Quizzes & Puzzles29 mins ago
What are the odds?
3 Answers
Hi - I originally posted this in Chatterbank as I didn't know where to put it so I will have another go here. My husband taught my 5 and 6 year olds to play clock patience/solitaire yesterday (where you have to get all the cards out before you get 4 kings). The 5 year old completed it on the second turn and the 6 year old on the first turn - they were using different packs of cards which had been thoroughly shuffled (bu being dropped on the floor and picked up again and well as the more traditional method). As there is no skill involved there must be a way to work out the odds of this happening (I don't particularly want to know how to work it out - just what the odds were of this happening). Thanks.
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For more on marking an answer as the "Best Answer", please visit our FAQ.According to the Wikipedia entry, the chances of winning are 1 in 13.
I am not sure how that figure is worked out, though.
http://en.wikipedia.org/wiki/Clock_patience
I am not sure how that figure is worked out, though.
http://en.wikipedia.org/wiki/Clock_patience
The chance of winning is 1/13 and of losing is 12/13.
If they play three games, the chances of winning two out of the three are 36/2197, or about 1 in 61.
This comes from 3 x (12/13) x (1/13) x (1/13), ie 3 x P(Lose) x P (Win) x P (Win).
The factor of 3 is because the losing game could be first, second or third in the sequence.
If, however, you discount the case where the first two games were won, so the third game would never be played, then the chance is only 2 x (12/13) x (1/13) x (1/13), or about 1 in 91.5.
I think.
Someone will correct me if I am wrong.
If they play three games, the chances of winning two out of the three are 36/2197, or about 1 in 61.
This comes from 3 x (12/13) x (1/13) x (1/13), ie 3 x P(Lose) x P (Win) x P (Win).
The factor of 3 is because the losing game could be first, second or third in the sequence.
If, however, you discount the case where the first two games were won, so the third game would never be played, then the chance is only 2 x (12/13) x (1/13) x (1/13), or about 1 in 91.5.
I think.
Someone will correct me if I am wrong.