Maths problem: Almost blinded out - make a stand!

[molotom]

You're in a poker tournament with many players on one big table. You get 1-outed by someone who you just have covered, and you are left with only 1BB. You only have a few more hands until it is your big blind, and decide to choose for a good hand then call/push with it. You assign every starting hand a number x between 0 and 1, based on its overall equity. Initially, we'll assume that for a random hand, x is uniformly distributed between 0 and 1.

You decide you want to maximise the expected value of x when you go in.

What is your optimal strategy?
Following this strategy, what is your expected value of x when you have N hands left?

(We'll say that when it's your blind, N=0).

[jay_shark]

Here are some good guide lines you can use which gives you an approximate idea of the hands you should be shoving with when you're in position x relative to the BB .

If you're utg with 1BB then you should shove with the top 50% of hands . If you're utg+1 then you should shove with the top 33% of all hands . In general if you're utg+x then you should shove with the top 1(x+1) of all hands .

I've computed the exact EV for a two player game and you may refer to this post for the exact answer but it's much more practical to use my approximation .

http://taigabridge.com/poker/articles/he_lastallin.htm

[molotom]

I think that the intuitive answer, that we push the top third with 3 hands left and so on is wrong.

Let x_n denote the expected value of x for the hand we go in with when we apply our optimal strategy to the case with n hands remaining.

If n=0, ie it's our big blind, then we're in with any hand no matter what. So x_0 = 0.5.

If n=1, now we make a decision: if our hand is better than the expected hand we'd end up with if we folded we go in. Otherwise we fold.

So we go in when x > x_0 = 0.5. The chance this happens is 0.5, and the expected value x when it happens is 0.75 (a random hand from the top 50%).

We fold when x < x_0. The chance this happens is 0.5, and the expectance is now just equal to x_0.

Therefore x_1 = 0.5*0.75 + 0.5*0.5 = 0.625.

In general,
x_(k+1) = (1 - x_k)*[0.5*(1+x_k)] + (x_k)*(x_k)
= [1+(x_k)²]/2

This means that in the case where we see three hands, we go in if our hand is in the top 37.5%, not just the the top third. This reflects the fact that while we might see three hands, we will sometimes end up folding the best one.

In practice the optimal strategy is more complicated, because the equity of a random poker hand is not uniformly distributed. However I hope this shows why the intuitive approach is incorrect.