Running It Twice - Page 2

TomCollins · #11 06-16-2005, 05:30 PM

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 · #12 06-16-2005, 07:13 PM

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 · #13 06-16-2005, 07:16 PM

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 · #14 06-16-2005, 08:14 PM

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 · #15 06-16-2005, 08:29 PM

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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.

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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.

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

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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.

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I think textbook writers agree with you. Problems like this are very common.

MikeL05 · #16 06-17-2005, 11:51 AM

[ 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 · #17 06-17-2005, 04:11 PM

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 · #18 06-17-2005, 04:23 PM

[ 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.

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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 · #19 06-17-2005, 04:44 PM

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/ [img]/images/graemlins/diamond.gif[/img] 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/ [img]/images/graemlins/diamond.gif[/img] 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 · #20 06-17-2005, 05:02 PM

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?