Can "doing business" without reshuffling influence EV?

[hairy]

In undergroud clubs "doing business" has become relatively common; i.e. both players agree to run the remaining cards multiple times, often 3 times, and typically without reshuffling. If you win 2 of 3 times you take 2/3 of the pot.

A common situation might be an all-in on the flop with top pair vs a flush draw where the players agree to run it three times so as to avoid both going broke.

I realize that "doing business" in the above situation primarily influences the variance, however I'm unsure to what degree the expected value may change if at all.

Any thoughts?

Also, I'm aware there may be legitimate non-EV related reasons for doing business such as if you are simply enjoying playing in a particular game but don't have the cash for a rebuy. I'm merely concerned with the EV issue.

[SamIAm]

Welcome to the forum. This is usually called "running it twice", and we get this question a LOT. Try searching for it; if you don't find anything, post here and we'll help you search.

The short answer is: EV is linear. EV[X+Y] = EV[X]+EV[Y]. Running it twice lowers variance and doesn't change EV.

Search and ye shall find.
-Sam

[rufus]

Running multiple times is EV neutral. Notably, if you can make descisions about play (e.g. whether or not to call an all-in), then you may be able to gain information about your opponents hand from his attitude about 'doing buisness'.

[hairy]

Thanks for the help. I understand that if you reshuffle the cards back into the deck the EV remains the same, but are you sure it makes no difference if you continue to deal off of the deck without reshuffling the cards once you've ran it once?

Imagine a simplified scenario where both players are all-in on the turn and the pot is 2 dollars. Lets say there are 10 cards left in the deck and one player has only 1 card that can make him a winner. If we run it once the player with one out's EV is 10% of the pot of 2 dollars which is 20 cents.

If we decide to run it twice without reshuffling there are two scenerios: 10% of the time the guy with 1 out hits his card the first time and then loses the second time. He gets 1 dollar 10% of the time under this scenario for an EV of 10 cents.

The other 90% of the time he loses the first round and is now 1/9 (>10%) to win half the pot (1 dollar) in the second round for an EV of slightly more than 10 cents.

Thus it seems to me, using my rudimentary and likely flawed statistics, that the player with 1 out actually increases his EV by running it twice here assuming the cards are not reshuffled into the deck before running it a second time.

Am I correct and if so what are the implications of this?

P.S. I tried searching the forum without luck.

[punter11235]

We had good thread about it which contained three diffrent proofs that dealing it X times doesnt change EV. Reshuffling cards back into the deck doesnt matter here - EV stays the same.

[SamIAm]

[ QUOTE ]
P.S. I tried searching the forum without luck.

[/ QUOTE ]I just simply don't believe you.
-Sam

[Siegmund]

Perhaps returning to your "10 cards left with 1 out" simplified example will help you to see why doing business without reshuffing doesn't change EV but reduces variance (reduces it more than business with reshuffling would, I might add.)

Running it once:
If your card is on top of the deck (1/10) you win.
If not, you lose. Expectation, 1/10 of the pot.

Running it twice:
If your card is among the top two (2/10), you will win one and lose one. if not you lose. Expectation, 2/10 * 1/2 = 1/10.

Running it three times:
If your card is among the top three (3/10), you will win one and lose two. If not you lose. Expectation, 3/10 * 1/3 = 1/10.

on up to...
Running it ten times:
You are absolutely guaranteed to win one and lose nine times. Expectation, 1/10.

Really, the cases add up to exactly what they should, no matter how many outs there are - it's just slightly harder to do the math when you have more than one out.

You have done it the hard way for running it twice: the expectation of the first card winning for you is 1/10; then, if you won on the first card, the second card is a guaranteed loser, but if you lost on the first card (as you will 9/10ths of the time) you have a 1/10 chance to win on the second, for a 1/10 + 9/10*1/9 = 1/10 + 1/10 = 2/10 chance of one win and one loss, and 8/10 chance of two losses.