Question

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?

Accepted Answer

The value of p+q is 55.

Answer

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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