PLO math question

[kirisim]

Help me out here:

Let’s say two players dump $1000 into an Omaha pot on the turn. They both have a made straight, but one guy has a redraw to a better hand, with a 25% shot of making that better hand. So if they run the river 1 time, the EV of the players is +$250 per hand and -$250 per hand, respectively. (Correct?)

Now, say they agree to run the river 2 times. 1) What is the EV of the respective players and 2) For the guy with no redraws, is he better off running it once, or twice?

[kirisim]

I think I got it. The guy with no redraws is better off running it once.

I get his EV as approximately -$312 per hand if he runs it twice.

[gaming_mouse]

The EV per hand does not change based on the number of times they go to the river. The EV is the EV -- that's it.

As for your calculation, I can't tell whether or not it's right because it's not clear what you are asking. Are you trying to find the EV of the turn bet, of a future river bet (if so, how much?), or of the hand in total (if so, how much was in the pot before the turn bet?)

gm

[kirisim]

I am saying that each player has put in $500 each. They are both all in on the turn. The cards are turned up. One guy has the 25% redraw. They are discussing whether to run it once, twice, or three times. What is better for the guy with no redraws??

[gaming_mouse]

[ QUOTE ]
I am saying that each player has put in $500 each. They are both all in on the turn. The cards are turned up. One guy has the 25% redraw. They are discussing whether to run it once, twice, or three times. What is better for the guy with no redraws??

[/ QUOTE ]

1 time is the best for him, because it's a losing proposition.

His EV per hand is a constant value. However, his EV for N hands is his EV per hand multiplied by N. Therefore, to miminize his overall expected loss, he should play as few hands as possible.

gm

[ThinkQuick]

I think that the answer you are looking for depends on your definition of 'better' i guess. There is no change in the Expected Value of the hands, but running it more times decreases the variance. This could be better in a tournament situation.

Your qestion reads, 25% of the time the river is run, player B (with no redraws) gt 0% of the pot, and 75% of the time the river is run, B gets 50% of the pot.

Therefore B 'owns' 37.5% of the pot and this is the percentage he will get on average. That is to say, if the river is run more and more times, the percentage of the pot going to B gets closer and closer to 37.5%.

If the $1000 pot is run once, B may get $0 or may get $500.
But if river is run a million times, B will get ~ $375. This significantly decreases B's variance, and the more times the river is run, the less chance B has of leaving with nothing (Better for B).

[edtost]

doesn't the ev change depending on which cards come up the first time it is run? not each run is with the full stub.

[ThinkQuick]

[ QUOTE ]

doesn't the ev change depending on which cards come up the first time it is run?

[/ QUOTE ]

In this case, the EV of the next run would change after the first run if you knew the result of the first run.

However, the EV as observed by the player in question is before any running, which is when he has to make his decision. Therefore the only EV of importance is that one., which is invariant.

[gaming_mouse]

[ QUOTE ]
doesn't the ev change depending on which cards come up the first time it is run? not each run is with the full stub.

[/ QUOTE ]

I don't know what you mean by full stub.

I will just say I assumed that by "running the river 2 times" the OP meant you would lay down 1 river card, do the payouts for it, then replace the money back in the pot (so that the pot was identical), put the river card back in with the remaining cards in the deck, re-shuffle, and go again.

That is, the two runs are independent of each other.

If that is not what he meant, then I don't what he meant, and my comments probably don't apply anymore.

gm

[ThinkQuick]

[ QUOTE ]

I don't know what you mean by full stub.

I will just say I assumed that by "running the river 2 times" the OP meant you would lay down 1 river card, do the payouts for it, then replace the money back in the pot (so that the pot was identical), put the river card back in with the remaining cards in the deck, re-shuffle, and go again.

That is, the two runs are independent of each other.

If that is not what he meant, then I don't what he meant, and my comments probably don't apply anymore.

gm


[/ QUOTE ]

The practice of "running it twice" in Omaha is not as you described.

It means that two sets of the remaining board cards will be dealt out, one on top of the other, and that the pot is split, with half going to the winner in the top hand, and half to the winner of the bottom hand (may be the same player).

That is, not independent.
By saying "not each run is with the full stub", the poster is saying that after the first river card is dealt, the remainder of the deck (stub = portion remaining) is no longer as it was before said card was dealt. I.e. since the 2nd river card is dealt after the first river card, (and that first card is NOT replaced and reshuffled), the odds regarding that second card are slightly different than what they were for the first river card, when one accounts for the removal of the first river card.

[kirisim]

Running the river twice means: The dealer burns & turns the river, then burns & turns the river again. So they are not independent. If a guy makes a flush the first time, there is one less of that suit remaining.

However, that changes the odds only a small amount, maybe a few percent.

[gaming_mouse]

[ QUOTE ]
Running the river twice means: The dealer burns & turns the river, then burns & turns the river again. So they are not independent. If a guy makes a flush the first time, there is one less of that suit remaining.

However, that changes the odds only a small amount, maybe a few percent.

[/ QUOTE ]

I see. Sorry, I was not familiar with Omaha.

In any case, assuming that the guy with the redraw can hit his hand twice (even if the second redraw is slightly less than 25%) you are still better having him redraw only once.

Do you want to take only 1 bet where you have the worst of it, or two?

gm

EDIT: Actually, let's do the math.

One run:

EV = 500*.75 + 0*.25 = 375

Two runs (assuming 25% redraw on each):

EV = 500*(.75)^2 + 250*2*.75*.25 + 0*.25^2 = 375

Interesting, it is actually the same. Now if the second redraw had only a 20% chance of success if the first redraw hit, say, but a 30% chance if the first one missed:

EV = 500*.75*.70 + 250*.75*.30 + 250*.25*.80 + 0 = 368.75

So two redraws is worse.

[pzhon]

E(A+B) = E(A)+E(B) even if A and B are not independent.

This means it doesn't affect the EV if you decide to run it once, twice, or any number of times. All that happens is the variance decreases.

Suppose the probability of winning one time is 25%=10/40. Here are the probabilities of the possible outcomes:

Win, Win: 10/40*9/39.
Win, Tie: 10/40*30/39.
Tie, Win: 30/40*10/39.
Tie, Tie: 30/40*29/39.

The total number of wins when you run it twice is 2*10*9+10*30+30*10 out of 2*40*39, or 780 out of 3120, 25% of the trials.

Hitting the draw decreases the probability of hitting the second time. Missing the first time increases the probability of hitting the second time. The weighted average of these effects is 0.