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When Mr and Mrs Clinton’s twin sons Ben and Bill had their tenth birthday. There were 6 kids at the party, including Ben and Bill. Ronald McDonald had made 3 hamburgers and 3 cheeseburgers and when he served the kids he started with the girl directly sitting left of Bill. Ben was sitting to the right of Bill. Ronald flipped a (fair) coin to decide if the girl should have a hamburger or a cheeseburger, head for a hamburger, tail for a cheeseburger. He repeated this procedure with all the other 3 kids before serving Ben and Bill last. Though, when coming to Ben he didn’t have to flip the coin anymore because there were no cheeseburgers left, only 2 hamburgers.
Ronald McDonald was quite surprised this happened, so he would like to know what the probability is of this kind of events. Calculate the probability that Ben and Bill will get the same type of burger using the procedure described above.

Anyone can help finding the probability?

 Jun 12, 2021
 #1
avatar+114919 
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There is a lot of irrelevant info here.

Lets forget the kids.

there are 3 cheesburgers and 3 hamburgers placed in a row.   

there will be  6!/(3!3!) possible orders = 20 altogether.

If the last two are hamburgers then on of the others is also a hamburger, 4 possible places in the line

If the last 2 are cheesburgers then again 4 possible ways.

That is 8 ways that ben and bill can get the same

 

Prob = 8/20 = 0.4

 Jun 13, 2021

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