My uncle’s ritual for dressing each morning except Sunday includes a trip to the sock drawer, where he:
(1) picks out three socks at random, then
(2) wears any matching pair and returns the odd sock to the drawer or
(3) returns the three socks to the drawer if he has no matching pair and repeats steps (1) and (3) until he completes step (2).

The drawer starts with 16 socks each Monday morning (eight blue, six black, two brown) and ends up with four socks each Saturday evening.

(a) On which day of the week does he average the longest time to dress?
(b) On which day of the week is he least likely to get a pair from the first three socks chosen?

Source: manchi tutorials

I don't anticipate going further.  (a) and (b) are asking for which day is is most/least likely to get 3 unmatched socks.  I figured it would be best to find the probability of every combination every day.

Day 1:
blue pair = 8*7*6+3*8*7*6+3*8*7*2=1680
black pair = 6*5*4+3*6*5*8+3*6*5*2=1020
brown pair = 2*1*0+3*2*1*8+3*2*1*6=84
no pair = 6*8*6*2=576
So the chance of no pair is 576/3360 ≈.17143
the chance of ending up with each color are the above counts out of 2784

If the brown pair is ever chosen, there will always be matching pairs for the rest of the week.

Day 2 - we need to consider the conditional probabilities based on a black pair or a brown pair
If day 1 was black, no pair = 6*6*6*2/14*13*12 = 432/2184
If day 1 was blue, no pair = 6*8*4*2/14*13*12 = 384/2184
Total probability of no pair =
1680/2784*432/2184+1020/2784*384/2184 = 485/2639 ≈ .18378

To continue, I'd need to find the probability of each possible remaining sock drawer (5 possibilities) and the day 3 solution would be a bit uglier.

Posted by Jer on 2013-06-29 15:03:34