-=Attempts added=-

A Pinochle deck is a special deck of cards with 48 cards in total. it consists of two copies of each of the 9, 10, J, Q, K and Ace of all four suits (so there are 2 nine of clubs, 2 nine of diamonds, 2 nine of hearts, two nine of spades, and so on for ever other denomination). Poker can be played with the Pinochle deck, but the probabilities are different from poker with a standard deck. Calculate the following from the Pinochle deck:

a) Total number of 5-card Poker hands possible (order does not matter)

Attempt: 48C5 = 1712304 (I believe this number includes repetitions, how do I get rid of those?)

b) Probability of a four of a kind (4 cards of the same denomination plus one card of a different denomination)

For this one, I know how to do it with a standard deck, as following, but not the Pinochle one. How would the numbers be changed? I know the technique is the same, but which numbers would end up changing?

Attempt: P(4 of a kind):_

Denominations: 13 C 1 x 12 C 1

Suits: 4C4 x 4C1

Total: 13C1 x 12C1 x 4C4 x 4C1 = 624

P(4 of a kind 52 cards) = 624/52C5 = 1/4165

So would the Pinochle probability be = 624/48C5 ?

c) Probability of a royal flush (A,K,J,Q, 10 all of the same suit)

Attempt: P(Royal Flush) = 8/(48 C 5) = 1/214038


1 Answer 1


I am not familiar with Pinochle decks. So for the count I will assume that for example the two $\heartsuit$ Queens are identical.

Then there are $3$ types of $5$-card hands: (i) all cards distinct; (ii) there is $1$ pair of identicals; (ii) there are $2$ pairs of identicals. For the count, we calculate the number of each type, and add up.

(i) This is easiest. There are $24$ different cards, and there are $\binom{24}{5}$ ways to choose $5$.

(ii) There are $24$ different cards. We choose $1$ of the kinds to have $2$ of, and then $3$ different cards for the rest. That gives total $\binom{24}{1}\binom{23}{3}$.

(iii) This can be tricky. We pick $2$ of the kinds to have $2$ each of, and then an odd card. That gives $\binom{24}{2}\binom{22}{1}$.

Now for poker hand probabilities, we have to forget about the count we just made. For the different hands that we counted above are not equally likely. The Type (i) jands are all equally likely, as are the type (ii) hands, as are the type (iii) hands, but we do not have equal likelihood between different types.

There are two ways to proceed. We can take account of the different probabilities for each type. But my inclination is to imagine that the "repeats" are coloured red and blue, making the $48$ cards distinct. Then there are $\binom{48}{5}$ equally likely hands.

For each type of poker hand, count how many ways there are to produce it, taking colouring into account. I will leave the (unpleasant) details to you. We need to define carefully what we mean by each poker hand. For example, five of a kind is now possible.

Let's do a relatively straightforward one, $1$ pair. The type of card we have $2$ of can be chosen in $6$ ways, $9$ to Ace. Suppose for example it is $2$ Kings. They can be either $2$ of the same suit ($4$ choices) or of different suits. If so, the different suits can be chosen in $\binom{4}{2}$ ways, and then the actual cards in $2^2$ ways (all combinations of red and blue), for a total of $4+(6)(4)=28$ ways. Now we choose $3$ types from the remaining $5$, and for each choose a colour, red or blue. This gives a total of $(6)(28)(80)$. Check it. Now divide by $\binom{48}{5}$ for the probability.


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