Quote:
Originally Posted by plexiq
The easiest way is to write a small program that canonicalizes flops/boards and then simply counts the resulting unique flops.
I never bothered to do it manually yet, but lets give it a try:- There are the 13 different trip-flops
- Paired flops have 13*12 unique rank structures, with 2 different suit structures each (two-suited or rainbow)
- Unpaired flops have 13C3 unique rank structures, with 5 different suit structures (rainbow, monotone, 3 two-suited)
> 13 + 13*12*2 + choose(13,3)*5
[1] 1755
Thank you, looks much easier then my own answer I would still like to contribute:
Why are there 1,755 distinct flops, if considering color isomorphism?
Total numer of flops possible: Draw any 3 of 52 → 52! / 3! * (52-3)! = 52*51*50/6 = 22.100
Probability of a monotone flop: Draw any first card. There are 12 other cards of the same suit. Draw a second card of the same suit from the 51 remaining deck → (12/51). Multiply with the probability to draw a third card of the 11 identical suited cards from the remaining 50 card deck (11/50). 0,05176. Multiply with total number of flops → there are 1.144 monotone flops.
Probability of a rainbow flop: Draw any first card. Draw a second card of the 3*13 cards of a different suit from the 51 cards left in the deck → (39/51). Multiply with the probability to draw a third card from the 2*13 cards of a different suit from the remaining 50 cards → (26/50). 0,39765. Multiply with total number of flops → there are 8.788 rainbow flops.
Probability of a two suits flop: Draw a second card of the same suit → (12/51). Multiply with probability to draw a third card of different suit (3*13/50). Each of these flops can occur in three suit variations, so multiply accordingly → 3*(12/51)*(39/50) = 0,5506. Multiply probability with total number of flops → there are 12.168 two suited flops.
Test: all probabilities add up to 1, and all flops counts add up to 22.100.
Now color isomorphism:
Each of the monotone flops can occur in four different suits. Devide by 4 = 286 different flops.
Two suited flop: every combo of two cards of the same suit occurs once for every color, i.e. four times. And each occurence is accompied by a third card of a different suit of which there are three. So devide 12.168 by (4*3) to get the count of unique two suited flops = 1.014.
To receive the count of unique rainbow boards, count the number of unique flops possible, if there were no color distinctions at all. It's like variations of drawing 3 cards from only 13 differently ranked cards, putting them back to the deck. (n-1+k)! / (k!(n-1)!) = 15! / 6*12! = 455
Add up 455 + 1.014 + 286 = 1.755 different flops.
Your answer is much more to the point. You were thinking from the other side, i.e. rank strunc → suit struc, while I am coming from the suit struct → ranks. I alreay realized that 455 is 5*91, but I couldn't figure why 5 until your answer.
Last edited by ZorroZock; 01-18-2017 at 12:03 PM.