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!!
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Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20...

This is not a puzzle. So, for those of you who follow this puzzle blog, please bear with me for just one post. Interesting Math in this art...

Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20...

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 tos...
4
ReplyDeleteRoll 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
Nice and elegant solution..
ReplyDeleteI don't how I got this wrong, but this is not the correct solution.
ReplyDeleteThe 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 :)