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

Factual Questions
1

This goes back to an earlier thread

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
(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?

2

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.