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(…

is the answer sqrt(10) inches?

ReplyDeleteyes suyash.. correct answer..

ReplyDeletecan someone prove that this is the best one can do :)

for the circumference to lie entirely on the black squares, it should not intersect any edge of the squares at points other than the vertices. so the question changes to 'what is the radius of the largest circle that can be drawn on a chessboard such that it never intersects any edge of the squares at points other than the vertices'.

ReplyDeletewe know that three non-collinear points in a 2-D space determine a unique circle. Since the circumference lies entirely on black squares, if one black square is containing the circumference, one of its diagonally adjacent black square has to contain it as well (excluding the case where the circle completely lies inside one black square, because that is clearly not the largest radius case).

now we have seven points (of the two black squares combines) out of which we have to choose 3 non-collinear ones such that the radius of the circle so formed is the largest. the common point is always included. for the remaining two points we consider the following cases:

1)point adjacent to common point in both black squares

2)point diagonally opposite to common point in both black squares

3)point adjacent to common point in one black square and diagonally opposite to it in another.

in case 2 and a subcase of 1, a circle is not formed because the points are collinear. out of the remaining subcases of 1 and case 3, it is easily seen that the radius is largest in case 3.

this take cares of a quarter of the circumference. the symmetry of the circle will make sure that the other 3 quarters also lie entirely on black squares.

By simple geometry (perpendicular bisector, pythagoras) we can prove that the radius of the circle so obtained is sqrt(10) inches.

nice explanation.. I actually got the solution by "drawing" circles on a chessboard in bitmap. :) Thanx

ReplyDelete