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.

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.

Re: Theoretical problem about coinflips
 

[ QUOTE ]

In the specified scenario it's never -ev to call when you have more chips than your opponent.
I thought that was interesting.


[/ QUOTE ]

This may be wrong.

Re: Theoretical problem about coinflips
 

when you have no FE and a coinflip gives you FE if you double up

Re: Theoretical problem about coinflips
 

[ QUOTE ]
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.

[/ QUOTE ]

Yes there are. But they all involve meta-game and EV past this hand. If you demonstrate that you will make calls, even coinflip and slightly the worst of it then you may prevent people from stealing/pushing in front of you.

Re: Theoretical problem about coinflips
 

[ QUOTE ]
when you have no FE and a coinflip gives you FE if you double up

[/ QUOTE ]

This is outside the strict mathematical aspects of this single hand, but actually, yeah, good example.

This hits to the heart of some issues that I have with SNG mathematical methods, especially concerning the conversion of chip values to their dollar equivalent. ICM.

On the one hand, we can say 'that's not fair, you are introducing other factors which do not really fit into the model which we are assuming is correct'

Lets face it though, Chips do NOT have a dollar equivalent in tournaments, even if that is what ICM tries to approximate for us.

The reason is simple - You don't get paid until the tourney is over, and so what happens on the next hand really matters. If your chip stack grants you less mathematically describable advantages (like FE) then these things cannot be ignored.

On that same line of thinking, even a very very big stack might have slightly more value than we suppose because it may actually tighten the calling ranges of opponents we have covered. This matters in future hands, and so despite how ICM may approximate our momentary equity, we cannot ignore hands that are yet to come in which our equity will be modified for the better by tighter calling ranges.

Now, this is not original and many people have been saying these kinds of things for a while, but still, I find myself more intrigued with these issues lately. In the past, I have implied that ICM doesn't ignore future hands, because it is based on probability calculations about what those future hands will result in.

Let me make clear, that I am happy with the ICM. I think it does what it is supposed to. It gives us a reasonable and functionally accurate estimate of what our equity in a tourney prize pool is, based on each player's respective chip position. nothing more.

What matters, I think, is that we get better able to recognize where future hands will modify what ICM calculations (when extended to hand analysis) are telling us, because there are factors outside the scope of the model.

Fold equity is a good example. Table image is another good example, and already mentioned in this thread. A well timed, slightly -$EV play when you are big stack can reap long term rewards (not considered in the usual ICM hand analysis) if it tightens up the range of hands that opponents will push into you with. Another example is having a shortstack at the table, and our occasional desire to keep a bubble going if the presence of the shortstack reduces the number of hands that our own all-ins will be called with.

I do not think that we need to modify ICM to account for these things. I have thought in the past about 'weighted' models which slightly overvalued/undervalued stacks, but I think this is a bad approach.

What we need to do is change the extension of our reasoning when we use the results of the model in hand analysis. For example, even if we find a situation where we can push with 74o for a slight gain in equity, this slight gain may not be worth the future ramifications of (possibly) having to show a 74o push.

To make what I mean clear, consider good old fashioned pot odds calculations. Nobody ever felt the need to alter our value of money when implied odds were brought into the picture. We simply realized that the potential for winning future bets meant that those pot odds calculations were incomplete, and that we could go against these calculations when we took this into account. We know we can call with a flush draw even if we aren't immediately getting proper odds, so long as we suspect that we have the potential to win future bets which make the play profitable.

In the same respect, I think ICM only provides us with immediate equity estimates, and can be overriden by future +$EV opportunities in certain situations. Rather than having this equity come from future bets in the hand, it now comes from future hands within the tournament. I think this was once suggested as 'implied equity'.

I am just thinking out loud here, and as I say, I'm sure this isn't news to many of you. This is especially obvious to me now that I am looking over this thread another time. Oh well...

Any other thoughts?

Regards
Brad S

Re: Theoretical problem about coinflips
 

[ QUOTE ]

[ 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 ]

I have dusted off my algebra and managed to derive this result.
(basically the +chips and the -chips cancel each other out.)

[ QUOTE ]

[ 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 ]

I also managed to derive a similar, although less elegant formula. Which shows the same thing.
Your mathematics is obviously much better than mine.

I assert that for tournaments that pay out more than 2 places further placings may slightly alter the curve they will not fundamentally alter it's shape.
Therefore the result above holds for all (regular) payout structures.
I don't have the mathematical ability to prove this, so I challenge anyone to disprove it. [img]/images/graemlins/smile.gif[/img]

In summary:

I now agree with NoSelfControl and believe his proofs and mathematics to be good.

It is always -ev (or at best breakeven) to call.
Even if you have the bigger stack..

Insty
who needs to study some more mathematics.

Re: Theoretical problem about coinflips
 

Good post.

Re: Theoretical problem about coinflips
 

If forced to GUESS, I would agree with Insty that it will always be negative EV when more than two players remain in the tournament and only one player is in the pot. However, it is impossible to use a term-by-term approach to prove it.

On the other hand, it MAY be positive EV to call a fair bet in a multiplayer pot because your being in the pot may affect some players' probabilities of winning more than others.

Interesting problems; I'll think about them some more over the next few days.

Re: Theoretical problem about coinflips
 

[ QUOTE ]
I am just thinking out loud here, and as I say, I'm sure this isn't news to many of you. This is especially obvious to me now that I am looking over this thread another time. Oh well...

[/ QUOTE ]

Brad, allow me, as somebody who's still coming to grips with ICM in many ways, to ask you to never stop "thinking aloud" on these forums. [img]/images/graemlins/smile.gif[/img]

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 ]

Re: Theoretical problem about coinflips
 

No because of the declining value of chips in a tournament.

Re: Theoretical problem about coinflips
 

[ QUOTE ]
No because of the declining value of chips in a tournament.

[/ QUOTE ]

This is a pretty vague statement. For instance, it's easy to come up with an example where, in a single hand, your chips have increased AND your tournament $ EV per chip has increased, i.e. the average value of your chips has gone up as your stack has increased. (My example is an all-in where you are short-stacked, win the main pot while the big stack busts several others out.) Can you define "value" and "declining value" precisely? (Even better would be a reference that has a proof.)

In the context of the OP, I think I can now show that $EV can only go down in tournaments that pay up to 4 palces. The jump from 3 to 4 wasn't as easy as I had hoped at the time of my earlier posts.