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Originally Posted by vixticator
Somewhat unrelated... let's say you go into final jeopardy with only 2 players because one guy was eliminated by having less than zero (right?) and the other two players have $1,001 (p1) and $1,000 (p2)? If they both play perfectly and let's say they somehow know each have 50% chance of getting the answer right how much should they bet? I assume there is a right answer... or is there a huge continuum of best responses?
This is an interesting question.
Say you are the 1001 guy and you know the other guy bets x. The correct move is to wager something between x+1 and x-1. If you're right then we will win. If you and your opponent are wrong then you will still win. The only way you lose is to be wrong and have him be right, which happens 25% of the time. Betting more causes you to lose even when you both are wrong while not gaining you any win equity, betting less is similar but you lose when you are both right. Since betting any amount is EV neutral ignoring winning, you can't gain from betting more or less, but it does cost you.
So your best response is to be in that (x-1,x+1) interval.
Say you bet y<999. If your opponent wagers x < y -1 then he will win whenever you are wrong and lose when you are right, making him win half the time. If he wagers x > y + 1 then he will win whenever he's right and lose whenever he's wrong, giving himself a 50/50 shot. If he wagers in between those, then he'll have only a 25% chance for the reasons I gave above when discussing what you should wager when he wagers x.
That makes his best response to be either greater than x+1 or less than x-1.
Because he can either go big or small given a fixed strategy for you, if he knows what number you are putting down then he can guarantee himself a 50% chance. Similarly, because an optimal wager for you if you know what he's putting down can be above or below it, you can always guarantee yourself a 3:1 edge if you cheat and know what he's putting down.
That means that there's no fixed equilibrium. The solution has to be something involving randomness.
I haven't actually studied FJ wagering, so I'm not sure if you can get an equilibrium by reducing the strategy space to just two possible wager sizes but it seems reasonable. That's a good place to start though.
Let's say the large bet is 1,000 and the small bet is 0. We'll say I'm the guy with 1,000 and you're the guy with 1,001. If I bet 1,000 with probability p and 0 with probability 1-p then you will win by betting 0 if either I'm wrong or I'm betting 0. This happens with probability .5p + 1 -p = 1-.5p. If you bet 1,000 you win if you are right, or if you're wrong but so am I and I bet 1,000 (you would win 1-0 in this case). The chances of that happening are .5 + .5*.5*p = .5 + .25p. As I showed above, it's not an equilibrium if either player is playing a fixed strategy. It only makes sense for you to randomize over the bet sizes if you are indifferent between playing either option, which only makes sense if the win% from betting 1,000 is the same as betting 0. That means that I must set p so that 1 - .5p = .5 + .25p or p = 2/3.
Say you bet 1,000 with probability q and 0 with probability 1-q. If I bet 0 then I win if you are wrong and bet 1,000. That probability is .5*q. If I bet 1,000 then I win if I am right and you are either wrong or bet 0. That probability is .5(1-q + .5q) = .5 - .25q. For the reasoning given above, we set these two equal and solve for q giving us q = 2/3.
So in this case, the unique equilibrium would be for each of us to wager 1,000 2/3 of the time and to wager 0 the other 1/3.
Actually, now that I think about it, this should be an equilibrium of the actual game. If you are doing this 2/3, 1/3 randomness then altering my wagers either doesn't matter or screws me. As I showed, I can't improve by wagering 0 or 1,000 all the time. I also can't improve by wagering anything in between. Any number less than 1 gives me a win in the same scenarios as betting 0 and anything 1 or greater gives me the same win scenarios as betting 1000. So I would have the same payoff across the whole spectrum of numbers, meaning that I can't do better by deviating whether that be playing a pure or mixed strategy.
For you, if I'm doing the 2/3, 1/3 mixing then betting less than 1 gives you the same win scenarios as betting 0. Putting anything between 1 and 999 is worse for you than betting 0 (and hence the mixed strategy) because you never win when you would have lost by putting 0 and you lose instead of win when you are wrong and I bet 0. Putting anything > 999 puts you in the same situation as wagering 1,000.
So yeah, that is an equilibrium for the game. There are definitely others, probably some where you could possibly bet any number, but that is the simplest IMO. A key to this is that each play is 50/50 to get it right meaning that the only thing that matters is winning or losing. If that weren't the case then my argument for it being an equilibrium would break down. For example, betting 100 might be better than betting 0 if you are a 3:1 favorite to be right because you gain $50 in equity which might be worth the hit you take in chances of being champion.
edit: I wasn't grunching but made such a long post that I missed several responses. Apparently second doesn't get to keep their money, which changes things a bit, though I suspect not much.