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?
