Hello all,
I was playing a game of Three card poker at the casino the other night (fyi it is a game where the player is dealt 3 cards and the dealer is dealt 3 Cards from a 52 deck shoe.) one hand was dealt where we ended up with the same hand.
My hand was Q , 7 , 3
And the dealer hand was also Q , 7, 3
so I didn't win or lose it was 'stand off'
it got me think what the odds are of this happening and I thought I had it...
my first calculation was to assume the dealer had 3 separate cards ( I hadn't considered him having a pair which would reduce the odds of me matching him or the small chance of him having a 3 of a kind which would obvs make my chances to match him nil.
so if the dealer has 3 separate cards ( no pair) the calculation would be
My first card 9/49. 1/5.44r
Second card.6/48. - 1/
Third card. 3/47
162/110544.=. 1 in every 682.37037 hands ( I think)
but I had forgotten about the pair & 3 of kind for the dealer which would make having the same hand less likely.
Dealer having pair would be some thing like
player first card. 5/49
Second.
So my question is if you help so I can sleep is what's the ultimate odds of the dealer and player having the same hand :
thanks for your reply's in advance
I was playing a game of Three card poker at the casino the other night (fyi it is a game where the player is dealt 3 cards and the dealer is dealt 3 Cards from a 52 deck shoe.) one hand was dealt where we ended up with the same hand.
My hand was Q , 7 , 3
And the dealer hand was also Q , 7, 3
so I didn't win or lose it was 'stand off'
it got me think what the odds are of this happening and I thought I had it...
my first calculation was to assume the dealer had 3 separate cards ( I hadn't considered him having a pair which would reduce the odds of me matching him or the small chance of him having a 3 of a kind which would obvs make my chances to match him nil.
so if the dealer has 3 separate cards ( no pair) the calculation would be
My first card 9/49. 1/5.44r
Second card.6/48. - 1/
Third card. 3/47
162/110544.=. 1 in every 682.37037 hands ( I think)
but I had forgotten about the pair & 3 of kind for the dealer which would make having the same hand less likely.
Dealer having pair would be some thing like
player first card. 5/49
Second.
So my question is if you help so I can sleep is what's the ultimate odds of the dealer and player having the same hand :
thanks for your reply's in advance