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There are 26 people lettered A to Z each has randomly been allocated a random unique integer from 1 to 26. There are 26 seats numbered 1 to 26.

The people enter in alphabetical order and sit on their corresponding seat if possible.

The first 4 people forgot their numbers and just sat in a random seat. All the people after the initial 4 remember their number but if their seat is taken they randomly sit in a seat available.

What is the probability that Z sits in his allocated seat?

Obviously Z sits in correct seat if A to D randomly sat in the 4 seats allocated to them which is; $$\frac{2}{13}\times\frac{3}{25}\times\frac{1}{12}\times\frac{1}{23}=\frac{1}{14950}$$

My issue from here is how do I calculate the probability that Z sits in correct seat if one of the initial 4 people doesn't sit in their allocated seats?

Asaf Karagila
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Ben Franks
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  • Note: the variant wherein only the first person has forgotten their seat is discussed in many places, e.g. here. This differs from that in that more than one person has forgotten their seat. – lulu Dec 29 '17 at 13:09
  • Hint: the logic here is pretty much the same as in the standard variant. We only care about the selection of seats $A,B,C,D,Z$. Most of the people just go to their assigned seats and are therefore irrelevant. If someone chooses a non-$Z$ letter below them in the alphabet then the important decision is merely deferred. Each "live" choice is a choice between $A,B,C,D,Z$ and, as these are effectively indistinguishable, the answer is $\frac 15$. – lulu Dec 29 '17 at 13:20

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