Ok here's the scenario.

Player #1 is dealt AA. Player #2 is dealt KK.

Player #3 has a range of {QQ-22,AKs,AKo}.

What is the percentage chance of Player #3 flopping a set?

I'm not sure how to approach this... I tried using combinations but I ended up with 18 combos out of 69 (3 AK, 6 combos of 11 pocket pairs) for a 26% chance of flopping a set (3 cards flop giving 18 combos sets)... but this doesnt account for paired boards etc.

I will assume that by a set, you mean a PP that hits the board. That is, we exclude AK that hits a paired board.

There are 6*11 combos of pocket pairs, and 4 combos of AK, for a total of 70 combos. So, the chances of player #3 holding a pair is 66/70.

When he holds a pair, he has 2 outs from the remaining 46 cards, so the chance he at least hits one of them is 1 - (44/46)^5.

So the combined probability of him (1) holding a pair and (2) hitting a set with it, is 66/70 * (1 - (44/46)^5) ~ 18.8%

If you want to account for trips as well, the probability is slightly higher.

Thanks for the response. 66/70 sounds good. I dont know why I thought only 3 combos of AK, my bad. I understand that so far.

Can you explain why the the formula is 1 - (44/46)^5? Wouldn't it be:

dealing 3 flops cards with 2 outs out of 46 unknown... be the addition of:

1) hitting the set in the first card of the flop (2/46) +
2) missing the first card, but hitting on the second (44/46 * 2/45) +
3) missing the first and second card, but hitting the third (44/46 * 43/45 * 2/44)

= 0.1275.

66/70 * .1275 = 12.02%

I don't get where your power of 5 comes from... is that to calculate for 5 board cards?

If we do 1 - (44/46)^3 we get 12.48% * 66/70 = for a 11.77% chance of flopping a set.

The power of 5 is supposed to be 3, I calculated for 5 cards. My bad.

But also, that whole approach was wrong, I don't know why I did it that way. Your method is correct. So it's 66/70 * .1275 = 12.02% like you say.

Edit:

Instead of 1 - (44/46)^3, the correct formula is 1 - (44/46)*(43/45)*(42/44) ~ 12.75%.

EDIT: Klyka has already corrected the above in a subsequent post.

This is incorrect. The probability of missing is 44/46 only on first card of the board, not every card, so we cannot simply raise this to the 5th power. The probability of missing on the 2nd card after missing on the 1st card is 43/45, and the probability of missing on the 3rd card after missing on the first 2 is 42/44. The OP was interested only in the flop, so the correct probability is

66/70 * [1 - (44/46 * 43/45 * 42/44) ]

=~ 12.0%


We can do this with combinations as

66/70 * [1 - C(44,3)/ C(46,3) ]

=~ 12.0%

where C(46,3) is the total number of flops, and C(44,3) is the number of flops which do not make a set. Note that this 12.0% will also include quads and full houses.

There is no way that player #3 can flop trips with his range since the first 2 players hold AA and KK.

I didn't get much right in that post..

That's okay Kykka is was early in the morning, and if you pulled an all-nighter like me, it's expected. Your response actually triggered enough for me to figure it out.