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Problem: Jak rolls two die and wants the probability of rolling at least a 5 to be $\frac{1}{2}$. How many should Jak roll?

basically, I got two answers (same), but different approaches. Can someone tell me why one of them could be wrong?

Solution 1

The chance of getting a 5 is $\frac{1}{6}$ and rolling n dies or n times (same meaning in this situation) is $\frac{n}{6}$. Now I want the probability to be at least $\frac{1}{2}$

So I have $\frac{n}{6} > \frac{1}{2} \implies n > 3 \implies n = 4$

That is I want to roll four dices.

Solution 2

This one is a bit more formal.

It basically thinks about the probability of NOT getting that 5. So I have $\frac{5}{6}$. Now rolling n times, I get $(\frac{5}{6})^n$

Thus, the event of getting at least one five would be $1 - (\frac{5}{6})^n$ and we want this to be greater than 1/2.

$$1 - (\frac{5}{6})^n > \frac{1}{2} \implies n > 3.80 \implies n = 4 $$

Both gives me n = 4, but clearly the first one is informal. Could someone tell me why the first one could be wrong?

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can "a chance" be $n/6$? – Artem Jul 18 '12 at 1:40
Oh I see. I made a mistake, it should've been $\frac{1}{6^n}$ and I wouldn't get n = 4 at all. – Hawk Jul 18 '12 at 1:44
up vote 4 down vote accepted

Unfortunately, the first one is in fact wrong because you can't add probabilities in this way. If you could, we could set up a ridiculous situation.

Say that, instead of a die, you are using a fair coin and you want to guarantee getting heads. If probabilities simply added, that means that the probability of getting heads after two trials would be $2\cdot\frac{1}{2}=1$. This is clearly not true. Your first method would be correct, however, if there was something to prevent the same event from happening twice, but this isn't the case in the problem.

Your second method is correct. You are finding the probability $P$ that the opposite of the event happens every time. $P$ is the complement of the event happening at least once, and therefore $1-P$ is the probability you are looking for.

Edit: I should have been clearer about this. $n=4$ is the right answer, and you justified it perfectly the second time. The first time just happens to be right; it's not a valid method of finding the answer to this problem.

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Oh I see. I made a mistake, it should've been $\frac{1}{6^n}$ and I wouldn't get n = 4 at all. – Hawk Jul 18 '12 at 1:44
That would be the probability of rolling a $5$ every time. An event of probability $P$ occurring $n$ times in a row has probability $P^n$, and the probability of the complement (opposite) of $P$ occurring during $n$ trials is $1-P^n$. – David Spencer Jul 18 '12 at 1:47

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