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Showing posts from August, 2013

The Careless Receptionist and Derangements

Follow @ProbabilityPuz Q: A large number of drunk guests arrive at a hotel where they have booked specific rooms. A careless receptionists hands over keys at random. What is the probability that at least one guest ends up in a room she booked? Kindle, 6" E Ink Display, Wi-Fi - Includes Special Offers (Black) A: A useful concept to understand is that of Derangement . A derangement is the number of ways a set can be permuted such that none of the elements are in their respective positions. For example if \(\{a,b,c\}\) is a set, then the derangements of the set are \(\{b,c,a\}\) \(\{c,a,b\}\) The above two are the only derangements of the set. The number of possible derangements of a set is usually specified as \(!n\) (note, the exclamation is before the \(n\)). There exists an expansion of \(!n\) which is $$ !n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!} $$ Coming back to the probability question at hand, the number of ways by which all guests land up in different

A Bayesian Treasure Hunt

Follow @ProbabilityPuz Q: Four friends A,B,C,D learn that a square tract of land they own has a treasure buried somewhere. They divide the tract of land into four equal quadrants, estimate that the probability in each of those square tracts as \(\{r,r,r,p\}\). A starts digging his quadrant first. The probability that A might miss the treasure given the dig is \(q\). Having dug out his quadrant, A exclaims that he still hasn't found the treasure. What is the updated probabilities of A's quadrant has the treasure? B's quadrant has the treasure? Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics) A: This is a classic Bayesian puzzle. The fact that A has dug up his quadrant and revealed that there is no treasure leaves open the possibility that he might have missed the treasure. Yet it is information that must be worth something which will help B,C,D update their belief that the treasure is present in their quadrant. First o