Bol Baby Bol

Bol Baby Bol

Source: P. Winkler

You have an opportunity to make one bid on an object, whose value to its owner is, as far as you know, uniformly random between $0 and $100. What you do know is that you are so much better at operating the widget than he is, that its value to you is 80% greater than its value to him.

If you offer more than the widget is worth to the owner, he will sell it. But you get only one shot. How much should you bid?

Update (11/12/09)
Solution: (First solved by Jaadu)
Highlight the part between the * symbols for the answer.
* He should not bid (or bid 0$) !! Explanation in comments!! :) *

Comments

  1. Anonymous4:16 PM, October 07, 2009

    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)
    Suppose you bid b.
    Prob(you get it)=b/100
    Prob(you don't pay more| you get it)=((b - b/1.8)/b)
    Therefore we need to maximise (b/100)*(b - b/1.8)/b)
    which is same as maximising b.
    So you should bid 100+epsilon.

    ReplyDelete
  2. Unknown4:59 PM, October 07, 2009

    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.

    ReplyDelete
  3. Pratik Poddar5:30 PM, October 07, 2009

    yo jaadu!!!

    correct answer!!
    Explanation:
    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

    So, if I get, expected value of the widget for me is 0.9x$ paying x$.

    If x is less, i.e I am not getting it, I did not gain/lose anything.

    So, overall I am losing. So, I should not bid.

    For the question posed by Jaadu, I dont have any idea. Someone please help!!!

    ReplyDelete
    Replies
    1. Unknown1:26 AM, September 23, 2016

      I think here we're supposed to calculate expected payoff rather than expected value of Y. Please have a look at my approach
      E[Payoff] = P(Y exp payoff is -ve => dont buy

      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

      Delete
  4. Pratik Poddar5:38 PM, October 07, 2009

    @anonymous,
    who are you? :D

    You have a good approach.

    I am not able to see a mistake here!! Someone please find a mistake..
    :)

    ReplyDelete
  5. Unknown6:11 PM, October 07, 2009

    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 thing

    ReplyDelete
  6. Pratik Poddar6:25 PM, October 07, 2009

    Right..

    He maximized the wrong thing..

    I dont care about what's the maximum probabality...

    My expected profit has to be maximized...

    Good attempt though.. :)

    ReplyDelete
  7. Pratik Poddar10:18 PM, October 07, 2009

    Pondering over Anonymous' comment more:

    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

    P2 = Probability(you get the widget and you pay more than its worth to you) = b/100 - b/225 = b/180

    What we want to maximize is P1 - P2
    which is maximized when b = 0 :)

    Now I am happy... :)

    ReplyDelete
  8. scadza12:07 PM, October 08, 2009

    I am also happy....great insight

    ReplyDelete

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