A spider eats 3 flies a day. Until the spider fills his quota a fly has a 50% chance of survival if he attempts to pass the web.

Assuming 5 flies have already made the attempt to pass, what is the probability that the 6th fly will survive the attempt?

(In reply to

simpler computation by Charlie)

Charlie... I thought of that when I first looked at the problem.... but it is misleading.

Though your calculation 'works out', the spider NEVER eats 5 or 4 flies in the first 5 flies. Once he's eaten the first 3, he stops capturing them.

If you want to use combinatorial mathematics (and I purposefully, avoided the combination function for ease of understanding), you should total up the likelihood of getting 3 eaten if the THIRD fly is eaten, the likelihood of getting 3 eaten if the FOURTH fly is eaten, and the likelihood of getting eaten if the FIFTH fly is eaten....

This would correspond to:

1/8 chance:

CCC

1/16 chance (each):

CCNC

CNCC

NCCC

1/32 chance (each):

CCNNC

CNCNC

CNNCC

NCCNC

NCNCC

NNCCC

total: 1/2

_____________

All the 16 remaining possibilities don't have 3 flies eaten (and all have 1/32 chance of occuring)

NNNNN

NNNNC

NNNCN

NNCNN

NCNNN

CNNNN

CNNNC

CNNCN

CNCNN

CCNNN

NCNNC

NCNCN

NCCNN

NNCNC

NNCCN

NNNCC

total: 1/2