Re: The maths of miniatures distribution (split from Sales and events)

[KingOfVrock]

Math is for losers!

[MyriadTabletop]

Math is for losers!

Lol. Roll initiative, you punk! Oh, I'm sorry. That is if you know how . Bwahaha.
(Aaaallll in good fun).

[Jandar]

I listed every possible combination (not permutation, since I'm not lining them up in order to display), and one can argue about which combinations are more likely and weight them; however, when you don't know, they're supposed to be weighted equally for the estimate.

I don't know what to tell you. Combinations are still the wrong tool because that's not how the samples (bricks) are organized during their selection. Consider using your method with only two-bricks. The combinations are:

Same Brick - 4 instances
AA, BB, CC, DD

Different Brick - 6 instances
AB, AC, AD, BC, BD, CD

Using this method, the probability of getting two of the same brick is 40%. And granted, this WOULD be true, but ONLY if every case was broken down, the bricks identified, and piles of pre-made pairs of bricks were assembled, each containing one of the possible combinations, from which the fulfillment worker randomly selected a pair from one of the ten piles of pre-arranged combinations. So yes, IF the samples are identified and pre-organized into "every possible combination", you can THEN calculate the probability of randomly selecting a combination with a desired trait FROM the pool all known combinations. But that doesn't reflect reality. Your method ignores the uncertainty of the value inherent to each brick by identifying them in advance, grouping them, and only -then- applying randomness by selecting from amongst the known, sorted combinations. Except the randomness occurs at the selection of each individual brick, not in the random selection from combinations of multiple -known- bricks. After all, we know definitively by the simplest math that the odds of getting two of the same selected brick is only 25%, not 40% if each is selected randomly.

Further, it's possible that the four "Same Brick" combinations could be further randomized into additional mismatched pairings and not occur at all. This is exactly why we use permutations and the sampling method. It accounts for -every- possible arrangement, and why order -does- matter.

By contrast, my method assumes only that when a customer orders two or more bricks, that a worker grabs a brick without knowing it's value, randomly, one at a time from a sufficiently randomized stock. I know this isnt perfectly accurate, but its the worst case scenario (more on that below). But even if a worker has only 1 loose brick and needs to open a new case for the rest of the order, it's purely random which brick from the next case they choose.

This all said, even this method doesnt perfectly match reality. Its not truly random, and there are confounding factors that impact the math that I alluded to above, I won't deny that. But they all serve to improve the chances of fewer duplicates. For example, for every brick removed from the sample pool it decreases the stock levels of that one brick while increasing the likelihood that one of the other three bricks will be selected since the stock is finite and shipped in perfect cases. So that makes duplicates less likely. Alternatively, if the cases are kept -as- cases and only broken as-needed (most likely), it pushes all of the math towards an even stronger likelihood of lower chances of duplicates while completely eliminating triple-repeats from even being possiblem. After all, if you order 4 bricks, the worker needs at most two-partial cases, meaning the worse possible case is two pairs of matching bricks, never three.

The problem is, we don't know their organization system; but it's fair to say they either keep bricks in sealed cases (skews the data towards less-duplication), or break them down into bricks (totally random, and math is as-above, adjusted slightly as stock cycles). It's another reason why using the "combinations" math is inappropriate.

Edit: I know this seems like off-topic, but I dont want people avoiding Bricks on sale because of improper statistics.

It really doesn't matter when or how one buys 4 bricks or in what order they're selected or opened; in the end, the stack of minis from those four bricks is exactly the same.
My table lists every possible outcome exactly once.

Your table repeats outcomes that are identical; for example, opening brick A then brick B gives the same minis collection as opening the bricks in the reverse order.

One could use your table as a way to weight the outcomes, but that makes assumptions.
Once could also weight the outcome of 4 different bricks close to 100% based on other assumptions (assuming they ship a case).
Making no assumptions leads to the odds I listed (and just to be fair, making no assumptions is also an assumption, but it's formally considered the default case).

BTW, these are probabilities not statistics.
Statistics would be the actual results of opening the 4 bricks, similar to the thread on this forum that shows the results for opening cases.
Probabilities always have assumptions, statistics are just data.
Tossing a coin many times and seeing 60% heads and 40% tails is a statistic.
Assuming the coin is fair and that it will give 50% heads and 50% tails is a probability.

[MyriadTabletop]

Your table repeats outcomes that are identical; for example, opening brick A then brick B gives the same minis collection as opening the bricks in the reverse order.

They're not identical. They're unique, because you must account for the frequency that each combination occurs.

Consider rolling a pair of pip-dice. We know there are 36 permutations and 21 combinations. Of the permutations, 15 appear inverted (30), with the remaining 6 being rolling "doubles". But we account for frequency when we determine the odds of what the result will be on any given "throw". We know that "rolling doubles" occurs with a 6/36 (or 1/6) probability, not 6/21. And the odds of rolling a 3 + 4 are 2/36 because we do count 3+4 and 4+3. It's not 1/21. Order absolutely matters. It doesn't matter that you "end up with the same two values", it's the random selection of a sample from a population with uniform distribution of finite values (equal chance of any one brick) that is analogous with the rolling of the dice. We are sampling a population, we aren't sampling combinations.

I think this is what you're missing. Dice (brick) #1 can be any one of a series of values; and Dice (brick) #2 can -also- be any one of a series of values; so we use the Product Rule of calculating probabilities. This can then be -represented- as a table, which let's us -visualize- the permutations that are accounted for in the Product Rule.

But here is an even easier analogy. We have a sack with an infinite number of marbles equally divided among red and white (infinite only to avoid the annoying fractional discrepancies introduced by removing one marble from the sack). When we choose one, the marble will always have been the color it is (in the same way the brick -has- definitive, but unknown contents), but it's the selecting of the marble from a 50/50 bag that introduces the randomness, and why we factor in order. We don't say "The combinations of any two marbles are RW, RR, and WW, so there is a 33% chance of drawing one of those three combinations when you pull two marbles from the bag," that would be incorrect. We know this because this experiment is functionally identical to flipping two coins, in which HH, HT, TH, and HH are the possible permutations, and the odds of getting two heads (two red marbles) is universally understood to be 25%. And the odds of getting 'one of each', is 50%. Neither relies on Combinations, and the odds would not be 33% for either. Now, all we have to do is add two more colors of marbles into the sack in equal quantities, call them Bricks, and we use the same method. Probabilities via Permutations.

As I said before, doing it with Combinations simply calculates 'the odds of selecting one or many combinations from a list of all possible combinations', which is not the same as selecting two random samples from an evenly distributed population.

To put a cap in this; If the contents of each brick (the result of the dice roll, the flip of the coin, the marble in the sack) is selected randomly, and the odds of the resulting value is equal across all possible values, then this is the method.

Now, I recognize that in application the probabilities will only apply for as long as the distribution remains equal, but for the sake of any method we have to assuming perfect randomness at easy stage, otherwise using any method would be a waste of time since we don't actually know how they manage their stock.

Edit: I know this seems like off-topic, but I dont want people avoiding Bricks on sale because of improper statistics.

BTW, these are probabilities not statistics.

I think this is just equivocation (or pedantry).

While statistics (little 's') does commonly refer to either a series of output values derived from manipulating the data (4 out of 5 dentists recommend....), it can also refer more generally to a collection of methodologies as a discipline, like taking a Statistics class (big 'S') where they teach how to calculate probabilities. I meant the latter. You are using Statistics inappropriately, in the same way someone might use Algebra incorrectly resulting in the wrong solution. Perhaps that's clearer, but if we're just going to mince on language I'm not really interested.

Either way, that was my last chance to explain it. I'm not sure what else I can do. I love math, but I'm not a fan of going in circles. So I'm good.

[KingOfVrock]

Math is for losers!

Lol. Roll initiative, you punk! Oh, I'm sorry. That is if you know how . Bwahaha.
(Aaaallll in good fun).

d10 or d20???

and isn't this whole argument a bit moot, as the bricks themselves are not identical (at least not anymore that I can remember?) - just that you would get a distribution of all minis (basically rares) in the set within a case?

Anyway just for fun, if there were 4 unique bricks, wouldn't the calculation be:
First brick - Let's call it brick A
Second brick - 75% chance of getting a different brick (ie, not A)...if it is different we'll now call this brick B
Third brick - 50% chance of getting a different brick (ie not A or B)...if it different we'll now call this brick C
Fourth brick - 25% chance of getting a different brick (not A or B or C) ...we're calling this brick D

75% * 50% * 25% = 9.375% chance of getting all 4 unique bricks in a 4 brick purchase

[MyriadTabletop]

...as the bricks themselves are not identical (at least not anymore that I can remember?) - just that you would get a distribution of all minis (basically rares) in the set within a case?

Honestly, I haven't followed the Case Results thread enough to really say. I -thought- that was true for at least the Rare distribution, and I did assume that in my math (as you did), but I have no clue.

The original question was posed by jcg, "I wonder what the odds are of getting a full case (vs 4 random bricks) if you order 4…."

So I think that was just the premise. You used the same math and method as I did, for what it's worth.

[Jandar]

Your table repeats outcomes that are identical; for example, opening brick A then brick B gives the same minis collection as opening the bricks in the reverse order.

They're not identical. They're unique, because you must account for the frequency that each combination occurs.

Consider rolling a pair of pip-dice. We know there are 36 permutations and 21 combinations. Of the permutations, 15 appear inverted (30), with the remaining 6 being rolling "doubles". But we account for frequency when we determine the odds of what the result will be on any given "throw". We know that "rolling doubles" occurs with a 6/36 (or 1/6) probability, not 6/21. And the odds of rolling a 3 + 4 are 2/36 because we do count 3+4 and 4+3. It's not 1/21. Order absolutely matters. It doesn't matter that you "end up with the same two values", it's the random selection of a sample from a population with uniform distribution of finite values (equal chance of any one brick) that is analogous with the rolling of the dice. We are sampling a population, we aren't sampling combinations.

I think this is what you're missing. Dice (brick) #1 can be any one of a series of values; and Dice (brick) #2 can -also- be any one of a series of values; so we use the Product Rule of calculating probabilities. This can then be -represented- as a table, which let's us -visualize- the permutations that are accounted for in the Product Rule.

But here is an even easier analogy. We have a sack with an infinite number of marbles equally divided among red and white (infinite only to avoid the annoying fractional discrepancies introduced by removing one marble from the sack). When we choose one, the marble will always have been the color it is (in the same way the brick -has- definitive, but unknown contents), but it's the selecting of the marble from a 50/50 bag that introduces the randomness, and why we factor in order. We don't say "The combinations of any two marbles are RW, RR, and WW, so there is a 33% chance of drawing one of those three combinations when you pull two marbles from the bag," that would be incorrect. We know this because this experiment is functionally identical to flipping two coins, in which HH, HT, TH, and HH are the possible permutations, and the odds of getting two heads (two red marbles) is universally understood to be 25%. And the odds of getting 'one of each', is 50%. Neither relies on Combinations, and the odds would not be 33% for either. Now, all we have to do is add two more colors of marbles into the sack in equal quantities, call them Bricks, and we use the same method. Probabilities via Permutations.

As I said before, doing it with Combinations simply calculates 'the odds of selecting one or many combinations from a list of all possible combinations', which is not the same as selecting two random samples from an evenly distributed population.

To put a cap in this; If the contents of each brick (the result of the dice roll, the flip of the coin, the marble in the sack) is selected randomly, and the odds of the resulting value is equal across all possible values, then this is the method.

Now, I recognize that in application the probabilities will only apply for as long as the distribution remains equal, but for the sake of any method we have to assuming perfect randomness at easy stage, otherwise using any method would be a waste of time since we don't actually know how they manage their stock.

Edit: I know this seems like off-topic, but I dont want people avoiding Bricks on sale because of improper statistics.

BTW, these are probabilities not statistics.

I think this is just equivocation (or pedantry).

While statistics (little 's') does commonly refer to either a series of output values derived from manipulating the data (4 out of 5 dentists recommend....), it can also refer more generally to a collection of methodologies as a discipline, like taking a Statistics class (big 'S') where they teach how to calculate probabilities. I meant the latter. You are using Statistics inappropriately, in the same way someone might use Algebra incorrectly resulting in the wrong solution. Perhaps that's clearer, but if we're just going to mince on language I'm not really interested.

Either way, that was my last chance to explain it. I'm not sure what else I can do. I love math, but I'm not a fan of going in circles. So I'm good.

It's sounding more and more like you took the basic class,
(for example: Math 70: Introduction to Probability at https://ww3.math.ucla.edu/courses/),
but not the upper division one,
(for example: Math 184: Enumerative Combinatorics at https://ww3.math.ucla.edu/courses/).

At least that would explain the insistence on the correctness of an un-nuanced approach.

Your table repeats outcomes that are identical; for example, opening brick A then brick B gives the same minis collection as opening the bricks in the reverse order.

They're not identical. They're unique, because you must account for the frequency that each combination occurs.

So if I grab a hamburger, fries, and a coke and you grab fries, a coke, and a hamburger, we're having a different lunch?

Math is for losers!

Lol. Roll initiative, you punk! Oh, I'm sorry. That is if you know how . Bwahaha.
(Aaaallll in good fun).

and isn't this whole argument a bit moot, as the bricks themselves are not identical (at least not anymore that I can remember?) - just that you would get a distribution of all minis (basically rares) in the set within a case?

I'm unaware of anyone missing any mini, given that they got every rare.
I believe that the sub-case variations don't vary enough to allow that to happen.

EXCEPT, that is, when they have those AB variants of a non-rare mini; then, I have seen it happen.

[MyriadTabletop]

So if I grab a hamburger, fries, and a coke and you grab fries, a coke, and a hamburger, we're having a different lunch?

Of course you're having the same lunch. But you keep making the same mistake, over and over and over (and not even trying to engage with any of my explanations).

The question is not, "How many different meals can you make from a menu of three items, and from that list, what are the chances of getting a combination that has one of each item?".

The question is, "If the waiter brings me three items from a menu randomly, what are the odds I get one of each menu item?"

(If Wizkids ships me four bricks from their inventory randomly, what are the odds you get one of each of the possible four bricks = the original question.)

This is honestly exhausting. I feel I've gone through lengths to help elucidate the issue, and I don't think you've returned the same effort in return. So at this point if you want to make any progress I would recommend that you go to one of the statistics subreddits that I'm sure exist, pose the problem to them (as you understand it), and find a different way explain why we would use your method. Because just repeating, "Yeah, but it's still the same bricks/meal," isn't getting us anywhere.

As for course work, I've taken statistics classes throughout my diploma program, and more during my bachelor's degree, and even more during my masters level courses. Setting aside that this problem only actually requires a "basics" level understanding to address, I'm quite comfortable in this space, even if it has been a while. I've appreciated the opportunity to refresh though.

So yeah... if you don't have anything new to share, or a new way to explain why you think you're right, probably just drop it. But if you do choose to continue, moving to PMs would be best. I'm sure by now folks here know our two positions and will make up their own minds without any more from us. We're officially way off topic anyways.

[BBShockwave]

This one surprised me considering how new it is.

I wonder what the odds are of getting a full case (vs 4 random bricks) if you order 4….

DnDMinis says in their description they will ship you a full unopened case if you buy 4 bricks.
Other shops, often even when I asked them several times in mail, could not assure me they would send a case. (Jedishop being one, after they stopped listing cases).

That's interesting because even the case doesn't guarantee 4 unique bricks (I've been burned on that).

Really? I have been buying cases since ToD and that never happened to me. Well, happened once, but that was an admitted fault of Tritex, who told me they had no full case of F&T for me, and the last 2 boosters of the 4th brick were replaced by DiA boosters. However, it turned out the 4th brick was also from another case, too. It was not identical to any other brick in the case, but I got 2 of the same Medium Rares in it (Green Hag and Tabaxi Fighter) but did manage to get all 4 Huge Rares in the case even with the mix-up.

Math is for losers!

The only thing I learned from combinatorics in secondary school was that
A: Combinatorics sounds a lot like Combaticons, so I kept thinking of how cool it would be to get a Bruticus figure
B: I was convinced by the extremely small chance of a string of numbers to appear in that order, meaning I have never played the lottery since then (not that I have ever done it before) - the only sort of game of chance I engage in is buying Mage Knight boosters.

and isn't this whole argument a bit moot, as the bricks themselves are not identical (at least not anymore that I can remember?) - just that you would get a distribution of all minis (basically rares) in the set within a case?

Well, no, based on case results I have seen, mine and others, you are guaranteed of 1 of each Rare in a case but beyond that your results will vary wildly.
Depending on how many Huge/Large Uncommons are in a set, you will get 3-4 of each (if there are 7-8 of them in the set) or 2-4 (if there are 9-10).
Medium Uncommons, you are probably getting 2-4 of each, with most of them being 3, and only 1-2 being 2 or 4.
Commons you are getting 3 to 5 of each, but this is the most varied. I got 6 Ghouls in Boneyard for example.
Of course this is for normal sets. Superboosters means you get less of everything.

[MyriadTabletop]

I would love to see how Boosters are assembled at the factory. Like... somewhere there is a stock flow diagram, lol.

[Jandar]

I would love to see how Boosters are assembled at the factory. Like... somewhere there is a stock flow diagram, lol.

That we can agree upon.

[Jandar]

So if I grab a hamburger, fries, and a coke and you grab fries, a coke, and a hamburger, we're having a different lunch?

I feel I've gone through lengths to help elucidate the issue, and I don't think you've returned the same effort in return.

I absolutely did.

Case (b) above in my earlier posts is rolling a 4 sided die (exactly the method you describe, sampling from a distribution).
It leads to 3!/4^3, which is your claim (KingofVrock's math too).

I just disagree that one can assume that it's done this way.

[MyriadTabletop]

I just disagree that one can assume that it's done this way.

All good. I think we've settled on it not actually mattering at all. I think the human element would confound any maths based best-guess anyways.