## Saturday, March 8, 2008

### Final Jeopardy Wagering Strategy--Part 1

I glanced at a couple of web sites on this topic, but the analysis seemed so lame that I decided to tackle it without studying what anybody else has said about it.

This is an important topic in the field of game theory, which deals with situations in which one must make decisions based on what one guesses others are going to do. The Prisoner's Dilemma is of course the major problem in the field of game theory, and has been the subject of much wonderful analysis over the years.

In order to arrive at a situation simple enough that we can get a handle on it, I will make the following assumptions for the purpose of this initial discussion.

1. Let's say the lowest player has a negative score, so that there are only two players playing Final Jeopardy.

2. Let's further assume that the outcome is purely random; i.e., each player will write down "heads" or "tails", and when the music stops Alex will flip a coin to determine the right answer.

3. Let's also assume that the goal is to win the game; i.c., the opportunity to come back the next day and play again trumps the opportunity to win a few more thousand dollars in a losing cause.

With these assumptions stated, let us give Player A \$10,000, and look at different levels for Player B and discuss correct wagering strategies for each.

The lowest non-trivial amount for B would be \$5,000. Since a tie is the same as a win, i.e., if you tie both players get to come back the next day, it is correct here for A to wager zero. If he wages a dollar, and B wagers everything, A's winning chances drop to 75% given the 4 equally possible results which flow from assumption #2. By risking nothing, A is guaranteed a return visit. Similarly, by risking \$5,000, B is ensured a 25% chance of coming back, assuming A correctly wagers nothing.

Now give B \$6,000. Normal wagering strategy here for A is to risk \$2,001, guaranteeing him the win if he gets the answer, even if B risks everything. This is such a common bet here for A that I think it is safe to say B should bet based on this assumption about what A is doing. If B now bets everything, it can be easily seen that he wins only 25% of the time (when he is right and A is wrong). However, B can risk down to 2K and still ensure that same 25% chance of winning. Correct strategy then for B would be to risk between 2K & 6K, depending on how he feels about the category of the question, and it would be a gross blunder for B to risk anything less than \$1,999, since this would give him a 0% chance of winning.

Now let's give B \$7,000, and here things start to get more interesting. A's normal bet here would be \$4,001. Assuming that bet on A's part, which is a pretty safe assumption, what should B's strategy be here? Well, B has a strategy that will jump his winning expectation clear up to 50%! By betting \$1,000 or less, he wins whenever A is incorrect.

So is A wrong for making the standard \$4,001 wager? Well, if he assumes B is sophisticated enough to go through this analysis and limit his wager to the 1K, then A can adjust his wager down to 2K or less and raise his winning expectation up to 75%! Given how innumerate most people are, it is hard to say if this assumption is valid. I would say, however, that it is pretty easy for B to see that if he risks everything, he can win only 25% of the time against the normal \$4,001 wager. Surely most B's who have given it any thought (which if you're going on the show you surely would do) can then do the simple math and realize the proper bet would be the 1K or less.

Now let's give B 8K. The "normal" bet here for A now is \$6,0001. This makes B's bet in response 4K or less, giving him a 50% shot. In response, if A anticipates the 4K bet, he can limit his bet to 2K (or less) and up his winning chances to 75%.

Finally, let's give B 9K. All B has to do now to assure himself a 50% chance against A's normal \$8,001 bet is to risk 7K or less. Surely an A who gives it any thought must realize B is more likely to risk 7K or less compared to risking everything, and so A should adjust his strategy here to allow for this. A bet of 8K would give him 50% against B's 7K bet, and A would have to lower his bet dollar-for-dollar to keep it at 50% as B lowers his bets.

I invite responses on the following issues:

1) What would a general formula be for A and B's betting strategy?

2) How would the strategies change when the 3rd player is introduced?

3) How would the strategies change if you allow for variances in difficulty of the questions; i.e., assume it is not strictly 50% right and wrong, but assume that if I get the right answer, my opponent is 75% likely to get it, and similarly if I get the wrong answer, my opponent is 75% chance of getting it wrong also. This is more in conformance with real life experience.

4) What would a survey of actual bets of Jeopardy contestants reveal?

5) What would a computer simulation reveal?

#### 8 comments:

chessart said...

Worked through some more issues last night in bed. If B has exactly 2/3 of A's total, say A has 7.5K & B has 5K, then A's usual strategy is to bet \$2,501.

However, this seems to be a gross blunder, since B can then get a 50% win rate by simply betting nothing! (He wins every time A is incorrect, which is 50% by our assumption). However, if A leaves off the extra dollar and simply bets \$2,500, then he will always come back against a zero wager by B. Any other wager by B gives A a 75% success rate, so it looks like zero is B's best wager. Granted, he can also get to 50% by wagering everything vs. a \$2.5K wager by A, but he has no way of knowing this will be A's wager!

Another principle flowing from the above is that B has no real chance to do anything creative in Final Jeopardy unless he is within 2/3 of A's total. This gives B guidance in his decisions near the end of Double Jeopardy, i.e., he needs to position himself to get within 2/3 of A.

chessart said...

I now have a handle on B's correct wagering strategy, after looking at charts of bets and results. What B needs to do is to "create separation", to use a basketball term, referring to the offensive player's attempt to put distance between himself and the guy guarding him, which will allow him to get a good shot.

The separation required is that B needs to have a difference between his bet and A's of at least the amount of the difference between their scores going into Final Jeopardy. Thus, if it is \$10K vs. \$8K, B should bet so as to have a difference of at least \$2K between the two bets. If he can do this, his winning percentage is then at 50%.

A's task, then, is to try to make for no separation, just like a good basketball defender; in fact, if he bets the exact amount as B, he then has a straight 75% shot.

Philip Weaver said...

I discussed this problem with some friends from work at lunch today. They both have backgrounds in high assurance and formal methods (so quite a bit of mathematics). One of them seemed more interested in criticizing the simplification of the problem than actually thinking about it, which was rather annoying. But, the other one thought it was quite interested, especially the fact that the players' relative probabilities of getting the question right are not independent.

Anonymous said...

Why do douchey Contestant As always bet \$xxx1 dollars, often beating contestant B by \$1 when they both get the question right? I mean, if you tie for the win, you both get the money, right? So all you're doing is screwing the other guy out of his money. Plus, you'd let a contestant come back whom you've already out-scored through two rounds, which is probably better than a random contestant.

Ben Rodriguez said...

Leading people should not be so greedy and be willing to play for a tie unless they have far more than double the score of second place or the contestants are in tournament play. A tie ensures that the player that you've already beat, in a way, comes back. This way one can increase their chances of coming back longer and longer. The problem is, most of the people on Jeopardy are so greedy that a tie is not in their dictionary. In fact, one lady had exactly half the score of the leader and refused to risk everything, assuming that a tie is a defeat. In fact, its gotten to the point where every time someone loses with a XXX1 wager I think "That's what you get for being greedy."

chessart said...

Ben, I understand your point about the extra dollar. I would not characterize it as being "greedy"; that seems too harsh. It's in our psyches to play for a win, not a tie. Also, the show might discourage people from playing for a tie, as it makes for a better show if there is a clear winner, rather than 2 of the 3 coming back the next day. Certainly if the leader has exactly double the 2nd-place player, than it would be a huge blunder for the leader to bet a dollar, as that risks the game.

Jon said...

I have always wondered about the possibility of collusion in these circumstances.

Player in the lead: "I will bet an amount such that if you bet all your money we will tie and both go home with money (and come back next game). I will only ask for of your winnings."

The other player would say 'yes' assuming he is getting more than \$2,000, which is the default amount given to the contestant who finishes in second place.

Anonymous said...

The other reason to play for a win instead of a tie is that, having played once (or more) already, you have an advantage against a new contestant who doesn't know the buzzer timing. Most people who get on Jeopardy in the first place already know most of the answers. But look at what a difference buzzer timing made in the Watson games.