What is the expected probability of a basket to be frequent on average? I was reading on frequent pattern mining algorithm and came up with the following question. Let a company have $10^4$ different products and there are $10^9$ transactions each containing exactly 10 different products. If for each transaction a product is uniformly chosen, then what is the probability that a given fixed transaction of size 10 is chosen for $10^3$ times among $10^9$ transactions?

This is self study and the problem is stated in Slide 9 in here
P(10)  9990! *  10! / 1000!

= 10! / 10000 x 9999 x 9998 x 9997 x 9996 x 9995 x 9994 x 9993 x 9992 x 9991   = 3.645 * 10 ^ (-34)

with 10^9 transactions, you have a poisson distribution with parameter 10 ^(-25) so the chances of 1 particular transaction is extremely negligible.  The chance of getting 1000 occurrences of a transaction is 

P(1000) = $\lambda ^ {1000} / 1000! e^{-\lambda} $

log(P(1000)) = 1000 log(3.645 * 10 ^ (-34)) - (1000 log 1000 - 1000) = - 76994 - 5907 = -82901

that's natural log, as log 10 it is -36003

so my answer is $P(1000) = 10 ^ {-36003}$
P(10)  9990! *  10! / 1000!

= 10! / 10000 x 9999 x 9998 x 9997 x 9996 x 9995 x 9994 x 9993 x 9992 x 9991   = 3.645 * 10 ^ (-34)

with 10^9 transactions, you have a poisson distribution with parameter 10 ^(-25) so the chances of 1 particular transaction is extremely negligible.  The chance of getting 1000 occurrences of a transaction is 

P(1000) = $\lambda ^ {1000} / 1000! e^{-\lambda} $

log(P(1000)) = 1000 log(3.645 * 10 ^ (-34)) - (1000 log 1000 - 1000) = - 76994 - 5907 = -82901

that's natural log, as log 10 it is -36003

so my answer is $P(1000) = 10 ^ {-36003}$