Given 3 equally likely outcomes - H, T, and B) - why is P(exactly 1 H in 3 tries) = 0.4444.... and not 0.381? I have a three-sided dice, with three letters: $H$, $T$ and $B$. I want to figure out the probability that, out of 3 rolls, I get exactly 1 H. I figured I'd start by finding the probability that the 3 rolls do not all result in the same outcome (I don't get three H's, three T's, or three B's.) That probability, by my calculation, = $1 - 3/27 = 0.8889$.
I figured that in $3$ rolls of the dice, given that they do not result in $3$ $H$'s, $3$ $T$'s or $3$ $B$'s, then these are the only possible outcomes:
$2$ $H$'s and a $T$
$2$ $H$'s and a $B$
$2$ $T$'s and a $B$
$2$ $T$'s and an $H$ -
$2$ $B$'s and a $T$
$2$ $B$'s and an $H$ -
$1$ $H$, $1$ $B$, and $1$ $T$ -
So, there are seven outcomes and only three ways to then get only $1$ $H$
Since there are only two ways to get $1$ $H$, therefore $P(1H$|not $3H$, $3T$, or $3B) = 3/7$.
Multiplying the two together, we get $0.381$, when my binomial calculator says the real answer is $0.4444....$ I feel as though I'm missing something here.
 A: Let $p=1/3$ be probability of H and since you don't care about the other two symbols you can replace T and B by X. This is now a biased coin with H occuring with probability $1/3$ and X occuring with probability $2/3$. So now your answer is by the binomial distribution
$$\binom{3}{1} (1/3) (2/3)^2=\frac{3\times 4}{3^3}=\frac{4}{9}$$ as required.
A: The question in your text is not the same as the question in your title.  In the title you do not exclude the rolls of three the same.  Assuming you do exclude those, there are $24$ sets of rolls $(3^3-3)$ where the $3^3$ is the total number of rolls and $3$ have them all the same.  There are $12$ ways to get exactly one $H$, so given the rolls are not all the same the chance of exactly one H is $\frac 12$, not $\frac 49$.  The value $\frac 49$ is correct if you do not exclude the case of all the same, because it is $\frac {12}{27}$
A: These are the probabilities:


*

*$P_0=P(H=\color\red0,T=\color\green0,B=\color\orange3)=\dfrac{3!}{\color\red0!\cdot\color\green0!\cdot\color\orange3!}\cdot\dfrac{1}{3^3}=\dfrac{1}{27}$

*$P_1=P(H=\color\red0,T=\color\green1,B=\color\orange2)=\dfrac{3!}{\color\red0!\cdot\color\green1!\cdot\color\orange2!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$

*$P_2=P(H=\color\red0,T=\color\green2,B=\color\orange1)=\dfrac{3!}{\color\red0!\cdot\color\green2!\cdot\color\orange1!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$

*$P_3=P(H=\color\red0,T=\color\green3,B=\color\orange0)=\dfrac{3!}{\color\red0!\cdot\color\green3!\cdot\color\orange0!}\cdot\dfrac{1}{3^3}=\dfrac{1}{27}$

*$P_4=P(H=\color\red1,T=\color\green0,B=\color\orange2)=\dfrac{3!}{\color\red1!\cdot\color\green0!\cdot\color\orange2!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$

*$P_5=P(H=\color\red1,T=\color\green1,B=\color\orange1)=\dfrac{3!}{\color\red1!\cdot\color\green1!\cdot\color\orange1!}\cdot\dfrac{1}{3^3}=\dfrac{6}{27}$

*$P_6=P(H=\color\red1,T=\color\green2,B=\color\orange0)=\dfrac{3!}{\color\red1!\cdot\color\green2!\cdot\color\orange0!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$

*$P_7=P(H=\color\red2,T=\color\green0,B=\color\orange1)=\dfrac{3!}{\color\red2!\cdot\color\green0!\cdot\color\orange1!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$

*$P_8=P(H=\color\red2,T=\color\green1,B=\color\orange0)=\dfrac{3!}{\color\red2!\cdot\color\green1!\cdot\color\orange0!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$

*$P_9=P(H=\color\red3,T=\color\green0,B=\color\orange0)=\dfrac{3!}{\color\red3!\cdot\color\green0!\cdot\color\orange0!}\cdot\dfrac{1}{3^3}=\dfrac{1}{27}$



Hence the requested probability is $\dfrac{P_4+P_5+P_6}{1-(P_0+P_3+P_9)}=\dfrac{12}{24}$
