Expected Value

Roll a die, and you get paid what the dice shows. if you want, but you don't have to, you can roll the die again and get paid what the second roll shows instead of the first. What is the expected value?

Update (11/12/09): Solution by Pintu Kumar (CSE, IITB) in comments!!
Update (21/01/10): Solution posted by Pintu was wrong. Correct solution posted in comments by me!!

Comments

  1. Pintu10:58 AM, October 28, 2009

    4
    Roll it once .. if the value is more than or equals to 3, value of the first roll in counted else we roll again.
    So expected value ..
    4.5*4/6 + 2/6*3 = 4
    Also .. it doesnt make a difference if we take the values strictly greater than 3 on the first roll

    ReplyDelete
  2. Pratik Poddar11:27 AM, October 28, 2009

    Nice and elegant solution..

    ReplyDelete
  3. Pratik Poddar4:15 PM, January 21, 2010

    I don't how I got this wrong, but this is not the correct solution.

    The strategy is roll it once. If the value is more than 3.5, value of the roll in first roll is counted. Else we roll again.

    So, expected value = 1/6*(6+5+4) + 1/2(3.5) = 4.25

    In Pintu's solution, he miscalculated the expected value of a roll of dice as 3. And hence his solution.

    The correct strategy and calculation gives expected value = 4.25 :)

    ReplyDelete

Post a Comment

Popular posts from this blog

Fraction Brainteaser

Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20 out of 28 Geometry questions.  In total, Siddhant scores 25 out of 34.  Vaibhav writes another Maths test and correctly answers 20 out of 25 Arithmetic questions and 6 out of 9 Geometry questions. in total, Vaibhav scores 26 out of 34. Note that a) Vaibhav scores more than Siddhant b) Siddhant score better than Vaibhav in both individual topics -  5/6 > 20/25 and 20/28 > 6/9 How is it possible? 
Read more

Buying Dimsums

Source: Alok Goyal (Stellaris VP, Ex-Helion VC) puzzle blog Problem: A fast food restaurant sells dimsums in boxes of 7 and 3. What’s the greatest number of dimsums a person cannot buy. Generalize it for p and q where p and q are relatively prime. I loved the puzzle. Hope you enjoy it too.
Read more

Consecutive Heads

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. 1) Calculate P{X>Y} 2) What's the expected value of X 3) What's the expected value of Y This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment. Solution: 1) (Solved by me finally after 13 months :)) Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently. So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0" f(State) = P(X>Y | "State" configuration initially)
Read more