Expected Value: Coin flip not same as previous flip

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DerekJ
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Homework Statement


Given a sequence of Heads and Tails, let’s say that the sequence has a switch each time one toss is different than the toss before it. For instance, the sequence HHTHTTTHTH has 6 switches.

Suppose you toss a fair coin N times and record the resulting sequence of Heads and Tails. If X is the number of switches in the sequence, find E(X).

Homework Equations


We are assuming that the expected value, E(X) = ∑x_i*P(X=x_i) (the sum of each element of X, multiplied by the respective probability of that element occurring)

The Attempt at a Solution


I understand that this is going to have something to do with the number of heads and tails in the sequence and the relationship between the two. Clearly 2/n of the sequences have 0 switches, and 2/n of the sequences have n-1 switches, however I'm unsure of how to work out the values in between in general.
 
on Phys.org
it's rather simple. use E(X+Y)=E(X)+E(Y)
 
DerekJ said:

Homework Statement


Given a sequence of Heads and Tails, let’s say that the sequence has a switch each time one toss is different than the toss before it. For instance, the sequence HHTHTTTHTH has 6 switches.

Suppose you toss a fair coin N times and record the resulting sequence of Heads and Tails. If X is the number of switches in the sequence, find E(X).

Homework Equations


We are assuming that the expected value, E(X) = ∑x_i*P(X=x_i) (the sum of each element of X, multiplied by the respective probability of that element occurring)

The Attempt at a Solution


I understand that this is going to have something to do with the number of heads and tails in the sequence and the relationship between the two. Clearly 2/n of the sequences have 0 switches, and 2/n of the sequences have n-1 switches, however I'm unsure of how to work out the values in between in general.

Can you see how to do it without doing any calculations? Hint: what is the expected number of Heads in a sequence of tosses?