A mathy ICM question Xpost in STT

[nomadtla]

On the page of the standard 2+2 linked ICM calculator there is a Detailed example to explain the math of ICM.

The detailed Example looks like
[ QUOTE ]
All players are equally likely to win any hand, regardless of skill or position.
The probability of player A placing first is equal to the fraction of the total chips in play held by A.
Given that player B places first, the probability of player A placing second is equal to the fraction of total chips, not counting player B's chips, that are held by player A. So the probability of player A placing second is equal to the sum of these probabilities for each player B, C, D... that places first, weighted by the chance of that player placing first.
Given that player B places first and player C places second, the probability of player A placing third is equal to the fraction of total chips, not counting player B or player C's chips, that are held by player A...
Etcetera.

This is best explained with an example: suppose players A, B, and C have 3, 2, and 1 chips respectively.

Player A's chance of 1st is 3/(3+2+1) = 3/6 = 1/2
Player B's chance of 1st is 2/(3+2+1) = 2/6 = 1/3
Player C's chance of 1st is 1/(3+2+1) = 1/6


Given B gets 1st, A's chance of 2nd is 3/(3+1) = 3/4
Given C gets 1st, A's chance of 2nd is 3/(3+2) = 3/5
So A's chance of 2nd is (3/4 * 1/3) + (3/5 * 1/6) = 1/4 + <u>1/10</u> = 7/20

Given A gets 1st, B's chance of 2nd is 2/(2+1) = 2/3
Given C gets 1st, B's chance of 2nd is 2/(2+3) = 2/5
So B's chance of 2nd is (2/3 * 1/2) + (2/5 * 1/6) = 1/3 + <u>1/15</u> = 2/5

Given A gets 1st, C's chance of 2nd is 1/(1+2) = 1/3
Given B gets 1st, C's chance of 2nd is 1/(1+3) = 1/4
So A's chance of 2nd is (1/3 * 1/2) + (1/4 * 1/3) = 1/6 + <u>1/12</u> = 1/4


Given B gets 1st and C gets 2nd, A's chance of 3rd is 1
Given C gets 1st and B gets 2nd, A's chance of 3rd is 1
So A's chance of 3rd is (1 * 1/3 * 1/4) + (1 * 1/6 * 2/5) = 1/12 + <u>1/15</u> = 3/20

Given A gets 1st and C gets 2nd, B's chance of 3rd is 1
Given C gets 1st and A gets 2nd, B's chance of 3rd is 1
So B's chance of 3rd is (1 * 1/2 * 1/3) + (1 * 1/6 * 3/5) = 1/6 + <u>1/10</u> = 4/15

Given A gets 1st and B gets 2nd, C's chance of 3rd is 1
Given B gets 1st and A gets 2nd, C's chance of 3rd is 1
So C's chance of 3rd is (1 * 1/2 * 2/3) + (1 * 1/3 * 3/4) = 1/3 + <u>1/4 </u>= 7/12
1st 2nd 3rd
Player A 1/2 7/20 3/20
Player B 1/3 2/5 4/15
Player C 1/6 1/4 7/12


[/ QUOTE ]

I understand the math behind all of it but I do not understand where the values I have underlined come from. Could someone who understands the math behind ICM please explain.

I know it's not dire important but I like to know all the math that goes behind everything and I can't figure out where those values come from.

[Siegmund]

Do you understand the other lines right above that? The underlined terms are just showing an intermediate step in the algebra to the final answer.

Taking the first one,
[ QUOTE ]

Given B gets 1st, A's chance of 2nd is 3/(3+1) = 3/4
Given C gets 1st, A's chance of 2nd is 3/(3+2) = 3/5
So A's chance of 2nd is (3/4 * 1/3) + (3/5 * 1/6) = 1/4 + 1/10 = 7/20


[/ QUOTE ]

A can come in 2nd either if the rankings are B-A-C or C-A-B. The probability of a B-A-C finish is P(B wins)xP(A beats C, given that B wins) = 1/3 * 3/4 = 1/4.
The probability of a C-A-B finish is P(C wins)xP(A beats B, given that C wins) = 1/6 * 3/5 = 1/10.
The final answer is the sum of these.

In general, what ICM does is list every possible permutation of players, calculate the probability that that sequence occurs by multiplying (winner's chipstack)/(total chips) * (2nd place chips)/(total - winner) * (3rd)/(total-winner-2nd) * ... on down to last, and then add these up to see how often you ought to finish in each place.

[nomadtla]

got it. So he's just showing his work entirely. Makes perfect sense now I was trying to make things way more complicated then need be.