Please explain whether we have to find the expectancy value or the number of tosses? Also,how to approach this problem.
closed as off-topic by Austin Mohr, Goos, no identity, TZakrevskiy, Shuchang Dec 17 '13 at 10:04
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Hint: Let random variable $X$ be the number of tosses. Yes, you are asked to find $E(X)$, the expectation of $X$.
The possible outcomes of the game are H; TH; TTH; TTTH; TTTT. In the first $3$ cases, we have $X=1$, $X=2$, and $X=3$ respectively. In the last two cases, we have $X=4$.
Now use this analysis to find $\Pr(X=1)$, $\Pr(X=2)$, $\Pr(X=3)$, and $\Pr(X=4)$.
Then find $E(X)$ in the usual way.
In 50% of all cases, the coin ends up head, and you finish.
In 50% of all remaining cases, 25% of total, the coin ends up head on the second toss.
Similarly for the third toss and fourth toss.
In all the remaining trials, four tails will have been tossed.
Can you make a weighted average of all five cases?