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If there are $x$ (where $x$ is even) people in an office and each person calls out name what is the probability that every worker calls the name of the worker that calls him.

Thanks.

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    All of the options I suggested involve calling randomly. Even if you meant "uniformly random", that doesn't answer my remaining questions.2012-12-11

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$(x-1)\times(x-3)\times(x-5)\times...\times3\times1$ distinct pairs of people.

In each one of these pair instances, each specific pair call one another's name by the probability of $\frac {1}{(x-1)^x}$ assuming no one calls his own name.

$\frac{(x-1)\times(x-3)\times(x-5)\times...\times3\times1}{(x-1)^x}$

EDIT: adding some explanations.

$x$ people ($x$ is even) forming $(x-1)\times(x-3)\times(x-5)\times...\times3\times1$ distinct pairs: In these pairs, any one of the $x$ people could have paired with $(x-1)$ others. An arbitrary other among those remaining can appear in this set of pairs with $(x-3)$ of the remaining persons and so forth.

Assuming each can call one another's name with equal probability, a person can call out anyone of the $(x-1)$ names except his own. The probability of each person calling the name of the person he paired up with is $1/(x-1)$. Each one of the $x$ people calling out the "right" names in a set of pairs is $1/(x-1)^x$. Multiply this by the number of such pair instances.

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    yup, figured trying n works. Thx again.2012-12-11