Calculating the probability of being dealt a straight in the Phase 10 card game

Question

Assuming there are 96 basic numbered cards, two of each rank from 1 through 12 in each of four colors in the deck. What is the probability of being dealt a 9 card straight, if I am to be dealt 10 cards in total? The problem for me is that I am over counting. I only know how to solve the problem assuming each card is unique, this is not the case as there is two of every card

Answer 1

There are $48 \choose 10$ hands you can draw. There are two complications to counting the number of straight hands. One is that you can get two cards of the same rank. It is easy to count each of the two straights. The second is that you can get a ten card straight and count the top straight and the bottom straight.

Hands with a ten card straight: there are three ranks that can be the lowest. For each rank you have four cards, so there are $3 \cdot 4^{10}$ of these.

Hands with a nine card straight and a second card that matches one of the nine in rank: four ranks that can be the lowest, nine ways to choose the rank with two cards, six ways to choose the two cards of that rank, eight ranks where you choose one of four cards, so $4 \cdot 9 \cdot 6 \cdot 4^8$ of these.

Hands with a nine card straight and an unconnected card. There are six choices for the ranks of the cards: $1-9+11, 1-9+12, 2-10+12, 1+3-11, 1+4-12, 2+4-12$. For each you have four choices of each rank, so there are $6\cdot 4^{10}$ of these.

The total number of hands with a nine card straight is $3 \cdot 4^{10}+4 \cdot 9 \cdot 6 \cdot 4^8+6\cdot 4^{10}=23592960$ The probability is $\frac {23592960}{48 \choose 10}\approx 0.0036,$ a little more than one in $300$