Pokermath problem

Reply Tue 28 Feb, 2012 03:51 pm

iam trying to learn poker and i have a mathematical problem.
you dont need to know poker for this problem

There is a model called "Independent Chip Model" and i cant reproduce the correct results.

Basicly it calcutes your win % in a tourney based on your chips.
Payout structure
1st 50% , 2nd 30%, 3rd 20%

if there are 10 players and each player paid 10$
the prizepool would be 100$
1st would get 50$, 2nd 30$, 3rd20$.

To calculate the first place is very easy.

Player A = 10000 Chips ( 0.327)
Player B = 15000 Chips (0.491)
Player C = 5000 Chips (0.163)
Player D = 500 Chips (0.0163)

Player A would win the touney in the first place in ( playerchips / maxchips)

Player A = 10000 / 30500 = 0.3278 => ~33%

I also figured out how i can calculate the 2nd place for each player ( not sure if this is correct)

To calc the 2nd place for player A i thought the following:

if player B wins which hapens 49% of the time, than player a would face player c and player d.

if player C wins than he would face player b and d

if player d wins than he would face player b and c.

so i came up with the this formula :

Player A = sum += WinchanceOf1st * (PlayerAchips / (TotalChips - WinnerChips(1st) ))

Player A =
0.491 * (10000 / (30500 - 15000)) +
0.163 * (10000 / (30500 - 5000)) +
0.016 * (10000 / (30500 - 500))

Player A would be 2nd in ~ 32%

i dont know if this approach is correct because im a mathnoob.
And i couldn't find the formula for the ICM. i dont know if this is public.

If this would be correct, how do i get the winchance of the 3rd place ?
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Reply Wed 29 Feb, 2012 01:50 am
Check out these sites:

Personally, I think the premise behind the probability calculation is wrong. In a tournament where players are eliminated one at a time (at least when you get down to the last table this is how it's done), why would you compute a player's probability of taking second place by removing all of the other players (one by one) and summing the probabilities of that player beating the remaining players based on their current chip counts?

Since the tournament is an elimination process, I'd approach it from the other direction and compute the probabilities of each player being eliminated in the next round. After each elimination, I'd distribute the eliminated player's chips among the remaining players based on the sizes of their stacks.
Reply Wed 29 Feb, 2012 04:44 am

but i just dont get your idea. so u want to calcualte the chance of losing for each player ? but u dont know what will happen ? lets say
Player A 1500 Chips
Player B 1200 Chips
Player C 1100 Chips
Player D 1400 Chips
If Player A goes allin , you dont know his chance of being eliminated because the chance is dynamic, he could win the pot right away, or he could get called by one player and if more than one player call his allin his chance of losing would increase.
So the chance of losing is based on the actions from the other players.
And what if player A doesnt play a single hand ?
Or if player A goes All in and gets called by player B than he wouldn't lose all of his chips, but he would be in a very bad situation.

but maybe im retarded and i missed something.
Reply Wed 29 Feb, 2012 05:21 am
If this would be correct, how do i get the winchance of the 3rd place ?

Iteratively, by removing the chips that first and second place hold to obtain a notional estimate of the third place expected value (as you did in your first calculation and so on for all possible payouts).

It is only a model, ie. it attempts to describe what happens based on an existing trend.

The modelled trend results are then multiplied by the known probabilitities of initial hands winning individual pots to mathematically escamotage decision making.
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Reply Wed 29 Feb, 2012 06:34 am
I am a poker player who is also math savvy. I read this explanation for ICM and it this particular calculation is a bit of hocus pocus. The poker books acknowledge as much.

The idea is simple enough and it is certainly correct. In a tournament the chips you lose are worth more (in dollar value) then the chips you win. If you understand this the math doesn't matter so much -- especially since even if the calculation were valid you aren't going to be doing this at a poker table.

But yeah, the math they give for this is bogus.

Reply Wed 29 Feb, 2012 10:35 am
sry i thought i got it , but are you sure that the formaula provided at http://www.pokerjunkie.com/icm-explaining-the-independent-chip-model => P(A3) = P(A3|B2)*P(B2) + P(A3|C2)*P(C2) + P(A3|D2)*P(D2) is correct ?

im sitting here since 3 hours :/

ill give you my calculations for player a

A 6737 chips 1st= 0.49 2nd=0.37
B 185 chips 1st= 0.01 2nd=0.02
C 1347 chips 1st= 0.09 2nd=0.16
D 5231 chips 1st=0.38 2nd=0.43

the calculation above is correct, i verfied these number with a working icm calculator.

Player A:
( (6737 / 8079) * (6737 / 11968) )* 0.02 +
( (6737 / 6922) * (6737 / 11968) )* 0.16 +
( (6737 / 6922) * (6737 / 8079) )* 0.43
= 0.4487 => which is wrong

the correct result should be 0.11
Reply Wed 29 Feb, 2012 02:23 pm
i made a mistake
( (6737 / 8079) + (6737 / 11968) )* 0.02 +
( (6737 / 6922) + (6737 / 11968) )* 0.16 +
( (6737 / 6922) +(6737 / 8079) )* 0.43
= 1.67

but its still wrong. Sad i'll test the other formula tomorrow
Reply Thu 1 Mar, 2012 07:01 am
got it, thank you guys
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