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QUESTION: In a three handed pinochle game last night, in eleven hands dealt, one person had 3 double pinochles.

Do you know the odds of this happening.

(and no, it wasn't me......)

Thanks.

Mike

This happened last night, and I was curious what the odds were of this happening......

ANSWER: I am not an expert at pinochle, so I will try to make clear what I think is happening.

48 cards are dealt -- the highest 24 cards in the deck, from two decks.

Double pinochle is both Jacks of diamonds and both Queens of spades.

The deal is 12 cards to each of 4 players, so a single person receives 1/4 of the cards.

There are 48 choose 12 ways to deal all the cards, and only 48 choose 8 ways of getting double pinochle, which we can call the probability p:

p = (48c8) / (48c12) ≈ 0.00541635

That's about half a percent or 1 in 200. However, to do this three times in a row is:

p^3 ≈ 0.000000158898

which is virtually impossible. Of course, you didn't say three times in a row. There are really two things you could be asking. Either you want to know the probability getting double pinochle

If you want exactly 3 out of 11, you need:

(11c3) p^3 (1-p)^8 ≈ 0.0000251035

and those chances are about 1 in 39,836.

The most reasonable interpretation, though, is "what are the chances of somebody getting three [or more] double pinochles" which is slightly more likely (you could get 3 d.p.s, but you could get the even more rare 4 or 5 or ... ). That is:

(11c3) p^3 (1-p)^8 + (11c4)p^4 (1-p)^7 + (11c5)p^5 (1-p)^6 + ... + (11c10)p^10 (1-p) + (11c11) p^11 ≈ 0.000025379

and those chances are about 1 in 39,403.

Note: For the purposes of this analysis, I've considered all cards unique -- even though the two decks might be indistinguishable to the naked eye (assume one has red back, the other blue). This makes the analysis easier. We could do it as if the two decks were indistinguishable, and that would affect the numerator and denominator in the final outcome above by the same fraction basically.

So, in short, the odds are nearly one in 40 thousand. But that doesn't mean impossible!

---------- FOLLOW-UP ----------

QUESTION: Thanks for the great analysis. In this case is was exactly 3 occurrences out of eleven hands dealt.

However, could I ask you to run the probability when the game is between three, not four, players?

Thanks again for your entertaining site.

Again, I will try to explain what I am assuming, since I don't know much about pinochle.

If there are three players and all cards are dealt, each player gets 16 cards. (In some variations this is not the case however!) The only thing that changes is p:

p = (48c12) / (48c16) ≈ 0.0308972

Note this is much more likely! You can still compute:

p^3 ≈ 0.0000294956

(11c3) p^3 (1-p)^8 ≈ 0.00378617

(11c3) p^3 (1-p)^8 + (11c4)p^4 (1-p)^7 + (11c5)p^5 (1-p)^6 + ... + (11c10)p^10 (1-p) + (11c11) p^11 ≈ 0.00403872

That last one is about 1 in 250 (more like 1 in 248 I guess).

I suggest you consider the "answer" to be the last figure. I know what you observed seems like it was

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