Forum:Ratios: need help

I've got a problem. I want to prove to a duelist that's complaining about Konami "exploiting rich duelists" by making Secret Rares, that his thinking is faulty by proving that you have MORE of a chance of getting the Secret Rare card you want from a box, than you would getting the Super Rare card you want from that same box, but I suck at math, and dunno how to calculate ratios of ratios.

Here's my math problem: there are 24 packs in a box, with the ratios of getting a Secret and Super Rare from that box being 1:24 and 1:5, respectively. There are 8 Secrets and 14 Supers in a set. If I've done my calculations correctly, that means I've got a 12.5% chance of getting the Secret Rare I want from a box, and a 7.14% chance of getting the Super Rare I want out of any 1 of the 5 or so Supers I can get in a box. Now, I know that duplicates means that technically, I'd have a much greater chance of getting the Super Rare I need (if 5 Supers are the average you can get in a box, that means I have up to a 35.71% chance of getting the Super Rare I need), but in practice, I'm betting that actually means I have a much LOWER chance of getting it, considering there's an equal chance that any one of the OTHER 13 Supers in the set would have duplicates in the box.

So, my question is thus: is this correct? Is my math right, and the percentages for getting the Super Rare you want to pull from a single box much lower than the chance of getting the Secret Rare you want from that same box? Any math wizards here want to correct me? Please do; I'm poor at math, so I'd probably miss something. 66.228.109.2 22:56, November 1, 2009 (UTC)

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• Ooh, math!

Ratios: 1:24 (ScR), 1:5 (SR)

Numbers: 8 (ScR), 14 (SR)

P(ScR,desired)=P(ScR)*P(desired)=(1/24)*(1/8)=1/192=0.0052=0.52%.

P(SR,desired)=P(SR)*P(desired)=(5/24)*(1/14)=1/70=0.0142=1.48% (about 3 times more than the above P).

These Ps are for 1 booster pack.

For a box, multiply by 24:

P(ScR,desired,box)=0.52%*24=12,48%.

P(SR,desired,box)=1.48%*24=35,71%.

Now, you mention possible duplicates for the SR cards, so that the P of pulling a specific one is less, since you may have more copies of an undesired card. However, in the box, you either have the desired card (in x quantity), or you don't (x=0). In case 1, multiple copies are irrelevant, and the same goes for case 2. So, multiple copies do not matter! You either get the card or you don't, and the amount of duplicates has no impact on the actual desired-card-pulling-P.

Let me elaborate a bit on the above, so that it is understood.

We randomly buy a 24 booster pack, untampered, box.

Opening it, we pull a desired SR card in 1 copy, 4 copies of the same undesired SR card, and an undesired ScR card.

What is important here is to illustrate the fact that the additional copies of the undesired SR card do not change the P of pulling the desired one.

We have here a case of sampling with replacement. Essentially, you "pick" a sample, record it, then put it back into the "pool". So, you can "pick" the same sample more than once, since it still is in the "pool". This displays the possibility of having multiple copies.

However, the same possibility is true for all the SR cards in the set (the possibility of pulling multiple copies of card 1 is equal to that of card 2 and so on), so there are no differences in the possibilities of pulling multiple copies of card "x" compared to card "y".

Also, (key point ahead), the possibility of "picking" the desired card is the same each time you "pick" a sample, being an independent event. And that possibility is hence equal to the possibility of "picking" any of the undesired cards, collectively equal to 1/8 (remember, we are studying the SR cards, so we omit the others for now).

You have a ratio of 1:5 for SR cards. So, you have 5 chances in which to "pick" the card you want.

P(picking the desired card)=(1/14)+(1/14)+(1/14)+(1/14)+(1/14)=5/14=35,71%, same as the P for the desired card being in the box.

So, since we have proven that the possibility of unwanted SR duplicates do not change the possibility of pulling the desired SR card, your original statement of "if 5 Supers are the average you can get in a box, that means I have up to a 35.71% chance of getting the Super Rare I need, but in practice, I'm betting that actually means I have a much LOWER chance of getting it, considering there's an equal chance that any one of the OTHER 13 Supers in the set would have duplicates in the box." is false.

Direct any questions to me, should any arise. --Darth Covah (Talk | Contribs) 14:08, November 2, 2009 (UTC)

Wait, what? That doesn't make much sense. Like I said, I'm not good at math, but that seems to fly in the face of conventional wisdom. There's more Super Rare cards you could pull than Secret Rare cards, so the greater chance of pulling a Super Rare should mean a greater chance of NOT pulling the card you want from them, than the lesser chance you get of pulling a Secret Rare. It just seems like the greater number of Supers available for pulling should offset the number of chances of pulling a Super in a box. Isn't this what people were arguing about, when Konami/UDE decided to whittle down the Ultimate Rares to JUST Ultras? That the loss of the Super Rare Ultimates would make getting a desired Super Rare harder to get than most of the other rarities? 66.228.109.2 20:27, November 2, 2009 (UTC)

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• I assure you the logic of the above calculations and results is sound. However, where do think it makes no sense? Does all of it make no sense to you, or is it a specific point you think is logically false?

"There's more Super Rare cards you could pull than Secret Rare cards, so the greater chance of pulling a Super Rare should mean a greater chance of NOT pulling the card you want from them, than the lesser chance you get of pulling a Secret Rare".

Now, let me refine the statement.

"There's more Super Rare cards you could pull than Secret Rare cards, so a greater chance of pulling a Super Rare should mean a greater chance of NOT pulling the card you want from them than the chance of pulling a Secret Rare"

This is, as I see it, a rephrasing of your original query.

I will repeat that this is a false statement, and proceed to explain, logically, why that is so.

(Deviations from previous calculations are due to rounding.)

The chance of pulling a SR in a booster is 1:5, so it is 20%. The chance of a pulled SR card being the one you want is 1:14, so it is 7.14%.

The combined chance of pulling a SR in a booster and that SR being the desired one is equal to the product of the individual chances. So, that chance is 1,42%.

Similarly, the chance of pulling a ScR in a booster is 1:24, so it is 4,16%. The chance of a pulled ScR card being the one you want is 1:8, so it is 12,5%. The combined chance, as stated above, is the product of the individual chances, which is 0,52%.

To sum it up, you are 2,73 times more likely to pull the desired SR than the desired ScR from 1 booster.

To see the chances for a whole box, just multiply each chance by 24 (the amount of boosters):

SR: 1,42*24= 34,08%

ScR: 0,52%*24= 12,48%

It really can't get any simpler. For a whole box, you have about 3 times more chances of pulling the desired SR than the desired ScR.

The increased amount of SR cards is relevant, but the deciding factor are the ratios. You would have to have about 3 times more SR cards than the current amount for the chance of pulling the desired SR being equal to the chance of pulling the desired ScR, and even more for it to be less.

But, since the ratios are fixed, and the amount of SR/ScR cards remains relatively stable, it is a fact that for a whole box, you have 2,73 times more chances of pulling the desired SR than the desired ScR.

I believe that this is clear now. If not, ask me to clarify the points you do not think are sound. --Darth Covah (Talk | Contribs) 00:37, November 3, 2009 (UTC)

Then why were people complaining about how impossible pulling the Super Rare cards they want, after Konami/UDE restricted the Ultimate Rare variants to Ultra Rares? I distinctly remember people complaining that Super Rares were harder to get than Secret Rares, after Konami/UDE stopped making Ultimate Rare versions of the former? I'm sorry, but that just flies in the face of conventional logic, and MUST be wrong. There's just no way that you'd have that big of a margin between the chances of pulling the Super Rare that you want, and the chances of pulling the Secret Rare that you want. The possibility of duplicates HAS to factor in SOMEWHERE, and the higher amount of SR cards compared to the ScR cards HAS to skew towards NOT getting the card you want, not towards GETTING the one you want. 66.228.109.2 00:52, November 3, 2009 (UTC)

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Excuse me, but if you've made up your mind that you are correct, what's the point of asking if you are?

To the point, then. UtR cards are not included in the ScR/UR/SR/R card ratios - they are independent. And of course removing all UtR cards from SR down decreases the chance of pulling a desired SR, but it is not enough to make it less than the chance of pulling a desired ScR. The calculations I performed above also do not include UtR cards which are derived from the set's SR cards. Hence, they are not influenced by them.

All in all, the people were right to complain, because they were right about the cards being harder to pull. But the SR cards didn't all of a sudden become harder to pull than ScR cards, just because they lost their UtR versions.

Finally, I will repeat that UtR cards are independent, they are not included in the ratio of the card they are derived from. So, my calculations are sound, and you are about 3 times more likely to pull a wanted SR than a wanted ScR. That chance may be in itself small, but your question was about SR rarity in comparison to ScR rarity.

Last, duplicates. If you read my first answer more carefully, you'll see that I have already examined that. Duplicates do not factor into the chance of pulling the desired SR card.

Despite my belief that the question has been answered, I shall leave the possibility of further questions open. Hope that your question has been sufficiently clarified. --Darth Covah (Talk | Contribs) 01:06, November 3, 2009 (UTC)

tldr; You have a 5/18 (approx. 28% chance) of getting your Super Rare card.

Basically, you can get the wanted card zero times, once, twice, ..., or five times. It's easiest to just find the chance of getting none, and subtracting it from one. The problem is that order doesn't matter, and you can select the same card more than once.

To solve this, draw it out this way: give each SR card a number, so that you have Super Rare #1 (SR1), SR2, ..., SR14. Then, put a "wall" between each card. You'll get something like this:

| | | | | | | | | | | | |

Where the pipes represent walls (the | thing is called a pipe). The space before the first "wall" represents the number of times that you pull SR1. The space between the first and second "walls" represents the number of times that you pull SR2. And so on. So, if you pulled SR13, then the picture would look like,

| | | | | | | | | | | | X |

Now, for your question, you pull five cards at random. For example, you might get one SR1, one SR4, one SR7, and two SR11. The diagramme would look like this:

X | | | X | | | X | | | | X X | | |

We have to count the total number of ways that you can pull cards. Using the above diagramme, you can think of it this way: you have 19 "slots" which start out as all "walls", and you choose five of them and turn them into "Xs". This will always give you a diagramme like the one above. There are a total number of bin(18, 5) ways of doing this, where bin() represents the binomial coefficient.

Now, let's say that you don't pull the Super Rare card you want. Arbitrarily, let's say that you wanted to pull SR1 (else, we can just re-order them). Using the above diagramme, this means that the first "slot" will always be a "wall". Thus, when you're choosing what "slots", you only have 18 to choose from. This gives you a total of bin(17, 5) ways of not getting what you want.

Now, you have a bin(17, 5)/bin(18, 5) = 13/18 chance of not getting the card, so you have a 5/18 chance of getting the card you want.

The problem with Darth Havoc's method is that he's double-counting. Let's say you get your wanted card on pulls #1 and #2. He would count that as both the first 1/14 and the second 1/14.

--Deus Ex Machina (Talk) 01:10, November 3, 2009 (UTC)

Here's my argument: since each Super Rare has an equal chance of being pulled for each Super in the box, and you have more than 1 chance to pull A Super Rare in a box, logically those chances SHOULD add up by the amount of chances of pulling a Super Rare in practice, meaning you should ACTUALLY have a LESS chance of pulling the card you want. Comparatively, since you only have ONE chance of pulling a Secret Rare in a box, the equal chance of pulling each Secret Rare is only applied ONCE, increasing the chance of pulling the card you want, in practice.

And why discount Ultimate Rares? Yes, they have their own ratio for being pulled in a box, but they are just alternate rarity versions of the cards in the set, so their ratios should be added to the ratios of the cards they're duplicating. 66.228.109.2 01:28, November 3, 2009 (UTC)

Unfortunately, probability isn't always intuitive. The birthday problem is probably the best example.

Let's say there are 8 Secret Rares and 14 Super Rares. If you only pulled one Super Rare and one Secret Rare per box, then obviously you have a lower chance of getting your desired Super Rare. However, what if you pulled 100 Super Rares per box, but only pulled one Secret Rare? Obviously, you'd have a better chance of getting your Super Rare than your Secret Rare.

We don't even have to go to 100 to reach this point. 5 is sufficient.

Limiting Ultimate Rares did decrease your chances of getting a Super Rare, but that doesn't affect the original question.

--Deus Ex Machina (Talk) 02:34, November 3, 2009 (UTC)

Crap. I was hoping on using this as a way of shutting up everyone on the Shriek boards who were complaining about how Konami is supposedly scalping duelists by making the OCG cards they want Secrets, since I know those same people complained about the difficulty of getting the cards they wanted if they were Super Rare when Konami limited the Ultimate Rares, and their argument does seem valid. It just seems right: more Supers x more chances of getting a Super = more chances of NOT getting the Super you want. Now all I have to argue against them is the fact that the demand for the card and whether it's been reprinted do a lot more to determine the difficulty and price of getting the card you want, be it from box-pulling or trading, than its rarity does, and no one's listening to that argument.66.228.109.2 03:01, November 3, 2009 (UTC)

Ah, here's your problem:

...more Supers x more chances of getting a Super = more chances of NOT getting the Super you want

That is correct. However, the chances of getting the Super you want also increases, and the chance of getting the Super increases at a much faster rate than the chance of not getting it.

Let's say that you only got one pull. Then there is one scenario where you get the card, and 13 situations where you don't get the card, for a total of 14 scenarios. You have a 1/14 chance of getting the card you want.

Now, let's say that you have two pulls, and let's say that you don't care about the order in which you pull the cards. You have a total of 91 scenarios (pair up each of the 14 cards with another one, and get rid of duplicates). Of these, you have 14 ways of pulling the card you want (pair up the card you want with one other card, possibly itself). This leaves 77 ways of not pulling the card you want. The number of situations where you don't get the card you want has obviously gone up. There are six times the number of situations where you don't get the card you want. However, the number of situations where you do get the card you want has also gone up, and much higher - it's fourteen times higher.

It doesn't matter how many situations there are. What matters is the number of situations in question divided by the total number of situations.

--Deus Ex Machina (Talk) 03:39, November 4, 2009 (UTC)