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#1
10-31-2005, 10:51 AM
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Theoretical problem about coinflips
Maybe easy to answer, and I certainly think I know the answer to this, but I'd like to see a definitive closure to the following question:
Assuming Independent chip model is accurate, opponents are equal in ability, and blind sizes are negligible to the problem, is there any situation where a known coinflip should be taken when an opponent pushes all in before you? Also assume that no matter how many players are at the table, showdown will only involve you and the lone pusher. Put another way, if I know for CERTAIN that my odds of beating an opponent in a showdown are exactly 50/50, and this opponent pushes all in ahead of me, are there any conditions under which I can call profitably. So just pull out the ICM calculator and find me one example. Some more complicated questions to answer if you are feeling really ambitious: If there are circumstances where you can call an opponent profitably with a true coinflip, what is special about those circumstances in general that makes this so? Conversely, if there are no circumstances under which this is the case, I'd like to see a proof. Regards Brad S As a corrollary to this problem, does the answer to this make sense given what we know about the relative value of chips in different sized stacks? Ie - common tourney knowledge often suggests that calling with a suspected coinflip is wise if we have a huge stack in comparison. This is becasue the small stack's chips are worth more than ours and we may actually have pot odds to do so when considering the 'extra' value of shorty's chips. 
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#2
10-31-2005, 10:58 AM
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Re: Theoretical problem about coinflips
I'm sure you know this already but yes, there certainly are situations when you should call as a 50/50, and even as a 45% dog and worse. They're rare but they do occur.
To give an example let's say you're in the BB with A8s and the SB has pushed. If you fold your EV is 0.1556 (and the SB's goes up to 0.1556 also). If you call and win your EV goes up to 0.2716, and the SB busts, thus 0. If you call and lose, your EV goes down to 0.0475, and the SB's up to 0.2430. In this case, you only have to be a 49.6% favourite, and can call as a coin flip, even assuming you want a 0.3% edge.
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#5
11-01-2005, 11:34 PM
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Re: Theoretical problem about coinflips
Here's what I know so far.
1. Assume you have the option to call (but not raise) the final bet by the only other player still in the pot. Imagine a bet putting one of you all-in or a bet on the river. 2. There may be dead money in the pot from blinds or earlier bets. However, assume the bet is "fair," i.e. the pot odds equal the odds against your winning the bet. 3. And so far, unfortunately, assume the tournament pays only two places with at least three still in it. The ICM then implies that calling the bet has negative EV. Sketch of the proof. Your total EV is comprised of several terms. It is fairly easy to check that your expected value for finishing first or for finishing second when anyone except the bettor finishes first is the same whether or not you call the bet. The only remaining term is when the bettor finishes first and you finish second. Here your EV decreases if you call the bet. In a nutshell, this is because f(z)=(y+z)(x-z)/(1-y-z) is concave down (x represents your percentage of the total chips, y the bettor's, and z your loss, which could be negative). This same proof shows that raising on the river if you are certain your opponent will call is worse than calling.
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#6
11-02-2005, 04:26 AM
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Re: Theoretical problem about coinflips
[ QUOTE ]
Here's what I know so far. 1. Assume you have the option to call (but not raise) the final bet by the only other player still in the pot. Imagine a bet putting one of you all-in or a bet on the river. 2. There may be dead money in the pot from blinds or earlier bets. However, assume the bet is "fair," i.e. the pot odds equal the odds against your winning the bet. 3. And so far, unfortunately, assume the tournament pays only two places with at least three still in it. The ICM then implies that calling the bet has negative EV. [/ QUOTE ] I disagree. With dead money in the pot it should be possible to construct a situation where calling has a positive ev. Unless the stuff about it being "fair" means this isn't possible, in which case just simplify the problem by assuming 0 dead chips. [ QUOTE ] Sketch of the proof. Your total EV is comprised of several terms. [/ QUOTE ] Ok. [ QUOTE ] It is fairly easy to check that your expected value for finishing first or for finishing second when anyone except the bettor finishes first is the same whether or not you call the bet. [/ QUOTE ] I don't see how this works, can you explain this further? [ QUOTE ] The only remaining term is when the bettor finishes first and you finish second. Here your EV decreases if you call the bet. In a nutshell, this is because f(z)=(y+z)(x-z)/(1-y-z) is concave down (x represents your percentage of the total chips, y the bettor's, and z your loss, which could be negative). [/ QUOTE ] Where does this function come from? I agree that it is -ev if you might be knocked out, but am I mistaken when I say that it is never -ev to call if you have more chips than the villain? [ QUOTE ] This same proof shows that raising on the river if you are certain your opponent will call is worse than calling. [/ QUOTE ] I'll have to think about this.
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#7
11-03-2005, 02:55 PM
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Re: Theoretical problem about coinflips
By grouping terms in just the right way, I was able to extend my proof to cover tournaments that pay 3 places. Under the independent chip model, when at least 3 remain in the tournament, it is ALWAYS negative EV in terms of tournament winnings to take part in a bet with one other player for which your chip EV is 0 (or less). Without mathematical typesetting available, I don't think I can convey any details in this forum. (The algebra is sufficiently involved that I used a symbolic manipulation package to double-check my work.)
Hopefully, more to come. If successful, I'll probably eventually write it all up in a format I could send out at some point, although 1st semester calculus and the idea of partial fractions might be a prerequisite for reading it. [ QUOTE ] Here's what I know so far. 1. Assume you have the option to call (but not raise) the final bet by the only other player still in the pot. Imagine a bet putting one of you all-in or a bet on the river. 2. There may be dead money in the pot from blinds or earlier bets. However, assume the bet is "fair," i.e. the pot odds equal the odds against your winning the bet. 3. And so far, unfortunately, assume the tournament pays only two places with at least three still in it. The ICM then implies that calling the bet has negative EV. Sketch of the proof. Your total EV is comprised of several terms. It is fairly easy to check that your expected value for finishing first or for finishing second when anyone except the bettor finishes first is the same whether or not you call the bet. The only remaining term is when the bettor finishes first and you finish second. Here your EV decreases if you call the bet. In a nutshell, this is because f(z)=(y+z)(x-z)/(1-y-z) is concave down (x represents your percentage of the total chips, y the bettor's, and z your loss, which could be negative). This same proof shows that raising on the river if you are certain your opponent will call is worse than calling. [/ QUOTE ]
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#9
10-31-2005, 11:09 AM
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Re: Theoretical problem about coinflips
Unless I misread that post, you make no assumptions about equal stack sizes. So I can think of two spots.
1) You have the pusher well covered. As an extreme, say you're 5 way and he has one chip. Or you have 7k and four people have t250 and one pushes. In these cases you benefit from busting a guy and really can't hurt your stack. I'm not going to sit on the ICM calculator but I think most people in the forum would make that call. 2) You're down to 2 players. Kind of a trivial example but hey, it works.
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#10
10-31-2005, 11:13 AM
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Re: Theoretical problem about coinflips
[ QUOTE ]
1) You have the pusher well covered. As an extreme, say you're 5 way and he has one chip. Or you have 7k and four people have t250 and one pushes. In these cases you benefit from busting a guy and really can't hurt your stack. I'm not going to sit on the ICM calculator but I think most people in the forum would make that call. [/ QUOTE ] It's extremely close, but ICM calc disagrees if blind sizes are negligible. At least as far as I have checked. That said, I too make these calls all the time, though I think it's just because usually the blinds really matter. [ QUOTE ] 2) You're down to 2 players. Kind of a trivial example but hey, it works. [/ QUOTE ] I suppose when down to 2, blind sizes can never really be negligible, so you are right I guess I am thinking more about situations with 4+ players, (and maybe with 3, though I am less sure about that) Regards Brad S
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