# Thread: Dice question

1. ## Dice question

Suppose that a 6 sided die has been altered so that the faces are 1,2,3,4,5,and 5. If the die is tossed 5 times, what is the probability that the numbers recorded are 1,2,3,4 and 5 in any order?

Thanks!

2. Well let's think.

We have a die of 6 sides.. but the sides are labeled 1,2,3,4,5,5.

We're tossing the die 5 times and trying to figure out the probability of getting 1,2,3,4,5 in any order.

First Roll:
To start off, the first number we roll can be either 1,2,3,4, or 5 to satisfy what we're trying to do.
What are the chances we roll 1,2,3,4, or 5 the first time? Well the die only has sides 1,2,3,4,5 and 5, so you will HAVE to get either 1,2,3,4 or 5 the first time you roll. So, the probability for the first roll is 6/6 or 1.

Second Roll:
Now, the first time we rolled, we either rolled a 1,2,3,4 or 5. So, since we want to roll one of each number, whatever we rolled the first time, we cannot roll again. For example-sake, let's say that the first roll, we rolled a 1. We cannot roll a 1 again, because then we will have two 1s, but we need to roll one of each number.
So, now we need to roll either a 2,3,4, or 5. Our chances of rolling one of these are 5/6.

Third Roll:
Let's say for simplicity that we rolled a 2 on the second roll. Now we have a 1 and a 2. We need either a 3,4, or 5. What are the chances of rolling one of these? 4/6.

Fourth Roll:
Let's say, again for simplicity, we rolled a 3 on the third roll. Now we have a 1,2, and 3. We need a 4 or 5. What are the chances of rolling one of these? 3/6.

Fifth Roll:
Now we have, let's say, 1,2,3, and 4. We just need the 5. Since there are two sides on the die labeled 5, our chances of rolling a 5, is 2/6.

Now we just multiply all the probabilities together.
$\displaystyle \frac{6}{6} * \frac{5}{6} * \frac{4}{6} * \frac{3}{6} * \frac{2}{6}$

We end up with $\displaystyle \frac{5}{54}$ or $\displaystyle \approx 0.0926$.

3. Hello, Dolvich!

Suppose that a 6-sided die has been altered so that the faces are 1,2,3,4,5, and 5.
If the die is tossed 5 times, what is the probability that the numbers
recorded are 1,2,3,4 and 5 in any order?

We know that: .$\displaystyle P(1) = P(2) = P(3) = P(4) = \frac{1}{6}$ .and .$\displaystyle P(5) = \frac{2}{6}$

The probability of getting a particular order, say, 1-2-3-4-5, is:
. . $\displaystyle \left(\frac{1}{6}\right)^4\left(\frac{2}{6}\right) \:=\:\frac{2}{6^5}$

There are 5! possible orders of the numbers.

Therefore: .$\displaystyle \text{Prob}\;=\;5! \times \frac{2}{6^5} \;=\;\frac{5}{162}$

4. Ahh, I get it now. Thank you!