Taking balls from a box

There are 100 balls in a black box: 50 white and 50 black. Take out 1 ball each time randomly until the box is empty (without replacement). Define status number as (1 + the number of times the colour of the ball drawn changes). For examply, status number equals 2 if we first take out 50 white balls continuously and then 50 black ball continuously. And status number = 100 if the series of ball taken out are white, black,white, black....white and black.
1/ What is the probability that status number = K (K is a given positive integer).
2/ What is the expectation of the status number?
 
(1) For the status number to be (K), the sequence of draws must have the form (A_1A_2...A_K), where each (A_i) is a homogeneous block of draws -- all white or all black -- and these blocks alternate in type. There must be (a=\lfloor \frac{K}{2} \rfloor) of one type of block and (b=K-\lfloor \frac{K}{2}\rfloor = \lceil\frac{K}{2}\rceil) of the other. So we need to find the number of (a)-tuples of positive integers ((x_1,...,x_a)), as well as the number of (b)-tuples of positive integers ((y_1,...,y_b)), that satisfy (x_1+\cdots+x_a=50) and (y_1+\cdots+y_b=50). But this is easy (bars and shorts method, balls and boxes method, whatever you wanna call it): In general, the number of positive (k)-tuples satisfying (x_1+\cdots+x_k=N) is (\binom{N-k+k-1}{k-1}=\binom{N-1}{k-1}). So the numbers of (a)- and (b)-tuples are, respectively, (\binom{49}{a-1}) and (\binom{49}{b-1}). These give the number of elements in the individual blocks, but we must also choose whether we start with a black block or a white block, and this can be done in 2 ways. The number of ways in which the status number can be (K) is therefore (2\binom{49}{a-1}\binom{49}{b-1}). The probability that the status number is (K) is then (\frac{2\binom{49}{a-1}\binom{49}{b-1}}{\binom{100}{50}}=\frac{2\binom{49}{\lfloor\frac{K}{2}\rfloor}\binom{49}{\lceil\frac{K}{2}\rceil}}{\binom{100}{50}}).

(2) We could use the result in (1), but it's much easier to use indicator r.v.'s. For each (i=1,...,99), let (X_i=1) if draws (i,i+1) are different, (0) otherwise. Then the e.v. of the status number is (1+E[X_1+\cdots+X_{99}]=1+99E[X_1]=1+99P(X_1=1)=1+99\cdot\frac{50}{99}=51).
 
(1) For the status number to be (K), the sequence of draws must have the form (A_1A_2...A_K), where each (A_i) is a homogeneous block of draws -- all white or all black -- and these blocks alternate in type. There must be (a=\lfloor \frac{K}{2} \rfloor) of one type of block and (b=K-\lfloor \frac{K}{2}\rfloor = \lceil\frac{K}{2}\rceil) of the other. So we need to find the number of (a)-tuples of positive integers ((x_1,...,x_a)), as well as the number of (b)-tuples of positive integers ((y_1,...,y_b)), that satisfy (x_1+\cdots+x_a=50) and (y_1+\cdots+y_b=50). But this is easy (bars and shorts method, balls and boxes method, whatever you wanna call it): In general, the number of positive (k)-tuples satisfying (x_1+\cdots+x_k=N) is (\binom{N-k+k-1}{k-1}=\binom{N-1}{k-1}). So the numbers of (a)- and (b)-tuples are, respectively, (\binom{49}{a-1}) and (\binom{49}{b-1}). These give the number of elements in the individual blocks, but we must also choose whether we start with a black block or a white block, and this can be done in 2 ways. The number of ways in which the status number can be (K) is therefore (2\binom{49}{a-1}\binom{49}{b-1}). The probability that the status number is (K) is then (\frac{2\binom{49}{a-1}\binom{49}{b-1}}{\binom{100}{50}}=\frac{2\binom{49}{\lfloor\frac{K}{2}\rfloor}\binom{49}{\lceil\frac{K}{2}\rceil}}{\binom{100}{50}}).

(2) We could use the result in (1), but it's much easier to use indicator r.v.'s. For each (i=1,...,99), let (X_i=1) if draws (i,i+1) are different, (0) otherwise. Then the e.v. of the status number is (1+E[X_1+\cdots+X_{99}]=1+99E[X_{99}]=1+99P(X_1=1)=1+\frac{99}{2}=50.5).


What am I missing? (2) doesn't seem right to me.

E[X1] != .5 ...the probability that the second ball is not the same as the first is 50/99, not .5.
 
What am I missing? (2) doesn't seem right to me.

E[X1] != .5 ...the probability that the second ball is not the same as the first is 50/99, not .5.

You're absolutely right.

It's still true, though, that (P(X_1=1)=P(X_2=1)=\cdots=P(X_{99}=1)), by symmetry, so it should actually be (1+99\cdot\frac{50}{99}=51).
 
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