tag:blogger.com,1999:blog-4115025577315673827.post4118319344877022623..comments2020-05-25T13:34:36.365+05:30Comments on CSE Blog - quant, math, computer science puzzles: Bol Baby BolPratik Poddarhttp://www.blogger.com/profile/11577606981573330954noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-4115025577315673827.post-58718578439894583542016-09-23T01:26:57.534+05:302016-09-23T01:26:57.534+05:30I think here we're supposed to calculate expec...I think here we're supposed to calculate expected payoff rather than expected value of Y. Please have a look at my approach<br />E[Payoff] = P(Y exp payoff is -ve => dont buy<br /><br />I dont think calculating expected value of Y is a right approach (u're missing the x/100 part), kindly correct me if I'm wrong <br /><br />Anonymoushttps://www.blogger.com/profile/15391484841580161769noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-70424894726058764762009-10-08T12:07:43.441+05:302009-10-08T12:07:43.441+05:30I am also happy....great insightI am also happy....great insightscadzahttps://www.blogger.com/profile/04733361617492050871noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-77137285449225159662009-10-07T22:18:30.632+05:302009-10-07T22:18:30.632+05:30Pondering over Anonymous' comment more:
P1 = ...Pondering over Anonymous' comment more:<br /><br />P1 = Probability(you get the widget and you don't pay more than its worth to you)= (b/100)*(b - b/1.8)/b) = b/225<br /><br />P2 = Probability(you get the widget and you pay more than its worth to you) = b/100 - b/225 = b/180<br /><br />What we want to maximize is P1 - P2<br />which is maximized when b = 0 :)<br /><br />Now I am happy... :)Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-22970231906496026082009-10-07T18:25:13.301+05:302009-10-07T18:25:13.301+05:30Right..
He maximized the wrong thing..
I dont ca...Right..<br /><br />He maximized the wrong thing..<br /><br />I dont care about what's the maximum probabality...<br /><br />My expected profit has to be maximized...<br /><br />Good attempt though.. :)Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-39383148356016078422009-10-07T18:11:56.086+05:302009-10-07T18:11:56.086+05:30the problem with Anonymous solution is the assumpt...the problem with Anonymous solution is the assumption that the buyer needs to bye the widget even in case it is not worth it.He maximized the wrong thingUnknownhttps://www.blogger.com/profile/17923995765018410791noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-58721904633518283352009-10-07T17:38:07.166+05:302009-10-07T17:38:07.166+05:30@anonymous,
who are you? :D
You have a good appro...@anonymous,<br />who are you? :D<br /><br />You have a good approach.<br /><br />I am not able to see a mistake here!! Someone please find a mistake.. <br />:)Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-7955871546702882042009-10-07T17:30:49.129+05:302009-10-07T17:30:49.129+05:30yo jaadu!!!
correct answer!!
Explanation:
Suppose...yo jaadu!!!<br /><br />correct answer!!<br />Explanation:<br />Suppose I bet $x and get the widget. So, the value of it for the owner would be $y, uniformly distributed between 0 and x. So, its value for me is $1.8y. Expected value for me is 1.8* Expected value of y = 1.8*x/2= 0.9x<br /><br />So, if I get, expected value of the widget for me is 0.9x$ paying x$.<br /><br />If x is less, i.e I am not getting it, I did not gain/lose anything.<br /><br />So, overall I am losing. So, I should not bid.<br /><br />For the question posed by Jaadu, I dont have any idea. Someone please help!!!Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-54628324775282029232009-10-07T16:59:36.939+05:302009-10-07T16:59:36.939+05:30Answer is 0 for this case.A much better question i...Answer is 0 for this case.A much better question is what must be the value of bidding if two chance is allowed.In this case answer is 50 for first bidding and then 100 for the second bidding.Unknownhttps://www.blogger.com/profile/17923995765018410791noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-39430787956310559432009-10-07T16:16:09.355+05:302009-10-07T16:16:09.355+05:30You want to maximise Probability(you get the widge...You want to maximise Probability(you get the widget,you don't pay more than its worth to you)= Prob(you get it)*Prob(you don't pay more| you get it)<br />Suppose you bid b.<br />Prob(you get it)=b/100<br />Prob(you don't pay more| you get it)=((b - b/1.8)/b)<br />Therefore we need to maximise (b/100)*(b - b/1.8)/b)<br />which is same as maximising b.<br />So you should bid 100+epsilon.Anonymousnoreply@blogger.com