#21
|
|||
|
|||
Re: A Similar Problem
[ QUOTE ]
What is the probability that you will get through the whole deck without ever dealing the same 2 cards simultaneously? Then for an encore what is the probability for n Decks (1 < n < 52) [/ QUOTE ] I'm not sure what your encore question means. Do you want the probability that on all 52 cards, every deck has a different card, or do you want the probability that not all decks coincide? If the former, it is the number of 52xn Latin rectangles. |
#22
|
|||
|
|||
Re: A Similar Problem
[ QUOTE ]
[ QUOTE ] What is the probability that you will get through the whole deck without ever dealing the same 2 cards simultaneously? Then for an encore what is the probability for n Decks (1 < n < 52) [/ QUOTE ] I'm not sure what your encore question means. Do you want the probability that on all 52 cards, every deck has a different card, or do you want the probability that not all decks coincide? If the former, it is the number of 52xn Latin rectangles. [/ QUOTE ] I assumed that we have 2 stacks with n decks in each stack, and that we want to go through all the cards, dealing a card from each stack, without ever dealing a pair of the same card. |
#23
|
|||
|
|||
Re: A Similar Problem
For the initial problem...here's an approach to the entire rank-derangement problem. I haven't thoroughly looked at it yet, but it seemed sound. Font is a bit ugly and hard to read, however.
http://math.dartmouth.edu/~doyle/doc...rank/rank.html |
#24
|
|||
|
|||
Re: A Similar Problem
Great link LetYouDown. I enjoyed reading the historical background. I guess I'm not the only one who likes this problem. Funny thing too is that I've considered pitching this game to a casino. I've come up with several ideas for it's presentation.
Anyway, Marilyn is a smart woman and I concur that the expected number of matches in 52 is 4. That was another thing I did with the program along with the earlier post I had with probability of no matches for each card. It goes like this Matches % 0 = 1.6% 1 = 6.89% 2= 14.44% 3 = 19.8% 4 = 20.1% 5 = 16.1% 6 = 10.5% 7 = 5.8% 8 = 2.7% 9 = 1.1% 10 = 0.4% 11 = .14% 12 = .04% |
#25
|
|||
|
|||
Re: A Similar Problem
[ QUOTE ]
Anyway, Marilyn is a smart woman and I concur that the expected number of matches in 52 is 4. [/ QUOTE ] That's obvious since the probability of a match on each card is 1/13, the average number of matches is 52*1/13 = 4. |
#26
|
|||
|
|||
Re: A Similar Problem
[ QUOTE ]
I assumed that we have 2 stacks with n decks in each stack, and that we want to go through all the cards, dealing a card from each stack, without ever dealing a pair of the same card. [/ QUOTE ]As the decks get large, doesn't this approach 1-1/e? (Assuming that "same card" means same rank and suit.) |
Thread Tools | |
Display Modes | |
|
|