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Let's say I have a deck of custom playing cards. In this deck there are 75 cards made up of 5 ranks (Ten, Jack, Queen, King, Ace) and 5 colors (Red, Orange, Green, Blue, Purple). There are 3 of each rank in each color (3 Red Tens, 3 Red Jacks, etc.).

During a hand of this game each player will be dealt 3 cards. Hands are only "valued" if all 3 cards are the same color. In other words a straight has to essentially be a straight flush to count. There are 4 possible "valued hands":

  • High Card
  • Pair
  • Straight
  • 3-of-a-kind

I am trying to calculate the odds of the pair, straight, and 3-of-a-kind. My calculations and thoughts are below but I am not a mathematician by any stretch of the imagination and I am unsure how close I am to the right track. Can anyone out there verify my calculations or point me in the right direction if I'm off?

Total number of possible 3-card hands:

$\binom{75}{3}$ = 67,525

3-of-a-kind with all 3 cards being the same color (example: 3 Purple Aces):

$\binom{5}{1}$ selects the color - out of the 5 possible colors

$\binom{5}{1}$ selects the rank - out of the 5 possible ranks (10, J, Q, K, A)

$\binom{3}{3}$ selects all 3 cards of that rank and color

...which gives me $$\frac{\binom{5}{1}\binom{5}{1}\binom{3}{3}}{\binom{75}{3}}$$

I calculate this to be $\frac{25}{67,525}$ = .037% or 1 in 2,701 as the odds of getting dealt 3-of-a-kind in the same color.

Straight - with all 3 cards being the same color (example: Red Ten, Jack, Queen):

$\binom{5}{1}$ selects the color - out of the 5 possible colors (let's say Red)

$\binom{3}{1}$ selects one of the ranks that can start a 3-card straight (10, J, or Q) - (let's say we selected Ten)

$\binom{3}{1}$ selects one of the 3 available Red Tens

$\binom{3}{1}$ selects one of the next ranks that would make a straight (for example if we chosen a Red 10 first we are looking for 1 of the 3 red Jacks)

$\binom{3}{1}$ selects one of the ranks that could finish the straight (for example if we chose Red 10 and Red Jack so far we would now need one of the 3 Red Queens)

...which gives me $$\frac{\binom{5}{1}\binom{3}{1}\binom{3}{1}\binom{3}{1}\binom{3}{1}}{\binom{75}{3}}$$

...which comes out to $\frac{405}{67,525}$ = .59% or 1 in 167 as the odds of getting dealt a 3-card straight in the same color.

Pair - with all 3 cards being the same color (example: Orange Jack and 2 Orange Kings):

$\binom{5}{1}$ selects the color - out of the 5 possible colors (Orange)

$\binom{5}{1}$ selects any one of the 5 possible ranks (King)

$\binom{3}{2}$ selects 2 of the 3 possible cards that are this color and rank (2 Orange Kings)

$\binom{12}{1}$ selects the last card from one of the remaining cards of this color that isn't the same as the pair

...which gives me $$\frac{\binom{5}{1}\binom{5}{1}\binom{3}{2}\binom{12}{1}}{\binom{75}{3}}$$

...which comes out to $\frac{900}{67,525}$ = 1.33% or 1 in 75 as the odds of getting dealt a pair where all 3 cards are the same color.

Are my calculations correct? Am I on the right track?

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  • $\begingroup$ They all look good to me. Incidentally, if you take away the ${5 \choose 1}$ for the color, and change the denominator to ${15 \choose 3}$, that would be equivalent to dealing from a sub-deck all of one color. The probabilities would become bigger and should be quite easy to estimate accurately by simulations. (Whereas a small prob value like $1/2701$ would be very hard to estimate accurately by simulations.) In fact, there are only ${15 \choose 3} = 455$ hands (of the same color) so it's not hard to list all of them in a small program. $\endgroup$ – antkam Apr 9 at 22:34
  • $\begingroup$ Awesome. Thanks antkam. Now you gave me something new to learn: simulation :) $\endgroup$ – Danger Games Apr 10 at 14:13
  • $\begingroup$ In my very limited experience, it is very easy to design a (card / dice / etc) game where the relevant probabilities are very hard to calculate precisely. The answer is almost always: simulate it 10000 times. :) However, small probs are hard to estimate because they only happen a few times, so the error is large. Luckily in your case the prob of having 3 cards of the same color is easily calculated to be ${5 \choose 1} {15 \choose 3} / {75 \choose 3} = {14 \over 74} {13 \over 73}$, so you can rescale later. $\endgroup$ – antkam Apr 10 at 14:21

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