LEGO Collectable Minifigures Series 6 Discussion

[KyleJohnson11]

Does anyone have the item number for Toys R Us in the UK?

[1980-Something-Space-Guy]

This works only if every fig has 1/16 chance to pop out, which is incorrect.

Bytheway I see two major lacks in the argument, probably I failed at reading the whole discussion but:

1) Nobody mentions that with 58 consecutive picks from the same box, you get 100%. Your magic number is 58, provided they come out of the same box (and it's easy to be you the first buyer if you really care).

2) Cheating is not forbidden. Feeling the minifigs is SUPER EASY. I got 0% errors on series 4 and 5. I collected 20 dwarves this way. So, for me, 16 will be my magic number. But I don't care since I booked a box two months ago :)

Of course, that is assuming figs have equal distribution (which they don't) and are completely random (which they aren't either, since people usually take them out of boxes that have 3 complete sets each). That's what why I said completely random. It gives you an estimate though.

And about 2), come on , we said random choice which means no feeling.

Edited December 19, 2011 by johnnyvgoode

[UsernameMDM]

Anyone seen these in North America yet?

[surrideo]

Truly loving the maths, takes me back to University.

Ill try and take a pic of the bag instruction/poster insert later and upload it.

A mixture of feeling the bags and the extraordinarily talented random pickings from the girl in WH Smith has limited me to buying only 28 to get the full set and the doubles Ill be keeping. Only one minotaur, glad of that.

[Brickmamba]

jealous much, has anyone seen these in oz yet?

[Dervish]

Just got some price information in Finland.

Atleast Lähikauppa company that includes atleast Siwa, Valintatalo and Euromarket, will be pricing series 6 for 2,29 euros per bag. So that's a 0,30 eur increase per bag :/

Will be avaible for orders in mid january...

[SilentMode]

Has anyone got a picture (or seen one online somewhere) of the 16 figures (i.e. the mini-poster that's inside each bag).

Here you go:

[NiceBloke]

Just found some of these in WH Smiths, Brentwood, Essex (that's in the UK)

Guessing I'll be a popular Dad when I get home tonight :D

[vexorian]

If my calculations are correct, 80 won't make that much of a difference.

The probability of getting a complete set out of 60 completely random bags is around 1.7%

With 80, it's approximately 4.9%

If the bags were completely random, you would need to buy at least 347 to get a chance greater than 50% to get a complete set!

For anyone interested, according to my calculations, the probability of getting at least one complete set in a random set of n bags of collectible minifigs is [(n-1)!*n!]/[(n-16)!*(n+15)!]. This is a function that grows very slowly for our purposes

Edit: by the way, to get a more complete picture, the probability of hitting all 16 figs just by buying 16 is something like 3.3*10^(-7)%.

This is a lot easier and less error prone by using dynamic programming. I have a program ready for that in my machine. If I remember correctly the uneven distribution of figures actually made it harder to complete the set. I also remember that the expected number of random purchases before you collect the entire set was just barely smaller than 60 in the previous series.

I'll try to code a new program soon .

Edited December 19, 2011 by vexorian

[1980-Something-Space-Guy]

Here you go:

Sweet. Thanks!

This is a lot easier and less error prone by using dynamic programming. I have a program ready for that in my machine. If I remember correctly the uneven distribution of figures actually made it harder to complete the set. I also remember that the expected number of random purchases before you collect the entire set was just barely smaller than 60 in the previous series.

I'll try to code a new program soon .

That would be very interesting! If your program arrives at a formula that considers the uneven distribution, I'd definitely like to see it. It would be interesting to evaluate the probability at certain numbers.

[lukeyboymac]

Managed to get 5 bags from WH Smith in Woking. Got 5 different ones too which is good considering they are behind the counter and you get what you are given. On the back of the receipt there is a five pound off voucher with a spend of twenty quid so that will make them a bit cheaper between Christmas and NYE which is when the voucher is valid.

I got the mechanic, the patient, braveheart, minotaur and clockwork robot. Think this may well be my favourite series so far.

Collecting them blind is much more fun as well imo.

[Zeya]

All right, I remade the calculations, now considering each permutation as a separate case, as you suggested, which I think is the right approach.

I used the inclusion-exclusion principle to calculate the amount of permutations that won't give us all figs, and arrived at %29*%2816-i%29^n*%28-1%29^%28i%2B1%29]]%2F%2816^n%29"]this formula for the probability of getting all 16 figs after buying n bags.

%29*%2816-i%29^n*%28-1%29^%28i%2B1%29]]%2F%2816^n%29"]

It might look ugly, but you'll notice that it yields 0 for values of n lesser than 16, and yields the same answer that you got before for n=16.

Thanks!

With this info, the probability with 60 bags is 70.4%, and with 80, 91,1%. It's probably worth the extra money.

I buy complete sets off Bricklink anyways.

Your math is great. Reminds me of my probability and statistics class in college. That course was cool for the first half and kind of scary in the second half, lol.

Practically speaking, here's what I think happens at the TRU shipping center for these large orders. Something like this... There's a guy with boxes of CMFs next to him. He fills up the order with loose packets (most people have this, sometimes they get a full box, but I think that's rarer because packets are smaller and lighter - less shipping cost for them.) So if I order 60 packets, the dude is likely somewhere in the middle of a box. So say he puts in the last 20 packets from his current box. Then he opens a new one. He grabs from anywhere in there and continues with the rest of the 40. He's not going in the same order, and he's likely grabbing at random, maybe even from the same area as the previous box he discarded. So then my order ends up with more repeats than expected and isn't the full set of what one might expect from a batch of 60 packets. He might be messing up the expected distribution of figs by grabbing "at random", thus screwing even an 80 packet order out of a certain fig type.

So to help combat this next time, I'll just order 20 more. It sucks because that means a 33% increase in time spent bag-fondling. On the plus side, it means more potential duplicates of army builder types that I like to stash away to sell at a later date. (I only keep something like 4 dupes over my full set of 16, FYI, so I'm not robbing the market too badly.)

Actually I met some people at work that also collect the CMFs, so I sell the dupes that identfy to them at cost, that way I get real money "back" instead of TRU gift card credit. Also one of them has kids and he wants certain ones, so I feel a little happy about that too. :)

[1980-Something-Space-Guy]

Your math is great. Reminds me of my probability and statistics class in college. That course was cool for the first half and kind of scary in the second half, lol.

Practically speaking, here's what I think happens at the TRU shipping center for these large orders. Something like this... There's a guy with boxes of CMFs next to him. He fills up the order with loose packets (most people have this, sometimes they get a full box, but I think that's rarer because packets are smaller and lighter - less shipping cost for them.) So if I order 60 packets, the dude is likely somewhere in the middle of a box. So say he puts in the last 20 packets from his current box. Then he opens a new one. He grabs from anywhere in there and continues with the rest of the 40. He's not going in the same order, and he's likely grabbing at random, maybe even from the same area as the previous box he discarded. So then my order ends up with more repeats than expected and isn't the full set of what one might expect from a batch of 60 packets. He might be messing up the expected distribution of figs by grabbing "at random", thus screwing even an 80 packet order out of a certain fig type.

So to help combat this next time, I'll just order 20 more. It sucks because that means a 33% increase in time spent bag-fondling. On the plus side, it means more potential duplicates of army builder types that I like to stash away to sell at a later date. (I only keep something like 4 dupes over my full set of 16, FYI, so I'm not robbing the market too badly.)

Actually I met some people at work that also collect the CMFs, so I sell the dupes that identfy to them at cost, that way I get real money "back" instead of TRU gift card credit. Also one of them has kids and he wants certain ones, so I feel a little happy about that too. :)

Thanks.

By the way, are the figs in a box stored in a certain order? I mean, for example, are all genies grouped together, or are they distributed throghout the box? If they are grouped, are the order of the figs the same in all boxes? I've never wondered that until now.

Why don't you buy a box? I don't know how people manage to buy boxes anyways, but it seems like a good deal if you've got the time to sell the rest.

[vexorian]

Sweet. Thanks!

That would be very interesting! If your program arrives at a formula that considers the uneven distribution, I'd definitely like to see it. It would be interesting to evaluate the probability at certain numbers.

I remade my program, because my program was first getting the expected number of bags you need to complete a specific set of minifigs, whilst you wanted the probability to get the set given a number of bags.

Anyway, according to my new program the expected number of random bags you will need to buy without cheating before you get a complete set is 58.15. The probability to get a complete set after buying 60 random bags is 63.06%. And after 80 bags is 86.44% The probability to get a complete set with 16 bags is 0.00008 %. Another important fact: You will need to buy a expected amount of 200 random before you get 10 Romans. And 90.5266297782 if you want 5 aliens, 3 robots and a space girl (Which is my current wish).

Would you like to know what happens if the minifigs were evenly distributed? Probability to get a complete set with 60 bags is 70.46%. With 80 bags it is 91.11% (So in fact, your math is correct)... And the expected number of bags before you get a complete set is 54.09.

Results for the previous series were similar, my conclusion is the same as always, there is absolutely no economic sense in playing fair and getting random bags. In order to get a complete set by that method you'll spend the same as buying a whole box and a whole box comes with 3 full sets and plenty of other figures. Buy boxes or feel the bags, your wallet will be happier.

Sorry, there is no formula (or rather, the formula is a simple recurrence relation which can't be calculated easily and thus needs a program) If you really, really want to see the code, here it is: http://vexorian.blogspot.com/2011/12/minifig-problem.html I know it seems difficult, but it is actually very simple once you get used to dynamic programming as it doesn't require a lot of analysis to get a specific formula. My first program was a lot faster because it considered directly that there are three or four 'classes' of minifig weights, but this new program can solve some special combinations like 5 and 6 figures of the same weight class but different actual types.

Edit: Table of probabilities for bag counts from 1 to 91: http://pastebin.com/eMY9dcJy

Edited December 19, 2011 by vexorian

[1980-Something-Space-Guy]

I remade my program, because my program was first getting the expected number of bags you need to complete a specific set of minifigs, whilst you wanted the probability to get the set given a number of bags.

Anyway, according to my new program the expected number of random bags you will need to buy without cheating before you get a complete set is 58.15. The probability to get a complete set after buying 60 random bags is 63.06%. And after 80 bags is 86.44% The probability to get a complete set with 16 bags is 0.00008 %. Another important fact: You will need to buy a expected amount of 200 random before you get 10 Romans. And 90.5266297782 if you want 5 aliens, 3 robots and a space girl (Which is my current wish).

Would you like to know what happens if the minifigs were evenly distributed? Probability to get a complete set with 60 bags is 70.46%. With 80 bags it is 91.11% (So in fact, your math is correct)... And the expected number of bags before you get a complete set is 54.09.

Results for the previous series were similar, my conclusion is the same as always, there is absolutely no economic sense in playing fair and getting random bags. In order to get a complete set by that method you'll spend the same as buying a whole box and a whole box comes with 3 full sets and plenty of other figures. Buy boxes or feel the bags, your wallet will be happier.

Sorry, there is no formula (or rather, the formula is a simple recurrence relation which can't be calculated easily and thus needs a program) If you really, really want to see the code, here it is: http://vexorian.blogspot.com/2011/12/minifig-problem.html I know it seems difficult, but it is actually very simple once you get used to dynamic programming as it doesn't require a lot of analysis to get a specific formula. My first program was a lot faster because it considered directly that there are three or four 'classes' of minifig weights, but this new program can solve some special combinations like 5 and 6 figures of the same weight class but different actual types.

Edit: Table of probabilities for bag counts from 1 to 91: http://pastebin.com/eMY9dcJy

Your program is great! I haven't programmed anything in years, but I sort of understood what you explained in your blog post. Kudos!

Thanks for the table too. These are some interesting numbers.

I think that an even distribution optimizes the expected number of bags needed to buy a complete set. Hmmm, suddenly I remembered the Arithmetic Mean-Geometric Mean inequality.

[Weil]

Good work on the maths guys!

I also wrote a program for this (also in python!) a couple of days ago when we first started talking about the probability but I haven't had time over the weekend to look at the results other than to say that they looked reasonable. I'll have a look and compare them to the results from the proposed analytical formula and the different progam.

I think it would be nice if an admin split this topic to a new one discussing the probabilities and perhaps make it a sticky as I think it will be very useful knowledge for a lot of CMF fans.

[1980-Something-Space-Guy]

I think it would be nice if an admin split this topic to a new one discussing the probabilities and perhaps make it a sticky as I think it will be very useful knowledge for a lot of CMF fans.

I second that! Will Lego hate us for spilling the beans?

And it's true, this info is useful for all series!

Ahhh, I love combining Lego and Math. Isn't it funny how we take our collecting seriously, calculating formulas and all?

[Weil]

Just shows what a bunch of nerds we all are

I can confirm that using my program for the case that all minifigures are equally likely I get the same numbers as you guys.

My program uses a completely different method to vexorian's. I used a Monte Carlo Method where conceptually I just got the computer to randomly generate a minifigure for each additional bag and recorded how many bags it took before there was a complete set. I then got the computer to run this a million times and divided the number of times I got each number of bags by a million to give the probabilities.

Edited December 20, 2011 by Weil

[1980-Something-Space-Guy]

Just shows what a bunch of nerds we all are

[Blondie-Wan]

It's so disappointing but not surprising to see three Roman soldiers per box. At least there are 5 highlanders. I really wish they would put 5 of the obvious army builders in these boxes. Hint to lego: if a mini figure is a soldier, it is probably a good candidate for a larger apportionment.

Of course I know it wasn't an accident but still kind of unfortunate in my opinion.

Still, I can't wait to get these in the NW US. I'm sure I'll get all the Roman soldiers I need, it will just take a little longer.

Why only three Romams per box? Didn't TLG think there would be army-building demand for them?

I myself see the minotaur as more of a unique monster than a army-builder, though I understand why others see it differently.

Remember, the target market for these is still overwhelmingly made up of kids, not AFOLs, and children are likely to have very different ideas about what minifigures are most desirable. Most kids probably don't have any reasonable hope or expectation of army-building, either; indeed, I suspect the average kid buying these gets only a couple. If you were a ten-year-old getting just a few minifigures out of a given series, would historical army-builders be your first choice?

That's why LEGO included so few. They want him to be rare.

That's the answer I didn't want to hear. Damn the marketing/sales schemers! (semi-sarcastic)

I really doubt that's it. TLG has no reason to care whether a given figure is slightly more rare than others (if they really wanted one to be rare, they'd have just one per box, or even one per case or fewer, as is common with other blind-boxed toys, trading cards, etc.; three in a box isn't really "rare" at all). I think they're just marketing their toys to children, rather than AFOLs, and skewing the box mixes accordingly (that is, I think they're trying to balance the overall supply so that it matches the overall demand well, with their predictions of what the demand will be coming from their research into what kids and maybe a few AFOLs want). Heck, this could even be an "early" box, and the proportions might be tweaked a little at some point during the production run, so that "later" boxes might have more Romans and less of something else (as has apparently happened a couple times with previous series).

Just shows what a bunch of nerds we all are

Yes, this is the tip-off - no one would ever have considered AFOLs to be nerds otherwise...

(But seriously, that's some very cool and impressive work some of you have done with your statistics. Bravo! )

[surrideo]

I have two six year old boys and their "army builder" appears to be the bandit and the space woman/barbarella, as all the double space aliens and minotaurs went to school with them to be traded. Maybe they will be lego "mates" for all the ADU agents they are getting at xmas

[Markypops]

Another UK retailer has series 6 in. The Tesco that I work in has just received an OFD that holds 4 boxes. Price point is £1.97.

Happy hunting!

[Mr Benn]

By the way, are the figs in a box stored in a certain order? I mean, for example, are all genies grouped together, or are they distributed throghout the box? If they are grouped, are the order of the figs the same in all boxes? I've never wondered that until now.

For the last 3 series I've just gone and bought a box - it just makes life easier (yes, and more expensive I suppose, but beats trawling around trying to get full sets)...

Anyway, in my experience there do seem to be 'clusters' in the boxes - for example I got through two rows of the box without getting a single one of three of the figures, at least 3 of each of which then cropped up in the next 20.

So I don't think it's entirely random - the boxes must have to be packed in a certain way anyway to ensure that the right quantity of each figure is in the box, there's probably a very secret formula to finding that fig you want!

[Weil]

For anyone that is interested I've made a plot of the results that me, johnnyvgoode and vexorian obtained each using a different method for the case of a uniform distribution of minifigures and for the case that there is a distribution of minifigures in series 6 as reported by SilentMode respectively.

The plot is available HERE

The fact that our results agree while using different methods validates the results shown as correct.

Obviously the results are only valid if you buy the packets without any knowledge of what's inside ie you don't feel them or buy a complete box containing known amounts of figures.

[1980-Something-Space-Guy]

For anyone that is interested I've made a plot of the results that me, johnnyvgoode and vexorian obtained each using a different method for the case of a uniform distribution of minifigures and for the case that there is a distribution of minifigures in series 6 as reported by SilentMode respectively.

The plot is available HERE

The fact that our results agree while using different methods validates the results shown as correct.

Obviously the results are only valid if you buy the packets without any knowledge of what's inside ie you don't feel them or buy a complete box containing known amounts of figures.

Cool! I wanted to see that formula plotted, but Wolfram Alpha didn't do it like this. Thanks!