imadork
all you chumps play solitare to find the fastest time that you can beat the game, as did i for the longest time. however, thinking about it, we are going about it all wrong. i say, find the game that is absolutly un-playable. not one play in the entire game. think about it scientifically:
there are 3 different things to think about,
1. no aces can be in the fifteen playable cards,
2. none of the seven playable cards in the row-stacks can be moved to a different row-stack.
3. none of the eight playable cards in the deck can be moved to any of the seven row-stacks.
the probability of satisfying only the first condition can be easily calculated. this probability is the number of all possible arrangements (combinations) of 15 cards taken from a deck with no aces, divided by the number of all possible arrangements of 15 cards taken from a deck that includes aces. by using binomial equations you find that on average 24.39 percent of solitare games start with no aces amongst the initially playable cards. equivalently, 75.61 percent of solitare games have at least one ace in the 15 initially playable cards. this still leaves us with a problem becuse you need to simultaneously calculate for all three situations. even considering only the games where no aces occur in the 15 playable cards, the number of all possible solitare games is quite large. under the previous scenario, the number of games to be examined totals about 7,035,128,610,578,640. well no one is calculting that so I found a website where someone simulated 10 million games. a total of 24,893 unplayable games were reported giving 0.0025 (rounded to two significant figures) as an estimate of the probability of occurrence of unplayable games. this estimate suggests that on average 0.25 percent (one out of 400) of all games are unplayable. now using the same simulation, it was found that that anywhere from 2.5 to 10 percent of all solitare games are unsolvable. So if you find yourself loosing a lot, you suck because 90 percent of the time you should be at least finishing. imadork
there are 3 different things to think about,
1. no aces can be in the fifteen playable cards,
2. none of the seven playable cards in the row-stacks can be moved to a different row-stack.
3. none of the eight playable cards in the deck can be moved to any of the seven row-stacks.
the probability of satisfying only the first condition can be easily calculated. this probability is the number of all possible arrangements (combinations) of 15 cards taken from a deck with no aces, divided by the number of all possible arrangements of 15 cards taken from a deck that includes aces. by using binomial equations you find that on average 24.39 percent of solitare games start with no aces amongst the initially playable cards. equivalently, 75.61 percent of solitare games have at least one ace in the 15 initially playable cards. this still leaves us with a problem becuse you need to simultaneously calculate for all three situations. even considering only the games where no aces occur in the 15 playable cards, the number of all possible solitare games is quite large. under the previous scenario, the number of games to be examined totals about 7,035,128,610,578,640. well no one is calculting that so I found a website where someone simulated 10 million games. a total of 24,893 unplayable games were reported giving 0.0025 (rounded to two significant figures) as an estimate of the probability of occurrence of unplayable games. this estimate suggests that on average 0.25 percent (one out of 400) of all games are unplayable. now using the same simulation, it was found that that anywhere from 2.5 to 10 percent of all solitare games are unsolvable. So if you find yourself loosing a lot, you suck because 90 percent of the time you should be at least finishing. imadork
posted by josh at 3:02 PM
2 Comments:
it's kelly. can't remember my gd username again.
you have way too much time on your hands. do some work damnit.
wow
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