Busted Aces -- twice in one hand.
 

(WARNING -- A bad beat story is included in this question)

Last night, I was playing $1/$2 NL and am dealt AA in the BB. Folded to the button who goes all-in. SB calls. I make a significant raise. SB thinks and goes all-in (about double my raise) and I call.

Button shows 99
SB has A2s

Flop: Q72
Turn: 9
River: 2

So I lose the main pot to 9s full of 2s and the side pot to trip 2s. From the calculators online, I know that I was about an 81% favorite over the 9s and a nearly 88% favorite over the A2.

How would I calculate the odds of BOTH hands beating me? Do I need to take into account that the two events are related (i.e., if a card helps one hand it is not going to help the other).

Thanks.

Re: Busted Aces -- twice in one hand.
 

AsAd vs 9s9d vs Ah2h
69.2% 19.0% 11.8%

As you can see, I made both of your Aces the same suits as the nines, and gave the A2 different suits altogether.

The percentages would change (but only slightly) depending on what suits there were. (E.G., in my example it is impossible for the 9s to make a flush that beats you, but the A2 hearts obviously could.)

Re: Busted Aces -- twice in one hand.
 

Oh, and by the way, on how to calculate it... lotsa folks on 2+2 use twodimes.net. I use this because I just like it better.

If you mean how do you actually do the math rather than just type the cards in somewhere, lol, I dunno. But plenty of people here do.

Re: Busted Aces -- twice in one hand.
 

[ QUOTE ]
Oh, and by the way, on how to calculate it... lotsa folks on 2+2 use twodimes.net. I use this because I just like it better.

If you mean how do you actually do the math rather than just type the cards in somewhere, lol, I dunno. But plenty of people here do.

[/ QUOTE ]

Sorry -- maybe the questions wasn't clear. I do use twodimes -- that is where the percentages I quoted came from.

I wanted to know the chance of being beaten by BOTH hands -- not losing to either of the hands (which is what twodimes or the cardplayer calculator gives you when you put in all three hands). In the example above, I would have still been up on the hand if only the 99 had beaten me and I won the larger side pot. My AA had to be cracked twice to lose this hand.

Assuming these were independent events, I think I would multiply the probability of losing one hand by the probability of losing the other hand in order to calculate the probability of losing both (2.28% in this case, I believe). But since these events are interrelated, I assume the probability of losing to both hands is even less -- maybe its not significantly less...

Re: Busted Aces -- twice in one hand.
 

To do this, put 99 as dead cards and try against A2.

Then do the same but A2 as dead cards. Multiply the two together.

Re: Busted Aces -- twice in one hand.
 

[ QUOTE ]
To do this, put 99 as dead cards and try against A2.

Then do the same but A2 as dead cards. Multiply the two together.

[/ QUOTE ]
Because the probabilities are related, you can't just multiply them to get the probability that they both occur. Some boards are particularly dangerous for AA, e.g., with a 4-flush on the board. They make it easy for AA to lose against many hands. Some boards are safer for AA, e.g., where AA is the nuts. AA simultaneously beats all hands. Opposing this, if A2s wins because two 2s come, it is much less likely than normal that there is also a 9 on the board. That A2s and that 99 beat AA are not independent. This would be much more clear if the matchups were AK vs. QQ and JJ or AA vs. A2o and A3o, but it is still true for AA vs. A2s and 99.