[jason1990]

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just like in a poker game you don't compare your current hand with past or future ones to get the EV of different actions. Rather, you *simulate* the exact hand a large number of times and look at what your average profit or loss is for each decision. That doesn't mean that that exact hand actually has to happen a large number of times

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Suppose you play 5/10 limit. On the turn, you have a flush draw and are faced with calling a $10 bet in a $60 pot. You calculate that you have +EV, so you call. If you continue to play 5/10 limit, then you are going to face this same situation many, many times in the future. It really will happen a large number of times. Your +EV really is going to turn into long term profit by the LLN.

Now suppose differently. Suppose you are very foolish. Every time you sit down to play poker, you play the highest stakes NL you can, and place your entire bankroll on the table. On the turn, you have a flush draw and are faced with calling all-in for your entire bankroll with 6 to 1 pot odds. If this situation repeats many times and you call every time, then you will eventually go broke. The EV of calling is still positive, but this +EV will not turn into long term profit by the LLN. So calling every time is no longer in your best interest.

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Say that instead of wagering money on each game, you play the following game against a friend:

Neither of you can know what your opponent is doing or how many points they have until you both decide you're finished and compare points to see who won.
You each start with one point, and each round you can decide to either wager all of your points in the same manner as the first example or finish playing. If you lose a wager you're forced to finish playing. What is the optimal strategy to win this game?

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This is interesting. Let me add to some things to it, so that we may analyze it. At the end, the person with fewer points pays the person with more points $1, with no money changing hands on a tie. The game will be repeated many times, so that our objective is to maximize EV.

I will choose to play n flips; n can be 0, in which case I just stand with my 1 point. My friend will choose m flips. Here is one possibility. I will choose n = 1, 2, or 3, with probabilities 33/117, 4/117, and 80/117, respectively. If my friend chooses m = 0, then I win if I do not bust and I lose if I bust. So my conditional EV, given that I chose n, is

p^n - (1 - p^n) = 2p^n - 1,

where p = 3/4. Hence, my overall EV is

(33/117)*(1/2) + (4/117)*(1/8) + (80/117)*(-5/32) = 1/26.

If my friend chooses m = 1, 2, or 3, then a tedious calculation will show that my EV is 0. If my friend chooses m > 3, then I win if he busts and I do not, and I lose if he does not bust. So my conditional EV is

p^n(1 - p^m) - p^m
= p^n - p^m(1 + p^n)
>= p^n - p^4(1 + p^n)

Evaluating this gives

n = 1 --> 201/1024
n = 2 --> 279/4096
n = 3 --> -459/16384.

Hence, my overall EV is greater than or equal to

(33/117)*(201/1024) + (4/117)*(279/4096) + (80/117)*(-459/16384) = 513/13312.

So we see that if my friend chooses m = 0 or m > 3, then I have positive EV; and if he chooses m = 1, 2, or 3, I have 0 EV.

I believe I have a proof that this is the only equilibrium solution, but my calculations are very tedious. Perhaps there is a simpler analysis.