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Le 18/12/2014 09:07, scott a écrit :
>>> 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.
Oups, yes, I forgot that shitty version of sharing cards from the table.
> 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.
>
And this assumes of course that the shared card are neither four aces or
the royal flush.
From wikipedia, in texas hold'em, there is 4,324 royal flush (from the
133,784,560 combinations of one hand in a seven-cards poker).
Not all of them would allow a four-aces hand for another player: you
need 2 or more aces in the shared cards, and none in the royal-flush's
player's hand.
Interestingly, the 3 shared aces would allow 2 players to go 4-aces and
a third with a royal flush.
--
Just because nobody complains does not mean all parachutes are perfect.
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