• Jan 3rd 2009, 04:33 PM
sdunkley
I play a card game called Spider Solitaire. For purposes of my question we can ignore the differences between suits but you still have the standard 13 cards per "set" (One through King). There are 8 sets total for a total of 104 cards. On one game I played I had gotten 76 cards face up and only one card was a Ten card. Ignoring order, I am wondering what the probability is for this result. (BTW, I had lost the game so I could not play further to get any of the remaining 28 cards face up.)
Thanks for any help.
• Jan 4th 2009, 06:02 AM
awkward
Quote:

Originally Posted by sdunkley
I play a card game called Spider Solitaire. For purposes of my question we can ignore the differences between suits but you still have the standard 13 cards per "set" (One through King). There are 8 sets total for a total of 104 cards. On one game I played I had gotten 76 cards face up and only one card was a Ten card. Ignoring order, I am wondering what the probability is for this result. (BTW, I had lost the game so I could not play further to get any of the remaining 28 cards face up.)
Thanks for any help.

Hi Sdunkley,

To answer your question directly, there are $\displaystyle N = \binom{104}{76}$ possibilities for the deal of 76 cards from the 104 card deck (disregarding the order of the cards). Of these, $\displaystyle 8 \binom{96}{75}$ have exactly one Ten. So the probability of having exactly one Ten in the deal is $\displaystyle 8 \binom{96}{75} / N = 3.494 \times 10^{-4}$.

However, I don't think that is really the question we should ask. There is nothing special about the rank Ten-- surely you would have been just as perplexed to have only one card of any given rank, and even worse, you could have zero cards of that rank. So I think a better question is this: What is the probability of dealing one or fewer cards of some rank?

If we draw zero cards of some rank, there are 13 ways to choose the rank and then $\displaystyle \binom{96}{76}$ to choose the remaining cards, so the total number of ways is $\displaystyle 13 \cdot \binom{96}{76}$, with probability $\displaystyle 13 \cdot \binom{96}{76} / N = 1.569 \times 10^{-4}$.

If we draw exactly one card of some rank, there are 13 ways to choose the rank, 8 ways to choose the particular card, and then $\displaystyle \binom{96}{75}$ ways to choose the remaining cards, so the total number of ways is $\displaystyle 13 \cdot 8 \binom{96}{75}$, with probability $\displaystyle 13 \cdot 8 \binom{96}{75} / N = 4.542 \times 10^{-3}$.

Finally, the probability of drawing zero or one cards of some rank is

$\displaystyle 1.569 \times 10^{-4} + 4.542 \times 10^{-3} = 4.699 \times 10^{-3}$.
• Jan 4th 2009, 02:20 PM
sdunkley
Thanks
Thanks awkward. It's been 35 yrs since I've done math of that sort. You are right that I was interested in your second question.