RE: Probability question: names in hats
March 15, 2016 at 12:00 pm
(This post was last modified: March 15, 2016 at 12:02 pm by robvalue.)
There are only 3 possible sequences which don't violate the rules that you can't pick your own number (except for the last guy):
2, 1, 3: probability 1/4
2, 3, 1: probability 1/4
3, 1, 2: probability 1/2
That's it. That's all the ways it can happen. They don't all have equal probabilities. And only the first one satisfies the criteria of player 3 drawing his name.
All I've done is extracted these 3 sequences from the possible 6 permutations of 1, 2, 3 by excluding the ones where player 1 or player 2 take their own number.
I've excluded 1, 2, 3
1, 3, 2
And
3, 2, 1
These three can't happen.
2, 1, 3: probability 1/4
2, 3, 1: probability 1/4
3, 1, 2: probability 1/2
That's it. That's all the ways it can happen. They don't all have equal probabilities. And only the first one satisfies the criteria of player 3 drawing his name.
All I've done is extracted these 3 sequences from the possible 6 permutations of 1, 2, 3 by excluding the ones where player 1 or player 2 take their own number.
I've excluded 1, 2, 3
1, 3, 2
And
3, 2, 1
These three can't happen.
Feel free to send me a private message.
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