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 Comparing two events (Posted on 2016-06-15)
Compare probability of getting at least one "6" in four rolls of a single 6-sided die with the probability of at least one double-six in 24 throws of two dice,

Will the preference change if the two dice are thrown 25 times instead of 24 ?

Traced to: The French nobleman and gambler Chevalier de Méré.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 spoiler -- try it yourself first | Comment 2 of 3 |
The probability of not getting a 6 in four rolls is (5/6)^4 = 625/1296 ~=  .4822530864197532, making the probability of actually getting a 6 approximately  .5177469135802468.

The probability of not getting a double six in 24 throws of two dice is (35/36)^24 ~= .508596123869094, making the probability of actually getting a double six approximately .491403876130906.

Given 25 throws, that latter probability becomes 1-(35/36)^25 ~= .505531546238381, which is still not as high as the single six in four rolls of one die.

You would have to go to 26 times to exceed the single-six probability: 1-(35/36)^26 ~= .519266781065093.

 Posted by Charlie on 2016-06-16 09:43:00

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