Thursday, October 30, 2008

Chances of quad aces against royal flush?

In the WSOP 2008, a royal flush beat out quad aces. What are the chances of that happening? I know they gave a result and I'm not sure how that was calculated, but I did my own calculation with my own handy online odds calculator. Let me know if I screwed it up :)

Here is a link to the calculation:


I'll step through the process I took:

The first player needs pocket aces. So of the 52 cards, we have 4 outs, we need 2, and we are drawing 2 cards.

For the second player, first he needs any one of the KQJT that is not the same suit of the other players hole cards. 50 cards left, there are 8 outs in the deck, we are drawing 1 card and we need 1 of the outs.

For the other hole card , we need one of the remaining KQJT of the same suit. 49 cards left, there are 3 outs, we are drawing 1 and need 1.

On the board (don't care what order or where), we need exactly 4 outs from the 5 cards: the other two aces, and the other two cards KQJT of the same suit. 48 cards left, drawing 5, 4 outs, we need all 4.

Answer, this happens in 877,963,124:1, or 0.000001139% of the time.

I think they came up with 2.x billion? I'm curious how they figured it.

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