Odds that someone will pick up top 10% behind you.

Hello,

I was just wondering how to calculate the percentage chance that at least one person will have a certain hand range behind you in Texas Holdem.

For example, if I am utg at a 9 person table, what are the odds that at least one person has a top ten % hand behind me. Or if I am on the button, what are the chances that the sb or bb has a top 15% hand.

Thanks in advance,
JonnyBGood

I don't know how exact you want it, but one way to approximate it is as follows. Assume the hands behind you each independently have a 10% chance of a top-10% hand. In reality, these are not completely independent. But for an approximation, it should be alright.

Calculate the chance that NO ONE behind you has a top 10% hand, then subtract that from 1.

For example, say there are four players behind you. Then each has a 90% chance of NOT having a top-10% hand. So the chance that all four have a non-top-10% hand is: (.90)*(.90)*(.90)*(.90) = .6561

.6561 is the probability no one has a top-10% hand. Therefore, the probability that someone has a top-10% hand is: 1-.6561 = .3439

You can see for N players, it would be 1-(.90)^N and you could use a different percentage as well.

For your specific button question, it's 1-(.85)^2 = .2775