# 13 Die rolls - odds \>50% all six numbers show?

**URL:** <https://boards.straightdope.com/t/13-die-rolls-odds-50-all-six-numbers-show/339517>\
**Category:** Factual Questions\
**Created:** [January 11, 2006, 5:23pm UTC](https://boards.straightdope.com/t/13-die-rolls-odds-50-all-six-numbers-show/339517 "2006-01-11T17:23:49Z")\
**Posts on this page:** 2\
**Page:** 1

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**Author:** ![wolf\_meister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/wolf_meister/32/15202_2.png) [@wolf\_meister](https://boards.straightdope.com/u/wolf_meister)\
**Post date:** [January 11, 2006, 5:23pm UTC](https://boards.straightdope.com/t/13-die-rolls-odds-50-all-six-numbers-show/339517/1 "2006-01-11T17:23:49Z")

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This goes back to an earlier thread

> **[Probabbility question](https://boards.straightdope.com/sdmb/showthread.php?t=351814)**
>
> I have a more complicated situation that the question below, but I will do that one with a computer program. I’d like to verify the program with the example below: Given one 6-sided die, how many rolls on average before every number is rolled?...

in which the question was asked about the _average_ number of throws it takes for a 6 sided die to show all six numbers. Several cites were posted and it was generally concluded that the _average_ number of throws would be 14.7.

Still, if this were a bet _or_ if you were designing a game, wouldn’t the 50% mark be just as important (if not more so) than the average? I surfed a great many sites and even wrote my own probability simulator (link is at the above cited thread) and have concluded that 51.3858194 % of the time, all 6 numbers will show with 13 throws or less. If you want to verify the calculation, the site I used was:  
[http://mathforum.org/library/drmath/view/56679.html](http://mathforum.org/library/drmath/view/56679.html)  
(Dr Math has it calculated for 12 throws but by substitution you will be able to calculate for 13 throws).

So, perhaps some clarification is required here. If I said, for an even money bet, that I could roll all 6 numbers in 13 tries (or even 14), you might conclude that you’ve got a sure thing because on _average_, it takes 14.7 throws. However, I would eventually come out “ahead of the game” so to speak.

So, the question is, which is more applicable - the _average_ or the 50% dstribution point?

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**Author:** ![CJJ](https://avatars.discourse-cdn.com/v4/letter/c/ecc23a/32.png) [@CJJ](https://boards.straightdope.com/u/CJJ)\
**Post date:** [January 11, 2006, 6:54pm UTC](https://boards.straightdope.com/t/13-die-rolls-odds-50-all-six-numbers-show/339517/2 "2006-01-11T18:54:28Z")

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> [@](#):
>
> So, the question is, which is more applicable - the average or the 50% distribution point?

For the purposes of your game, the 50% distribution point is more important. The average is slightly skewed by those rare instances where it could take say \>100 attempts before you roll the missing number. But in your case the game presumably ends after the 13th roll, so these \>100 outliers have no more weight in the calculation that, say, those attempts that would have succeeded on the 14th roll.