RE: Probability question: names in hats
March 15, 2016 at 11:55 am
(This post was last modified: March 15, 2016 at 11:57 am by Whateverist.)
(March 15, 2016 at 11:35 am)robvalue Wrote: The remaining sequences just have to bring the total up to 1, they don't have to all be the same probability. Let's see:
2, 3, 1 = 1/2 * 1/2 * 1 = 1/4
3, 1, 2 = 1/2 * 1 * 1 = 1/2
Note how the second line here happens half the time. If player one takes number 3, the other two picks are forced.
So these are the remaining two legal combinations. Total probability, along with the "winning" one: 1/4 + 1/4 + 1/2 = 1
I don't follow what you're doing here. But before we jump to probabilities can you show me explicitly what drawing sequences you think satisfy the problem? What is the sample space? (I assume you do not include sequences in which any but the last keeps his own name as they aren't allowed to keep them anyway, right?)
Fill in the blanks please: { _____ , _____ , _____ , _____ }