When six dice are rolled, the number of different numbers which can appear range from 1 to 6. Suppose that once every minute, the croupier rolls six dice and you bet $1, at even odds, that the number of different numbers which appear will be exactly 4.
If you start with $10, roughly how long will it be, on average before you are wiped out?
:)
Update (11/12/09): Solution in comments!!
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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...
12 minutes.....
ReplyDeleteabout 22 minutes
ReplyDeleteabout 22 minutes
ReplyDeletesorry guys...
ReplyDelete:(
Not the correct answer :(
eternity :P
ReplyDeletesahi hai jaadu...
ReplyDelete:)
badhiyaan :)
@junta... probability that the even occurs is > 0.5, and so, he would expect his money to grow :)
I am not able to figure out how many cases would it take to arrange  123412 and 123411 types of cases (this no*360) should be the total possible cases
ReplyDeleteHmm, Got it  6C4(4C1 * 6!/3! + 4C2 * 6!/(2!*2!))/6^6
ReplyDeleteHmm good one.
:)
ReplyDelete@ranka... jee times... huh!! :)