4
$\begingroup$

The problem I'm trying to solve goes like this: Suppose that you play a game by rolling three dice, and your score is the highest number that appears on any of the dice. Suppose your opponent's score is $4$. What is your probability of winning the game?

The solution to the problem goes like this: the required probability is easier to compute as $1$ minus the probability that our score is $\leq 4$, which works out to be $1 - (\frac{2}{3})^3$.

Now, when I worked out this problem, my reasoning was like this: We only need to work out the probability that one of the die returns a $5$ or $6$, and then the rest of the two can be anything. So, if I fix one of the dice as $5$, the total outcomes for this case are $6^2$ (because anything can appear on the other two dice), and the same goes for the case when I fix one of the dice as $6$.

By my reasoning, the answer is $\frac{2 \times 6^2}{6^3} = \frac{1}{3}$.

I have two questions here:

  1. Why is my reasoning wrong?
  2. Why does the solution say it's easier to compute it the other way?
$\endgroup$
7
  • 2
    $\begingroup$ You are over counting. Your count the combination $5,5,5$ three times, for instance. $\endgroup$
    – lulu
    Commented May 25, 2018 at 20:47
  • $\begingroup$ @lulu Three times? How's that? $\endgroup$
    – ankush981
    Commented May 25, 2018 at 20:49
  • 4
    $\begingroup$ Overcounting? It seems more like he is undercounting. The probability $\frac{2\times 6^2}{6^3}=\frac{1}{3}$ is the probability that the first die specifically is what allows him to win (among possibly others). It is possible for the first die to be low while the second or third die are what trigger the win instead. $\endgroup$
    – JMoravitz
    Commented May 25, 2018 at 20:49
  • $\begingroup$ You count it once as "first die is $5$, second two are $5,5$" then again as "second die is $5$, first and third are $5,5$" and yet again as "third die is $5$, first two are $5,5$". $\endgroup$
    – lulu
    Commented May 25, 2018 at 20:50
  • $\begingroup$ The other way is better because otherwise you need to use the inclusion-exclusion principle. $\endgroup$
    – Alex R.
    Commented May 25, 2018 at 20:51

4 Answers 4

3
$\begingroup$

Because you are considering only the cases with 5 and 6 in one die but you should consider all the possible cases that is

  • one die 5 or 6 and the others 1 to 4 that is $\frac{3\times2 \times 4^2}{6^3} = \frac{12}{27}$

  • two dice 5 or 6 and one 1 to 4 that is $\frac{3\times 2^2 \times 4}{6^3} = \frac{6}{27}$

  • three dice 5 or 6 that is $\frac{2^3}{6^3} = \frac{1}{27}$

that is

$$\frac{12}{27}+\frac{6}{27}+\frac{1}{27}=\frac{19}{27}$$

For that the way proposed in the book is convenient since counting the no winning cases is simpler.

$\endgroup$
2
$\begingroup$

Suppose that the dice are different colors, say red blue and green.

Your calculation is for the probability that specifically the red die showed a $5$ or a $6$ and the blue and green die both showed anything:

$$\frac{\color{red}{2}\times\color{blue}{6}\times\color{green}{6}}{\color{red}{6}\times\color{blue}{6}\times\color{green}{6}}$$

Your error was in forgetting that this is not the only way in which you win. You could also have won in the event that the red did not show a $5$ or $6$ but the blue did and the green shows anything. Further you could have won in the event that the red and blue both don't show a $5$ or $6$ but the green does which occur with the following probabilities:

$$\frac{\color{red}{4}\times\color{blue}{2}\times\color{green}{6}}{\color{red}{6}\times\color{blue}{6}\times\color{green}{6}},~~~~\frac{\color{red}{4}\times\color{blue}{4}\times\color{green}{2}}{\color{red}{6}\times\color{blue}{6}\times\color{green}{6}}$$

We get as a result, the total probability that you win as the sum of these three:

$$\frac{2\times 6\times 6 + 4\times 2\times 6+4\times 4\times 2}{6\times 6\times 6}$$


Alternate methods of coming up with the final answer would be instead of breaking it into cases based on which die was the "first" to show a winning number, you could break it down into the total number of 5's/6's rolled. You could also approach this via inclusion-exclusion. Or you could do as the book suggested and look at the complementary event's probability. All of these will arrive at the same final numerical answer though the way in which they are written may vary slightly.

$\endgroup$
1
$\begingroup$

A couple of comments and answers have mentioned inclusion/exclusion but no one’s illustrated its application here, so here goes...

The probability that a particular die shows a 5 or 6 is $1/3$, so one might think naively that the total probability of at least one of the three dice is showing a 5 or 6 would be $p = 1/3+1/3+1/3 = 1$. This is a patently absurd result. The problem is that this counted each of the cases in which two dice show a 5 or 6 twice. We can try to adjust this by subtracting the probabilities of these combination events: $$ p = \left(\frac13+\frac13+\frac13\right)-\left(\frac1{3^2}+\frac1{3^2}+\frac1{3^2}\right) = \frac23.$$ This isn’t right, either, though. Each of the events in the above sum also includes the possibility that all three dice are showing 5 or 6, but we’ve now completely eliminated that from the sum, so we have to add it back: $$p = \left(\frac13+\frac13+\frac13\right)-\left(\frac1{3^2}+\frac1{3^2}+\frac1{3^2}\right) + \frac1{3^3}= \frac{19}{27}.$$ This, finally, is the correct answer.

More generally, if you have three events $E_1$, $E_2$ and $E_3$, the inclusion-exclusion principle says that $$\Pr(E_1 \cup E_2 \cup E_3) = \Pr(E_1)+\Pr(E_2)+\Pr(E_3) - Pr(E_1\cap E_2) - \Pr(E_1\cap E_3) - \Pr(E_2\cap E_3) + Pr(E_1\cap E_2\cap E_3).$$ This generalizes to larger unions: you add up the individual probabilities of the events, subtract the probabilities of two of them occurring at a time, add back the probabilities of three at a time, subtract the probabilities of four at a time, and so on. It’s pretty clear that this is a lot more work than the suggested method, which turns the “or” represented by the union of events into an “and” by looking at their complements instead.

$\endgroup$
0
$\begingroup$

Your reasoning isn't so much wrong as far as it goes, it's just doesn't capture the total distribution of outcomes. You only counted one way of getting a 5 or 6. You've ignored the fact that you could get a 5 or 6 on any of the three dice and even more than one 5 or 6.

Considering all these things in the way you tackled the problem, the answer is:

$$^3C_1(\frac{1}{3})^1\cdot(\frac{2}{3})^2 + ^3C_2(\frac{1}{3})^2\cdot (\frac{2}{3})^1 + ^3C_3(\frac{1}{3})^3 = .7037$$

It's easier to calculate:$$1 - (\frac{2}{3})^3 = .7037$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .