I wife and I have recently been watching Jeopardy!, which can be done quickly with a DVR. I have been interested in the betting strategies. It has seemed to me that some players don't use optimal strategies. For example, there was an episode where a player who was leading correctly answered the Final Jeopardy! question, but lost because she failed to wager enough money. I dug a little deeper to see what others had to say.
In Final Jeopardy!, the players each make a wager before answering a question. They may wager any dollar amount between $0 and the total they have gained through the episode. I want to know if there is a sure strategy for how much one should wager.
If one player has won more than double the other opponents, then he is assured of winning as long as he picks a wager so that even if he gets the question wrong, he would still have more than double what the other opponents started with. If he wants to maximize his expected winnings, then he should bet as much as safely possible if he thinks he has a better than 50/50 shot at answering correctly, and he should bet 0 otherwise.
If one player does not have more than double the other opponents, then she may possibly lose. I found through my Google search (I'm not including links, but there are many sites that talk about this) the 2/3 rule. Suppose that the player in the lead has more than 1.5 times her nearest opponent, but less than or equal to 2 times her nearest opponent. Then the best play is to bet so that if she gets it right, she will have more than double what the nearest opponent started with. In mathematical formula terms, suppose the leader has $A and the 2nd-place player has $B, and that 2 B >= A > 1.5 B. Then the leader should bet an amount $x so that A-B > x > 2B-A, which is possible since A-B > 2B-A is equivalent to A > 1.5 B. If the leader gets it right, she wins. If she gets it wrong and the 2nd-place player gets it right and bets big, she loses, but that would have happened regardless of what she had bet. If both players get it wrong, then the leader wins because she still has more than what the 2nd-place player started with. In this situation, the 2nd-place player might as well bet as big as possible, unless he is concerned about the difference in prize between him and the 3rd-place player.
If the leader is ahead by less than 1.5 times his nearest opponent, then the situation is trickier. My search said that the leader will always have the same betting strategy: bet so that he has barely more than double of what the 2nd place person started with. This is not because it is the most proven strategy per se, it is just because that is what you see on TV. The case then is that the 2nd-place player needs to assume this and bet so that she just beats out the leader in the case that he gets it wrong and she gets it right. One link for this his here.
This however doesn't sit well with me because the proper strategy for the 1st place player can't be to always bet big. For example, in the scenario that the 2nd-place wagers a very small amount, if the 1st place player bet $0, he would beat the 2nd-place player no matter what. In this scenario, there is no fool-proof bet that either player can make (i.e. there is no Nash Equilibrium, as named for the guy from A Beautiful Mind.) If the 1st-place player doesn't like his chances in getting the Final Jeopardy! question right, why not bet small instead of the standard leader bet? Why is that a bad strategy? (I do note, however, that it is most often the case that some or all of the contestants get Final Jeopardy! right.)
No comments:
Post a Comment