**Source:**Taken from the book "Heard on The Street" (Problem 4.2 in Revised 9th Edition) by Timothy Falcon Crack. Saw a very similar problem in the placement test of a hedge fund coming to campus. (Cannot disclose the name. Policy signed :P)

**Problem:**I will roll a single die not more than three times. You can stop me immediately after the first roll, or immediately after the second, or you can wait for the third. I will pay you the same number of dollars as there are dots on the single upturned face on my last roll (roll number three unless you stop me sooner). What is your playing strategy?

Update(06/02/10):

**Solution:**Posted by Vivek Kumar Jha (Elec IITB 3rd yr B.Tech) in comments!! Generalized solution by Aman !!

Basically we need to work backwards.... if you were given single throw then u dont have ny choice n expected outcome is 3.5 . Now for two rolls you can prove that quitting after 1st roll iff outcome is more than 3 for which expected value is 4.25(5/2 + 3.5/2).Now for three throws expected outcome is maximized (5.5/3 + 4.25*2/3 = 4.67) if you choose any value greater than 4 in first throw and greater than 3 in 2nd case.

ReplyDelete@viki (vivek).. nice solution

ReplyDeleteTry generalising it for N throws.. :)

For 4 throws, if we look for a 5,6 in the first throw and subsequently as before, we get the expected value as 4.94.

ReplyDeleteSince this is still less than 5,in case of 5 throws, we look for 5,6 again in the first throw, and proceeding as before get the expected value to be 5.13 (> 5).

Hence for all N>5, the optimum strategy should be to : Look out for 6 for N-5 throws; 5,6 for next 3 throws and 4,5,6 for the 2nd last throw.

One more interesting observation that I made was that the series of Expectations converges to 5.5 as N-> infinity.

@Pratik and viki : can any one of you generalise it for a hypothetical die with values from 1 to M?

A small correction : The expectations converge to 6 (as is seen directly).

ReplyDeletePosting the mathematical proof: for N throws (writing Aman's solution with explanation)

ReplyDeleteExpected return from the last throw = 3.5

So, in the last throw, if the output is > 3.5, stop.. else check the last throw.

So, expected value from the second last throw = 3/6*5 + 3/6*3.5 = 4.25

Continuing the same way in the third last throw, expected value = 2/6*5.5 + 4/6*4.25 = 5.5/3 + 8.5/3 = 14/3= 4.66

Expected value in the fourth last throw = 2/6*5.5 + 4/6*4.66 = (5.5+9.32)/3 = 4.94

Expected value in the fifth last throw = 2/6*5.5 + 4/6*4.94 = 5.13

The optimum strategy should be to : Look out for 6 for N-5 throws; 5,6 for next 3 throws and 4,5,6 for the 2nd last throw.

And of course for N->infinity, expected value turns out to be infinity. :)

Thanx a lot for the solution Aman.

As far as the mathematical formula for a hypothetical die with values 1 to M is concerned..

It get messy with cases concerning whether M is divisible by 2 and 3 and 8 or not.

Assuming M = 6k

Expectation from first throw = (M+1)/2

Expectation from second throw = 1/2*((M+2+2M)/4) + 1/2*(M+1)/2 = 5M/8 + 1/2

Expectation from third throw = .. and so on..

Result would be analogous. Thanx a lot for your solution.