Probability Puzzles

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 ...

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...

Follow @ProbabilityPuz Q: An urn contains 20 uniquely identifiable balls. How many draws with replacement needs to be done before you are \(95\%\) sure that you have seen all of them? A Book on Statistical Inference A: A benign looking question, but its closed form solution is fairly complex and quite unlike a typical counting puzzle. First lets learn a useful tool: Stirling numbers of the second kind .  It is a complete name and they all go together. It takes in two arguments say \((k,n)\). This is represented as \(S(k,n)\) or also as \({k\brace n }\). It represents the number of subsets of a set of size \(k\) which has a size of \(n\). For example, if a set is given as \({a,b,c}\) then the number of sets of size 2 is \(\{\{a,b\},\{b,c\},\{a,c\}\}\). So we phrase this as $$ S(3,2) = 3 $$ The generic expression for \(S(k,n)\) is given by $$ S(k,n) = \frac{1}{n!}\sum_{i=0}^{n}(-1)^{i}{{n}\choose{i}}(n-i)^{n} $$ The present problem is similar to starting with \(k\) b...

Q:You have two dice. One of them is a regular one numbered from 1 to 6. The other one is blank on all sides. How do you number the blank one with two distinct numbers such that when you roll both dice, the sum of the number that shows up is almost evenly distributed? A Book on Statistical Inference A: If you assume that the blank dice was a regular one, then it is easy to see why the distribution of the sum would look like a triangular wave. There are more combinations of numbers between 1 and 6 that sum to 6 than there are to 2 and 12. For example, {1+5, 2+4, 3+3} all add up to 6 whereas only {1+1} add up to 2 and {6,6} add up to 12 respectively. Interestingly, if you insert the numbers 1 and 2 on three faces each you will get a fairly even distribution of the numbers. A comprehensive case by case enumeration is shown below. First Roll Total @ Second Roll = 1 Total @ Second Roll = 2 1 2 3 2 3 4 3 4 5 4 5 6 5 6 7 6 7 8 Notice, the distribut...

Follow @ProbabilityPuz Q: You have a large number of coins out of which some small fraction \(p\) are faulty ones. The faulty ones have slightly different weight (maybe more or less) and for some reason the error-free-test to weigh the coins is expensive and you want to minimize the number of tests done. So you embark on weighing them in bulk lots. What is the optimal lot size so that you minimize the number of weighs per coin. Probability Theory: The Logic of Science A: Lets assume we went with a lot size of \(x\). If we can compute the expected number of weighs per coin needed for a lot size of \(x\) then this strategy must apply to the entire set of coins. So we need not know the entire lot size. Next, let us try and estimate the expected number of weighs needed. There are two scenarios. You weigh the entire lot and you find no discrepancy in the weight You weigh the entire lot and you find a discrepancy in the weight. Next we note the following The probability ...