tag:blogger.com,1999:blog-6309069204584200907.post3256028857260256064..comments2017-09-17T10:01:58.754-07:00Comments on Puzzles from real interviews: Another Game of DiceAdministratornoreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6309069204584200907.post-64761449997956640152012-08-13T14:32:10.462-07:002012-08-13T14:32:10.462-07:00That will be the case for your friend not you!That will be the case for your friend not you!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6309069204584200907.post-24273940726783768052012-08-02T19:32:00.768-07:002012-08-02T19:32:00.768-07:00i agree with the last comment...!!!
probability ...i agree with the last comment...!!!<br /> probability would be 20/36 and not 15/36<br /><br /><br />tie cases should be included...!!Antariksh Srivastavanoreply@blogger.comtag:blogger.com,1999:blog-6309069204584200907.post-84375377801583052252012-06-19T11:26:16.311-07:002012-06-19T11:26:16.311-07:00I think in the solution for winning in first roll ...I think in the solution for winning in first roll doesnt consider the winning in tie situation. 15 cases are there for 2,3,4,5,6 i.e<br />2 can beat 1<br />3 -- 2,1<br />4 -- 3,2,1<br />5 -- 4,3,2,1<br />6 -- 5,4,3,2,1<br /><br />so total adds to 15 cases as above but here tie cases are not included i..e 2 can beat 2 , 3 can beat 3 like that so there are 5 more such cases one each for 2,3,4,5,6<br /> so total probability will be 20/36..Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6309069204584200907.post-3563911787238523222011-10-25T21:15:46.030-07:002011-10-25T21:15:46.030-07:00In second game, probability that you win is:
5/12*...In second game, probability that you win is:<br />5/12*(1 + 1/6 + 1/6^2 + ...) = 5/12*6/5 = 1/2<br /><br />Does it answer your question?<br /><br />tmbtwAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6309069204584200907.post-34575105418412143312011-10-14T07:24:17.989-07:002011-10-14T07:24:17.989-07:00How is it equal in the second game? Isn't ther...How is it equal in the second game? Isn't there still the condition that if they tie the friend wins?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6309069204584200907.post-70475077406117297272011-08-08T11:38:00.545-07:002011-08-08T11:38:00.545-07:00The second time you roll a 1, you should keep it r...The second time you roll a 1, you should keep it right? Why do you calculate further after calculating the .486111...?Anonymousnoreply@blogger.com