If we roll a die 20 times what are the chances that all numbers 1 through 6 come up?

I thought the size of the space was 6^20 and then the size of the event would be

(6!*6^12) however this is obviously wrong.

Thanks for any help

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- Aug 21st 2011, 06:17 AMhmmmm20 rolls of a die chance all numbers come up?
If we roll a die 20 times what are the chances that all numbers 1 through 6 come up?

I thought the size of the space was 6^20 and then the size of the event would be

(6!*6^12) however this is obviously wrong.

Thanks for any help - Aug 21st 2011, 08:12 AMTravellerRe: 20 rolls of a die chance all numbers come up?
Can you compute the chances that at least one number does not come up, using derangements?

- Aug 21st 2011, 08:50 AMPlatoRe: 20 rolls of a die chance all numbers come up?
I think that reply #2 means use

*inclusion/exclusion*.

Lets play the "back-of-the-book" game.

You look up the answer and see $\displaystyle \sum\limits_{k = 0}^5 {\left( { - 1} \right)^k \binom{6}{k}\left( {\frac{{6 - k}}{6}} \right)^{20} } .$

WHAT THE HECK????

Now it is your responsibly the explain why that answer works.

Can you tell us all what is going on there? - Aug 21st 2011, 01:24 PMhmmmmRe: 20 rolls of a die chance all numbers come up?
Ah ok so If it as all the ways you can choose the 20 with all 6 die thrown then that is the number of surjections from the set of 6 elements to the set of 20 elements which is

$\displaystyle \Sigma^{5}_0 (-1)^k 6Ck (6-k)^{20}$.

We then simply divide by the total space 6^20 and arrive at what you have there.

Sorry quite simple I was just thinking of it in a completely and much less helpful way.

thanks a lot (sorry for so much editing I made loads of typo's) - Aug 21st 2011, 01:32 PMPlatoRe: 20 rolls of a die chance all numbers come up?
Very good: count the surjections. That is good.