Binary possibilities

[Pumpkinstew]

Originally Posted by kaiowas

Incidently if you had 11585 cards you'd be guaranteed that at least one pair would conflict.

Hmm, not sure about that. If you have 2^26 'bins' of 64 cards available to you and you assume that the cards are randomly and equally distributed through the entire range of values then you must need 2^26+1 cards to guarantee a duplicate.

You're right that I wrongly used factorial in my first post though. The wiki description of the birthday problem uses factorials in the soultion though and as best I can tell it's the same problem.

[Mark]

For 100% probability of a duplicate, you would indeed need 2^26+1 cards. I'll take a guess that 11585 is the 'on average' (i.e. 50%) probability.