Question

You meet a nasty dog in a dark alley. The dog has a bag and the bag has a pebble in it. The pebble is equally likely to be black or white. The dog then adds a white pebble and shakes up the bag. He then asks you to take a pebble from the bag without looking inside it. What are your chances of taking out a white pebble? (Taken from Richard Wiseman's blog). What if the odds of the original pebble being black or white weren't supplied?

Answer 1

0.75

Answer 2

Prove it!

Answer 3

There are two cases possible. First, the bag already had a white ball, in which the probability of picking a white ball is 1 or 100%.
The second case is when the bag has a black ball. So the probability of picking a white one is 0.5 (50%).
The combined probability is the sum divided by two. (1+0.5)/2 = 0.75

Answer 4

Nicely done! Much simpler than my proof:
P(picking white pebble) = P(initial pebble white) * P(pick white pebble in this case) + P(initial pebble not white) * P(pick white pebble in this case).

In shorthand:
\(P(W) = P(A) * P(W/A) + P(A’) * P(W/A’)\)

P(W/A) is always 1 (picking a white pebble from a bag of two white pebbles),
P(W/A’) is 0.5 (picking a white pebble from a bag of one white and one black),
P(A’) = 1 – P(A)

So we get:
\(P(W) = P(A) * 1 + (1 – P(A)) * 0.5;\)
\(P(W) = 1/2 (1 + P(A))\)
\(P(W) = 1/2 (1+0.5)\) = 0.75

Answer 5

Your answer is good even when we don't know the probability of the initial ball being white or black.

Answer 6

Yeah, I must have intentionally left the P(A) in until the end. Did it a few weeks ago, and don't really remember it!