Have you ever been punished for being too good at spider solitaire? I mean, have you ever been stuck because you collected too many suits? Many versions of the game don’t allow you to deal from the deck if you have empty columns, nor do they allow you to get back a completed suit. If the number of cards left on the table in the middle of the game is less than ten — the number of columns — you are stuck. I always wondered what the probability is of being stuck. This probability is difficult to calculate because it depends on your strategy. So I invented a boring version of spider solitaire for the sake of creating a math problem. Here it goes:
You start with two full decks of 104 cards. Initially you take 54 cards. At each turn you take all full suits out of your hand. If you have less than ten cards left in your hand, you are stuck. If not, take ten more cards from the leftover deck and continue. What is the probability that you can be stuck during this game?
Let us simplify the game even more by playing the easy level of the boring spider solitaire in which you have only spades. So you have a total of eight full suits of spades. I leave it to my readers to calculate the total probability of being stuck. Here I would like to estimate the easiest case: the probability of being stuck before the last deal.
There are ten cards left in the deck. For you to be stuck, they all should have a different value. The total number of ways to choose ten cards is 104 choose 10. To calculate the number of ways in which these ten cards have different values we need to choose these ten values in 13 choose 10 ways, then multiply by the number of ways each card of a given value can be taken from the deck: 810. The probability is about 0.0117655.
I will leave it to my readers to calculate the probability of being stuck before the last deal at the medium level: when you play two suits, hearts and spades.
No, I will not tell you how many times I played spider solitaire.Share: