ChatterBank0 min ago
How to calculate......
Trivia really, but I was wondering how would you calculate the number of games of solitaire you would need to play before all the combinations of the cards were used?
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While there are the previously stated number of possible combinations for the order of 52 cards, the number of times one would actually have to shuffle the cards to arrive at each and every combination can not be determined. Due to the random nature of the shuffle it is far more likely that many of the possible combinations would be repeated before all possible combinations were manifested.
Ignoring the "lands on edge" scenario, one could in theory flip a coin and endless number of times before the other of the two possible outcomes became manifested.
Consider a �deck� of cards consisting of only the four aces. The odds of arriving at all of the 24 possible order arrangements without hitting on one of the previous outcomes is less than 1 in 2 billion!
80658175170943878571660636856403766975289505440883277824000000000000 (click on the link for the actual number)
is the minimum number of games you would have to play to arrive at all the possible combinations of cards but the odds of arriving at all possible combinations before repeating any of them are . . . Astronomical!
Ignoring the "lands on edge" scenario, one could in theory flip a coin and endless number of times before the other of the two possible outcomes became manifested.
Consider a �deck� of cards consisting of only the four aces. The odds of arriving at all of the 24 possible order arrangements without hitting on one of the previous outcomes is less than 1 in 2 billion!
80658175170943878571660636856403766975289505440883277824000000000000 (click on the link for the actual number)
is the minimum number of games you would have to play to arrive at all the possible combinations of cards but the odds of arriving at all possible combinations before repeating any of them are . . . Astronomical!
The answer is: you can't. You might equally ask: "How many times must I roll a die to be sure that every number, 1 through 6, has been rolled at least once?" Six times? Ten million times? There is always the chance that you never roll, for example, a three. You can, however, answer the question: "How many times must I roll a die in order to be at least [b]95%[/b] sure of rolling every number at least once?" Or any percentage you wish.
So, when playing solitaire, a card combination might be similarly missed, over and over, and never come up during play. So you can never calculate how many games it will take for every card comination to have definitely been played. You [i]can[/i] calculate the probability that every combination will come up after [i]x[/i] games. Or, conversely, you can find how many games you need to play to be [i]y[/i] per cent sure that every card combination will have been played by then!
The exact probabilities will depend on what you define as a "card combination". Is the number of card combinations the number of unique, winning games of solitaire? Or is a card combination the order in which you place all the cards onto the four ace piles? There are many interpretations of your question and I can only imagine the method of calculation to be lengthy and complicated, and the solution to be "a large number of games".
So, when playing solitaire, a card combination might be similarly missed, over and over, and never come up during play. So you can never calculate how many games it will take for every card comination to have definitely been played. You [i]can[/i] calculate the probability that every combination will come up after [i]x[/i] games. Or, conversely, you can find how many games you need to play to be [i]y[/i] per cent sure that every card combination will have been played by then!
The exact probabilities will depend on what you define as a "card combination". Is the number of card combinations the number of unique, winning games of solitaire? Or is a card combination the order in which you place all the cards onto the four ace piles? There are many interpretations of your question and I can only imagine the method of calculation to be lengthy and complicated, and the solution to be "a large number of games".
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