# A fair coin is tossed until either a head comes up or four tails are obtained.what is the expected no. of tosses? [closed]

Please explain whether we have to find the expectancy value or the number of tosses? Also,how to approach this problem.

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## closed as off-topic by Austin Mohr, Goos, userNaN, TZakrevskiy, ShuchangDec 17 '13 at 10:04

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Hint: You can't have more than 4 tosses! –  BRayhaun Dec 17 '13 at 8:34
Hmmm, just to have some laughs. HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTT THHH THHT THTH THTT TTHH TTHT TTTH TTTT –  Euler....IS_ALIVE Dec 17 '13 at 8:38

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.

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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?

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We never go beyond four tosses, as either four tails will have comme up, or three tails and a head. Either way, you're done. –  SQB Dec 17 '13 at 10:29
this is what I'm saying, isn't it? –  Jan Dvorak Dec 17 '13 at 10:52
Well, I interpreted your "[s]imilarly for the (...) fourth toss. In all the remaining cases (...)" as having anything beyond four tosses. Perhaps you can edit your answer to remove such ambiguity. –  SQB Dec 17 '13 at 10:57
is my answer clear now? I didn't want to give too much of a hint by using a singular there. –  Jan Dvorak Dec 17 '13 at 10:58
Yes (I've removed my -1). –  SQB Dec 17 '13 at 11:02