Special rule poker combinatorics: combos of cards so Andy, James both beat Ida's hand

swooper

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Hey! im currently stuck on a combinatorial question and would appreciate if someone could help ´ , basically its a special game of poker 3 cards are on the table and each player gets to take 2 cards , the 3 cards on the table can be counted in for everyone. So basically there are 3 people playing Ida, Andy and James we know that Ida has spades queen and spades 10 and on the table there is spades ace, diamonds queen and clubs ten so the question is:"how many different kind of combinations of cards can Andy and James have so BOTH of them has a better pokerhand than Ida" i have gotten that Andy or james can seperatly have 33 combinations of different cards to have a better pokerhand than ida (if it was to be a 1v1) but im not quite sure how to do it when both has to have better, doesnt it depend on which cards that Andy/James picks up?
 
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Hey! im currently stuck on a combinatorial question and would appreciate if someone could help ´ , basically its a special game of poker 3 cards are on the table and each player gets to take 2 cards , the 3 cards on the table can be counted in for everyone. So basically there are 3 people playing Ida, Andy and James we know that Ida has spades queen and spades 10 and on the table there is spades ace, diamonds queen and clubs ten so the question is:"how many different kind of combinations of cards can Andy and James have so BOTH of them has a better pokerhand than Ida" i have gotten that Andy or james can seperatly have 33 combinations of different cards to have a better pokerhand than ida (if it was to be a 1v1) but im not quite sure how to do it when both has to have better, doesnt it depend on which cards that Andy/James picks up?
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Do you see that as things now stand, Ida has two pair ?
I admit the word BOTH in this context could confusing. I suggest we take to mean each of Andy & James has a better hand than Ida.
If you are unsure of poker ranking look here.
But I suggest that you want to find the number of two card hands that are disjoint (no cards in common, taken in pairs) so that both beat Ida's two pairs.
Now you must show us some work.
 
Pka im not quite sure what you mean but i can explain what i did to find the amount of combinations one of them have in a 1v1

basically the only type of poker hands that will win/can be picked up is

ladder -> three of a kind > two pairs

now for the ladder to happen the only cards that we get to pick up is king and knight, we have 4 of each so this gives us 16 combinations
three of a kind we need either 2 Aces or 2 queens or 2 tens and since ida has one and on the table there is one (10 and queen) we only have 1 of each left this gives us 1 combination for each of those which is 2 in total and now we have the aces left and we have 3 aces left and we pick 2 this gives us 3 combinations and in total all the combinations we can have three of a kind is, 5.
now the only two pairs we can get that are higher than idas two pairs would be either a ten and an ace or a queen and an ace and we have 2 queens , 2 tens, 3 aces if you do the first one you get 6 combinations and same with the other one and in total this is 12 which gives us our result what one of them can have and that is, 33 combinations. Now what im confused over is that if we have 2 people against her doesnt the combinations that the other person can get chance depending on what the first one picks? im a bit confused

Ladder combinations = 16
three of a kind combinations = 5
two pair combinations = 12
total = 33
 
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