Don't know if it ok to open a new thread here, I just think people might not see my reply at the original thread.

First of all, I havent read all the replies, I just found Andy's calculation is logically unjustified.

Now, why did I say that?

There is one fact, that 'Andy' ignored, and he called it 'subtlety', that there is only 52 card in a deck. Personally, I don't think this is subtle at all, nor it is a minor issue. In fact, this is the ONLY issue that matters.

Ignoring this fact, which means the probability of drawing out as a underdog remains the same, the equity will OF COURSE, ALWAYS stays the same. You really dont need any calculation to draw that conclusion. Because all you did, is run the EXACTLY same hand a number of times, and each and everyone of them has the EXACTLY same probabilty of winning/losing.

Edit: My apologies to Andy, I was wrong at this point of implying the EV is going to change. But my point remains the same

Now why isnt it good to run it more times when you are a FAVOURITE?

NO, IT IS NOT THE VARIANCE

That is because, you are effectively giving more chances to get out DRAWN.

REMEMBER there are only 52 cards in a deck, and are only a few cards that can help the DOG. As a favourite, you DONT WANT TO SEE THEM. If you ran it twice, you are given out more chances to SEE them, that is IT.

The chances of hitting 3 cards when there is 35 cards left is much higher than there is 40 left, isn't that obvious?

I think this is very simple to understand?


Now because this is a 'mathematical' discussion, let's see some numbers.

Lets use a rather simple example (coz I am lazy)

AK vs KJ, and you are dead to a JACK on the flop:

you have two chances to hit remaining 3 jacks in the deck, the estimated probability is: (considering only you and your opponent's hands)

(3/45)*(39/44) = 5.9%
-> one jack on the turn, a card that is NOT an ACE on the river (Situation A)

(42/45)*(3/44) = 6.36%
-> one card that is NOT an ace on the turn, and a jack on the river (Situation B

(3/45)*(2/44) = 0.3%
-> hit runner runner Jacks (situation C)

all toghether the estimated probability that KJ will draw out on AK is:
12.56% (might be a different from the simulation results, this is just an estimate anyway)

So the probability of NOT drawn out is about 100% - 12.56% = 87.44% (Situation D)

Now what if we run it twice?

The first run obviously is identical to the above

now the second run:

provide that it is situation (D)

we had a new set of probabilities, similar to the above caculation only now there are only 43 cards left:

(3/43)*(37/42) + (40/43)*(3/42) + (3/43)*(2/42) = 13.12% (Situation E)

Notice 13.12% > 12.56%

This is the probability that KJ drawn out on AK PROVIDE THAT it didnt drawn out in the first run, the overall probability for this to happen is:

P(D) * P(E) = 87.44%*13.12% = 11.47% (Situation F)

THIS IS A SPLIT POT, but this is only one situation

IN total we have the following situations:

1) when we draw out on AK at the one time, but didn't draw out at the other time (this has eight combinations, 1st turn, 1st river, 2nd turn, 2nd river, 1st turn and river and NO JACKS in 2nd, 1st turn and river and one ace one jack, 2nd turn and river and no jacks in 1st run, 2nd turn and river with one ace and one jack in 1st run)


2) when we draw out on AK at the first time AND we draw out on AK at the second time (this has eight combinations, 1st turn+2nd turn, 1st turn+2nd river, 1st river+ 2nd turn, 1st river + 2nd river, 1st turn and river + 2nd turn and no ace on the river, 1st turn and river + 2nd river and no ace on the turn, 1st turn and no ace on the river + 2nd turn and river, 1st river and no ace on the turn + 2nd turn and river)

3) we didnt draw out on neither runs:
this is rather easy to calculate: 87.44%*(1-13.12%) = 75.98% (Situation G)

Notice the drop in probability compare to situation D

which means what? your AK is that 87.44% - 75.98% = 11.46% less likely to hold for the whole pot!

Since no one is paying me for doing this, Im not gonna bother about the details. But HOPEFULLY you can see why it is NOT GOOD to run more than once as a favourite.

that is a lot of math, but i think it is basically saying what always confused me about the situation.

by not shuffling the turn and river cards back into the deck after the first deal, the second deal has different odds than the first. i don't think it actually changes your outcome, but i don't have the math to explain why.

can someone give a non-math explanation as to why this does not effect the outcome if you are
a) about even (flush draw + overcards against pair) or
a big favourite (like sammy's kings vs. aces).

if the odds of the second deal are different than the first, why does the expected outcome not change?

As everyone has been saying. .

Running it twice makes no difference to the EV and it makes no difference whether you reshuffle the deck or not. If you miss the first time your odds of drawing out will increase the second time, but the reverse is also true and if you hit the first time you're less likely to hit the second time. It cancels out.

This is the simplest possible example:

Say you have a set of Kings against a set of Aces on the turn and the pot has $44. So the set of Kings is drawing to one out. If you run it once, he'll win 1/44 times, so his EV is $1. If you run it twice and don't reshuffle the card, he'll get half the pot 1/22 times (he can never win the whole pot) so his EV is again $1. If you run it twice and do reshuffle the card, he'll win the whole pot 1/1936, get half 86/1936, and get jack squat 1892/1936, so his EV is:

44 * 1/1936 + 22 * 86/1936 = 1.

bolded word should be edited.

otherwise, thanks - exactly the explanation i was looking for.

Sigh, Please understand the example in my post:

run it once, you have 87.44% to hold and win the whole pot

run it tiwce, you have 75.98% to hold and win the whole pot

Tell me which situation you rather be in?

OF COURSE the EXPECTED VALUE stays the same if you put every card on the deck out, the payout would be much closer to the EV isn't it?

BUT thats not the point, why would you given out on the situation that you are have a HUGE edge to win the whole pot, and give your opponent a FAIR shot at his money?

JUST FEAR the BAD BEATS?

this can only be justified if the pots are so HUGE you are becoming risk averse as the favourite , which is not a case on day to day plays.

Okay I understand your point. By agreeing to run it twice you do give yourself less chance to win the whole pot, but also less chance to come up empty. It's like the difference between a 10% chance to win $5000 and a 50% chance to win $1000 -- mathematically it's the same thing, but psychologically there's a big difference.

There are tons of reasons why you might want to decrease your variance. For example, say you feel you are the best player at the table and get it all in as a 2/1 favourite against a weak player. Now it would obviously be to your advantage to run it twice here, since you want to limit the role of random luck, and even if you only split it, you should be able to exploit your greater skill later in the game. By the same token, if you were in the same situation against a better player, you would only want to run it once, because you want to give yourself the best possible chance of breaking him.

In most situations, in cash games, I don't care in any one situation whether or not I'm sucked out on. This is because I'm not overplaying my bankroll, and over time, so long as I know I get my money in as the favorite, I will come out ahead. When I'm ahead, all in, online where you can only run it once (though I think an ability to run it twice, if implemented correctly, could be a nice software addition to online poker sites), and I suffer a bad beat, I just smile and rebuy.

I would still choose to run it twice in several situations despite the fact that I don't "fear the bad beat". Dingas brings most of them up (which, in turn, I brought up in the other thread). I'd rather run it twice against a bad player that will leave if I bust him, because bad players leaving the table is -EV for me. Vice versa for a strong player that I want to bust. Additionally, there's the tilt factor: if I think I have a chance at getting a player to tilt, I will gladly do whatever I can to bump up the variance.

But this is not a mathematical issue. Yes, we all get that by running it twice, we are decreasing the chance that we have of scooping the whole pot as a favorite, and we are also decreasing the chance of us being sucked out on. None of that matters in the long term, so I am concerned only with the metagame effects of variance in the short term.

Now, what you write here is nice and I appreaciate your thinking compeletely.

As you might noticed the reason for me to write this thread is what Andy says in the original thread:" I would run it as many times as you want as a favourite" and that is just not correct.

Here is what I have to say after your reply in the original thread:

If you are the favourite, your best chance to win the whole pot is to run it once. When you run multiple times, your chance of winning that proportion of the pot is going to be reduced.


Sure it doesn't matter if you add them all up and talked about EV, by my point is very simple, if run it multiple times your are getting less in your chance of winning the whole pot.

Think it this way, if you deal every card in the deck out run by run, your opponent will almost always get a proportion of the pot. (apart from rare situations where your killer card presents itself in the same run with his) In this situation, he is getting a fair shot at 'his' equity.

If now, everybody plays this way, all you have to do is to pay the exact money to get your equity and that would be a 'ultimate fair game' Because everyone makes the correct decisions and everyone gets his/her deserved equity (which is now the money they are paying), in a game like this everyone would go home with the money they brought in.

But thats not poker. That would make the game pointless.

Now, I think I have explained my point fairly simply, whether to appreciate it or not, it's up to you.

That is exactly what I was talking about.

Here your 'better player' is a risk-averse person and the 'worse' player is a risk-seeking person.

Now I don't understand the reason that why they are taking such an attitude, especially for the worse player you just described.

The better player I kind of understand because you don't want a bad beat. So in that situation you would agreed to run it twice with better cards. (Still not justified IMO)

The worse player would get a LARGE increase of probability to win some money, and I think that is a huge gain. Because as the underdog your chances of breaking the better card, by definition is not good. personally I would always agree to run it more times as the underdog. In fact, here is the place I would ask to run it 'as many times as you like'.

Op:

Your math is wrong. You are wrong. In your discussion above, you don't condition the second run through based on every outcome from the first run through. You should have 16 different calculations for the two run through and compare those to the 4 calculations for only 1 run through. You have to play out each scenerio with both runs through. Yes, they may hit on the first run through, but they may miss on the second run through and will only get half the pot. The pot is divided evenly by the amount of times they run the hand, remember? It is possible to hit on the first run through, miss on the second run through, and only get half the pot. Your calculations do not take into account these scenerios, and therefore are incomplete and misleading.

The expectation is exactly the same. Reread the post where Andy Beal gives the proof. It is correct. It is based on the independence of expectations and the linearity of expectations. The expectation is precisely and exactly THE SAME no matter how many times you run it. In fact, you can run it in many strange and odd ways and the EV will be exactly the same. Sorry, but you're not right.

http://www.fullcontactpoker.com/poker-foru...showtopic=54819


If you want a more concrete example, I have a nice example in the same post. I did what you tried to do above, but I did it right. It took me a bit to get the calculations right, and I don't include them all in my post, but I guarantee you that they are correct and demonstrate clearly that both cases have the same EV.

Seriously, read this thread again and again until you get it.

When did I EVER say that the EV is changed?

Yes on one category of situations I didn't seperate if the ACE hit on the river or on the turn in some of the runs. But it doesn't matter.

No I am NOT wrong by saying you are more likely to get outdrawn in the 2nd run once you hold up in the 1st run. Yes I didn't metion that you have already secure the other half of the pot.

But YES you are less likely to hold up in the 2nd run if you did hold up in the 1st one.

I will have a look at your proof and get back to you, yes I will appreciate all the things you have to say.

Please believe me if your calculation make sense, I am going to get it.

Also, if you like to really check my calculations, I 'translated' my example into a similar format of yours:

AK versus KJ, on the flop

flop: BBB ( B = BLANK, here it represents every card other than A or J )

Situations when KJ draw out on AK:

A)BBBJB
B)BBBBJ
C)BBBJJ

There ARE no other situations when KJ can win this hand, these are all covered by my calculation.

IF that is the first run, AK needs to avoid these situations to hold up, so none of these situation has happened.

NOTE: There are three situations that AK can hold up but diminish the chance to hold up in the second run:

1) BBBAB
2) BBBBA
3) BBBAA

I DIDNT consider them because in those situations it will further decrease the chance for AK hold up in the 2nd run.

Now the 2nd run, NOT considering situations 1),2),3):

The situations will be identical to above apart from the fact that there are less cards in the deck that is BLANK.

All those situations that KJ can outdrawn AK in the second run, when AK hold up in the 1st run IS covered by my calculations.

No my calculation is NOT wrong when I calculate the probability of KJ drawn out in Neither runs. You don't need to do enumeration for this.

IF you can have a look at this and prove me wrong, by all means.

Thank you.

You didn't understand what I was doing with my calculations. You don't need to lay out the flop as all blanks.

Also, your discussion uses counter outs and is much more complicated of an example than mine is. Try to use an example where there are no "counter outs" at first. Something like Ak vs AJ is a bit easier since there are A LOT fewer combinations to explicitly enumerate.

When considering running it twice, you must lay out every different combination of hitting outs. Here's what mine meant:

HM HM <-- For example meant hitting an out on the first turn and missing on the first river and hitting again on the second turn and hitting again on the second river.

There are 16 of these combinations that you have to write out and calculate the odds for. Let's do this one for practice: Let's assume we have 5 total outs (just so we don't "run out of outs" in the HH HH scenarios. I assure you this won't chance weather the EV is altered but rather only makes for an annoying example)

(5/45)(40/44) * (4/43)(39/42)

Now, in the above example, the person gets the entire pot, so we award them 1.

In an example like:

HM MM <-- The person would only get half the pot, so we give them 1/2.


To do the calculation correctly, you have to write out every one of these probabilities and multiply them by the amount of the pot they win (0, 1/2, or 1) and add all these numbers up.

You then have to show that this number is the same as if you only ran it once.

I've done this, though. I've actually done it in terms of x "outs" which makes it slightly more general. Doing it, I get a rather ugly equation in terms of x. Shockingly, though, it simplifies to the same equation that I get when I run it only once.

This is where you go wrong. In your analysis above, you consider an ace as a counter out. But you don't enumerate the percentages when one or two aces hit the first time, reducing the counter outs, or when two jacks hit, giving us only one more out. Your example is very complicated and for your discussion to be complete, you have to go into A LOT of permutations.


So you should agree that running the hand any number of times produces the same EV. The only thing that is reduced is the VARIANCE. Think about it. If you're a 66% favorite to win and you run it twice and you win one and you lose one, the value you get from the hand is closer to the true ev than if you run it once and win. Your variance about the true EV is reduced (ie 66-50 is 16, 100=66 is 34).

The increased chance that you will be outdrawn on the second run conditioned on being outdrawn on the first run only drives the most probable result from two runs closer to 50%. Variance is reduced.

In the long run over many hands, it doesn't matter how many times you run it. The only case where I can see it mattering is if a person has their entire bankroll on the line and they are only playing this one hand ever. But this example requires a consideration of Utility, which is beyond what I'm prepared to discuss in this post. I hope this helps.

LongLiveYorke,

I am very pleased that there are people out there who actually understand the difference between a real mathematical proof and an ad hoc argument or "proof by example" or argument by intuition. Neither of the latter are proofs, as I know you are well aware.

That being said, I am still not sure what the creator of this thread's point was. EV is the same. Period. It is. I will be willing to bet any sum of money you want and we will test it against a million, a billion, a trillion number of simulatations on a computer if anyone would like. Just post here, we can name the stakes, get the money in escrow, and set up a independent third party for the simulations.

Now, if his point was an argument for or against running it once or twice, well , there is no one right answer since it depends on each individual's inherent tolerance for risk. And I thought I made that point loud and clear long ago in my post by arguing way one player might very well want to run it twice as a fav/dog and another might not want to run it twice as a fav/dog.

NOW, on to responding to parts of the original poster's comments...



EV does not change. I am befuddled as to what your point IS.

[quote name='Mriya' date='Friday, April 7th, 2006, 5:02 AM' post='972659']
Now why isnt it good to run it more times when you are a FAVOURITE?

NO, IT IS NOT THE VARIANCE

That is because, you are effectively giving more chances to get out DRAWN.
[\quote]

Sorry, but it has everything to do with the variance. Either one chooses to enter into a transaction with probability p (where maybe p>.5 or p<.5). AFTER that choice, a player may or may not choose to reduce his variance. There is not "right" answer since in the end the EV is invariant. It is a matter of taste and as such I don't know how one can "prove" which choice is "correct".

[quote name='Mriya' date='Friday, April 7th, 2006, 5:02 AM' post='972659']
Since no one is paying me for doing this, Im not gonna bother about the details. But HOPEFULLY you can see why it is NOT GOOD to run more than once as a favourite.
[\quote]

LOL, again, if EV is the same (as you yourself agree above) how can either decision be termed "good" or "bad"? The expectation is the same! This is why there is no one answer: variance is a matter of choice when EV is invariant.

Running it once provides the best chance to scoop the whole pot for BOTH players.

Someone who's 5% to win a pot will be .25% to scoop if it's run twice.
Someone who's 95% to win a pot will be 90.25% to scoop if it's run twice.

Who cares?

Increasing the probability of scooping shouldnt be of any concern if there's no difference in the average value of what you'll win.

I think we have settled everything in this thread:

http://www.fullcontactpoker.com/poker-foru...pic=54819&st=20

I take full responsibility for not made myself clear to everyone, especially to Andy whose original point stands completely as it is.

My point never directly contradict to his original point so as far as I am concern the title is inappropriate.

Abba is right, the chance of scooping matters only at the particular pot, and not if you consider all the hands you play all toghether.

I thank everyone who has been helpful to me. Particularly Shaffer.

However longliveyorke, If you read my post carefully, you will notice I have already explained your concerns on my calculations. Yes I understand your enumeration (and your format please don't assume I can not understand a simple enumeraion exercise, I take no offense nonetheless), and yes that is a correct way of doing it. But NO my calculations does not contradict yours and NO that difference in probabilities are NOT changes in EV.

Thank you for your help though, and the same goes to everyone.

Interesting.

I might be dumber for having read this--then again, I might be smarter. All the while, I can't help but feel like the Math Nerds in high school just got into a bit of a scuffle only to put their compasses and protractors down and shake hands.

Nice ending, fellas.