Calculating Split Percentages for Double Deck Pinochle - Expert Help Needed

[rakbeater]

I am writing a book on double deck pinochle and need to get accurate percentages. I am hoping this forum is willing and able to help me! If I am in the wrong forum, and someone could direct me to a more appropriate forum, that would be greatly appreciated.

Knowing what double deck pinochle is, is irrelevant to solving the problem. The problem is this:

There are 80 cards in a pinochle deck with four suits: Hearts, Spades, Clubs, and Diamonds. 20 cards per suit. It is a 4 player game, each player gets dealt 20 random cards.

My question is: In my hand I have X number of cards in one suit, I want to know the percentage of times (20-X) will split between the other 3 players.

For example:

If I have 8 cards in my hand, leaving 12 cards to be split among the other 3 players:

12-0-0 ?%
11-1-0 ?%
10-1-1 ?%
9-3-0 ?%
9-2-1 ?%
etc, etc, etc

I need to know the split percentages when I have 5 cards in my hand all the way up to 18 cards in my hand.

Hopefully this makes sense. If it doesn't please reply and I will clarify. I have no background in Prob/Stat so I am completely at the mercy of others. Thanks in advance for the help.

My email is rakbeater at gmail.com if that is needed for any reason. Thanks again!

 

[rakbeater]

http://talkstats.com/showthread.php?t=12966

I was wondering if this response I received in this other forum is correct? I appreciate any help in figuring this out. I don't want to have incorrect information in my book. Thanks again!

 

[moonman239]

I wouldn't post your e-mail if I were you. I would post it as rakbeater at gmail dot com. Otherwise when the email collector bots come to this thread, they will see your email address and add it to their list of people to spam.
Anyways, the answer you received makes sense to me. You're dealing with cards that can have any suit, and that can be drawn in any order.

 

[Fredrik]

I don't understand what he's doing in step 3, so I didn't read on after that. Note that I'm not saying he's wrong. I often make mistakes in problems like this, so don't assume that my attempt is correct either. I'm just telling you what I'm thinking.

There are $${60\choose 12}$$ ways to distribute 12 hearts over 60 positions. Now consider the possibility that player A gets 7 hearts, player B gets 3 hearts and player C gets 2 hearts. There are $${20\choose 7}$$ ways to distribute 7 hearts over the 20 positions of player A's hand, and we calculate the corresponding numbers for players B and C the same way. So the probability of this particular distribution of the hearts should be

$$\frac{{20\choose 7}{20\choose 3}{20\choose 2}}{{60\choose 12}}$$

But what we really want is the probability that one of the players gets 7 hearts, another gets 3 and another gets 2. To get that we just need to multiply the above by 6, because there are 6 ways to distribute the values 7,3,2 over 3 players. If the distribution had been e.g. 8,2,2, the number would have been 3, because there are only 3 ways to choose who gets the 8 hearts. And if the distribution had been 4,4,4 the number would have been 1, because these numbers can't be rearranged. So you multiply by 6,3 or 1, depending on the number of players that get the same number of hearts.

Hm, interesting. I entered the numbers for the 9-2-1 distribution into his formula and mine in Mathematica, and the results differ by a factor of 6. If I leave out the 3!=6="number of ways to distribute the numbers 9,2,1 over three players", I get his result.

 

[rakbeater]

I wouldn't post your e-mail if I were you. I would post it as rakbeater at gmail dot com. Otherwise when the email collector bots come to this thread, they will see your email address and add it to their list of people to spam.
Anyways, the answer you received makes sense to me. You're dealing with cards that can have any suit, and that can be drawn in any order.

Thanks for the advice re: my email address. Thank you everyone else for your assistance.

 

[Fredrik]

Rakbeater: The last paragraph in my post and the last thing BGM says in point 4 of his post imply that one of us is wrong by a factor of 6 (or that we're both wrong). So you can't take my calculation as a verification of his. The fact that we're so close probably means that one of us is making a relatively minor mistake, but there's still a mistake in (at least) one of our answers. I don't know if it's in mine or his.

So I hope someone else will take a look at this. If I did something wrong, I'd like to know it.

 

[rakbeater]

ok I will have to check it out. Thanks for the concern. I do want to get this right.

 

[rakbeater]

Rakbeater: The last paragraph in my post and the last thing BGM says in point 4 of his post imply that one of us is wrong by a factor of 6 (or that we're both wrong). So you can't take my calculation as a verification of his. The fact that we're so close probably means that one of us is making a relatively minor mistake, but there's still a mistake in (at least) one of our answers. I don't know if it's in mine or his.

So I hope someone else will take a look at this. If I did something wrong, I'd like to know it.

Take a look at the other thread, where someone commented and said that you are both right. One is for specific distribution and the other formual is for generic distribution. Let me know if that makes sense.

 

[Fredrik]

Rakbeater, you haven't made it clear if you're looking for the probability that one of the players have 9 of the suit, another has 2 and another has 1, or the probability that player A has 9 of the suit, player B has 2 and player C has 1. These probabilities differ by a factor of 6.

If I understand BGM correctly, he made a small mistake in his first post by suggesting that he was calculating the first of these probabilities, when in fact he was calculating the second. In that case, the calculation of the first probability by his method, in Mathematica, is

6 Binomial[12, 9] Binomial[48, 11] Binomial[3, 2] Binomial[37,18] Binomial[1,1] Binomial[19,19]/(Binomial[60, 20] Binomial[40, 20] Binomial[20, 20]) = 12160/4443467 = 0.0027366

And the calculation by my method is

6 Binomial[20, 9] Binomial[20, 2] Binomial[20, 1]/(Binomial[60, 12]) = the same result

So it looks like you have your answer. Just remember that if it's the first probability you're interested in, the 6 at the start of both calculations needs to be replaced by 3 when two of the other three players have the same number of cards of that suit, and by 1 when they all have the same number, e.g. a 4-4-4 distribution.

 

[Petr Mugver]

I think the formula above is wrong. If you put easy numbers like (20,0,0,0) or (19,1,0,0) you don't get the expected results (1 and 1/3 in the previous examples). It gives wrong results by up to 10^50!

I have a formula that looks different. Let $$N_A$$, $$N_B$$, $$N_C$$, $$N_D$$, be the number of cards of one type in possession of player A, B, C and D.
A certain distribution has probability

$$P(N_A,N_B,N_C, N_D)=\frac{20!}{N_A!N_B!N_C!N_D!}\left(\frac{1}{4}\right)^{20}$$

The probability tha A has $$N_A$$ cards (and the other players $$_{20-N_A}$$) is

$$P(N_A)=\sum_{N_B+N_C+N_D=20-N_A}P(N_A,N_B,N_C, N_D)=\frac{20!}{N_A!(20-N_A)!}\left(\frac{1}{4}\right)^{N_A}\left(\frac{3}{4}\right)^{20-N_A}$$

And the probability you are looking for should be

$$P_{N_A}(N_B,N_C,N_D)=\frac{P(N_A,N_B,N_C, N_D)}{P(N_A)}=\left(\frac{1}{3}\right)^{20-N_A}\frac{(20-N_A)!}{N_B!N_C!N_D!}$$

Combinatoric is always a mess...

Checks:

$$P_{20}(0,0,0)=1$$
$$P_{19}(1,0,0)=P_{19}(0,1,0)=P_{19}(0,0,1)=1/3$$
$$P_{18}(2,0,0)=P_{18}(0,2,0)=P_{18}(0,0,2)=1/9$$
$$P_{18}(1,1,0)=P_{18}(0,1,1)=P_{18}(1,0,1)=2/9$$

And so on...

$$P_{N_A}(20-N_A,0,0)=P_{N_A}(0,20-N_A,0)=P_{N_A}(0,0,20-N_A)=\left(\frac{1}{3}\right)^{20-N_A}$$

Seems reasonable! In particular:

$$P_{8}(7,3,2)=P_{8}(7,2,3)=P_{8}(3,7,2)=P_{8}(3,2,7)=P_{8}(2,7,3)=P_{8}(2,3,7)=1.5\%$$

 

[Fredrik]

I think the formula above is wrong. If you put easy numbers like (20,0,0,0) or (19,1,0,0) you don't get the expected results (1 and 1/3 in the previous examples). It gives wrong results by up to 10^50!

For the distribution (20,0,0,0), my formula is (in Mathematica syntax):

Binomial[20, 0] Binomial[20, 0] Binomial[20, 0]/(Binomial[60, 0])

and Mathematica tells me this is =1. For (19,1,0,0), it's

3 Binomial[20, 1] Binomial[20, 0] Binomial[20, 0]/(Binomial[60, 1])

which is also =1. (The factor of 3 is there because I don't care which one of the other three players has the one heart that I didn't get).

My formula yields different results than yours for the (18,2,0,0) and (18,1,1,0) distributions. I get 19/59 and 40/59 respectively. For (8,7,3,2), I get roughly 7.2% Hm, that sounds like a lot. (When you compare yours to mine, keep in mind that I always include a factor of 1,3 or 6 that you don't, because I don't care how the numbers 1,1,0 or 7,3,2 are distributed among the three players).

Edit: I just calculated the result for (18,2,0,0) by a different method and got the result 19/59 again. This is the method: We need to find the probability that neither player A nor player B gets any of the 2 hearts among the 60 cards. The probability that the first card is not a heart is 58/60. For the second card, it's 57/59. We need to multiply these numbers for 40 cards. The product can be expressed as (58!/(18!))/(60!/(20!)). Multiply this by 3 because we don't care which two players don't get a heart. Result: 19/59.