Post by jabberwok on Apr 13, 2016 17:00:45 GMT 8
I had some spare time so I calculated some poker odds. leaving it here for recording,sharing,and for someone to double check my math. I didn't really like statistics in math so anyone feel free to check on the calculations and point errors out. I will be writing the combination as C[n r] since I can't find a pretty way to do it on this board.
[Odds of Holding one pair]
♡◇♧♤
Lets say the hand is A◇,A♧,5,7,9
You hold the two aces and drop the rest. What are the odds of getting a two pair, three of a kind or stronger?
52 cards a deck(13*4) + wildcard means 53 cards a deck. since 5 cards are drawn 48 cards are left
since 2 cards are being held and 3 redrawn, the number of possible hands are C[48 3] = 48! / 45! / 3! = 17,296
1. Total number of hands = 17,296
Five of a kind requires you draw 2 aces A♡,A♤, and the wild card joker.
2. Five of a kind = 1
Four of a kind requires you draw 2 aces or one ace and a joker and any one card(excluding aces and joker)
so C[3 2]*C[45 1] = 3 * 45 = 135
3. Four of a kind = 135
There are 2 ways to make a Full house in this situation. make a three of a kind with the aces and getting one pair or drawing a three of a kind.
since we discarded 3 cards of one rank each, 3 ranks have 3cards left and 9 ranks have 4cards left.
pick 1 of 3 cards from the aces and joker, pick 1 rank from the 3 ranks and pick 2 cards from the 3 cards in that rank
C[3 1]*C[3 1]*C[3 2] = 3*3*3 = 27
pick 1 of 3 cards from the aces and joker, pick 1 rank from the 9 ranks and pick 2 cards from the 4cards in that rank
C[3 1]*C[9 1]*C[4 2] = 3*9*6 = 162
Ace full house = 27+162 = 243
getting the other full house requires drawing 3 cards of the same rank so
pick 1 rank from the 3 ranks and pick 3 cards from the 3 cards in that rank
C[3 1]*C[3 3] = 3
pick 1 rank from the 9 ranks and pick 3 cards from the 4 cards in that rank
C[9 1]*C[4 3] = 9*4 = 36
Other full house = 3+36=39
4. Full house = 243 + 39 = 288
Three of a kind requires a ace or a joker so
pick 1 of 3 cards from the aces and joker, pick 1 card out of 45 cards, pick 1 card out of 44 cards
C[3 1]*C[45 1]*C[44 1] = 3*45*44 = 5940
this also contains the ace full houses so we subtract that amount 243
5. Three of a kind = 5940 - 243 = 5697
two pair requires 2 cards of the same rank so
pick 1 rank from the 3 ranks and pick 2 cards from the 3 cards in that rank and pick 1 card out of 42 cards
48 - (2 aces and joker, 3cards from this rank = 6) = 42
C[3 1]*C[3 2]*C[42 1] = 3*3*42 = 378
pick 1 rank from the 9 ranks and pick 2 cards from the 4 cards in that rank and pick 1 card out of 41 cards
48 - (2 aces and joker, 4cards from this rank = 7) = 41
C[9 1]*C[4 2]*C[41 1] = 9*6*41 = 2214
6. Two pairs = 2214 + 378 = 2592
temp saved
[Odds of Holding one pair]
♡◇♧♤
Lets say the hand is A◇,A♧,5,7,9
You hold the two aces and drop the rest. What are the odds of getting a two pair, three of a kind or stronger?
52 cards a deck(13*4) + wildcard means 53 cards a deck. since 5 cards are drawn 48 cards are left
since 2 cards are being held and 3 redrawn, the number of possible hands are C[48 3] = 48! / 45! / 3! = 17,296
1. Total number of hands = 17,296
Five of a kind requires you draw 2 aces A♡,A♤, and the wild card joker.
2. Five of a kind = 1
Four of a kind requires you draw 2 aces or one ace and a joker and any one card(excluding aces and joker)
so C[3 2]*C[45 1] = 3 * 45 = 135
3. Four of a kind = 135
There are 2 ways to make a Full house in this situation. make a three of a kind with the aces and getting one pair or drawing a three of a kind.
since we discarded 3 cards of one rank each, 3 ranks have 3cards left and 9 ranks have 4cards left.
pick 1 of 3 cards from the aces and joker, pick 1 rank from the 3 ranks and pick 2 cards from the 3 cards in that rank
C[3 1]*C[3 1]*C[3 2] = 3*3*3 = 27
pick 1 of 3 cards from the aces and joker, pick 1 rank from the 9 ranks and pick 2 cards from the 4cards in that rank
C[3 1]*C[9 1]*C[4 2] = 3*9*6 = 162
Ace full house = 27+162 = 243
getting the other full house requires drawing 3 cards of the same rank so
pick 1 rank from the 3 ranks and pick 3 cards from the 3 cards in that rank
C[3 1]*C[3 3] = 3
pick 1 rank from the 9 ranks and pick 3 cards from the 4 cards in that rank
C[9 1]*C[4 3] = 9*4 = 36
Other full house = 3+36=39
4. Full house = 243 + 39 = 288
Three of a kind requires a ace or a joker so
pick 1 of 3 cards from the aces and joker, pick 1 card out of 45 cards, pick 1 card out of 44 cards
C[3 1]*C[45 1]*C[44 1] = 3*45*44 = 5940
this also contains the ace full houses so we subtract that amount 243
5. Three of a kind = 5940 - 243 = 5697
two pair requires 2 cards of the same rank so
pick 1 rank from the 3 ranks and pick 2 cards from the 3 cards in that rank and pick 1 card out of 42 cards
48 - (2 aces and joker, 3cards from this rank = 6) = 42
C[3 1]*C[3 2]*C[42 1] = 3*3*42 = 378
pick 1 rank from the 9 ranks and pick 2 cards from the 4 cards in that rank and pick 1 card out of 41 cards
48 - (2 aces and joker, 4cards from this rank = 7) = 41
C[9 1]*C[4 2]*C[41 1] = 9*6*41 = 2214
6. Two pairs = 2214 + 378 = 2592
temp saved
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(for me now at least)
