More random maths

[manuel]

So a quick maths problem, which on the surface seems very easy, but I can’t seem to figure out how to get at a solution....

 

you are collecting Lego mini figures. There are 12 unique figures to collect. They come in sealed mystery bags of 1, and let’s say there is an equal probability of any of them being in that bag. 
 

what is the probability of getting a full set of 12 unique figures if you buy 20 bags.

 

*genuinely asking for a friend 

[monkeyseemonkeydo]

And this friend... does his name start with G and end in f?! Sorry, I've never been any good with probabilities! Obviously the first one has a one in one chance of being 'new' so that leaves you with 11 to go with a 1 in 12 probability of the next 19 being 'new' but beyond that I couldn't tell you. Gut feeling is the odds are against you getting the remaining 11 in 19 shots though!! 

[aener]

Would it not be:
(12/12) * (11/12) * (10/12) * (9/12) * (8/12) * (7/12) * (6/12) * (5/12) * (4/12) * (3/12) * (2/12) * ((1/12) * 8) ?

......................................................................................................................................................^^^^ The eight remaining chances, assuming you got all the others when you needed them. 1/12 is the lowest probablility, so I think this should give you the worst-case scenario where you still collect all 12?

(That gives 4.29e-4, which seems appropriately low )

 

[manuel]

I don’t think so, as there are significantly more than 8 ways that those last 8 can go down, and even more ways that you could get the 12 not in the first 12 picks. How many more is what I’m struggling with. 
 

I feel like it is a bit like the birthdays problem and the probability is surprisingly high, but no method to get there...

[aener]

I thought that as I was typing it, but then my instinct overruled my second-guessing.
My head is telling me that's a worst-case scenario, mathematically, because it has the biggest reduction combination whilst still getting all 12 figures.
If the first two you opened had the same figure, the #/12 probabilities would all get shifted down the orderings by one opening, so every other route would have more of the larger probabilities. I might amend my proposal to be eight individual " * (1/12)" s, as I'd forgotten they're in sequence rather than in parallel - as with rolling two sixes being a 1/36 chance rather than 1/12.

I'm perfectly happy to be proven wrong as I always hated statistics, but the logic in my head is telling me this way.

[JT!]

This is called the Coupon Collector's Problem:

Answer = 5.14%

[Davetrials]

well, maths is ace.

[monkeyseemonkeydo]

2 hours ago, JT! said:

This is called the Coupon Collector's Problem:

Answer = 5.14%

Can we have some more steps please? My head just exploded.

[JT!]

38 minutes ago, monkeyseemonkeydo said:

Can we have some more steps please? My head just exploded.

I failed a-level maths, so I'm in the same boat, I asked a friend and that's what they came back with. Look into Coupon Collector's Problem if you want to get deeper into the math.

[manuel]

Awesome - I haven’t found that method for finding a probability anywhere but will take your word for it as the expected number is something like 33 to be likely to have a full set (not sure what the variance is but expect it’s quite high)

the equation for expected number is pretty cool. 

[JT!]

"More likely than not" looks to be 34.

If you wanted to be very likely, you're looking at around 65.

X axis is number of packages, Y axis is probability.

 

[Topsy]

I actually only just had all this at school.. And I've forgotten most of it already within two weeks.

[JT!]

On 5/3/2021 at 12:26 PM, manuel said:

So a quick maths problem...

Bit of a bump, but relevant.