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)

f(0,0) = 1/4[f(…

answer: 4

ReplyDeletesum all the digits of every exponent of 2 till you get a single digit. Observe that the pattern is 1,2,4,8,7,5,1,2,4,8,7,5(repeats in cycles of 6).

Thus the single digit for 2^29 will be 5. Now since we know that it consists of 9 distinct digits, try to find the digit which when removed from the number 0123456789 gives the single digit sum as 5.

:) yo asad!! fundoo!!

ReplyDeletethanx..

Correct Answer!

thanx.

ReplyDeleteDude u r doing a fundoo job of maintaining ur blog and updating the puzzles regularly.

Anyways try to post a proof for the solution I have given, like why is the cycle length 6.

thanx Asad!

ReplyDeleteActually there exists a simpler solution:

2^29 mod 9 = (8^9)*4 = -4 = 5

So, (sum of the digits of 2^29) mod 9 = 5 {This is true since 10 mod 9 = 100 mod 9 = 1}

Let the missing number be k

(Sum from 0 to 9) - k mod 9 = 5

k mod 9 = 40

k = 4

:)

I am happy! :D

I dont know abt the circle fundaa

ReplyDeletebut it is done by the mod method

it is simpler.

nice question, blows u at first

Proof of Asad's solution:

ReplyDelete2^6 mod 9 = 64 mod 9 = 1

So, the cycle repeats after every 6 steps

Effectively, Asad is also doing exactly the same thing!!

"sum all the digits of every exponent of 2 till you get a single digit. Observe that the pattern is 1,2,4,8,7,5,1,2,4,8,7,5(repeats in cycles of 6)."

i.e. find 2^i mod 9 for all i and see that since 2^6 mod 9 =1, the sequence repeats on length 6.

"Thus the single digit for 2^29 will be 5. Now since we know that it consists of 9 distinct digits, try to find the digit which when removed from the number 0123456789 gives the single digit sum as 5."

i.e. since 2^29 mod 9 = 5, find the sum of digits such that sum of digits mod 9 = 5, i.e. 45-k = 5 mod 9, i.e. k =4 :)