That is, during their time in the wings and the Crazy Chair.

I know that even many of those which were on for 25-30 games didn't have all 22 boxes selected up to and including their game.

Ray is the one that seems to be closest that I can recall with any certainty with 19 of the 22 boxes [Only ones he didn't have during his time were numbers 5, 7 and 8], and he only had two boxes [9 and 12] twice, with all of the others once, and he only appeared in 21 games.

I think Original Lucy should have had all 22 numbered boxes during her 50 game run, but I am not quite sure.

Original Lucy didn't (nor did she have all 22 amounts), which leads me to believe that nobody has...greeny will know for sure, though.

What did Lucy not have?

Erm, in the box amounts stakes I don't think she ever had 1p or £1,000...no idea on the numbers though. Greeny, where are you...

So I take it Sarah didnt do it in 33 shows either?

No, she never had 1p. (And probably others too, but I'm sure of that one.)

She had £1,000 twice I think.

I'll try and work out some odds:

If you do 22 shows, the chances of having every value, (or every box number) are:

(1/1) x (21/22) x (20/22) x (19/22) x (18/22) ....... x (1/22)

Which can be shortened to (21!/22^21)

Which equals 3.29 x 10^-9 or 1 in 303,761,260...

I'm trying to work out what happens at 23 boxes and so on and I think its to do with getting 22 more chances at the combination. e.g. 23 combinations of aaaaaaaaaabaaaaaaaaaaa where a is a different value and b is the same value.

So then it gets complicated with 24 and 2 'Bs' as theres even more combinations. If I understand my pascals triangle correctly thats 276 combinations.

So for 30 boxes, we have 22 'similar' values, and 8 'different' values so the amount of combinations is 58,529,925. This means the odds for having all 22 different values or box numbers in 30 games is:

This means the odds of having all 22 values in 30 games are about 1/5.

Damn those odds are too short lol. I've mucked up somewhere...
I only did this coz I was bored of some current maths I'm doing... so its probably not understandable anyway ^^

I think I'm vaguely on track, and I'm sure MisterAl will probably be here soon...

Lucy definately never had box 22, and I think there might be a couple of others that she didn't have. She certainly had 2, 5 and 11 a lot.

When I was looking at this topic yesterday, my initial idea for calculating this probability was pretty much the same as what you've done. However, it's not right because it assumes that only 1 of the combinations (of which there are 58,529,925 in your example) will give the 22 different values, when it could be that many of the combinations do.