Special Thread For Chen-Ankenman Mathematics of Poker

[Jerrod Ankenman]

[ QUOTE ]
Not satisfied with my solutions. Solved three one street
games. Then linked the solutions into a series. This is
one game with three streets, not three independent one
street games. The three streets are interdependent, not
independent streets. Therefore the methodology is
suspect.

Street one. The pot is 10. The bet is 60. From other
toy games X should call pot/(pot+bet) of the time.
X calls 1/7 of the time. Y maximized against this
strategy. Y should bet 12/77 of his other(non AK) hands.
Y was betting 26(14 AK) hands out of 91. This was
optimal strategy for the half-street game.
What if X maximized against Y's optimal strategy? X
should always fold. Whenever Y bets fewer than 50% of
the hands, X can guarantee plus EV by folding to all
bets. Y must bet at least .396 of his others before
X needs to call to improve on +0.5(the value of the
game in the 3 street solution).

Conclusion. There's more to this 3 street game than
just solving 3 single street games. Optimal strategies
for one individual street may not be optimal for
multiple streets. My solution for the street one game
is only correct for a street one game in isolation.

This chart is what's happening. Street two is a game
within street one. And street three is a game with
street two. The value of the top left hand cell of
the street one and two matrices is a variable. This
game must be solved by recursion.

jogs

[/ QUOTE ]

Yes, that's right. Here's what happens. (Ignore checking hands - assume that checking on any street is tantamount to giving up)

Suppose we have street 3. We know what the bet size is going to be on street 3, so Y ends up with a distribution of all his value bets and exactly enough bluffs to make his bluffing ratio on the river correct. Call his value bets V and his bluffs B = (alpha)V for the appropriate alpha = 1/(p+1). If the bets on the first two streets are both $60 and the initial pot is $10, then we have a pot size of $250 and a bet size of $60, so alpha is 60/(250+60) = 6/31.

So if we have V value bets, we need 6/31 bluffs. So the total number of hands we need in order to play the river correctly is (37/31)V - 31/31 of which are value bets and 6/31 of which are bluffs.

Now let's look at the 2nd street. The trick here is to use as your value bets the (37/31)V number and not V itself. So here we have a pot of $130, a bet of $60, and 37/31V value bets. alpha is 60/190 = 6/19. So now we need (6/19)(37/31)V hands to bluff with on the 2nd street (these hands will be given up on the river). So we have V(222/589) hands that are 2nd street bluffs only.

Now our total number of hands to play the 2nd street properly is 222V/589 + 37/31V, or 925V/589.

Now on the first street, we have an alpha of 60/(60+10) = 6/7. So 6/7 of the 2nd street hands will be bluffs on the first street. So that's (6/7)(925/589)V that are bluffs on the first street.

I attached a little Excel file to this post - it shows the relative frequencies of each strategy for X, and the value of each of Y's strategies. You can see that using this method, Y is indifferent between calling any number of streets and folding immediately, and the value of any strategy for him is -$19.17.

Hope that helps.

Edit: Uh, I guess you can't attach files or I am unable to find it anyway. You can grab this file here.