Running It Twice

[Jazza]

for those who don't know what it means i think it's easiest to show an example:

two players playing NL holdem get it all in on the flop, player A has K/images/graemlins/heart.gifK/images/graemlins/spade.gif, player B has
A/images/graemlins/diamond.gifJ/images/graemlins/diamond.gif, and the flop is 9/images/graemlins/diamond.gif4/images/graemlins/diamond.gif2/images/graemlins/spade.gif

after seeing each other's hand and realizing it is about a coin flip, they agree to 'run it twice'

the dealer now deals the turn and river, and a second turn and river:

first turn and river: 8/images/graemlins/diamond.gif5/images/graemlins/diamond.gif
second turn and river: T/images/graemlins/spade.gifJ/images/graemlins/club.gif

now, because they won one each, they split the pot

had one of the players won both boards, he would take the whole pot

running it twice reduces variance, which many players are happy to do, but my question is can running it twice change the EV for the players?

[Usual]

Assuming the turn and river cards are replaced and reshuffled, then EV remains the same. (Variance is halved)

[Nottom]

[ QUOTE ]
Assuming the turn and river cards are replaced and reshuffled, then EV remains the same. (Variance is halved)

[/ QUOTE ]

They aren't.

[irchans]

EV is still the same even if the cards are not shuffled. Let me think about it for a bit and try to come up with a proof.

[Derek in NYC]

Id like to see the proof, since this is counterintuitive to me.

[jason1990]

Call the players A and B. The key observation is this: the probability that player A wins the second run is the same as the probability that he wins the first run. What may confuse some people is this: the conditional probability that he wins the second run, given that he won the first or given that he lost the first, is something different. But before the runs take place, the probability he wins the second run is exactly the same as the probability he wins the first.

So let P be the size of the pot, let X=1 if player A wins the first run and 0 otherwise, and let Y=1 if player A wins the second run and 0 otherwise. Then the amount that player A wins is

X*P/2 + Y*P/2

and his EV is

E[X*P/2] + E[Y*P/2] = [P(X=1)+P(Y=1)]*P/2
= [2*P(X=1)]*P/2
= P(X=1)*P.

On the other hand, if they only run it once, he will win X*P and his EV will also be P(X=1)*P.

[pzhon]

[ QUOTE ]
The key observation is this: the probability that player A wins the second run is the same as the probability that he wins the first run.

[/ QUOTE ]
That's one of two basic facts that may be helpful. The other is that E(X+Y) = E(X) + E(Y), even when X and Y are not independent.

[TomCollins]

Here is an example showing how EV is the same even without reshuffling. This is not a proof that this always works. I'll try to get that in another post.

Suppose you have KK, opponent has 55. Board is K952.

He wins 1/44 times. EV = 1/44*pot for opponent.

Suppose you run this three times without reshuffling:
He has 2 possibilities:
Lose all 3:
(43/44)*(42/43)*(41/42) = 41/44. EV = 0

Win 1, lose 2:
3/44 (1- 41/44).
Also: (1/44)+(43/44*1/43)+(43/44*42/43*1/42) = 1/44+1/44+1/44 = 3/44.
His EV = 3/44 * 1/3 pot = 1/44*pot.

Total EV for taking 3 cards is 1/44+0 = 1/44. Exactly the same as taking 1 card. But there is less variance since he wins 1/3 of the pot more often.

[Jordan Olsommer]

So instead of playing say one $10 pot with a particular flop of XYZ , they're essentially just playing two $5 pots with XYZ flops, and asking the dealer to expedite the process? Makes sense.

[MikeL05]

Your original example might hint at why I think it does change the EV.

Basically, AJs needs an A or a diamond to win (lets exclude other more obscure cases, like JJ or some kind of diamond combo that gives KK a boat).

AJs needs only ONE diamond to win, but in the first turn and river, he used TWO of his outs. When running this once, that's irrelevant, and won't change anything. When running it twice, it slightly reduces his chances of winning the second run. Same if he got a diamond AND an A in the first run.

As I see it, the idea of running it twice is that doing so would maintain an "average" probability of each hand winning over the two runs equal to its chance of winning the first run. But I would think that using two outs in the first run would violatet this assumption. I'm going to take some time soon and do some math on this, it really interests me.

[TomCollins]

You may think it, but that doesn't make it so.

Suppose you have 2 cards to come. If the dealer burns 3 cards instead of 1, do your odds of winning change at all? Of course not.

So your odds of winning the first run are equal to the odds of winning the second run. Therefore, EV is the same.

[AaronBrown]

I think the disagreement here is symantic.

Before the players agree to this they each have an EV (K's is +7.48% of the pot, AJ is the opposite of that). After they agree, they still have the same EV's. That's what people mean by the EV doesn't change.

Once the first hand is dealt, obviously the EV changes for the first hand, it becomes 100% for one player (there is no possibility of a tie with this hand). But the EV of the second hand also changes at this point. I think the people who are arguing EV changes mean this.

K's chance of winning the first hand is 53.74%. If he does win, his probability of winning the second hand drops. If he loses, his probability of winning the second hand increases. These changes exactly offset. But they do mean the variance reduction of this procedure is even greater than two coin flips versus one. There is negative correlation between the two winners.

Without getting too fancy about it, there are 12 cards (three aces and nine diamonds) that will probably let AJ win. If they deal 22 hands from the remaining 45 cards, with average luck those 12 win cards will come up on 10 different hands of the 22. It won't be 12 because about 1.5 times 2 winners will come up at once, 12/45 of the time the last undealt card will be a winner, and some of the Aces might be paired with a King. However, it's very likely that AJ will win between 7 and 13 of the 22 hands. The EV is still the same, but most of the variation has been eliminated.

[Jordan Olsommer]

Deleted my earlier post, because I had mistaken the second-trial EV when TomCollins was talking about the overall EV, which indeed does stay the same.

For an illustration of why, just think of a computer program enumerating all possible river cards for a certain matchup and specified board. The way it does this is by basically dealing out a card, evaluating the hands to see who won, and then throwing that card out and dealing another card from the deck which not only has one card less than before, but like in the situation described by the OP, you know exactly what that card is.

That changes the EV from trial to trial, but it doesn't change the overall EV.

[PairTheBoard]

This is kind of amazing to me. I always considered the idea that E(X+Y)=E(X)+E(Y) even when not independent, to be sort of ho hum. And the observation that the apriori chances in the first and second run are the same regardless of the nonshuffle seems pretty obvious. Yet you apply a little math to this observation and you get a result that really makes you blink your eyes. Without the math you might ponder the thing for quite a while. I think this would be a great teaching tool to accompany the Expected Value Theorum when it's presented.

PairTheBoard

[jason1990]

[ QUOTE ]
[ QUOTE ]
The key observation is this: the probability that player A wins the second run is the same as the probability that he wins the first run.

[/ QUOTE ]
That's one of two basic facts that may be helpful. The other is that E(X+Y) = E(X) + E(Y), even when X and Y are not independent.

[/ QUOTE ]
Ah, yes. I should have mentioned that. (We definitely need a FAQ for this forum.)

[ QUOTE ]
PTB:

This is kind of amazing to me. I always considered the idea that E(X+Y)=E(X)+E(Y) even when not independent, to be sort of ho hum. And the observation that the apriori chances in the first and second run are the same regardless of the nonshuffle seems pretty obvious. Yet you apply a little math to this observation and you get a result that really makes you blink your eyes. Without the math you might ponder the thing for quite a while. I think this would be a great teaching tool to accompany the Expected Value Theorum when it's presented.

[/ QUOTE ]
I think textbook writers agree with you. Problems like this (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2427203&page=&view=&s b=5&o=) are very common.

[MikeL05]

[ QUOTE ]
You may think it, but that doesn't make it so.

Suppose you have 2 cards to come. If the dealer burns 3 cards instead of 1, do your odds of winning change at all? Of course not.

So your odds of winning the first run are equal to the odds of winning the second run. Therefore, EV is the same.

[/ QUOTE ]

I'm sorry, I'm new here, but this makes absolutely no sense.

First, your example of burning 3 cards is both (a) obvious, and (b) irrelevant. In that case, the burned cards are unknown. In my case, we KNOW that they are, for example, two diamonds.

As far as my odds of winning the first run being equal to my odds of winning the second run... this is also ridiculous. If I have AA and you have KK, and we're all in after a flop of 4-8-T rainbow and decide to run it twice... KK has perhaps a 10% chance of winning. Say the first run is a turn of K and a river of K. Now for the second run, KK has a 0% chance of winning.

[TomCollins]

You don't know when you make the deal.

Suppose you burn 3 cards instead of 1. Call the two boards X and Y.

Now, EV(X) = EV(Y). I think you realize that.

So EV(running it twice) = EV(X) + EV(Y) /2. EV(X) = EV(Y), so 2 EV(X) / 2 = EV(X) = The same.

Your "counter-example" doesn't really prove anything. Of course when you remove two Kings, your EV is exactly 0 on the run. But you didn't know that two kings were missing ahead of time.

Counterintuitive? Of course. False? No. Proof is stated above.

If you know that diamonds already came off, and you have the choice to make a deal, it obviously changes the situation. But at the point you make the deal, everything is unknown.

[Stephen H]

[ QUOTE ]
I'm sorry, I'm new here, but this makes absolutely no sense.

First, your example of burning 3 cards is both (a) obvious, and (b) irrelevant. In that case, the burned cards are unknown. In my case, we KNOW that they are, for example, two diamonds.

As far as my odds of winning the first run being equal to my odds of winning the second run... this is also ridiculous. If I have AA and you have KK, and we're all in after a flop of 4-8-T rainbow and decide to run it twice... KK has perhaps a 10% chance of winning. Say the first run is a turn of K and a river of K. Now for the second run, KK has a 0% chance of winning.

[/ QUOTE ]

I'm fairly new here as well, but it makes sense to me. The example given of burning 3 cards is extrememly relevant. We're talking about the odds of winning each trial before we run any trials; naturally the odds of winning the second trial are different once we know the outcome of the first trial. The deal to run it twice is done before we know what the first run is, so we can't base our odds of winning the second run on the outcome of the first run. Put it this way. What if the dealer was going to burn 3 cards, but would burn two of them face up, before dealing the turn and river? Do your odds of winning change? Of course not. What if we "run it twice" but don't count the first run? Are your odds different than if we just ran it once? Before we run either trial, the odds of winning each trial are equal.

Take your example. Sure, if the first run is KK then the second run has a 0% chance of winning. And the first run now has a 100% chance of winning, too.
What if the first run is T-7? Now the first run has a 0% chance of winning, and the second run has better odds. Now what are the odds of winning the second run if we don't know what the first run cards are?
Now, since the odds of winning the first run and the odds of winning the second run are not independant, you can't multiply the odds together to find out what the odds are of winning both runs...but we don't need to know the odds of winning both runs to calculate the EV, as it turns out.

[MikeL05]

Sigh... the AA vs. KK thing has nothing to do with my original point... it was just to show why Tom's comment ("So your odds of winning the first run are equal to the odds of winning the second run. Therefore, EV is the same.") just makes no sense.

My original point, which no one has really approached yet, was regarding how a player could use two of his outs in the first run in order to gain the same effect that would be gained by only hitting one of the outs.

I have no interest in listening to people just blindly tell me that this doesn't matter because of burning 3 cards vs. 1, or whatever else. I understand how that works, and bringing it up makes me think you're missing my point.

I'd like to see some math here. I'll try to do this myself if I ever get un-lazy enough to take the time to do it. Basically, you should be able to draw some kind of branching probability thing, in which you take the possibilities of the first run (neither player improving, one player improving, the other player improving, both improving, and the diamond flush player improving by using TWO diamonds instead of one to win), multiply them by their probabilities, and figure out the odds of the second run concurrent upon the results of the first run.

Let's go back to what I was saying before, but trying a different example. Let's say instead we run this 12 times. The A/ /images/graemlins/diamond.gif player uses up 2 of his outs to win in the first 6 trials; he now has a 0% chance of winning the last 6 trials. Conversely, if he uses just one of his A/ /images/graemlins/diamond.gif outs to win the first 6 trials in a row, he still has some chance of winning trials 7 through 12.

Now running it once, as normal, none of this matters. But when running it 2+ times, it DOES seem to matter to me how many outs someone uses up to win each individual trial. So it would appear to me that the player who is ahead going into the trials, and thus does not rely on outs, has an advantage in trials in which two outs can be used up.

[jason1990]

What exactly is your point? If we don't know the result of the first run, then the probability that a particular player wins the second run is the same as the probability he wins the first. Do you not accept that? Or do you accept it, but disagree with someone's proof? There has already been sufficient mathematics demonstrated in this thread. So there's no need for you to get "un-lazy." Perhaps you might benefit from trying this problem:

You deal 26 holdem hands (hole cards only, obviously) from a shuffled deck. What is the expected number of hands that are suited?

[moomoocow]

OK - let's do this for a really simple example -

You hold KK
Your mortal enemy holds QQ

The flop and turn come
[ K Q Q ] [5]

you're both all in on the turn - flip your cards and decide to run it twice.

Your EV =
P (1st run is a king) * P(2nd card is anything else) * 1/2pot +
P (1st run is not a king) * P(2nd card is a king) * 1/2Pot +
P(1st card not a king)*P(2nd card not a king) *0

= 1/44 * 1 * 1/2p + 43/44 * 1/43 * 1/2p
= 1/44p

where p is pot sized.

Just as the dealer is about the do the first run. You yell "Stop - we're jsut going to run it once because the eV is different."

Your opponent sighs and says no - the EV from running it once is :

EV = p(river is king) * Pot + p(river not king)*0
= 1/44p

You sheepishly sit down and the dealer proceeds to run it twice - bringing a 4 and a 7. Berating your bad beat, you realize EV is such a silly concept - anyone who cares about "expected" value over and above "realized" value should be worried about paying their "realized" nut - not their "expected" nut /images/graemlins/smile.gif

[MikeL05]

Interestingly, I had an epiphane (sp?) on the bus ride home. So while hitting two diamonds on the first run does slightly reduce the chances of hitting the outs on the second run, I failed to figure in the fact that hitting exactly one out in the first run actually INCREASES the chances of winning on the second run to slightly slightly above that of a single run.

I appreciate all of the sarcastic and/or caustic replies though. I'm not really sure what the point of those was. Do you want to prove how smart you are here? I thought this board was for asking questions and receiving some help or some interesting theory. Stick to that and maybe we'll all get a bit more accomplished here.

Thanks to all of the helpful responses, sorry to clutter the thread with my incorrect hunch.

[mmmmmbrother]

i dont think you understand that when you agree to run it twice, you dont know the outcome for the first run, so the odds are the same.

the way your thinking about it is if you make the decision after t he first run. when you see the first two diamons come up, then it does reduce the chances of winning again, but you didnt know that.

[TomCollins]

Maybe when you take the bus to the 10th grade, you can learn about probability and math and stop relying on intuition.

[jason1990]

[ QUOTE ]
I thought this board was for asking questions and receiving some help or some interesting theory.

[/ QUOTE ]
Believe it or not, I was actually trying to help you. But I guess that was a waste of time, since you don't appear to actually be reading this thread. You're just riding the bus and enjoying the thread in your head. I'm all for epiphanies and it's great that you had one, but if you would just once read this thread carefully, you would see that it's completely irrelevant.

[MikeL05]

And maybe when you take the bus to 5th grade, you can learn about maturity, and stop relying on an online forum to boost your ego and defy your self-esteem issues.

I didn't understand something, and I tried to work it out verbally with some help here. Sorry I'm rusty at math, and I'm sorry you failed to explain it to me in a way that I could immediately understand.

Jason, I wasn't referring to you. Were you being a jerk in your post? I didn't think so. I appreciate the help, what you were saying just wasn't "clicking" with me. Just because I don't get it, doesn't mean I'm not reading and not trying.

[TomCollins]

It seems you'd rather learn by intuition than theory. You aren't going to get much of that at 2+2. If you tried to say what part confused you, we all can explain more parts in more detail.

Was it the addition part or the division that confused you?