An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is:
A. 3
B. 4
C. 5
D. 6
GATE 2005 · Engineering Mathematics · Binomial Distribution · medium
Answer: Expected number of tosses = 3. Answer: A.
Set up expectation equation: Let E = total expected tosses. After toss 1, we need at least one more toss. Let E2 = expected tosses from toss 2 onward (given toss 1 known). At each step: with prob 1/2 the next toss matches the current anchor (done, 1 more toss used); with prob 1/2 it differs (1 toss used, new anchor, same situation). So E2 = (1/2)*1 + (1/2)*(1 + E2). Solving: E2 = 1/2 + 1/2 + E2/2, so E2/2 = 1, so E2 = 2.
Add the first toss: The first toss must always be made. E = 1 + E2 = 1 + 2 = 3.