Card holding probability in 4-handed games

In a 4-handed card game like Spades or Hearts, say that the first player leads a Club and everyone follows suit, then he leads another Club, accounting for 5 of the 13 Clubs, leaving 8 out. If the 2nd player is now out of Clubs, how do you compute the probability that at least one of the next 2 players is also out of CLubs?

I tried listing all of the possible distributions (8-0-0, 0-8-0, 0-0-8, 7-1-0, 7-0-1, etc.) and counted those with 0 in the 2nd or 3rd positions as a percent of the total, but came up with about 48%, which seems way too high.

I get 4.165833185332 %

Thanks, but how do you get that?

There are 13 clubs and 39 other cards in a deck. We remove the 5 clubs that have been played from the deck. We also know that player2 has 12 other cards.
This leaves 27 other cards and 8 clubs for a total of 35 cards.
The number of ways player3 can have no clubs is combin(27,12) out of a total of combin(35,12) ways. combin(27,12)/combin(35,12)=0.020832671221993
The same for player4
So we now have 2*0.020832671221993=0.041665342443985 as the probability of either player3 or 4 having no clubs except we are counting getting no clubs in either hand twice, so we must subtract that.

combin(27,24)/combin(35,24)=0.000007010590666
0.041665342443985 - 0.000007010590666 = 0.04165833185332

Hope that helps. Feel free to ask any more questions.

Thanks. That is the info I was looking for.