Originally Posted by Val
Last night DD had a probability question: There were five numbers in a bag (0-4), and three were picked and not replaced. Zero wasn't picked. What was the probability of picking a 2 on the next turn?

DD was flustered because she'd removed the 2 in our mockup of the problem. But she wasn't sure what to think. "Should I put back the 2???" I explained that it was a bad question and asked her to think about different possible scenarios. She ended up understanding that because the question didn't specify whether or not the 2 had been picked, there were two possible scenarios: it had been picked, and the probability was therefore zero, or it hadn't, and the probability was therefore
one-half.

Interesting, my take is I have a 50% chance of getting the known: 0. And a 50% chance of picking the unknown, and the unknown has a 25% chance of being 2. So .5 X .25 = 12.5% of picking 2. If that is the answer they want, it's a good problem.

You can get at it from the pick/not picked approach. Then you have to calculate the given x selections the probability of not picking a particular. Set aside 0. First pick 3/4 of not being 2. Second is 2/3 chance of not. Third is 1/2 chance of not. 1/2 * 3/4 * 2/3 = 1/4 (the same as simply picking 2 as the excluded number pick.)

So 25% of the time it's not picked. In that scenario, you have a 50% chance of picking 2. Which is 12.5%.
As noted the not picked case is 0%. So the total probability is still 12.5%.