*Originally posted by roma45*

**he has only 5 to choose from, we already know that he has been wrong all week, but it is still a 1 in 5 chance of being right. the only fact we need is the number of choices and hes been wrong on the other days. if he was correcy on every day then it would be a certainty he would be right on a friday as well. as it is its a 1 in 5 chance**

It's true that if you had correct pairs for the first four days, you'd know the fifth day was also correct, but that isn't the problem given. Here we're told a bit of information that on the surface seems not to affect the odds, but when you look deeper you see that it does (though to be fair, it only reduces the odds by a small amount).

I'll demonstrate any easy to visualize brute force method on a very similar problem. Consider wolfgang labels four kinds of socks for each day in his four day work week from Monday-Thursday. Wednesday night in the pub you hear that wolfgang picked the wrong sock pair for the first three nights and are asked about the odds that wolfgang is left with the correct sock pair for Thursday. 1 in 4 chance you say? Let's see.

I'll use M for Monday, T for Tuesday, W for Wednesday, R for Thursday. There are 4! = 24 permutations of sock pairs for the four day week.

MTWR

MTRW

MWTR

MWRT

MRTW

MRWT

TMWR

TMRW

TWMR

TWRM

TRMW

TRWM

WMTR

WMRT

WTMR

WTRM

WRMT

WRTM

RMTW

RMWT

RTMW

RTWM

RWMT

RWTM

Now go through and cross off all the ones that have a match on M, T, or W (i.e. M in the 1st slot OR T in the 2nd slot OR W in the 3rd slot). The information given to us rules out these permutations, leaving only the permutations that have been wrong all three days so far.

TMRW

TWMR

TWRM

TRMW

WMTR

WMRT

WRMT

WRTM

RMTW

RWMT

RWTM

We're left with only 11 permutations that are possible. This represents our remaining pool of all possible outcomes and the denominator in our probability. Now let's count how many of these 11 possible outcomes have a match on the last day meaning an R in the 4th slot.

TWMR

WMTR

Only 2 out of 11 or 18%, much less than the 25% you'd predict without being told that wolfgang picked wrong on Monday, Tuesday and Wednesday.

I'll leave the 5 day problem posed in the first post to you, but hopefully this example helps you visualize why it isn't correct to just divide 1 into the number of types of socks and get 1/5.