Probability of winning 3+/4 rolls

Question

I have a pretty basic probability question that's bugging me.

Say we have two dice, a and b. When rolling both dice together, dice a has a $\frac{1}{5}$ chance of winning a roll round and dice b has a $\frac{4}{5}$ chance of winning a roll round, and best $3$ out of $4$ rolls wins. What is the probability that dice a wins?

To me, that seems like the probability that dice a wins at least $3$ times, which is $\left(\frac15\right)^3$? Or is that missing something? Is $\left(\frac15\right)^3$ only the probability that he wins exactly $3$ times and I need to account for a possible $4$th win?

How would you do this?

Answer 1

To win exactly $3$ times out of $4$ rolls, is to win $3$ times and lose once.

Now, first of all, that one loss could be in the first roll, the second, third, or fourth. That is , in terms of wins ($W$) and losses ($L$), you have the following possible patterns:

$LWWW$

$WLWW$

$WWLW$

$WWWL$

The probability of getting the first pattern is $P(LWWW)=\frac{4}{5}\cdot\frac{1}{5}\cdot\frac{1}{5}\cdot\frac{1}{5}$

The probability of getting the second pattern is $P(WLWW)=\frac{1}{5}\cdot\frac{4}{5}\cdot\frac{1}{5}\cdot\frac{1}{5}$

The probability of getting the third pattern is $P(WWLW)=\frac{1}{5}\cdot\frac{1}{5}\cdot\frac{4}{5}\cdot\frac{1}{5}$

The probability of getting the fourth pattern is $P(WWWL)=\frac{1}{5}\cdot\frac{1}{5}\cdot\frac{1}{5}\cdot\frac{4}{5}$

So note that each case has a probability of $\frac{4}{5}\cdot\frac{1}{5}^3$

And so, since each of these patterns lead to $3$ out of $4$ wins, and since they are all mutually exclusive, we can just add their individual probabilities to get the probability of getting exactly $3$ wins out of $4$ rolls, which is $4 \cdot\frac{4}{5}\cdot\frac{1}{5}^3$

Once you understand this, you'll know to use the following formula:

$$P("k \ outcomes \ out \ of \ n \ tries")={n \choose k}\cdot p^k \cdot (1-p)^{n-k}$$

for getting $k$ outcomes with probability $p$ out of $n$ events. That is, in your case you have $4$ rolls, so $n=4$, and you wanted $3$ of them to be a win, i.e. $k=3$, where a win has probability $p=\frac{1}{5}$. So, applying this general formula to your case you get:

$${n \choose k}\cdot p^k \cdot (1-p)^{n-k}={4 \choose 3}\cdot \frac{1}{5}^3 \cdot (1-\frac{1}{5})^{4-3}=4\cdot \frac{1}{5}^3 \cdot \frac{4}{5}$$

Now, to win at least $3$ times, is to win exactly $3$ times or to win $4$ times, so we need to add the probabiliyy of wining all $4$ times, which of course is just $\frac{1}{5}^4$

But notice, even here the general formula works. To win $4$ out of $4$ times, you get $n=k=4$, and so,

$P(WWWW)={4 \choose 4}\cdot \frac{1}{5}^4 \cdot (1-\frac{1}{5})^{4-4}=\frac{1}{5}^4$

Finally, since getting exactly $3$ wins is mutually exclusive from getting exactly $4$ wins, we can add up the probabilities to get the probability of $a$ winning:

$$P(a \ wins)=P(3W)+P(4W)=4\cdot \frac{1}{5}^3 \cdot \frac{4}{5}+\frac{1}{5}^4$$

Answer 2

Here are all the ways the game can go.

$\begin{array}{}4 - 0 &{4\choose 0}(\frac 15)^4(\frac 45)^0&0.16\%\\ 3 - 1 &{4\choose 1}(\frac 15)^3(\frac45)^1&2.56\%\\ 2 - 2 &{4\choose 2}(\frac 15)^2(\frac45)^2&15.36\%\\ 1 - 3 &{4\choose 3}(\frac 15)^1(\frac45)^3&40.96\%\\ 0 - 4 &{4\choose 4}(\frac 15)^0(\frac45)^4&40.96\%\\ \end {array}$

$2.74\%$ A wins, $81.92\%$ B wins, $15.36\%$ draw.