Three Card Poker - combinations for Prime bets

Hi all,

Hoping I've posted in the right place and that I can get some help here. I've been looking at the maths behind three card poker and am hoping to get some clarification regarding the combinations of hands that can win on a prime bet.

Most (if not all) of my maths info comes from the , and generally can follow the majority of it however have been completely stumped by the combinations of prime bets. By my workings there are a total of 5200 hands that can be made of 3 cards of the same color, this being a product of 2 (number of colors) x 26C3 (or C(26,3) depending on the notation you are used to. However the site states that the total combination of hands for 3 of the same color as not quite sure what I'm overlooking.

If anyone is able to shed light on this I would really appreciate this. It will help me make sense of the calculation for 6 cards of the same color on the website too.

Many thanks!

You're math is correct, but you must subtract the probability of all 6 cards the same color.

P(6 cards same color) = 2*c(26,3)*c(23,3)/c(52,3)/c(49,3)
You can just calculate 2*c(26,6)/c(52,6), but the first calculation follows the game play, where there are 2 separately dealt hands.

P(3 cards same color)= 2*c(26,3)*c(49,3)/c(52,3)/c(49,3)-P(6 cards same color)

Quote: CrystalMath

You're math is correct, but you must subtract the probability of all 6 cards the same color.

P(6 cards same color) = 2*c(26,3)*c(23,3)/c(52,3)/c(49,3)
You can just calculate 2*c(26,6)/c(52,6), but the first calculation follows the game play, where there are 2 separately dealt hands.

P(3 cards same color)= 2*c(26,3)*c(49,3)/c(52,3)/c(49,3)-P(6 cards same color)

Thanks for that, just tried out those formulas and have some follow up questions if that's alright?

1. For the probabilities for P(6 same color) why is it we multiply by C(23,3) and then divide by c(49,3) as well? This is purely just to make sense of the formula.

2. For P(3 same color) why do we multiply by c(49,3) and then divide by it too? Will they not cancel out in this instance?

I got the following values too:
P(6 same colour) = 0.022617558
P(3 same colour) = 0.21267656 (this seems pretty high, hence why I wanted to check the formulas.

Sorry for all the questions, I just like to make sure I'm following the logic with such calculations and thank you for responding once again!

The format of these formulas is to have the same denominator as the hits calculated by the Wizard 3CP. To just calculate the probability of 3 cards matching, we really don't need to consider the outcome of the dealer hand, which is represented by c(49,3)/c(49,3).

As for a sanity check on the 3 card numbers, you have about a 50% chance of the second card matching the first and about 50% chance of the third card matching the second. This gives 25% overall, so 21.26% is certainly reasonable.