First of all, sorry for my bad English!
So I’m designing a game that has some similarities with Poker as you’re trying to get triples, straight (suite), flush (same color), etc. There are some differences though. The deck has 32 cards, cards are numbered from 1 (not an ace) to 8 and there are 4 colors (like real card deck). Combination are made out of 4 cards only (in my actual prototype, players has 2 cards in their hands and there’s 2 cards on the board).
As I am not really good with probabilities, I’m asking some help to calculate the probabilities to get these combinations. If someone has the kindness to help me figure out how to calculate these numbers, could it be possible to also include the formula (like if the game change and there’s is 10 values instead of 8 or 3 colors instead of 4).
- 1 pair
- 2 pair
- Straight (Like 3-4-5-6. The #1 does not act like an ace)
- Flush (4 cards of the same color)
- Straight Flush (4 cards of the same color that has following numbers)
I think there are 863,040 possible hands (32 x 31 x 30 x 29), but I’m not sure because I think the way I calculate takes in consideration the order and it's not necessary. I think there are 20 possible straight flushes (1-2-3-4 // 2-3-4-5 // 3-4-5-6 // 4-5-6-7 // 5-6-7-8 // so 5 straight x 4 different colors). There are 8 possible quads (1 per value).
With this, I’m trying to find out which combination beats which one by the probability to get each ones.
I don’t know if I have been clear enough!
Thanks you very much for your help!
There are 32 choose 4 = 35960 possible hands (you were right, you don't need the order so you have to divide your number by 24 (the number of possible orders of 4 cards)
You already calculated the number of possible quads and straight flushes.
Flush: There are 4 possible colours. For each colour there are 8 choose 4 = 70 possibilities: So there are 4*70=280 possibilities for a Flush. 20 of these are a straight flushes, so you have 260 true flushes remaining.
Straight: 5 possibilities for the lowest card. Each card has 4 possibilities for its colour. Gives us 5*4*4*4*4=1280 straights. 20 of these are straight flushes, so we have 1260 true straights remaining.
Triple: 8 possibilities for the number you have the triple of. 7 possibilities for the number of the remaining card. 4 possibilities for which colour is missing in the triple. 4 possibilities for the colour of the remaining card. Together these are 8*7*4*4=896 possibilities.
2 pair: 8 possibilities for the number of the first pair. 7 possibilities for the number of the second pair. Divided by two as you don't care which pair is the first, and which one is the second. 4 choose 2 = 6 possibilities for the colours of the first pair, 6 possibilities for the colours of the second pair. Together these are 8*7*6*6/2=1008 possibilities.
1 pair: 8 possibilities for the number of the pair. 7 choose 2 = 21 possibilities for the numbers of the 2 remaining cards. 6 possibilities for the colours of the pair. 4 possibilities for each of the remaining cards. Together we have 8*21*6*4*4=16128 possibilities
35960 possible hands, of which we have:
Straight Flush: 20
2 pair: 1008
Wow! Thank you so much! It will help me a lot!! Thank you very much for the method too!! Really appreciated!