Lucia Grimes

2022-07-03

A six sided die is rolled six times. What is the probability that each side appears exactly once?

Melina Richard

Beginner2022-07-04Added 14 answers

Step 1

On the first roll, there are no restrictions. The die is allowed to be any of the 6 equally likely values. Thus the probability of not duplicating any numbers so far after roll 1 is $\frac{6}{6}$, or $100\%.$.

For each subsequent roll, the number of "successful" rolls decreases by 1. For instance, if our first roll was a [3], then the second roll needs to be anything but [3], meaning there are 5 "successful" outcomes (out of the 6 possible) for roll 2. So, since each roll is independent of the previous rolls, we multiply their "success" probabilities together. The probability of rolling no repeats after two rolls is $\frac{6}{6}\times \frac{5}{6}=\frac{5}{6},$, which is about $83.3\%.$.

Step 2

Continuing this pattern, the third roll will have 4 "successful" outcomes out of 6, so we get

$Pr\left(\text{3 unique rolls}\right)=\frac{6}{6}\times \frac{5}{6}\times \frac{4}{6}=55.6\%$

and then $Pr\left(\text{4 unique rolls}\right)=\frac{6}{6}\times \frac{5}{6}\times \frac{4}{6}\times \frac{3}{6}=27.8\%$

$Pr\left(\text{5 unique rolls}\right)=\frac{6}{6}\times \frac{5}{6}\times \frac{4}{6}\times \frac{3}{6}\times \frac{2}{6}=9.26\%$

and finally $Pr\left(\text{6 unique rolls}\right)=\frac{6\times 5\times 4\times 3\times 2\times 1}{{6}^{6}}=1.54\%.$.

On the first roll, there are no restrictions. The die is allowed to be any of the 6 equally likely values. Thus the probability of not duplicating any numbers so far after roll 1 is $\frac{6}{6}$, or $100\%.$.

For each subsequent roll, the number of "successful" rolls decreases by 1. For instance, if our first roll was a [3], then the second roll needs to be anything but [3], meaning there are 5 "successful" outcomes (out of the 6 possible) for roll 2. So, since each roll is independent of the previous rolls, we multiply their "success" probabilities together. The probability of rolling no repeats after two rolls is $\frac{6}{6}\times \frac{5}{6}=\frac{5}{6},$, which is about $83.3\%.$.

Step 2

Continuing this pattern, the third roll will have 4 "successful" outcomes out of 6, so we get

$Pr\left(\text{3 unique rolls}\right)=\frac{6}{6}\times \frac{5}{6}\times \frac{4}{6}=55.6\%$

and then $Pr\left(\text{4 unique rolls}\right)=\frac{6}{6}\times \frac{5}{6}\times \frac{4}{6}\times \frac{3}{6}=27.8\%$

$Pr\left(\text{5 unique rolls}\right)=\frac{6}{6}\times \frac{5}{6}\times \frac{4}{6}\times \frac{3}{6}\times \frac{2}{6}=9.26\%$

and finally $Pr\left(\text{6 unique rolls}\right)=\frac{6\times 5\times 4\times 3\times 2\times 1}{{6}^{6}}=1.54\%.$.

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