### 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!! :) *

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!! :) *

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)

ReplyDeleteSuppose 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.

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.

ReplyDeleteyo jaadu!!!

ReplyDeletecorrect 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!!!

I think here we're supposed to calculate expected payoff rather than expected value of Y. Please have a look at my approach

DeleteE[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

@anonymous,

ReplyDeletewho are you? :D

You have a good approach.

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

:)

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

ReplyDeleteRight..

ReplyDeleteHe maximized the wrong thing..

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

My expected profit has to be maximized...

Good attempt though.. :)

Pondering over Anonymous' comment more:

ReplyDeleteP1 = 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... :)

I am also happy....great insight

ReplyDelete