No Limit Hold 'Em Theory And Practice

[rspaterson]

I have a question about blocking bets. It's quite possible that I'm just having a brain fart. The material I'm referring to is on page 139 of No Limit Hold 'Em Theory And Practice.

It appears to me that the expectation equation for the blocking bet is incorrect.

$123=(0.50)($200)+(0.50)(0.35)($280)+(0.50)(0.65)(-$80)

That is to say that 50% of the time you bet $80 into a $200 pot your opponent will fold. 35% of the other 50% of the time you bet $80 into a $200 pot your opponent will call, but you'll have the best hand. 65% of the other 50% of the time you bet $80 into a $200 pot your opponent will call, but they'll have te best hand. My question is as follows.
I undersdtand that your $80 bet is not included in your take in the first part of the equation because you bet, your opponent folded, and the money you bet is simply returned to you (it does not become part of the pot and is not subject to rake). Why isn't your bet included in your take in the second part of the equation. When you bet and are called your money becomes part of the pot and is subject to rake. Shouldn't the equation look like this:

$123=(0.50)($200)+(0.50)(0.35)($360)+(0.50)(0.65)(-$80)

I do pretty well with math, but am by no means a genius. I often overlook things, and try to make them too complicated. Please help.

[Albert Moulton]

Maybe a simpler example might help.

In a fair 1-coin-flip game of heads or tails, if you bet $10 on heads and get called (risking $10 to win $10 with a 50% to win, a 50% chance to lose, and a 1:1 pay off), then your EV will be the sum of
.5 (lose) * -$10 = -$5, and
.5 (win) * $10 = $5,
for a total EV of -$5 + $5 = $0, or "neutral" EV.

On the other hand, if you counted your "$10" wager in the "pot" when you win, then you'd essentially be double counting it (i.e. .5 (win) * $20 = $10, making the coin flip game incorrectly appear to be +5 EV per flip).

So, in your OP, you are risking $80 to win some amount $X, some percentage of the time. If you lose, then you lose $80. But if you win, then you win $X. You don't win $X+$80.

[ChrisV]

We're trying to determine expectation, not the size of the pot. Look at it this way: If you bet $80, the immediate effect is to subtract $80 from your expectation. If your opponent folds, the $80 is returned to you:

$200 - $80 + $80 = $200 expectation.

If your opponent calls and you win, $160 is added to your expectation:

$200 - $80 + $160 = $280 expectation.

If you are called and lose, the $80 is not returned to you:

$0 - $80 = -$80 expectation.

Note that the starting expectation is $200 in the cases where you have the best hand and $0 if you don't.

[rspaterson]

thank you very much. it's easily understood. i think i'd just been reading too long.

[rspaterson]

thank you.