ICM Question

[Ryski]

Grabbed Tai's calculation from another thread for the purpose of this example. I understand the theory of ICM but not necessarily the calculation.

OK, the ICM on this is as follows:

- If you fold you will have 2180 chips worth 17.5% of the prize pool
- If you call and win you will have 4960 chips worth 32.3% of the prize pool
- If you call and lose you are out.

Therefore, you need to be (17.5/32.3) = 54% to win against opp's range to make this a good call.

2 Questions here:

1) What is the (17.5/32.3) equation? I know where the numbers come from but I'd like the equation explained to me in english =p

2) What happens if the chipstacks are different and you call and lose but ARE NOT out - ie you have the other guy covered. Since a third number now appears ( only 2 in the example above ), where does this fit in the above equation?

Thanks.

[taipan168]

1) What is the (17.5/32.3) equation? I know where the numbers come from but I'd like the equation explained to me in english =p

What we are trying to figure out is the equity (% chance to win) our hand has against opp's range that will make this a good call. Since if we fold we will have 17.5% of the prize pool and if we win we will have 32.3%, we need to figure out what % of the time we need to win 32.3% to make that a better option than folding and having 100% chance of having 17.5%.

2) What happens if the chipstacks are different and you call and lose but ARE NOT out - ie you have the other guy covered. Since a third number now appears ( only 2 in the example above ), where does this fit in the above equation?

In that case the equation becomes (% fold - % lose)/(% win - % lose). So in the above example, if we wouldn't be totally out but we had 4% of the prize pool with a very short stack, it would be (17.5 - 4.0)/(32.3 - 4.0) = we need to be 47.7% to win.

Hope that helped!

[Ryski]

Thanks man. That is exactly what I was looking for.

[Sakuraba]

and thanks from me too, this definitely helped.