Expected number of failures when tossing coins I'm trying to solve a problem but I suspect my solution is incorrect. I'm hoping someone can verify and perhaps give me an idea about how to solve it correctly.
We have three fair coins and each is tossed until the first head appears on each coin. (So once one head appears on one of the three coins, we stop tossing that particular coin.) When we are finished, a total of $6$ tails have been obtained. What is the expected value of the number of tails on the first coin?
This is how I have tried to solve it:
If a total of $6$ tails have appeared the possible number of tails obtained with the first coin is 0, 1, 2, 3, 4, 5 or 6. I define a random variable $X \sim \mathrm{Geometric}(1/2)$ so that the value of $X$ is the number of tails (failures) until the first head (success) is obtained.
I know that the PMF of a random variable with Geometric distribution is $(1-p)^k*p$ and in this case $p = 1/2$
From here I went on to compute the expected value as
$$\mathrm{E}(X) = \sum_{k=0}^6 k*\mathrm{P}(X=k) = \sum_{k=0}^6 k*(\frac{1}{2})^k*\frac{1}{2}$$ 
and reached the result $15/16.$
As I said, I'm pretty certain this is not the right answer, but I'm not sure why and how to solve it correctly, so any help is appreciated.
 A: If by "first coin" you mean "first coin to show a head", here is my answer.
The probability that the each coin shows $k$ tails before the first head is $\frac1{2^{k+1}}$, so the probability for each arrangement that the sum of the number of tails is $6$ is $\frac1{2^9}$. That is, each such arrangement has equal probability.

One Method to Compute the Number of Arrangements
The number of such arrangements where the lowest number of tails is $k$ or greater is the coefficient of $x^6$ in $\left(\frac{x^k}{1-x}\right)^3$ which is the coefficient of $x^{6-3k}$ in $\left(\frac1{1-x}\right)^3$, which is 
$$
\begin{align}
(-1)^{6-3k}\binom{-3}{6-3k}
&=\binom{6-3k+2}{6-3k}\\
&=\binom{8-3k}2[k\le2]
\end{align}
$$

Another Method to Compute the Number of Arrangements
To count the number of ways to have $3$ non-negative integers that sum to $6$, we can use the Stars and Bars Method, which says the number is $\binom{6+3-1}{3-1}=\binom{8}{2}$ (the same as above).
To count the number of ways to have $3$ positive integers that sum to $6$, we can count the number of ways to have $3$ non-negative integers that sum to $3$ (by preloading each stars and bars partition with $1$). This gives $\binom{3+3-1}{3-1}=\binom{5}{2}$ (the same as above).
To count the number of ways to have $3$ integers, greater than or equal to $2$, that sum to $6$, we can count the number of ways to have $3$ non-negative integers that sum to $0$ (by preloading each stars and bars partition with $2$). This gives $\binom{0+3-1}{3-1}=\binom{2}{2}$ (the same as above).

Thus, the number of arrangements with exactly $0$ as the lowest number is $\binom{8}{2}-\binom{5}{2}=18$.
The number of arrangements with exactly $1$ as the lowest number is $\binom{5}{2}-\binom{2}{2}=9$.
The number of arrangements with exactly $2$ as the lowest number is $\binom{2}{2}-0=1$.
Thus, the expected value is
$$
0\cdot\frac{18}{28}+1\cdot\frac{9}{28}+2\cdot\frac1{28}=\frac{11}{28}
$$
A: Since the coins are fair, we can assume the tosses are independent, and the expected number of tails on each coin is the same. So by symmetry $E[X_1]=E[X_2]=E[X_3]$, and we are given that $E[X_1]+E[X_2]+E[X_3]=6$ for all combinations of coin tosses that meet the stipulations of the problem. Hence
$$E[X_1]=2$$
A: The answer is 2.
Treat them as iid random variables X1 X2 and X3. You have to essentially find E[X1 | X1+X2+X3] = ( X1 + X2+ X3 )/ 3.
