Jeopardy is a very popular game show on television. It has three parts: Jeopardy, Double Jeopardy, and Final Jeopardy. Final Jeopardy occurs at the end of the program. The three players have each accumulated a different amount of money during the earlier parts of the program. Let us say that the amounts of money they have accumulated are:
Player 1: A dollars
Player 2: B dollars
Player 3: C dollars
where A >= B >= C >= 0. (If a player has 0 dollars or a negative number of dollars, that player is not allowed to play in Final Jeopardy.)
Each of the players bets some amount of his money (in secret) and is then asked to answer a question. (All the players are asked to answer the same question, again in secret). If a player’s answer is correct, the bet is added to his accumulated money. If the answer is incorrect, the bet is subtracted from his accumulated money. The contestant with the largest amount of money wins, gets to take home his accumulated money, and comes back to play the next day. If the top two or three players are tied, they all win, go home with their money, and come back to play the next day. We assume this outcome is acceptable to all players. If the top player has 0 dollars, he does not win and does not come back the next day. The question is: how much should each contestant bet.
We recommend strategies for each of the three players in the first three sections and then summarize the strategies in Section IV. In Section V, we briefly discuss the situation from the viewpoint of mathematical Game Theory. In Section VI we make some final comments.
I. Player 1
Consider first Player 1, who has the largest amount of money.
First of all, if A = 2B, then Player 1 can bet 0. The worst thing that can happen is that Player 2 bets B and answers correctly. Then there will be a tie, and both players win. If Player 1 bets anything and answers incorrectly, he would lose if Player 2 bet B and answered correctly
If A > 2B, Player 1 can also bet 0 and win even if Player 2 bets B. However Player 1 can bet as much as A – 2B – 1 and still be guaranteed to win if Player 2 bets B and answers correctly and Player 1 answers incorrectly. We assume that Player 1 makes this bet. (After all, the goal of the game is to win as much money as possible, and if Player 1 answers correctly, he will have 2(A – B) – 1. Of course if he thinks the question will be very hard and he is likely to get it incorrect, he can bet 0)
Now suppose that A < 2B. Then Player 1’s goal should be: If Player 1 answers the question correctly, he will win no matter what the other players bet or whether or not they answer their questions correctly.
Since Player 2 might bet B and hence end up with 2B dollars, Player 1 must bet 2B – A + 1 to ensure he will win if both he and Player 2 get the correct answer.
A special case occurs when A = B (where Player 1 cannot bet 2B – A + 1). If both Player 1 and Player 2 were to bet 0, then there would be a tie and both players would win. But if Player 1 bet 0 and Player 2 bet B (or even 1) and answered the question correctly, then Player 1 would lose even if he answered correctly. That goes against the above goal and would be very embarrassing to Player 1 on national television. My guess is that Player 1 would bet A. We discuss this situation further in Section V, where we note that Game Theory suggests that both players would bet 0.
Summary of Players 1’s Betting Strategy
if A > 2B then bet A – 2B – 1 else
if A = 2B then bet 0 else
if A = B then bet A else
bet 2B – A + 1;
Examples:
A = 10, B = 8, and C = 0. Since Player 2 might bet 8 and, if he answers correctly, have 16, Player 1 must to bet 7, so that if he answers correctly, he will have 17 and hence win.
A = 10, B = 5, and C = 0. Player 1 will bet 0, and Player 2 will bet 5. If Player 2 gets the correct answer, Player 1 and Player 2 will be tied and both will win. If Player 2 gets the incorrect answer, Player 1 will win.
A = 10, B = 3, and C = 0. A > 2B, so Player 1 could bet 0. But since, even if Player 2 bets 3 and answers correctly he will have only 6, Player 1 can bet 3 and win even if he answers incorrectly.
II. Player 2
Assume that A <= 2B, so that Player 2 has a chance of winning. Player 2 should have two strategies, one against Player 1 and one against Player 3. Each strategy will lead to a bet. Player 2 should then bet the larger of the two amounts.
Player 2 against Player 3
If B >= 2C, then no matter what Player 3 bets, Player 3 cannot overtake Player 2, and hence Player 2 can bet 0.
If B < 2C, then Player 2’s goal should be: If Player 2 answers the question correctly, he beats Player 3 no matter what Player 3 bets and whether or not Player 3 answers his question correctly.
Since Player 3 might bet C and hence end up with 2C dollars, Player 2 must bet 2C – B + 1 to ensure he will beat Player 3 if both he and Player 3 get the correct answer. The situation is the same as with Player 1 and Player 2.
If B = C, the situation is also the same as with Player 1 and Player 2. Therefore, Player 2 should bet B.
As with Player 1, when B > 2C, Player 2 sometimes need not bet 0, but can bet some positive amount. We ignore this possibility in this and the following discussions because of its added complications (now the accumulations of all three players are involved).
Summary of Player 2’s Betting Strategy Against Player 3
if B >= 2C then bet 0 else
if B = C then bet B else
bet 2C – B + 1;
Examples:
B = 8 and C = 3. Since even if Player 3 bets 3 and answers correctly he will have only 6, Player 2 can bet 0.
B = 8 and C = 6. If Player 3 bets 6, and answers correctly, he will have 12. Hence Player 2 must bet 5, so that if he answers correctly, he will have 13.
Player 2 against Player 1
Player 2’s only chance of winning is if 2B >= A. We assume that Player 1 uses the strategy above.
If 2B = A, Player 1 bets 0. Player 2 should bet B, so that if he answers correctly, he ties and hence wins.
We do not discuss the situation where B = A, because in this situation Player 2 will think he is Player 1 and bet accordingly. (He will bet B.)
If 2B > A. Player 1 bets 2B – A + 1. Player 2 can win only if Player 1 answers incorrectly. If Player 1 answers incorrectly, he will have
A – (2B – A + 1) = 2(A – B) – 1
Suppose B is greater than or equal to this accumulation
B >= 2(A – B) – 1
Solving for B
B >= (2A – 1) / 3
In this case Player 2 can bet 0.
if B >= (2A – 1) / 3 then bet 0
Example:
A = 10 and B = 7. If Player 2 were to bet 7 and answer correctly, he would have 14, so Player 1 must bet 5. If Player 1 answers incorrectly, he will have 5. Hence Player 2 can bet 0 and beat Player 1 (assuming Player 1 answers incorrectly) even if Player 2 answers incorrectly.
Suppose B is not larger than the accumulation Player 1 will have if he answers the question incorrectly. In this case, Player 2’s goal should be: If Player 1 answers the question incorrectly and Player 2 answers the question correctly, Player 2 will win.
Player 2 should bet enough so that his accumulation will be larger than that of Player 1, assuming Player 1 answers incorrectly. If Player 1 answers his question incorrectly, he will have 2(A – B) – 1
Therefore Player 2 must bet.
(2(A – B) – 1) – B + 1 = 2A – 3B
In order to execute this goal, B must be larger enough to make this bet. It must be true that In other words, it must be true that 4B >= 2A. But we have already assumed that to be true, or else Player 2 could not possibly win
Example:
A = 10 and B = 6. If Player 2 were to bet 6 and answer correctly, he would have 12. Therefore Player 1 must bet 3. If Player 1 answers incorrectly, he will have 7. Therefore Player 2 must bet 2, so that if he answers correctly, he will have 8.
Summary of Player 2’s Betting Strategy against Player 1
if B >= (2A – 1) / 3 then bet 0 else
if 2B = A then bet B else
if 2B > A then bet 2A – 3B;
Putting everything together
Summary of Player 2’s Betting Strategy
Execute each of the following if statements, and make the larger of the two bets
if B >= (2A – 1)/3 then bet 0 else
if 2B = A then bet B else
if 2B > A then bet 2A – 3B;
if B >= 2C then bet 0 else
if B = C then bet B else
bet 2C – B + 1;
III. Player 3
Player 3 can win only if Player 1 and Player 2 both answer their questions incorrectly (or if A = B = C). As before, we consider the betting strategies against the two other players separately, and select the larger of the two bets, but as we shall see, the betting strategies are connected in interesting ways.
Player 3 against Player 1
First consider the special situation where A = B. In that case we recommended that both Player 1 and Player 2 bet B. Of course if A = B = C, Player 3 will think he is Player 1 and bet C. If A = B, but C is not equal to A, then if Player 3 bets 0, he will win if both Player 1 and Player 2 answer incorrectly.
Assuming A is not equal to B, then order to guard against Player 2 betting his entire accumulation of B, Player 1 bets 2B – A + 1. If Player 1 answers his question incorrectly, his accumulation will be 2(A – B) – 1.
If C >=2(A – B) – 1, then Player 3 can bet 0 and win if Player 1 answers incorrectly. Otherwise he must bet 2(A – B) – C. In order to execute this strategy, C must be larger enough to make this bet. It must be true that C >= 2(A – B) – C. In other words, it must be true that C >= (A – B). If that is not true, Player 3 cannot exceed the accumulation Player 1 has if he answers incorrectly.
We need not consider the special case where: 2C = A In that case there are two possibilities:
1. C = B, in which case Player 3’s bet is C based on his strategy against Player 2 discussed below
2. C <> A / 2 and hence Player 1 is forced to bet 2B – A + 1 (or B). Thus Player 3 can bet the amount in the preceding paragraph against Player 1.
Summary of Player 3’s Betting Strategy against Player 1
if C >= 2(A – B) – 1 then bet 0 else
if C > (A – B) then bet 2(A – B) – C;
Note that if A = B, then it follows that C >= 2(A – B) -1, so we do not need a separate test for A = B.
Example:
A = 10, B = 8, and C = 6. Player 1 must bet 7. If Player 1 answers incorrectly, he will have 3. Player 3 can bet 0 (against Player 1). We complete this example later.
Player 3 against Player 2
As we said before, Player 2 selects his bet based on the threat from Player 3 and the hope that Player 1 will answer his question incorrectly. Player 2 selects the larger of the two bets. Player 3’s strategy is based on both of Player 2’s possible strategies.
Player 3’s Response to Player 2’s Betting Strategy Against Player 3
As we said before, if A = B, we recommend that Player 1 and Player 2 both bet B, and (if C is not equal to A) that C bet 0. Player 2 did not select his strategy based on Player 3, but Player 3 will win if both Player 1 and Player 2 give the incorrect answer.
Assuming that A is not equal to B, then if 2C > B, Player 3 might bet C, and then Player 2 must bet 2C – B + 1. This is exactly the situation that Player 2 faced against Player 1. Player 3 can use the same strategy.
Summary of Player 3’s Betting Strategy in Response to Player 2’s Betting Strategy Against Player 3
if (A = B) OR (C >= (2B – 1) / 3) then bet 0 else
if (C = B) OR (2C = B) then bet C else
if 2C > B then bet 2B – 3C;
Examples:
A = 10, B = 8, and C = 6. Player 2 must bet 5 (to guard against Player 3 betting 6), and if he answers incorrectly, he will have 3. So Player 3 can bet 0 against Player 2. This completes the previous example, where A = 10, B = 8, and C = 6. Player 1 must bet 7 (against Player 2), so Player 3 can bet 0 (against Player 1). Hence Player 3 can bet 0 against both players and win if they both answer incorrectly.
A = 20, B = 16 and C = 9. Player 1 must bet 13 (to guard against Player 2 betting 16) and if he answers incorrectly, he will have 7. Player 2 must bet 3 (to guard against Player 3 betting 9), and if he answers incorrectly, he will have 13. Player 3 must bet 5 against Player 2 and can bet 0 against Player 1. Thus Player 3 wins if he bets 5, answers correctly, and the other two players answer incorrectly.
Player 3’s Response to Player 2’s Betting Strategy Against Player 1
As we said before, Player 2 might bet either B, 2A – 3B, or 0 against Player 1
If Player 2 bets B, we have already said that Player 3 should bet 0 (or C if A = B = C)
If Player 2 bets 2A – 3B, it is because he is trying to exceed the amount Player 1 would have if he answered his question incorrectly. So if Player 2 answers his question incorrectly, his final accumulation will be less than the accumulation Player 1 had when he answered his question incorrectly. But Player 3’s strategy against Player 1 already has gotten his accumulation greater than Player 1’s and hence greater than Player 2’s (assuming Player 3 answers correctly). Thus Player 3 can ignore this bet from Player 2
If Player 2 bets 0 (based on the strategy described above) and answers his question incorrectly his remaining accumulations would be B. But if C > B / 2, then Player 2 would bet 2C – B + 1 to guard against Player 3 betting C. We have already discussed this situation under “Player 3’s Response to Player 2’s Betting Strategy Against Player 3” and again Player 3 can ignore this bet from Player 2.
Therefore, we can ignore here any possible bet that Player 3 might make in response to Player 2’s betting strategy against Player 1.
Summary of Player 3’s Betting Strategy
Execute each of the following if statements, and make the larger of the two bets
if C >= 2(A – B) – 1 then bet 0 else
if C > (A – B) then bet 2(A – B) – C;
if (A = B) OR (C >= (2B – 1) / 3) then bet 0 else
if (C = B) OR (2C =B) then bet C else
if 2C > B then bet 2B – 3C;
IV. Summary
Summary of Players 1’s Betting Strategy
if A > 2B then bet A – 2B – 1 else
if A = 2B then bet 0 else
if A = B then bet A else
bet 2B – A + 1;
Summary of Player 2’s Betting Strategy
Execute each of the following if statements, and make the larger of the two bets
if B >= (2A – 1)/3 then bet 0 else
if 2B = A then bet B else
if 2B > A then bet 2A – 3B;
if B >= 2C then bet 0 else
if B = C then bet B else
bet 2C – B + 1;
Summary of Player 3’s Betting Strategy
Execute each of the following if statements, and make the larger of the two bets
if C >= 2(A – B) – 1 then bet 0 else
if C > (A – B) then bet 2(A – B) – C;
if (A = B) OR (C > 2(B – 1) / 3) then bet 0 else
if (C = B) OR (2C = B) then bet C else
if 2C > B then bet 2B – 3C;
If the betting strategy for some player is that he bets 0, then he is guaranteed to win (or tie) if the higher numbered players answer incorrectly.
If there is no bet specified for some player, then that player cannot possibly win even if the other players answer incorrectly.
V. Game Theory
The problem is more complicated than we have explained so far. For example, assume that A = 10, B = 7, and C = 0 (Player 3 is out of the game). According to our previous analysis, since Player 2 might bet 7 and if he wins, have 14, Player 1 should bet 5. Then since, if Player 1 answers incorrectly, he will have 5, Player 2 can bet 0 and hence win whenever Player 1 answers incorrectly.
But suppose Player 1 knew with 100% certainty that Player 2 would bet 0. Then Player 1 could bet 0 and win whether or not he answered correctly.
But if Player 2 knew with 100% certainty that Player 1 would bet 0, he would have to bet at least 3 to win if he answered correctly. But if he answered incorrectly, he would have 4. But 4 is less than the amount Player 1 would have had if he had made his original bet of 5 and answered incorrectly. Thus Player 2 loses in this situation, but would have won if he had bet 0. And if Player 1 had bet 0 and Player 2 had bet 7 and both had answered correctly, Player 1 would lose but would have won if he had made his original bet of 5.
And on and on.
So what are the players to do? There is a field of mathematics called Game Theory that deals with such situations. I do not know enough about Game Theory to discuss in detail what solution it would suggest. But I would guess that it would involve each player selecting his bet based on some probability distribution, with there being some probability that he would win and another probability that he would lose even if he answered correctly. (Note that neither player would know with 100% certainty what the other player would bet.)
My personal view is that no real Jeopardy player would adopt such a strategy. For example, Player 1 would not want to use any strategy in which there is some possibility that he answers the question correctly, but loses to Player 2. That would be too embarrassing on national television. Therefore, I believe that real players will use the strategies outlined in the previous sections.
A particularly interesting situation occurs when A = B. Suppose A = 10, B = 10, and C = 0. If both Player 1 and Player 2 bet 0, then they both win no matter whether they answer correctly or incorrectly. This is clearly the best outcome for both players. If either player makes any other bet and answers incorrectly, he loses (unless the other player makes the same bet and also answers incorrectly). This situation is called an equilibrium in Game Theory, and the assumption is that both players would bet the equilibrium bet---in this case, 0. As we said before, however, the question is, would a real player make such a bet on national television, where it is possible that he answers correctly and still loses (because the other player did not use that strategy, made some bet and answered correctly). Again, my opinion is that real players would not bet 0, but would bet B so that they are guaranteed not to lose if they answer correctly.
I have discussed this situation with my friend and colleague who is a game theorist. When I suggested that I thought Player 1 would always bet the high amount suggested by my goal because he didn't want to be embarrassed on national television if he answered correctly but lost, my friend replied.
If not getting embarrassed is all important, we call that changing Player 1's utility function. Then Game Theory says he should make the high bet almost all the time and we can forget the small probability he should make the low bet. Making the low bet might put too much stress on his heart.
VI. Some Final Comments
The situation is even more complicated. Before the players make their bets for Final Jeopardy, the category of the question is revealed: “American Presidents,” “Ancient Rome,” etc. Each player might try to assess the probability that he will answer a question in that category correctly and then use that probability to refine his bet. We ignore this possibility, again because I think the players would always want to follow the goals given above.
You might think that we are spending too much time on situations like A = B or A = 2B that are very unlikely to occur. But that’s the way computer scientists think. Computers have to act correctly in all situations, even those that are very unlikely. And, in fact, these situations have occurred a surprising number of times on Jeopardy.
There are many Web sites that also address the issue of betting in Final Jeopardy. I particularly recommend
http://www.j-archive.com/help.php#faithlove, which contains pointers to many other sites, including a “wagering calculator” that suggests bets for particular values of A, B, and C. Sometimes those suggested wagers are the same as recommended here. Sometimes they are different.
