tag:blogger.com,1999:blog-4115025577315673827.post8348526323197340376..comments2020-05-25T13:34:36.365+05:30Comments on CSE Blog - quant, math, computer science puzzles: Baseball PartyPratik Poddarhttp://www.blogger.com/profile/11577606981573330954noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-4115025577315673827.post-69543133354527611012010-11-20T22:56:25.655+05:302010-11-20T22:56:25.655+05:30Abhishek,
Given that person 1 is in team T, there...Abhishek,<br /><br />Given that person 1 is in team T, there are remaining N-1 persons, who are equally likely to be in any team. So, expected size of T = 1 + expected no. of people , other than person 1 , who are in Team T.<br />=1+(N-1)/s.cherahttps://www.blogger.com/profile/15883972329291461469noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-91876413211730174762010-11-20T16:26:46.646+05:302010-11-20T16:26:46.646+05:30I got my answer as n/s ... and I can't figure ...I got my answer as n/s ... and I can't figure out where I'm wrong:<br /><br />Let I(X_i) = 1 if person i gets his team cap, 0 otherwise<br /><br />Let X = nos of people that get their team caps<br /><br />Now , X = I(X_1) + I(X_2) + ... + I(X_N)<br /><br />therefore <br /><br />E[X] = E[I(X_1) + I(X_2) + ... + I(X_N)]<br /><br />=> E[X] = N * E[ I(X_1) ]<br /><br />=> E[X] = N * (Probability that person 1 gets his team cap)<br /><br />Let P = (Probability that person 1 gets his team cap)<br />therefore <br />P = (k sums over 1 to N) [ P(team size of person 1 = k) * (probability that person 1 gets his team cap given his team size is k) ]<br /><br />(probability that person 1 gets his team cap given his team size is k) = k/N<br /><br />therefore <br /><br />P = (k sums over 1 to n) [ P(team size of person 1 = k) * k/N ]<br /><br />=> P = (1/N) * (expected team size of person 1)<br /><br />since expected team size of person 1 = expected size of any team = N/s<br /><br />therefore<br /><br />P = 1/s<br /><br />therefore<br /><br />E[X] = N/sAAhttps://www.blogger.com/profile/10718585328710156757noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-67639448913660826492010-11-20T06:12:51.972+05:302010-11-20T06:12:51.972+05:30I think a more compact solution can be:
X_i: 1 if...I think a more compact solution can be:<br /><br />X_i: 1 if fan i picks his cap correctly, 0 otherwise<br /><br />E[X_i]=P(X_i=1)=Pr(he picks his own cap)+P(he picks j's cap and j is fan of the same team as i)<br />=1/n + (n-1)/n(1/s) = (n+s-1)/ns<br /><br />Now X= X_1+X_2...X_n<br /><br />By linearity of expectation:<br /><br />E[X]=E[X_1]+E[X_2]...E{X_n]<br />=n*(n+s-1)/ns<br />=(n+s-1)/sUnknownhttps://www.blogger.com/profile/03495672216634189831noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-41860665064568515362010-10-22T11:31:37.791+05:302010-10-22T11:31:37.791+05:30yes pratik
i am gaurav sinha, cse 1996 graduate o...yes pratik<br /><br />i am gaurav sinha, cse 1996 graduate of iitk,<br />now working in indian revenue service (Customs and Excise).<br /><br />its real fun and refreshing solving maths problems on ur blog. Thanx for same.cherahttps://www.blogger.com/profile/15883972329291461469noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-37989418054968950852010-10-22T01:18:18.227+05:302010-10-22T01:18:18.227+05:30Bingo! Correct solution. Thanks for the answer and...Bingo! Correct solution. Thanks for the answer and the suggestion :)<br /><br />@chera.. Are you Gaurav Sinha from IITK?Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-24799415545201907972010-10-21T21:23:13.509+05:302010-10-21T21:23:13.509+05:30Answer is (n+s-1)/s.
I think the problem state...Answer is (n+s-1)/s.<br /><br />I think the problem statement could be made more clear. For example, we could say that <br /><br />(i) after arrival of guests, the host had arranged the n caps and mixed them in a basket and the fans have randomly picked caps from this basket. <br /><br />or <br /><br />(ii) the guests already had their team caps and after their arrival they had put their caps in a basket. After the party is over, the guests pick cap randomly from basket.<br /><br />Solution: Let the teams be numbered 1,2,3,...s<br />Let X= total number of fans who pick cap of their team.<br />Let Xi= number of fans of team i who pick a cap of team i . [i.e. correct cap]<br />Then X=X1+X2+….+Xs.<br />E[X]=E[X1]+E[X2]+….+E[Xs]<br /><br />By symmetry , <br />E[X1]=E[X2]=….=E[Xs]<br /> Fix and Concentrate on any particular team, say team 1.<br />Then we may write E[X] = s*E[X1].<br /><br />To find E[X1], suppose that there are r fans of team 1. If r=0, then clearly E[X1]=0. So suppose r>0 and the fans of team 1 are F1,F2,…,Fr. {r can be any number from 1 to n.} <br />Let a random variable Yi {i from 1 to r} be defined as Yi=1 iff fan Fi chooses cap of team 1.<br /><br />Then , X1=Y1+…+Yr. <br />So E[X1]= E[Y1]+E[Y2]+….+E[Yr]<br /> =r * E[Y1] because E[Yi]=E[Yj] by symmetry for all i,j.<br /><br />Now E[Y1]=r/n because there are r caps of team 1 out of total n caps.<br />So given an r, E[X1]=r*E[Y1]= r *r/n=r^2/n.<br /><br />So E[X1]= sigma r^2/n * P(r).,<br /> Where p(r) is probability that team 1 has r fans and where summation sigma ranges from r= 0 to n.<br />It is readily seen that <br />P(r)= comb(n,r)*((s-1)^(n-r) )/ s^n.<br /> <br />So E[X1]= sigma r^2/n * comb(n,r)*((s-1)^(n-r) )/ s^n; {where summation sigma is from r = 0 to n.}<br /> = ((s-1)/s)^n * sigma r^2/n * comb(n,r)* (s-1)^(-r) <br /> =(n+s-1)/s^2. <br /> Where we use identity sigma r^2 * comb (n,r) *x^r = n*x*(1+n*x) * ((1+x) ^(n-2)) by putting x=1/ (s-1).<br /><br />This identity can be proved by differentiating w.r.t. x both sides in binomial expansion of (1+x)^n, multiplying the resultant equation with x and differentiating again.<br /><br />So required answer is E[X]= s*E[X1]= (n+s-1)/s.cherahttps://www.blogger.com/profile/15883972329291461469noreply@blogger.com