A Full Monty!

One of your sisters (Jane or Mary) is taking a shower. You hear some singing.
You know that Jane always sings in the shower but that Mary only sings during
one shower out of four. What is the probability that Jane is in the shower?

Assuming that they each shower daily, I would say, one chance in five. Out of the 8 showers in 4 days, 5 have someone singing. In only one of those 5 is the singer Jane.

Good thinking, Bob. Except that Jane is the one that always sings.
So the answer is 4/5. Well done.

You were referring to the Monty Hall
famous probability problem, right CCb ?

Yes, Denis. It was a problem I read (worded differently) in an article
about variations of the Monty Hall problem, Monty fall and Monty crawl.

The problem that made Marilyn Vos Savant famous...

Coincidentally, I had just watched this Numberphile video on the Monty Hall problem (there are many to choose from!) that makes the answer much more intuitive.

The solution is not clear when there are just three doors so instead, say that there are 100 doors. You pick a door. It's clear that your odds of having picked the correct door are 1 in 100. Now Monty Hall eliminates 98 of the other 99 doors, leaving only one other door remaining. It's easier to see in this example that you would obviously want to switch to the one door of the 99 that Monty did not eliminate. The odds of that door being correct are 99 in 100.

The same holds true for 3 doors but in this case the odds of your initial pick are 1 in 3 and the odds of the other non-eliminated door are 2 in 3.