Odds calculation help.....4 handed
January 4th, 2014 at 7:41:52 AM
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Hi guys. Really just after some conformation of my maths. I'm new at it, and going on help of others.
4 players at a hold em table. All 4 go all in preflop, and all turn AQ.
Odds of that happening in any given deal?
I get 0.000000000303734 from (16*4!*4!*44!)/52!
If that is correct, and can someone give a fraction estimation to 1?
As I said, newbie to it, and kinda going from help.
Great site btw!
4 players at a hold em table. All 4 go all in preflop, and all turn AQ.
Odds of that happening in any given deal?
I get 0.000000000303734 from (16*4!*4!*44!)/52!
If that is correct, and can someone give a fraction estimation to 1?
As I said, newbie to it, and kinda going from help.
Great site btw!
January 4th, 2014 at 8:03:58 AM
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Got to a decent calculator. So 3292354494 to 1?
Cheers
Cheers
January 4th, 2014 at 9:34:01 AM
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Player 1's first card has to be one of 8 cards (any ace, or any queen), and his second card has to be one of 4 cards (either an ace, if his first card was a queen, or a queen, if his first card was an ace)
Player 2's first card has to be one of 6 cards (any of the three remaining aces, or any of the three remaining queens), and his second card has to be one of 3 cards
Player 3's first card has to be one of 4 cards, and his second card has to be one of 2 cards
Player 4's first card has to be one of 2 cards, and his second card has to be a particular card.
That's (8 x 4 x 6 x 3 x 4 x 2 x 2 x 1) / (52 x 51 x 50 x 49 x 48 x 47 x 46 x 45) = 1 / 3292354406.25
Player 2's first card has to be one of 6 cards (any of the three remaining aces, or any of the three remaining queens), and his second card has to be one of 3 cards
Player 3's first card has to be one of 4 cards, and his second card has to be one of 2 cards
Player 4's first card has to be one of 2 cards, and his second card has to be a particular card.
That's (8 x 4 x 6 x 3 x 4 x 2 x 2 x 1) / (52 x 51 x 50 x 49 x 48 x 47 x 46 x 45) = 1 / 3292354406.25
