1980-Something-Space-Guy

Confirmed where?

The funniest part about this is the "VIP" in the set's name. Seriously, I can't believe this. Although I wasn't planning on buying this, I'm kind of disappointed. I always expect Lego to keep a high standard of satisfying consumer needs. I just would have never expected Lego to do something like this, which seems, and please excuse the word, mediocre. Oh well, I'll go grab my series 6 figs.

I'm loving the new R5 unit. :wub: Thanks for the pics and review!

Hahahaha, that is simply awesome! Good work, and thanks a lot for sharing.

The oddest part is you're right.

Awesome pics! I'm glad Indy's still having amazing adventures. I gotta go to Hawaii someday. Thanks again for sharing!

To me, it's much like the "To be continued..." at the end of the Back to the Future movies.

:thumbup:

Are you Latin American? I've noticed that each time there are more of us here, so I thought it'd me nice if we had a topic for us, given that we're such a tied community. If you are Latin American, let us all meet here!

Hello Gaby! It's always nice to see another fellow Latin American AFOL around here. ¡Pura vida desde Costa Rica, y bienvenida a Eurobricks!

I'd be happier with 5 stormtrooper sets, to tell you the truth, and perhaps you could get discounts if you bought a certain amount. I don't think Hasbro would allow all this though.

Thanks for another great review Whitefang! I always read your reviews before buying my minifigs. Thanks!

Well, here's something I don't understand. That thread makes me thing that the creature is custom made. Yet we've seen it on the prelims. Could anyone explain this to me?

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

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?

I like the new lagoon creature. I'm interested in some of these figs, although I hate the werewolf. I can imagine them reusing the CMFs though... And that just doesn't seem right, by the way CMFs are advertised. I don't care very much though.

More astromechs! I'll be very poor by the end of next year.

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.

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.

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.

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.

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.

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.

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.