Probability of losing with 4 aces

What are the odds of losing with 4 aces to any straight flush in Texas Holdem? Looking for scenario where player A must have 2 aces in hand and player B must use both hole cards to make the straight flush. I saw the odds where 4 aces lose to royal flush but not to any straight flush.

(Sorry, I don't want to steal Wizard's thunder, but I think my probability knowledge is good enough for this. EDIT: I calculated odds that are pretty close, but there's a sticky eventuality I didn't get right. It's really annoying to calculate -- this might be a case where it would be easier to just throw a computer sim at the problem, if anyone has one? EDIT again: I've just realized that my assumptions don't hold for poker with a single deck. Best to ignore this analysis and just write your own computer program to do the counting, I guess.)

Wizard claims to have calculated the odds of a straight flush here, but not the odds of four-of-a-kind on a specific rank: <oh hold up, I removed this link because I'm not allowed to post links>.

I couldn't find four-of-a-kind on a specific rank for seven cards in Wizard's poker table or in the post I found that included the straight flush figures, so I'm going to calculate it myself.


Odds of the opponent getting a straight flush: 0.00027851 according to Wizard.

Assuming just one opponent, then assuming that we're using a large enough number of decks that the odds of a card being an ace are always 1/13, odds of a card being one suit are 1/4, etc --

First, odds of the two cards you got both being aces:
- 1/13 * 1/13 = 1/(13 * 13) = 1/169
Odds of at least two cards in the middle being aces: 2 successes out of 5 trials at 1/13 odds:
- (5 choose 2) * (1/13) * (1/13) * (12/13) * (12/13) * (12/13) = (10 * 12 * 12 * 12)/(13 * 13 * 13 * 13 * 13) = 17280/371293
Summary: odds of getting a four of a kind, all aces, with both hole cards: 17280/62748517 = 0.00027538

So, odds of getting four aces, then having your one opponent get a straight flush: 0.00027851 * ~0.00027538 = about 0.00000008. With one opponent, this situation will happen to you about once in 13,048,475 games. Pretty unlikely! The opponents' odds get better if you have more than one opponent, of course.

This is probably pretty close as a back-of-the-envelope calculation, but I've just realized that one of your aces can't play into the opponent's straight flush, because otherwise he'd have a pair in his hand meaning he wouldn't have a straight flush. So you have to calculate straight-flush in six cards rather than straight flush in seven. That means the odds are probably slightly lower. However, I'm not sure how to do the straight flush in six cards where one is known to be an ace calculation. Can anyone else help?

Welcome, Zekka, the Wizard can't read, much less reply to, everything posted, so I have never heard him object to someone else answering, it is pretty much expected.

I like your avatar