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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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 solution | Comment 1 of 3
For all cases, the probability one "at least one success" is the complement of "no successes" and so can easily be computed by subtracting the latter probability from one.

In the first case
1-(5/6)^4≈.5177

In the second case
1-(35/36)^24≈.4192

So it is more likely to get at least one six in four rolls than to get at least one double-six in 24 rolls.

Increasing to 25 rolls does not change this
1-(35/36)^25≈.5055

However, with 26 rolls this does the more likely outcome
1=(35/36)^26≈.5193
 Posted by Jer on 2016-06-16 09:42:03

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