 $500000

$750000 $750000 $750000 $750000

$1000000 $1000000 $1000000 $1000000 $1000000 $1000000 $1000000 $1000000 $1000000

Expected Value $151608 $178784 $206977 $190709 $250096 $300110 $375063 $333417

Offer $9000 $28000 $58000 $75000 $113000 $199000 $275000 $267000

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 $9,000 to $13,000 in 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, they player should keep refusing banker offers until the offer 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, 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). You can play Deal or No Deal at NBC.com. Dec. 26, 2005

I disagree with your answer to the Monty Hall question in the November 19, 2004 column. Assuming the car is behind door one there are actually four possibilities as follows, where the prize is behind door 1.

Player picks door 1 --> shown 2 --> switch to 3, lose
Player picks door 1 --> shown 3 --> switch to 2, lose
Player picks door 2 --> shown 3 --> switch to 1, win
Player picks door 3 --> shown 2 --> switch to 1, win

As you can see the probability of winning is 50% whether you switch or not. Furthermore it just goes against common sense that switching would be better.

Your mistake is assuming that each of these events has a 25% possibility. Following is the correct probability of each event.

Player picks door 1 (1/3) * shown 2 (1/2) = player loses (1/6)

Player picks door 1 (1/3) * shown 3 (1/2) = player loses (1/6)

Player picks door 2 (1/3) * shown 3 (1/1) = player wins (1/3)

Player picks door 3 (1/3) * shown 2 (1/1) = player wins (1/3)

So, losing events have a total probability of 2*(1/6) = 1/3 and winning events have a total probability of 2*(1/3)=2/3. Dec. 5, 2004

On a game show there are three doors. Behind one door is a new car and behind the other two are goats. Every time the game is played the contestant first picks a door. Then the host will open one of the other two doors and always reveals a goat. Then the host gives the player the option to switch to the other unopened door. Should the player switch?

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 door the player picks, he always can reveal a goat behind another door. The reason there is much debate and misunderstanding about this problem is because those people who are too young to remember the show don't know that the host, Monty Hall, did this every time for years. This problem has become known as the Monty Hall paradox.
Let's assume that the prize is behind door 1. Following are what would happen if the player 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. Therefore, the player should definitely switch. Nov. 19, 2004

What is the optimal strategy for the Plinko game on the Price is Right?

From left to right the prizes are $100, $500, $1,000, $0, $10,000, $0, $1,000, $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 or deny my hypothesis I did a search and found that there are lots of links devoted to the study of Plinko. 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 $2,557.91, on either side of the middle is $2,265.92, and tapering off as you move away further from the center. Sept. 14, 2004

It's not exactly gambling but I always wondered on the Price is Right gameshow what the best strategy to take when spinning the big wheel when you are not last to spin. Assuming you can't control your spin (completely random outcome), 5 cent increments from $.05 to $1.00, you get one spin or two spins added together, you can't go over 1.00. At what amount should you not take your second spin so that you can have the best chance to beat the player who spins after you? - 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.

The third player should spin again if his score is less than the current highest score. If his first spin ties the highest score he should spin again if the tie is at 45 cents or less. If the tie is at 50 cents, he has an equal chance of winning by spinning again or not. If there is a three way tie after the first spin of the third player, he should spin again if the tie is at 65 cents or less. April 11, 2004

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