Under French law, the Beaujolais Nouveau
( wine ) is released at

**12:01 A.M. ** on the third Thursday in November every year.

Last year, prior to the above date** 11** wine amateurs deposited ** 12 ** small-size barrels at their merchantâ€™s store (**A,B,C,D,E,F,G - 7** liters each; ** S,T,U - 5** liters each and ** V ** two barrels: one** 7** liters and one** 5 ** liters).

When the wine arrived ** 70** liters were poured in the above barrels so that each barrel got an integer number of liters. The Customers were billed accordingly.

If every possible distribution of wine among the ** 12** barrels is equally likely, what is the possibility that V(ictor) had to pay for ** 11** liters of wine?

This is a little tricky to do exactly analytically.

The barrels add up to 76 liters, so we are 6 liters short. Instead of distributing the 70 liters, let's distribute the 6 liter shortage. The complication is that all 6 liters of shortage cannot be distributed to any of the 5 liter barrels, but even if they were 6 liters this would be extremely low probability. Let's ignore that for now.

Then, the probability that the 1st liter of shortage goes to V is 2/12 = 1/6. The probability that it does not go to V is 5/6.

The probability that exactly 1 liter of shortage goes to V = 6*(1/6)*((5/6)^5) = (5/6)^5 = 3125/7776 ~ .4019. The same probability as rolling 6 dice and having exactly one show a 1.

Now, let's adjust for the impossibility of a 5 liter barrel being 6 liters short. The probability of trying to put 6 liters in one specific 5 liter barrel = (1/12)^6 = 1/2,985,984. There are 4 such barrels, so multiply by 4 to get 1/746,496 ~ .00000134. Subtracting this from the numerator and the demoninator gives us (.4019 - .00000134)/(1-.00000134) ~ .4019. In other words, it does not make a significant difference.

The exact fraction, I think, is (3125/7776 - 1/2985984)/(1-1/2985984) = 1199999/2985983 ~ .4019

Of course, I could be wrong