Okay, so I was reviewing this thread from before: http://www.vbforums.com/showthread.p...light=pinochle and was trying to apply it to a new situation (and also the question asked there).
If I use Jemidiah's method in post 11, I'm not sure I'm getting the same numbers as he did.
I'm trying to figure out odds for Aces around and double Aces around.
I started with quadruple aces around (4,4) ^ 4 from each suit * (64,4) = 635376
Then I went to triple aces (4,3) ^ 4 * (68,8) - 4x Aces from above = 1,892,353,865,232
Then I went to double aces (4,2) ^ 4 * (72,12) - 3x Aces and 4x Aces from above = 19,908,924,100,186,400
Then I got single aces (4,1)^4 * (76,16) - 2x/3x/4x Aces from above = 2,752,579,046,460,370,000
With (80,20) hands = 3,535,316,142,212,180,000
With these numbers dividing the total number of hands by the Number of Aces around hands, I'm getting a result of 1 in 1.28 hands, and double aces as 1 in 177.57 hands.
I've calculated hands through Excel and Python programs that I've written, and python and perl programs that my sister has written and I'm getting numbers of 1 in 4.84 and 1 in 466.04.
Can someone help me determine where I'm going wrong? The reason I'm trying to figure this out is I'm comparing numbers from 4 player, double deck pinochle with 6 player triple deck pinochle (i.e. 6 of each card of each rank).
My numbers are showing me that it's easier to get anything around once, when using 2 decks, but it's easier to get things around 2 or more times when you're playing with 3 decks). I find it odd that you can get Aces around once, fewer times with 3 decks, but double/triple aces around more times with 3 decks. (I have over 400 million simulated hands with programs... and the program results all line up...)
Edit: I'm also trying to calculate the probability of a run. I saw this thread http://www.vbforums.com/showthread.p...un-in-pinochle, but it looks like the OP didn't finish clarifying the questions asked of him. I'm also basing it off of a double deck of pinochle cards (i.e. 4 of each card of each rank from Jack through Ace (tens are included, but they're ranked below Aces, above Kings), double deck Pinochle doesn't use the 9's).
So I'm thinking it should be:
Random suit (4,1)
Random ace of that suit (4,1)^5 (^5 does Ace, Ten, King, Queen, Jack).
Then there are 75 cards left, and you still need 15 in your hand, so that should be (75, 15)
so if I take (4,1) ^6 (Suit + 5 ranks) * (75,15), I'm getting a number of 9.33 E+18, yet (80,20) says there should only be 3.53E+18 total combinations.... so I'm off somewhere.
In regards to previous unanswered questions, yes having both other face cards is still a run. The only point at which it quits being a run is if you end up with 2 of every card... then it's a double run. I was thinking I'd figure out the number of ways to get a quad run (should only be 1...make that 4, since it could happen in any suit...), then the number of ways to get a triple run (minus the one(make that four) for the quad run), then the ways of getting a double run (minus the ways of getting a triple/quad run), and then the ways of getting a single run (minus the double/triple/quadruple runs). But even ignoring double/triple/quadruple runs when looking for a single run, my numbers still seem way off...
If I use Jemidiah's method in post 11, I'm not sure I'm getting the same numbers as he did.
I'm trying to figure out odds for Aces around and double Aces around.
I started with quadruple aces around (4,4) ^ 4 from each suit * (64,4) = 635376
Then I went to triple aces (4,3) ^ 4 * (68,8) - 4x Aces from above = 1,892,353,865,232
Then I went to double aces (4,2) ^ 4 * (72,12) - 3x Aces and 4x Aces from above = 19,908,924,100,186,400
Then I got single aces (4,1)^4 * (76,16) - 2x/3x/4x Aces from above = 2,752,579,046,460,370,000
With (80,20) hands = 3,535,316,142,212,180,000
With these numbers dividing the total number of hands by the Number of Aces around hands, I'm getting a result of 1 in 1.28 hands, and double aces as 1 in 177.57 hands.
I've calculated hands through Excel and Python programs that I've written, and python and perl programs that my sister has written and I'm getting numbers of 1 in 4.84 and 1 in 466.04.
Can someone help me determine where I'm going wrong? The reason I'm trying to figure this out is I'm comparing numbers from 4 player, double deck pinochle with 6 player triple deck pinochle (i.e. 6 of each card of each rank).
My numbers are showing me that it's easier to get anything around once, when using 2 decks, but it's easier to get things around 2 or more times when you're playing with 3 decks). I find it odd that you can get Aces around once, fewer times with 3 decks, but double/triple aces around more times with 3 decks. (I have over 400 million simulated hands with programs... and the program results all line up...)
Edit: I'm also trying to calculate the probability of a run. I saw this thread http://www.vbforums.com/showthread.p...un-in-pinochle, but it looks like the OP didn't finish clarifying the questions asked of him. I'm also basing it off of a double deck of pinochle cards (i.e. 4 of each card of each rank from Jack through Ace (tens are included, but they're ranked below Aces, above Kings), double deck Pinochle doesn't use the 9's).
So I'm thinking it should be:
Random suit (4,1)
Random ace of that suit (4,1)^5 (^5 does Ace, Ten, King, Queen, Jack).
Then there are 75 cards left, and you still need 15 in your hand, so that should be (75, 15)
so if I take (4,1) ^6 (Suit + 5 ranks) * (75,15), I'm getting a number of 9.33 E+18, yet (80,20) says there should only be 3.53E+18 total combinations.... so I'm off somewhere.
In regards to previous unanswered questions, yes having both other face cards is still a run. The only point at which it quits being a run is if you end up with 2 of every card... then it's a double run. I was thinking I'd figure out the number of ways to get a quad run (should only be 1...make that 4, since it could happen in any suit...), then the number of ways to get a triple run (minus the one(make that four) for the quad run), then the ways of getting a double run (minus the ways of getting a triple/quad run), and then the ways of getting a single run (minus the double/triple/quadruple runs). But even ignoring double/triple/quadruple runs when looking for a single run, my numbers still seem way off...