Three balls are randomly and independently tossed into bins numbered with positive integers so that for each ball, the probability that it is tossed into bin i is 2`^-i` for I =1, 2, 3, 4,... More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is p/q, where p and q are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins 3, 17, and 10.) What is p+q?

The value of p+q is 55.

Expert Answers

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Any triple may be described by its starting index `j ` and the distance between the closest balls `k , ` i.e. `j , j + k , j + 2k . ` Here `j = 1 , 2 , 3 ... ` and `k = 1 , 2 , 3 ...`

The probability of a triple `j , j + k , j + 2k ` in some fixed order of balls is the product of the corresponding probabilities, `2^( -j ) * 2^( -j - k ) * 2^( -j - 2k ) = 2^( -3j ) * 2^( -3k ) . ` Because the order is insignificant, the probability of each triple is 6 times greater.

The probability of even spacing is therefore

`6 sum_( j = 1 )^( oo ) ( sum_( k = 1 )^( oo ) ( 2^( -3j ) * 2^( -3k ) ) ) =` ` 6 sum_( j = 1 )^( oo ) ( 2^( -3j ) ) * sum_( k = 1 )^( oo ) ( 2^( -3k ) ) =`

`= 6 * ( 2^( -3 ) / ( 1 - 2 ^( -3 ) ) )^2 = 6 * ( 1 / ( 8 - 1 ) )^2 = 6 / 49 .`

This way, `p = 6 , q = 49 , p + q = 55 .`

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