Three card poker Stand off odds

Browncow

New member
Joined
Jan 17, 2020
Messages
2
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
 

Romsek

Full Member
Joined
Nov 16, 2013
Messages
784
two cases, one where there are 3 different ranks, one where each gets a pair of something.

case I: dealer gets dealt 3 different ranks with probability

\(\displaystyle p_D = 1 \cdot \dfrac{48}{51}\dfrac{44}{50}\).

Your probability of getting dealt same ranks is

\(\displaystyle p_P = \dfrac{3}{49}\dfrac{3}{48}\dfrac{3}{47}\)

The product of these is

\(\displaystyle p_1=\dfrac{48 \cdot 44 \cdot 3^3}{51\cdot 50 \cdot 49\cdot 48 \cdot 47} = \frac{198}{978775}\)

case II: dealter gets dealt a pair

\(\displaystyle p_D = 1 \cdot \dfrac{3}{51}\cdot \dfrac{48}{50}\)

Your probability of getting dealt same ranks is

\(\displaystyle p_P = \dfrac{2}{49}\dfrac{1}{48}\dfrac{3}{47}\)

and the product is

\(\displaystyle p_2 = \dfrac{3 \cdot 48\cdot 2 \cdot 3}{51\cdot 50\cdot 49\cdot 48\cdot 47} = \frac{3}{978775} \)

The sum of these is the probability of dealer and player getting dealt the same hand.

\(\displaystyle p = \dfrac{201}{978775} \approx 0.002\)
 

Browncow

New member
Joined
Jan 17, 2020
Messages
2
Thank you so much for your reply.

Could you tell me the chances then using your calculation as a fraction like eg. 1/800 ' 1 in every 800 hands will end up in a 'stand off' draw

Also does the calculation assume after the dealers hand has been turned that the players hand can match in any order. Surely the chance of the players first card to be turned would have a chance of 9 in 49 of matching one of the dealers

sorry if this has been calculated. I am as you can probably tell, not a mathematician :)

cheers
 

Romsek

Full Member
Joined
Nov 16, 2013
Messages
784
I seemed to have lost a 0 in there. The probability is roughly 0.0002, not 0.002.

This is 1 out of 5000 hands.

I verified this by simming a million plays.

The order of the cards in the hand doesn't matter. (a,b,c) would match (c,b,a)
 
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