I think that you need to take into account that if you want to toggle 2 bits, you can only do if you flip bits from position 0..30. Toggling bit 31 is only going to toggle this bit no matter what. Therefore, you need to multiply (33-N)/32 to your proposed result, to keep this into account.

@Mythreya's analysis is correct but incomplete. To get the expected value, you have to multiply the number of bits by their probability. Answer is Sigma{k/(2^k)} for k = 1 to 32.

Using this approach, the answer is 2 bits will change on an average: https://answers.yahoo.com/question/index?qid=20110413165236AAU9tmO

@Henry David Thoreau: The question is not asking for expected value, just the discrete density function. Also P(k=32) needs special handling.

@Henry, no ... probability that at least 1 bit will change is (k/2^k); but, probability for all k bits to change, I guess, would still be 1/2^k. Right?

And, k=0 to 31 !!

k=1-32 p=k/2^(k-1)

E(n) = 1 + E(n-1) / 2 where E(n) is the expected number of bit changes when 1 is added to an n-bit integer. E(1) = 1. E(2) = 1 + E(1) / 2 = 1.5. E(3) = 1 + E(2) / 2 = 1.75 We can verify it for n = 3. The possible values are as follows 000 -> 1 change 001 -> 2 change 010 -> 1 change 011-> 3 change 100-> 1 change 101-> 2 change 110-> 1 change 111-> 3 change Total changes: 14 Expected change = 14 / 8 = 1.75;

answer is (2^1+2^2+2^3....2^32)/(32*(2^32))

If we just look at two bits - (b1,b2) f(b1) = { 1 b1=0; f(b2) if b1=1}, the probability of one bit changing is half. The probability of two bits changing is half * f(b2), which is half*half if b2=0 and half*half*f(b3) if b2=1. Therefore, for k=1-31 p(k) = 1/2^k and for p(k=32) = 1/2^k because when the 32nd bit flips there is nothing more to be done and the recursion stops