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posted by [personal profile] atreic at 12:01pm on 19/11/2013
A-level probability question, to bring joy to children at Christmas:

Tesco have an offer on Octonauts figures at the moment (that's true, so if you were hoping to buy some, now you know). The parents of an adorable 2 year old want 5 out of the 8 characters (he owns 3 already). However, the figures are not listed separately, just as 'one supplied.'

How many should they order into store to have a good chance (say >90%) of getting the 5 they want out of the random selection that Tescos send? There's no issue with taking any surplus back, so they could order vast quantities... but that would seem a little absurd.

[I can see 3 similar problems. (a) the simple problem, where 'Tescos' is a box containing all 8 characters, once and once only (b) a finite problem, where Tescos is a finite but large box (say containing 64 octonauts, 8 of each) and (c) an infinite problem, where Tescos is an Octonaut producing machine that can produce as many Octonauts as are ordered and dispenses them at random. I think the actual problem is (c), but I'm quite interested in the solutions to (a) and (b) as well and how it makes the maths different...]

My vague thoughts...

The simple problem would be easy. If you only bought 7 octonauts, there'd be a 5/8 chance that the last octonaut was in the set 'octonauts I wanted' and a 3/8 chance it wasn't. So that's only a 37% chance of a happy toddler... so you buy all 8 and have 100% chance. (is that right? Have I done something wrong?)

I wondered if it's best tackled as 1 - the probability you don't get all the figures you want. I thought briefly this was 1 - (3/8)^n, where n is the number of figures you buy, but actually this is Wrong, because that is the probability you get at least one figure you want, not that you get all five. And you must have to do something a bit like permutations and combinations, because four Pesos is different to a Peso, a Kwazii, a Barnacles, and a Tweak, even if in both situations you have 'four things you wanted'

So the first octonaut comes out of the machine. You have a 5/8 chance it's one you want, and a 3/8 chance it isn't. Gah, I'm going to draw a huge probability tree, which is hard in text. If you got the one you wanted, you now have a 4/8 chance of getting another one you want, and a 4/8 chance of not doing so.

*draws tree diagram up to three Octonauts*

So the number of octonauts at each node of this tree is a binomial distribution ( minus 1). That is, after one octonaut, the nodes are 'have 1 wanted octonaut' or 'have 0 wanted octonauts' And after two octonauts, the nodes are 2, 1, 1, 0. And after 3 it's 3,2,2,1,2,1,1,0 = 1*3,3*2,3*1,1*0.

Oh, this is coin tosses! Every time you pull the octanaut lever it's like tossing a coin, and heads is 'octonaut I want' and tails is 'octonaut I didn't want'. EXCEPT it's a biased coin (the first time you have 5/8ths chance of wanted Octonaut, rather than 1/2) and EXCEPT the bias of the coin changes based on what you got... because if you already have one, you have 4/8th chance of a wanted Octonaut, and if you already have two you have a 3/8ths chance... So, err, not much like coin tosses.

So it doesn't matter how you got to the 'I have two octonauts I want' node, any future path from their is the same. But there are lots of different ways of getting to (eg) the 'I have two octonauts I want' node and their probabilities aren't the same - eg after 3 Octonauts, 1,1,0 is more likely (P = 0.20ish) than 0,1,1 (P = 0.12 ish). Which makes sense, because you are more likely to get an Octonaut you want when you don't have any octonauts you want at all.

Hmm. I could solve it by drawing a Huge Diagram, but I'd get something wrong. My brain is saying 'simulation!' but a) that's not Real Maths, and b) I don't know how to do it in a quick and dirty way. Not in Excel, I guess... ;-)

Any help?

ETA: I _think_ I've solved this now, although in a very cludgy way!

See comments, especially
There are 33 comments on this entry. (Reply.)
posted by [identity profile] at 12:13pm on 19/11/2013
There was a similar question on facebook a while ago:

I wonder how many facebook friends you need so that you are more like than not to have a friend with a birthday on every single day of the year?

This was solved neatly by Ben:

Under (presumably) the same assumptions as above (uniform distribution of birthdays, ignoring leap years):

Define P(N,D) to be the probability that N people have exactly D different birthdays. Then we want to find the smallest N* such that P(N*,365)>0.5.

I'm thinking induction is the way to go. Suppose we have P(N-1,d) for all d, i.e. we know the chance of N-1 people having exactly d birthdays (for d = 1, 2, ... N-1).

Then to get P(N,D) there are two ways:
1) The N-1 people had D-1 different birthdays and the Nth person has a different birthday. The chance of this is (365-(D-1))/365 = (366-D)/365.
2) The N-1 people had D different birthdays and the Nth person has the same birthday as one of them. The chance of this is D/365.

So the recursion is
P(N,D) = [ P(N-1,D) * D + P(N-1,D-1)*(366-D) ]/365
with initial conditions P(1,1) = 1 and P(1,D) = 0 for D>1.

You can probably solve this with generating functions, but to be honest I don't care that much! A simple bit of code gives N* = 2287. This is pretty close to Vincent's answer. I wonder what happens if Vincent computes the median, and not the mean? Of course, since I said the code was simple, it's also quite possible that I made a mistake!


I think this is probably the solution for this... define P (N,X) as the probability after buying N Octonauts you have X Octonauts that you want? (I did this as P(N,O) with O for Octonaut initially, but it's _awful_ notation, because O looks like 0) Then we could have the recursion as P (N,X) = P(N-1,X) * P(didn't get an octonaut you want, given you already have X) + P (N-1,X-1)* P (did get an octonaut you want, given you already have X-1)

P(didn't get an octonaut you want, given you already have X) is easy, it's (X+3)/8 (so as soon as you have all 5, you never get an octonaut you want). Likewise P (did get an octonaut you want, given you already have X-1) is (6-X)/8

But I don't know how to actually do the recursion and calculate the answer...
Edited Date: 2013-11-19 12:16 pm (UTC)
posted by [identity profile] at 12:17pm on 19/11/2013
Presumably if I could calculation the answer, the answer is to find the smallest N such that P(N,5) > 0.9 ...
posted by [identity profile] at 12:22pm on 19/11/2013
Also, what is P(10,5)? 'Hmm, get about 10, that'll probably do' seems to be most people's intuition...
posted by [identity profile] at 01:44pm on 19/11/2013
OK, it turns out you can do this in an excel spreadsheet (you just calculate P(1,0) and P(1,1) and then stick in the formula for the recursion.

P(10,5) is a miserable 16% (although you'll have a 44% chance of having that oh-so-frustrating 4)

To actually get P(N,5) > 0.90, you need to order 30 Octonauts.

On the other hand, a pile of 30 Octonauts would be great :-)
posted by [identity profile] at 02:45pm on 19/11/2013
I guess it depends what happens to the surplus. Do they just get returned to the warehouse? Or do they assemble into one giant robot Octonaut that saves the planet!?
posted by [identity profile] at 02:51pm on 19/11/2013
You are fabulous! (and I'm quite pleased that this turned out to be a not trivially easy problem, rather than just that my brain has turned to complete mush after having children, though I was never very good at thinking in the right way for probability questions)

I couldn't quite bring myself to order in 30 sets of figures, so have ordered 10 and will report back on Thursday when I pick them up about what I ended up with. My disappointment at the dismal success rate predicted is improved slightly by the thought that once I've bought the figures at offer price, I can then order more in if needed and exchange even after the offer has finished. I'm sure all this would be easier if any of the Tescos in a 20 mile radius actually stocked the darned things.
posted by [identity profile] at 02:58pm on 19/11/2013
On the bright side, if you do get the five you want, you can feel very smug about not only have the five you want, but also about knowing how lucky you were to manage it! :-)

We have a big tescos with lots of toys that is our local, so if you do end up trying to hunt down Just One, let me know... I haven't specifically looked for Octonauts though.
posted by [identity profile] at 01:50pm on 19/11/2013
Can't you have a stab at this with expected values?

When you've bought 0 Octonauts, the expected number that you've got that you want is 3.

When you've bought 1, it's 3 + 5/8

When you've bought 2, it's 3 + 5/8 + (4 3/8)/8

When you've bought 3 ... and then your criterion for stopping is something like an expectation value of 7.2. Maybe.
posted by [identity profile] at 12:26pm on 19/11/2013
From a helpful friend's facebook message:

It sounds like a variant on the coupon collector's problem: (the variant being you already have 3 of the `coupons'). I haven't read the wiki article in any detail; I just knew the name of the problem.
posted by [identity profile] at 01:02pm on 19/11/2013
Right, that sounds good to me. I assumed this would be easy, but now I think it isn't :)

An alternative way of putting it would be to say "I'm trying to collect N coupons, and every box I order has a 5/8 chance of containing a coupon". I'm not sure if that would be any easier to solve.
posted by [identity profile] at 01:04pm on 19/11/2013
I sort of assumed someone like you or Simon would come along and go 'but this is trivially equivalent to $thing' and solve it in three lines, so it's nice that no-one so far has pointed out anything Completely Obvious I'm missing...
posted by [identity profile] at 01:09pm on 19/11/2013
I feel unreasonably glowy to be included in a group with Simon :) You're the one who knows about stats :)

That's what I thought at first, but it seems statistics is full of things like the birthday paradox where it's not _that_ complicated, but "finding the proper name for it on wikipedia" is the only successful strategy to actually finding the solution, which is what someone did.

Addendum: Scratch my previous suggestion, all the formulas in wikipedia are expressed in terms of a sum from 1 to 8 of the time needed to get the N'th coupon, so you ought to be able to just drop the first few terms of the series. And it's short enough you might just be able to calculate a numerical value for 8... (Although the "how many to have a 90% chance of getting all" is a bit more complicated than "expect time to get all")
simont: (Default)
posted by [personal profile] simont at 09:34am on 20/11/2013
I've only just now seen this post :-) but here's an analytic solution of sorts.

If there's only one Octonaut you need, that's reasonably easy: ordering n Octonauts and finding that not one of them is the kind you need is equivalent to saying that all of them are chosen from the remaining 7 possibilities, so the chance of that happening is (7/8)n. Hence, getting at least one of the kind you're looking for will happen with probability (1 - (7/8)n).

And if you wanted at least one of the five kinds of Octonaut, the same reasoning would apply: you'll get at least one of them with probability (1 - (3/8)n), because the only way you can't is if all n of them are one of the 3 kinds you've got already.

(You've gone through that reasoning yourself in your original post, I know. I'm just going through it again myself as a warm-up exercise to get my head in gear.)

But if you want all of the five kinds to show up at once, that's more fiddly. I suspect an Inclusion-Exclusion Principle exercise will be required.

Start with probability 1 (all possibilities).

Subtract five lots of (7/8)n, indicating the five subsets of possibilities in which one or other of the Octonauts you want fail to show up.

Now we've double-subtracted every situation in which two desired Octonauts fail to show up. There are (5 choose 2) = 10 of those and each one has probability (6/8)n, so add 10 × (6/8)n back in.

But now we've got an extra copy of each case in which three fail to show up, so now subtract again, and so on.

So we end up with the probability of getting at least one of each of the five required Octonauts being

P = 1 - 5 × (7/8)n + 10 × (6/8)n - 10 × (5/8)n + 5 × (4/8)n - 1 × (3/8)n

And that formula gives me 0.1616 for n=10, and 0.9107 for n=30, so it agrees with your Excel solution.

I suspect there's no nice formula to invert this (i.e. to give you n when you put in P), but it's an analytic solution to half of it at least.
posted by [identity profile] at 12:15pm on 20/11/2013
Collating comments from email so I have them all in one place. You know, for all those other times I need to solve this problem ;-)

"I was briefly worried that my calculation didn't _exactly_ match
Pete's numerical Perl solution, until I realised he'd done it in a
Monte Carlo way so some deviation from the true figures is to be
posted by [identity profile] at 12:30pm on 20/11/2013
> I know nothing about monte carlo - is there a dummy's guide to what it is
> and why it's not exact?

What I meant by a Monte Carlo approach is a random approach. Pete's
program actually modelled the problem as if for real - it chose a
bunch of Octonauts _at random_, checked whether they included one of
each of the five desired kinds, and then did that over and over again
in a loop and tracked what proportion of the trials were successful.

So it's inexact simply because it depends on randomness - if he'd been
really unlucky, his program might have just happened to get one of
every kind of Octonaut in every trial by sheer luck, or get none of
them on any occasion. The best you can say about the accuracy of the
Monte Carlo approach is that there's a _high probability_ of the
experimental result being near to the exact value, and it clusters
more closely the more trials you do.

(But its great virtue is that if you can live with that uncertainty in
the output, it's able to model problems of arbitrary complexity, long
after it becomes intractable to do the maths analytically or to
iterate over absolutely all possibilities!)
posted by [identity profile] at 12:31pm on 20/11/2013

Oh, of course. He rolled lots of dice, and although that's a good way of
working out probability, it's not a great way, because you could just be
really lucky at rolling dice.
posted by [identity profile] pete stevens at 02:17pm on 20/11/2013
Hmm. My program isn't terminating - it will run until power failure if it keeps being unlucky.

But on the being lucky/unlucky. I ran 16k full trials. So the probability of coming out with the answer 5 is (5/8 . 4/8 . 3/8 . 2/8 . 1/8)^16384 which is 120 * 2^-245760 which is well below the Heisenberg limit of 6.6 * 10^-34 at which point you probably need to start worrying about the correct type of octonaut spontaneously popping into existence without having to visit Tescos.

Incidentally in 64k trails (65536) the worse results were 86, 83 and 81 purchases to get the full set.

posted by [identity profile] at 12:45pm on 19/11/2013
Are all the octonauts produced in equal quantities? Often with children's characters, some are more popular than others and those are produced in greater quantities.

Also, this reminds me of a time when I was small and wanted charmkins, and my mother and aunt, going out shopping, said they would buy me one charmkin, and which did I want (Blossom), and (in case they didn't have her) which would be my second choice (Brown-eyed Susan)?

But as a Lovely Suprise, they in fact bought me both. And although I was grateful, I was also really sad, because why on earth did they assume that the Charmkin I wanted should Blossom not be available was the same as the Charmkin I would want to go alongside Blossom? In fact, I definitely wanted a humanoid Charmkin followed by a non-humanoid Charmkin, and having two humanoid Charmkins was (for reasons I cannot now remember) no better than having one humanoid Charmkin.
posted by [identity profile] at 01:05pm on 19/11/2013
This story is sad.

Also, I had never heard of Charmkins, and have just looked them up and they're so cute!
aldabra: (ghost)
posted by [personal profile] aldabra at 12:46pm on 19/11/2013
If I were Tesco, I'd probably order in 100,000 of Kwazii and 100,000 of Barnacles and so on. And then depending on the eptitude and cost control of my distribution factory I might open the boxes one at a time in order of arrival and fill orders that way. So the answer might be several hundred thousand.

What's the deadline? Christmas? If there's time to iterate I'd order in a preliminary eight and see what I got. And then investigate eBay for low-faff swapping.

Although, last time I counted we had 17 Tescoi in Cambridge. What I'd actually do is cycle to the largest half dozen in order of accessibility and see what they had on the shelves, first.
posted by [identity profile] at 12:56pm on 19/11/2013
^^^ This.
posted by [identity profile] at 01:06pm on 19/11/2013
Yes, that was my worry. But then I became interested in the hypothetical question, rather than the actual problem, which I fear may be as you state it...

_17_ Tescoi? Wow.
posted by [identity profile] at 01:11pm on 19/11/2013
Many of them are "Tesco Express" and probably have few or no toys of this kind; I think I'd only bother trying Milton, Fulbourn, and Newmarket Road unless I'd seen evidence of the toys in the smaller shops. Bar Hill is a difficult cycle, but likely to have more stock.
posted by [identity profile] at 01:31pm on 19/11/2013
I think the back road path from Dry Drayton to Bar Hill is quite a good cycle, but it's ages since I've done it, so I could be wrong...
aldabra: (Default)
posted by [personal profile] aldabra at 03:59pm on 19/11/2013
Not all of them big enough to probably have what-were-they? Octonauts?
posted by (anonymous) at 11:13am on 21/11/2013
Toys don't come in boxes of one type, because otherwise a smaller shop which ordered only one or two boxes wouldn't have a good assortment. So opening boxes one at a time actually gives you a fairly good random assortment, depending on whether are are shortpacked.


posted by [identity profile] pete stevens at 10:36pm on 19/11/2013



my $tests = 16384;
my @trials;
for (my $i = 0; $i < $tests; $i++) {
my $trials = 0;
my @oct = ( 1,1,1,0,0,0,0,0);
while ((eval join '+', @oct) != 8) {
$oct[int(rand() * 8)] = 1; $trials++;
my $c = 0;
for (my $i = 0; $i < 50; $i++) {
if (defined($trials[$i])) { $c += $trials[$i]/$tests; }
print "$i : " . $trials[$i] . "(" . $trials[$i]/$tests . ") - " . $c . "\n";


Cumulative probability distribution

# ./
0 : (0) - 0
1 : (0) - 0
2 : (0) - 0
3 : (0) - 0
4 : (0) - 0
5 : 47(0.00286865234375) - 0.00286865234375
6 : 180(0.010986328125) - 0.01385498046875
7 : 350(0.0213623046875) - 0.03521728515625
8 : 545(0.03326416015625) - 0.0684814453125
9 : 729(0.04449462890625) - 0.11297607421875
10 : 840(0.05126953125) - 0.16424560546875
11 : 891(0.05438232421875) - 0.2186279296875
12 : 914(0.0557861328125) - 0.2744140625
13 : 952(0.05810546875) - 0.33251953125
14 : 929(0.05670166015625) - 0.38922119140625
15 : 913(0.05572509765625) - 0.4449462890625
16 : 870(0.0531005859375) - 0.498046875
17 : 791(0.04827880859375) - 0.54632568359375
18 : 800(0.048828125) - 0.59515380859375
19 : 744(0.04541015625) - 0.64056396484375
20 : 642(0.0391845703125) - 0.67974853515625
21 : 633(0.03863525390625) - 0.7183837890625
22 : 521(0.03179931640625) - 0.75018310546875
23 : 459(0.02801513671875) - 0.7781982421875
24 : 387(0.02362060546875) - 0.80181884765625
25 : 370(0.0225830078125) - 0.82440185546875
26 : 361(0.02203369140625) - 0.846435546875
27 : 331(0.02020263671875) - 0.86663818359375
28 : 263(0.01605224609375) - 0.8826904296875
29 : 228(0.013916015625) - 0.8966064453125
30 : 197(0.01202392578125) - 0.90863037109375
31 : 187(0.01141357421875) - 0.9200439453125
32 : 147(0.00897216796875) - 0.92901611328125
33 : 165(0.01007080078125) - 0.9390869140625
34 : 102(0.0062255859375) - 0.9453125
35 : 98(0.0059814453125) - 0.9512939453125
36 : 102(0.0062255859375) - 0.95751953125
37 : 76(0.004638671875) - 0.962158203125
38 : 85(0.00518798828125) - 0.96734619140625
39 : 72(0.00439453125) - 0.97174072265625
40 : 60(0.003662109375) - 0.97540283203125
41 : 53(0.00323486328125) - 0.9786376953125
42 : 41(0.00250244140625) - 0.98114013671875
43 : 47(0.00286865234375) - 0.9840087890625
44 : 33(0.00201416015625) - 0.98602294921875
45 : 30(0.0018310546875) - 0.98785400390625
46 : 25(0.00152587890625) - 0.9893798828125
47 : 25(0.00152587890625) - 0.99090576171875
48 : 24(0.00146484375) - 0.99237060546875
49 : 23(0.00140380859375) - 0.9937744140625
posted by [identity profile] pete stevens at 10:37pm on 19/11/2013
I realise the problem said 'don't sledgehammer it, it'd be boring' but I thought about the question for a while and concluding that it would be boring, but not as boring as not doing that :-)
posted by [identity profile] at 03:37am on 22/11/2013
Well, it was good that I was prepared for disappointment - I got 5 different types of Octonauts, but only 2 out of the 5 I wanted. I've now ordered 20 more sets in the hope of a better haul.

(I wanted Dashi, Shellington, Inkling, Tweak and Tunip. I got 4 Kwazii's, 3 Barnacles, Peso, Shellington and Inkling)
posted by [identity profile] at 08:01am on 22/11/2013
This is sad. Although at least it proves that they don't open one box of 5000 Tunips and fill the whole order from that.

[Also, I wonder if we should adjust our model given that both narratively and based on the data we have, we expect many more Kwazii's and Barnacles...]

Good luck! Do you want me to have a snoop in bar hill tesocs? I'm there most weekends, but maybe posting a Tweak is more faff than it's worth...
posted by [identity profile] at 03:20pm on 22/11/2013
Success! My second order produced:
7 Tweaks, 3 Tunips, 3 Pesos, 2 Kwaziis, 2 Dashis, 2 Shellingtons and an Inkling

Interestingly, it doesn't lend more weight to the hypothesis about there being more Barnacles and Kwaziis.
posted by [identity profile] at 03:22pm on 22/11/2013

I needed some good news, now I'm grinning :-D
posted by [identity profile] at 06:41pm on 22/11/2013
You really should come to MathsJam one of these days. (The next one is Tuesday December 17th...)


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