>> There's a video out there of a Texas Hold'em showdown between
>> two players where one of the players has four aces and the other
>> has a royal flush. What is the probability of this happening?
>> (Express the answer in the form "1 in x".)
>>
>
> Mu. "0 in infinity" (not 1, it's 0). Unless you play with more than 52
> cards (that's call cheating in the rural country-part, but may be it is
> the rule in the self-called civilised area such as Wall street ?)
>
> A royal flush has an ace (and king, queen...) , so that would make 5
> aces in the deck. Heart, spade, diamond, club, and... what your name for
> the fifth ?

In Texas Hold'em each player only has 2 cards, and must choose 3 from 
the 5 shared face-up cards on the table to make their hand. So it's 
possible by a few combinations. Off the top of my head there are only 
three "types" of possible hands that will work:

P1    P2    Shared cards
A A   Q K   A A 10 J x
A A   x K   A A 10 J Q
A x   Q K   A A A 10 J

Obviously the suit of P2's cards must match the relevant ones on the 
table, and P2 could have any combination on 10,J,Q,K so long as the 
others are on the table to make the royal flush. And the order of cards 
doesn't matter. That should be enough information to figure out the 
probability. If I get time later I'll give it a shot.

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