538 Riddler Express

The Sandbox Games
23

Bump:

Now I know there is something a bit anti-Monty-Hall-ish about this.

Bill has four opaque bags, each of which has three marbles inside. Three of the bags contain two white marbles and one red marble, while the last bag contains three white marbles. The bags are otherwise indistinguishable.

Ted watches as Bill randomly selects a bag and reaches in without looking to grab two marbles without replacement. It so happens that both marbles are white. Bill is about to reach in and grab the last marble in that bag.

What is the probability that this marble is red?

Thing is, if Bill has looked inside before consciously choosing two white marbles, then the answer will be 75%.
But that’s not what happens, so no Monty Hall shit.

So, we are essentially determining the probability that a 2w1r bag was chosen, given two white marbles were pulled.
So, Bayes’?

Summary

P(2w1r bag was chosen | two white marbles were pulled) =
P(two white marbles were pulled | 2w1r bag was chosen) *
P(2w1r bag was chosen) /
P(two white marbles were pulled)

24

IT’S 50/50

25

Not a riddler question, but a cute math problem I came across:

Emma and Lindsey are playing a game. First Emma chooses n integers, then tells Lindsey which she chose. Using each number exactly once, Lindsey must create a number divisible by 6 using addition, subtraction, multiplication, or division. Each operator can be used multiple times, in any order Lindsey wants. What is the smallest n such that Lindsey can always succeed?

Answer: I originally saw the question posed as to how can you always do it for n=4, but n=3 suffices.