What's Wrong With This Statement?

[Skeesix]

Yes, but let's say you caught a peek at the bottom of the deck or one of the burn cards and you saw a jack, just for the fun of it.

[David Sklansky]

It's slightly more. Now tell me why.

[Cerril]

I guess I don't get it...

20% you win $40, 80% you lose $10 (unless there's a runner-runner for a tie), that's obviously where the >20% comes in (equal you might as well checkfold).

There are no other variables in this example, being highly contrived. The chance to steal is hardwired so we don't have to figure out the probability of one of the players still in having a pocket pair with the if-then statement. I'm assuming we also don't figure in image or anything outside the hand - it's played in a vacuum.

So it seems like they need to call somewhere in the neighborhood of 19% for a bet to be profitable

[La Brujita]

I don't normally try to answer these questions because frankly I am dumber than many of you but don't you have to take into account the effect of the rake (and the dealer toke if a live game)?

[Skeesix]

There's no mathematical explanation for why it should be more, particularly if you assume that the Jack is dead(which is the only way you can assume that only pocket pairs will call you in the real world).

So it must be a psychological consideration. Unfortunately I'm not familiar with any way to quantify a psychological consideration as more significant than the 6-10 percent chance of a tie on that flop.

[pzhon]

If you don't bet, you aren't dead. That is a more important factor than the possibility that your bluff is called and you tie.

[David Sklansky]

"There's no mathematical explanation for why it should be more, "

Incorrect

[Gabe]

Because a Jack could come and you could steal then.

[Earthy Tones]

[ QUOTE ]
You then say to yourself the following. "It is correct for me to bet $10 all in, if there is a greater than 20% chance I will steal the pot and incorrect if the chances are smaller than 20%."

[/ QUOTE ]

You originally come to this conclusion by saying to yourself:
If I bet and take down the pot I will net $40, if I bet and lose, I lose $10.

So

X*$40 must be > Y*$10

for your bet to be correct, where X and Y are the probability of the outcome of each occurrence. (There is, however, the small chance of a split. This does not change the spirit of the problem though.)

So if you take down the pot 20% of the time, the values X=.2, Y=.8
make both of our expressions equal

.2*$40 > .8*$10 … $8 > $8

This means that 20% percent is the value where the bet is break even, therefore you would need a probability of taking down the pot to be more than 20% for your bet to have a positive EV.

Sorry for that long winded explanation of something simple.

What I think the question is really asking is: Do you make a bet with an EV of exactly the same value of the bet itself?

This is all I could come up with..
First, we assume you are a winning poker player. This $10 is putting you all in and we also assume you will be playing poker at another time in your life, probably in the very near future. Therefore this $10 is worth more to you at another time where your stake at the table is normal sized.

So the chance of taking the pot down must be slighty greater than 20% for it to be correct to bet here.

[Leavenfish]

Clearly there is no obvious answer here as someone would have gotten it by now...let me venture a guess, that I assure you is not me simply being a smart ass. I mean it.

I would say that if $10 will put you 'all in' in a 10/20 game, an incorrect decision ends your session and what can be worse than that?

However, you are clearly not a very good player (or on a BAD streak, or simply WAY out-matched) if you are down to $10 in a 20/40 providing you bought in with a decent bank roll.

I say: take your $10, go buy yourself a drink and ponder why you found yourself in such an ignominious situation to begin with.

Serious. [img]/images/graemlins/wink.gif[/img]

---Leavenfish