Question

A fair octahedral die is rolled - what is the probability that after five rolls at least one number has been obtained twice?

Answer 1

ok lets call
A = after five rolls at least one number has been obtained twice

and lets call

A' = after five rolls no numbers were repeated

P(A) = 1 - P(A')

do you agree?

Answer 2

Got that

Answer 3

ok so lets find P(A') as it might be easier

what is the probability i have no repeated numbers after 1 roll?
well its 1 as i cant have repeats on only one roll

what is the probability that i have no repeats after two rolls?

well whichever number we got on the first roll we cant have again, so this time round we have
\[\frac{7}{8}\]

remember this relies on the first case being true:

\[1 \times \frac{7}{8}\]


now we have rolled the die twice and have used up two numbers, we have 6 left that are allowed on the third roll:

\[1 \times \frac{7}{8} \times \frac{6}{8}\]
i'm going to change the 1 to 8/8 because it looks nicer and you can see the pattern emerge:

\[\frac{8}{8} \times \frac{7}{8} \times \frac{6}{8}\]

so thats after three rolls, can you tell me what P(A') is after five rolls?

Answer 4

8/8 * 7/8 * 6/8 * 1/8 * 1/8 ?

Answer 5

hmm not quite, i'll use a specific example



lets say i roll the die and get a 1 , then on the next go i am allowed any of the following: {2,3,4,5,6,7,8}


so the probability of no repeats on the second roll is 7/8 times the probability of no repeats on the first go

lets say i roll and get a 2, now we are allowed any of the following:
{3,4,5,6,7,8}
so the probability for no repeats on this go is 6/8 times the probability of no repeats by the second go: 6/8 x 7/8 x 1

so now i roll and get a 3, what numbers are left? well:
{4,5,6,7,8}

so 5/8 x 6/8 x 7/8 x 1

what numbers am i allowed if i now roll a 4 and dont want a repeated value?

Answer 6

5, 6, 7, 8

Answer 7

yep, so now we're on the fifth roll and the probability of no repeats so far is

4/8 x 5/8 x 6/8 x 7/8 x 8/8

Answer 8

so calculate this value, its P(A') and then P(A) = 1- P(A')

Answer 9

There are 8^5 outcomes when we throw a fair die having 8 sides.
There are
\[\frac {8!} {3! 5!}

\]
ways to choose five distinct numbers from 1 to 8. Since order is not important.
There are
\[5!\frac {8!} {3! 5!}=\frac {8!} {3! }
\]
The probability of no repeats is
\[
\frac {\frac {8!} {3! }}
{8^5}=\frac{105}{512}

\]
Which is exactly what you get by what @eigenschmeigen was trying to guide you through.