discerte math probability question Two teams play a series of four matches (the first team to win three matches wins the series); obviously, there is a possibility of a draw. Team I, when playing Team II, has only the probability 1/3 of winning any given match. However, Team I wins the first  match. Find the probability  of a draw
3 possible answers : (27% , 38% or 49%)
My formula: c(3,2) * 1/3 * 1/3 * 2/3  + c(3,3) * 1/3 * 1/3 * 1/3 = 25.7% which not exactly in the answer,
My other formula: 1 - ( c(3,2) * 2/3 * 1/3 * 2/3 + c(3,3) * 2/3 * 2/3 *  2/3) = 25.9%
I must did something wrong, please correct me, thank you.
 A: Team I only has to win one match. There are three matches to 'choose from', so you get a factor ${3\choose 1}$. Then, team I has to win one and lose two matches, so you get $\frac 13(\frac 23)^2$. All together, it is
$$
{3\choose 1}\frac 13\left(\frac 23\right)^2=\frac 49=0.44444
$$
I guess I did something wrong too...
I think the answers or the problem have an error in them somewhere.
A: Think of it this way. There are exactly three scenarios that aren't draws.


*

*Team I wins in 3

*Team I wins in 4

*Team II wins in 4.


Scenario 1 has a 1 in 9 chance of happening ($1/3$ for Team I to win match 2, then another $1/3$ for Team I to win match 3 as well).
For scenario 2, they must split matches 2 and 3, then have Team I win match 4. There is a $4/9$ chance that the teams split matches 2 and 3 (since the probability of Team I winning both is $1/9$ and the probability of Team II winning both is $4/9$), plus a $1/3$ chance of Team I winning Game 4, leaving a $4/27$ probability for scenario 2.
Lastly, for Team II to win, they must win the next three games, which has a probability of $(2/3)^3 = 8/27$.
That leaves a (3 + 4 + 8)/27 = 15/27 probability of not drawing. Thus, we get a probability of 12/27 = 4/9 of drawing.
