tag:blogger.com,1999:blog-4115025577315673827.post3837677200582238129..comments2019-05-18T16:40:30.584+05:30Comments on CSE Blog - quant, math, computer science puzzles: Mathematics of Housie/TambolaUnknownnoreply@blogger.comBlogger3125tag:blogger.com,1999:blog-4115025577315673827.post-27461541698059590372013-08-21T22:32:21.435+05:302013-08-21T22:32:21.435+05:30@ Pratik: I like this problem. I have a question r...@ Pratik: I like this problem. I have a question regarding your solution. I think in your solution you have implicitly assumed that K calls for each player are independent, only then you can write Probability that at least one number is left in all 100 tickets after k calls = (1-p(k))^100. I think all we can assume is that N tickets are independent. For example consider following 2 events:<br /><br />E1 = At least one number is left in ticket 1 after 15 calls<br />E2 = At least one number is left in ticket 2 after 15 calls<br />Since these 15 calls are &#39;same&#39; for player 1 and player 2, I think E1 and E2 are not independent. Please correct me if I am wrong.Somil Bansalhttps://www.blogger.com/profile/10169516350758924271noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-50193901621223511162010-06-18T00:28:12.776+05:302010-06-18T00:28:12.776+05:30Doing calculations using MAPLE Answer for Problem ...Doing calculations using MAPLE<br />Answer for Problem 1: 67.52<br />Answer for Problem 2: 53<br /><br />So, If 100 people are playing or 100 tickets are sold, we expect 67.52 numbers to be crossed.<br /><br />If 53 people would have played, we expect 70 numbers to be crossed.Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-64920479987008334682010-06-18T00:01:31.986+05:302010-06-18T00:01:31.986+05:30Not many people are interested in doing the calcul...Not many people are interested in doing the calculations.. :P<br />I hope my analysis is correct :)<br /><br /><b>Problem 1:</b><br />Probability that all the numbers on a given ticket are crossed in &lt;= k calls (k&gt;=15) is<br />binomial(90-15, k-15)/binomial(90,k) = {75!k!(90-k)!}/{(k-15)!(90-k)!90!} = {k.(k-1).(k-2) .. (k-14)}/{90.89.88. .. .76}<br />(say p(k))<br /><br />Probability that at least one number is left on a given ticket in k calls is 1 - p(k)<br /><br />Probability that at least one number is left in all 100 tickets after k calls = (1-p(k))^100<br /><br />Probability that the game finishes in greater than k calls = (1-p(k))^100<br /><br />Probability that the game finishes in greater than equal to k calls = (1-p(k-1))^100<br /><br />So, expected number of calls = sigma over k from 15 to 90 (1-p(k-1))^100<br />= sigma over k from 15 to 90 (1-[(k-1)(k-2)..(k-15)/(90.89..76)])^100<br /><br />After doing calculations for this expression in MAPLE, I will add the answer in the next comment.<br /><br />Observations: Clearly, the expected value is less than 76. When the number of tickets sold is exactly one, then also, we expect the number of numbers crossed to be &lt; 76. So, the game we played was indeed &quot;unexpectedly&quot; long :)<br /><br /><b>Problem 2:</b><br />The expression formed using logic as in the previous problem is as follows:<br />70 = sigma over k from 15 to 90 (1-[(k-1)(k-2)..(k-15)/(90.89..76)])^N<br />Solve for N.Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.com