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(…

divide the triangle into 16 congruent equilateral triangles (by the obvious method). Now number them 1-16.. as 1 for the top one, 2 3 4 from left to right of the next three and so on

ReplyDeleteSo we are removing the bottom left part i.e 5,10,11,12..

the 4 congruent parts are: (1,2,3),(6,7,13),(14,15,16) and (4,8,9)

Draw a triangle ABC with P, Q and R as mid-points of AB, BC and CA respectively. Remove the triangle PQB (as asked in the question).

ReplyDeleteLet mid-points of AR, PR, QR and RC be X, Y, Z and W respectively. Join XY, YZ and ZW.

Thus, the four identical shapes are:

AXYP, PYZQ, XYZW and ZWCQ :)

@Siddhant, Shaunak..

ReplyDeleteBoth the solutions are correct (and identical!)

@Shaunak.. Nice explanation. Thanks!

P is midpoint of AB, Q is midpoint of BC, R is midpoint of CA. Remove triangle PQB. Divide segment PQ into 4 parts of ratio(1:5:1:1)by points P1 PQ1 and Q1 . Divide AC into 4 segments of ratio(5:1:5:5) by points P2 PQ2 and Q2 [Note that PQ:AC=1:2]. 4equal parts are A P P1 P2, P1 P2 PQ1 PQ2, PQ1 PQ2 Q1 Q2 , Q1 Q2 C Q

ReplyDelete