ICM Question

[iMsoLucky0]

Noob question. I have seen this term used some, but I don't know wha tit means.

Anyone care to enlighten me?

[ChrisV]

The Independent Chip Model is a model that seeks to predict the percentage share in the prize pool of all remaining players, based on their chip amounts. Its limitations are that it assumes everyone plays the same and that it does not take into account how playing power varies with stack size (e.g. having a big stack on the bubble).

There's a calculator here (http://www.bol.ucla.edu/~sharnett/ICM/ICM.html).

[KJ o]

Is there a formula available anywhere? I would like to be able to calculate it myself. In particular, I am looking at combining it with the twodimes.net type calculations, to get $EV in addition to CEV.

[AleoMagus]

That seems to be the million dollar question.

I actually thought I had a pretty good handle on how exactly ICM calculations were achieved until I tried to answer a thread asking this very question.

After an hour of trying to derive an exact formula or prodedure to get ICM results, I gave up because mine never matched up with the ICM calculator.

I would also very much like to know the answer to this question. I've seen the links in the dethgrind thread and have seen the coding, but it all means little to me.

Regards
Brad S

[ilya]

I think it's fairly complex. My dad's a math Ph.D. and he doesn't seem to think that it's trivial.

[hansarnic]

The formula is basic, just very time-consuming to use if you have lots of players.

For example, for three players, A (5000 chips), B (3000), C (2000), prize money $250, $ 150, $100

Possible results:

ABC
ACB
BAC
BCA
CAB
CBA

Take A's chance of winning first (5000 / total chips in play = 50%).

He can come 1st one of two ways (either B or C comes second). B comes second 60% of the time (3000 / remaining chips (which is 3000 + 2000), meaning C comes second 40% of the time which is (2000 / (2000 + 3000)). So you then have a % chance of these two possibilities (ABC is 50% x 60% = 30% & ACB is 50% x 40% = 20%), then repeat for other possible outcomes. Total up the % chance of each player in each position & multiply by the prize money to calculate $EV.

Never done it 'manually' for four players & upwards though, which is where it gets time-consuming....

[The Yugoslavian]

[ QUOTE ]
That seems to be the million dollar question.


[/ QUOTE ]

Someone has to know this code, right? Who came up with the original ICM stuff, was it Bozeman? Eastbay?

What would be really cool is if there was an application that followed chip counts and your hand in real time from your table(s) and all you had to do was plug in the range you felt your opponent had in order to get CEV and $EV figures for calling allins. I suppose if you could hook that up you could probably also create an algorithm to use some sort of range of calling hands for opponents behind you when open pushing and it could calculate your $EV in that situation too based on how likely you'd run into calling hands and such.

Of course, this is really close to bot territory I suppose but it'd still be a pretty sweet tool for making your decision.

Does everyone just still use something like Pokercalc or Pokerstove after the fact to calculate CEV of your hand vs. range of hands, tally up the chip counts that result from a fold or push and then plug that into ICM?

Yugoslav

[eastbay]

[ QUOTE ]
[ QUOTE ]
That seems to be the million dollar question.


[/ QUOTE ]

Someone has to know this code, right? Who came up with the original ICM stuff, was it Bozeman? Eastbay?


[/ QUOTE ]

I wrote the code that drives dethgrind's web page. I'm not sure what more there is to say about it other than what's already been said in prior threads, or what the code itself says.

I certainly didn't invent the idea and neither did Bozeman.

eastbay

[tallstack]

I know that many people have written their own versions of these and I have one for 3 and 4 player scenarios that works pretty well for situations like the one dethgrind posted about link (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=1122239&page=&view=&s b=5&o=). If you, or anyone else, would like to have a look then let me know and I will forward it to you.

Regards,

Dave S

[AleoMagus]

Thanks, that's an excellent explanation

Regards
Brad S

[rachelwxm]

[ QUOTE ]
The formula is basic, just very time-consuming to use if you have lots of players.

For example, for three players, A (5000 chips), B (3000), C (2000), prize money $250, $ 150, $100

Possible results:

ABC
ACB
BAC
BCA
CAB
CBA

Take A's chance of winning first (5000 / total chips in play = 50%).

He can come 1st one of two ways (either B or C comes second). B comes second 60% of the time (3000 / remaining chips (which is 3000 + 2000), meaning C comes second 40% of the time which is (2000 / (2000 + 3000)). So you then have a % chance of these two possibilities (ABC is 50% x 60% = 30% & ACB is 50% x 40% = 20%), then repeat for other possible outcomes. Total up the % chance of each player in each position & multiply by the prize money to calculate $EV.

Never done it 'manually' for four players & upwards though, which is where it gets time-consuming....

[/ QUOTE ]

Good explaination. One assumption made by ICM is that while I understand your shot of first place is proportional to your stack, I still don't know why it distribute stack proportional to the rest of the people. In your case
A 1st =50% (easy to understand)
ABC =30% ACB=20% (not so trival to me)
/images/graemlins/smile.gif

[hansarnic]

[ QUOTE ]
[ QUOTE ]
The formula is basic, just very time-consuming to use if you have lots of players.

For example, for three players, A (5000 chips), B (3000), C (2000), prize money $250, $ 150, $100

Possible results:

ABC
ACB
BAC
BCA
CAB
CBA

Take A's chance of winning first (5000 / total chips in play = 50%).

He can come 1st one of two ways (either B or C comes second). B comes second 60% of the time (3000 / remaining chips (which is 3000 + 2000), meaning C comes second 40% of the time which is (2000 / (2000 + 3000)). So you then have a % chance of these two possibilities (ABC is 50% x 60% = 30% & ACB is 50% x 40% = 20%), then repeat for other possible outcomes. Total up the % chance of each player in each position & multiply by the prize money to calculate $EV.

Never done it 'manually' for four players & upwards though, which is where it gets time-consuming....

[/ QUOTE ]

Good explaination. One assumption made by ICM is that while I understand your shot of first place is proportional to your stack, I still don't know why it distribute stack proportional to the rest of the people. In your case
A 1st =50% (easy to understand)
ABC =30% ACB=20% (not so trival to me)
/images/graemlins/smile.gif

[/ QUOTE ]

ABC and ACB are the only two ways A can win, therefore must total to 50%.

The chance of results ABC and ACB are determined by B & C's stack size. The formula assumes that when A wins, B's chance of coming second is calculated by his relative stack size vs. C.

B has 3000 chips, C has 2000. Therefore B will come second 50% more than C as he has 50% more chips. Or expressed another way, B's chance of coming second (when A wins) is his stack size (3000) divided by the total remaining chips (B+C or 3000 + 2000).

[rachelwxm]

[ QUOTE ]

B has 3000 chips, C has 2000. Therefore B will come second 50% more than C as he has 50% more chips. Or expressed another way, B's chance of coming second (when A wins) is his stack size (3000) divided by the total remaining chips (B+C or 3000 + 2000).

[/ QUOTE ]
Why B comes 2nd 50% more than C? Any intuition beyond the math?

[eastbay]

[ QUOTE ]
[ QUOTE ]

B has 3000 chips, C has 2000. Therefore B will come second 50% more than C as he has 50% more chips. Or expressed another way, B's chance of coming second (when A wins) is his stack size (3000) divided by the total remaining chips (B+C or 3000 + 2000).

[/ QUOTE ]
Why B comes 2nd 50% more than C? Any intuition beyond the math?

[/ QUOTE ]

Same reason A came in 50%?

That actually always bothered me a little. It seems more intuitive to me to compute that as 55%: (3000+2500)/10k: A's stack equally distributed to the remaining players, rather than ignored. So there you have it: eastbay's modified ICM, ICMe.

eastbay

[rachelwxm]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

B has 3000 chips, C has 2000. Therefore B will come second 50% more than C as he has 50% more chips. Or expressed another way, B's chance of coming second (when A wins) is his stack size (3000) divided by the total remaining chips (B+C or 3000 + 2000).

[/ QUOTE ]
Why B comes 2nd 50% more than C? Any intuition beyond the math?

[/ QUOTE ]

Same reason A came in 50%?

That actually always bothered me a little. It seems more intuitive to me to compute that as 55%: (3000+2500)/10k: A's stack equally distributed to the remaining players, rather than ignored. So there you have it: eastbay's modified ICM, ICMe.

eastbay

[/ QUOTE ]

I am with you sir. What I think why A win 50% is very intuitive, I think given A win 50%, B win 50% more than C is not.

In real poker, I think if there is one short stack and two big stack, natually I think each of the big stack would have equal chance to take the smaller one out, it does not work out like since A has more chips, he should have greater share of C's chip. Therefore your version of ICM. /images/graemlins/smile.gif