calculating mean of coin toss

Question

i am having a problem solving this one:

we toss a fair coin twice, and then we toss a coin one more for each time we got a H. let X denote the number of total H obtained.

what is the mean?

from what i understand, $E[X] = \sum_i P(X=x_i)$, so it should be equal to $0*P{x=0} + ... +4*P{x=4}$. the result should be, according to the written answer, 1.5, but i don't get that. what i wrote is right?

Answer 1

• If 0 heads come up (w.p 0.25), $X = 0$ w.p. 0.25 (because no more coins are tossed)

• If 1 head comes up (w.p. 0.5), $X=2$ w.p. 0.25 and $X=1$ w.p. 0.25 (because the extra toss is equally likely to come up head or tail)

• If 2 heads come up (w.p. 0.25), $X=2$ w.p. 0.0625, $X=3$ w.p. 0.125, and $X=4$ w.p. 0.0625 (the two extra tosses can result in TT, HT, TH, or HH)

Combining it all, $P(X=0) = 0.25$, $P(X=1) = 0.25$, $P(X=2) = 0.3125$, $P(X=3) = 0.125$, and $P(X=4) = 0.0625$.

Probabilities add up to 1, so that's a good sign!

$\mathbb{E}[X] = 1 \times 0.25 + 2 \times 0.3125 + 3 \times 0.125 + 4 \times 0.0625 = 1.5$

Answer 2

If you toss $TT$, you get 0 heads for $\frac{1}{4}$ prob. If you toss TH or HT, you get to toss one more time which may be H or T. The sample space {THT, THH, HTT, HTH} out of $2^3=8$ ways it can play out. Thus it is for a prob of $\frac{2}{8}$, you get one head and for another $\frac{2}{8}$ prob, you get two heads. The last case is if you toss HH, you get two more tosses which gives you a favorable outcome of (HHTH, HHHT, HHTT, HHHH} out of $2^4=16$ ways it can play out. Thus it is for a prob of $\frac{1}{16}$ you get two heads (HHTT), and for a prob of $\frac{2}{16}$ you get three heads and for a prob of $\frac{1}{16}$, you get four heads.

Thus the required probability $=\frac{1}{4}\times 0+ \frac{2}{8}\times 1+\frac{2}{8}\times 2+\frac{1}{16}\times 2+\frac{2}{16}\times 3 +\frac{1}{16}\times 4$ $$ = \frac{3}{2}$$