![]() ![]() This is what I don't know how to account for. The actual number of unique hands is less than this because of the aforementioned fact that there are multiple skip cards, multiple wilds and multiple blue12's, red12's, etc. (Parenthetical note, indicated redundantly by the fact that this paragraph is in parentheses. Wolfram Alpha gives us this number: 38,722,819,230,810 different hands. (TotalToChooseFrom choose NumberNeeded) Since the deck has 108 cards, and I need to calculate (108 choose 10). The mathematical function that does this is the choose function. ![]() The first calculation I will find is how many different hands of phase 10 there are. If any of you readers want to instruct me, please leave a comment below. I hope that hasn't distorted things too horrendously, and perhaps I'll return and clean this post up if I ever figure out how to handle it. The fact that there are wild cards, and that there are two blue 12's and two green 12's etc is a fact that complicates things beyond what I know how to handle, so I have ignored that in these calculations. ![]() This sort of a question requires you to calculate how many different combinations of cards are possible for each hand, and divide that by how many different combinations of hands there are. What are the probabilities of being dealt one of the given phases right away? Yesterday I wrote about some of the probabilities associated with the game phase 10. ![]()
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