LEGO Collectable Minifigures Series 6 Discussion

[ruboand]

Great minifigures!

My top 3:

1.Roman

2.Sleepyhead

3.Robot

I would have preferred 5 of these in each box!

[Ardelon]

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.

The Roman was one of the most anticipated minifigs by me, but I'm pretty disappointed, for the following reasons:

- The helmet seems too bulky, a helmet the thickness of a classic knight/sports helmet would have been better. If it was possible to get that much detail on it, that is.

- As for the "thong", TLG should have dunked the top part of the Roman's legs into red dye, it's probably not a process they use, but it doesn't seem too complicated and it would improve the look of this and other figs with short pants/skirts. Or even all-red legs with a small yellow band below the "knees" would have been better.

- I would also have liked to see back printing, and preferably a red basic torso color rather than light bley; with silvery or even light blay armor printed on the red, just for authenticity.

- I'm also disappointed we didn't get a specialized pilum instead of a regular CMF spear (I don't mind we didn't get a gladius, though it would have been cool to get a gladius with a PotC-style sheath). I feel a pilum would have been more relevant for the Roman than the broadsword for the highlander (while a nice touch, the Scots used all kinds of swords, anyway).

That's not to say the Roman is a bad fig, I still see him as one of the most useful, just not as cool as he could be.

[Masked Builder]

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.

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

[Ardelon]

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)

[Dervish]

Any of the UK shops that have the series 6 in stock, do international mailorders?

I just visited my trusty supplier that checked out the status of the series 6 and said, in january :(

It's better to have them shipped now, and wait 3 weeks that just wait 3 weeks to see them in stores :I

EDIT: John Lewis seems to ship, but doesn't have the figs online yet..

Edited December 17, 2011 by Dervish

[Brickmamba]

Strange choice of language...."Conveniently left out".

To be pedantic they are £2.00 in TRU, £1.99 in WH Smith High street and £1.95 in John Lewis. It is good they haven't had a price hike though.

so have series 6 'officially' been released in UK? may have to get onto my sis in law regarding getting me a box if that is the price

[Clone gunner comander jedi]

Hey Brickset have updated there collectable minifigures galery with hi res images of series 5 and 6 look for yourself.

Love the robot

Edited December 17, 2011 by Clone gunner comander jedi

[surrideo]

That little robot is all that stands between me and full set of minifigures lol

[Zeya]

That little robot is all that stands between me and full set of minifigures lol

Ain't it always the way? Last round I bought 60 figs from TRU online, and of course they didn't send me an unopened box. I was still missing 3 figs after that batch and had to go hunting. Call me paranoid, but I suspected they had been picked through or something. I'm going to be ordering maybe 80 this time and hopefully that'll get me a better chance at a full set.

[mo123567]

Ain't it always the way? Last round I bought 60 figs from TRU online, and of course they didn't send me an unopened box. I was still missing 3 figs after that batch and had to go hunting. Call me paranoid, but I suspected they had been picked through or something. I'm going to be ordering maybe 80 this time and hopefully that'll get me a better chance at a full set.

...And that's why they use blind packaging.

[Etzel]

Ain't it always the way? Last round I bought 60 figs from TRU online, and of course they didn't send me an unopened box. I was still missing 3 figs after that batch and had to go hunting. Call me paranoid, but I suspected they had been picked through or something. I'm going to be ordering maybe 80 this time and hopefully that'll get me a better chance at a full set.

Wouldn't it be cheaper to just buy a full set from Bricklink?

I've bought boxes of the latest 3 series, but I'm not sure if I will do that this time as I'm a bit short on money. I think I will try and be patient and wait for them to get to the local stores and feel my way to a full set and some extras. In one of them that I visited earlier this week they had a huge amount of S4 on racks, very good for keeping track of which one you have already checked.

[1980-Something-Space-Guy]

Ain't it always the way? Last round I bought 60 figs from TRU online, and of course they didn't send me an unopened box. I was still missing 3 figs after that batch and had to go hunting. Call me paranoid, but I suspected they had been picked through or something. I'm going to be ordering maybe 80 this time and hopefully that'll get me a better chance at a full set.

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)%.

Edited December 18, 2011 by johnnyvgoode

[Weil]

I'm afraid that I don't think your calculations are correct.

This can be shown by calculating the (much simpler) probability that you get 16 different figures after buying 16 bags.

With your first bag the probability of getting a unique figure is 16/16.

With your second bag the probability of getting a unique figure is 15/16.

With your third bag the probability of getting a unique figure is 14/16.

.

.

.

With your sixteenth bag the probability of getting a unique figure is 1/16.

So the overall probability of this perfect outcome is (16/16)*(15/16)*(14/16)*...*(1/16)=16!/(16^16) which is approximately 1*10^(-6) or 0.0001%. Several orders of magnitude higher than you get with your formula!

Calculating the general formula for the probability of getting 16 unique figures after buying n bags is surprisingly difficult.

[1980-Something-Space-Guy]

I'm afraid that I don't think your calculations are correct.

This can be shown by calculating the (much simpler) probability that you get 16 different figures after buying 16 bags.

With your first bag the probability of getting a unique figure is 16/16.

With your second bag the probability of getting a unique figure is 15/16.

With your third bag the probability of getting a unique figure is 14/16.

.

.

.

With your sixteenth bag the probability of getting a unique figure is 1/16.

So the overall probability of this perfect outcome is (16/16)*(15/16)*(14/16)*...*(1/16)=16!/(16^16) which is approximately 1*10^(-6) or 0.0001%. Several orders of magnitude higher than you get with your formula!

Calculating the general formula for the probability of getting 16 unique figures after buying n bags is surprisingly difficult.

I think that you're not using the correct method, although I can't find anything wrong in your reasoning right now, but I'll give it a thought.

For example, let's say there were only three different figs.

There are 10 possible outcomes: (1A, 1B, 1C; 2A, 1B, 0C; ..., ;3A, 0B, 0C; ...)

So for 3, the probability of getting them all by only buying 3 is 1/10.

However, using your reasoning would yield 1*2/3*1/3=2/9.

I used balls and separators to calculate the number of total possible distributions of buying n figs, which is (n+15)!/[(15)!*(n!)], and the total possible distributions of n figs with at least one of each, which is (n-1)!/[(15)!*(n-16)!]

I can't see the mistake in either reasoning right now, but I'm too sleepy. This is why I sometimes hate combinatorics.

Edited December 18, 2011 by johnnyvgoode

[Weil]

I think that you're not using the correct method, although I can't find anything wrong in your reasoning right now, but I'll give it a thought.

For example, let's say there were only three different figs.

There are 10 possible outcomes: (1A, 1B, 1C; 2A, 1B, 0C; ..., ;3A, 0B, 0C; ...)

So for 3, the probability of getting them all by only buying 3 is 1/10.

But the probability of those 10 outcomes are not equal so you can't just divide it by 10. There's two outcomes as to whether or not I win the lottery, I do or I don't. That doesn't mean I have a 50/50 chance!

You have to take into account the different combinations with which you can get 1A, 1B, 1C compared to getting 3A, 0B, 0C.

A,A,A is the only combination of getting 3A,0B,0C.

A,B,C; B,C,A; C,B,A; B,A,C; A,C,B and C,A,B are all combinations of getting 1A,1B,1C.

Therefore one of each is 6 times as likely as getting all of one.

[Phyre]

^^I love all this LEGO math. ^^

But for me, I just use the "touch and feel" method. I usually can't afford to buy more than a few minifigs in one trip to the store, so I never buy a full series all at once. But when I completed my Series 5 collection, I went to the store to try and get the six figures I was missing plus an extra gladiator, and got them all right using the touch and feel method. Obviously if you're buying them online, like from LEGO directly, what you get is most likely completely random. But if you buy a full sealed box of 60, you are ensured to get three complete series. If you buy random figs out of a box in a store, then you have to take into account how many of each fig were originally in a sealed box, how many are missing from that box, and if there are any other open boxes nearby that figs could have been moved to or from. So the probabilities change depending on what method you use to get your figs. Let's see some calculations for that.

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

Yeah, all of the "army builders" that collectors labeled beforehand seem to be the ones they put three of in a box - Zombie, Spartan, Elf, Dwarf, Gladiator, and now Roman. It stinks, but that's the way it goes.

[AoNZ]

That robot is gorgeous

I wonder if he could have benefitted from different minifig hands or even just grey hands to match his body?

Regardless he is one of my favourites.

[CM4Sci]

No luck here in MN..

Has anyone created one of those Google Maps Minifigure Maps? It'd be splendid to have one right about now.

-Sci

[1980-Something-Space-Guy]

But the probability of those 10 outcomes are not equal so you can't just divide it by 10. There's two outcomes as to whether or not I win the lottery, I do or I don't. That doesn't mean I have a 50/50 chance!

You have to take into account the different combinations with which you can get 1A, 1B, 1C compared to getting 3A, 0B, 0C.

A,A,A is the only combination of getting 3A,0B,0C.

A,B,C; B,C,A; C,B,A; B,A,C; A,C,B and C,A,B are all combinations of getting 1A,1B,1C.

Therefore one of each is 6 times as likely as getting all of one.

Second thought, I think that you're right. I'll try to research a bit more into this and see if the probability for n can be calculated.

Edit: here's more info on the subject. Apparently the optimal amount of figs you should buy is 45. Whew, that's a relief. Sorry for scaring you all people.

Edited December 18, 2011 by johnnyvgoode

[Wardancer]

Hmm I am sceptical about the minotaur. At first I thought I would want 8. But now with LotR coming I will perhaps save that money. There is something about the minotaur that I do Not like 100 percent.

It would be most honorable if some of the early buyers compiled a dot code sheet.

[1980-Something-Space-Guy]

Calculating the general formula for the probability of getting 16 unique figures after buying n bags is surprisingly difficult.

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.

Edited December 19, 2011 by johnnyvgoode

[thefyfe]

Officially found in Glasgow uk today at whsmith!

[Roger Rabbit]

Got my first five here: Flamenco Dancer, Minotaur, Leprachrun, Space Girl (aka Barbarella/Blonde chick from Abba) and Lady Liberty. Couldn't find them in Smith's near work, but trotted along to Toy R Us and couldn't resist grabbing five.. although there wasn't that many left to be honest. Can't wait to complete my collection.

At the moment, the Statue of Liberty has to be my favourite (and the only one I'm buying more than one of, I want to use one as a minifigure fancy dress costume). although I have a soft spot for the Genie and Sleepyhead, so it could change once I get them. The Roman would rank higher if he wasn't wearing a thong..

[GregD]

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

That would be nice to see if anyone could upload it, thanks.

[Itaria No Shintaku]

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 :)