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 May 4th, 2017, 07:20 PM #1 Newbie   Joined: May 2017 From: Mandurah, Western Australia Posts: 3 Thanks: 0 Picking card chance, infinite set, non-standard example Howdy, this'll be my first post on this site. I believe this question is probably at a high-school level. Trying to work out percentages for an online application, but it's been years since I needed probability theory of this complexity. Let's say I have an infinite list of cards. The odds of me pulling one of the cards I want from this set is a constant 1.5% per card. Cards are provided in a pack of 8 at a time. There can be duplicates. I.e. I can get more than one of what I want per pack. If I have 10 copies of the card that I wanted presently, how many packs would I have had to open? I realise that there won't be a set number due to the nature of the question, but let's make it: "The odds that I'd have 10 copies after x packs is 50%" and "The odds that I'd have 10 after y packs is 90%" Solve for x and y. Working out would be appreciated so I can apply it to other questions that may come up later. I assume I have to use nCr somehow, but can't think of how to you apply it here.
 May 5th, 2017, 07:49 AM #2 Newbie   Joined: May 2017 From: Mandurah, Western Australia Posts: 3 Thanks: 0 Edit: One thing I think I might have misstated: It's only sort of an "infinite list of cards." Let's say there's ~500 unique cards. If I pull cards from a theoretical "deck", I could pull forever, but the cards I pull will be copies from these 500. So let's say I have something like 10 cards of these 500 that I want, but the odds that I pull a card from the top of the deck and it isn't one of these 10 is 98.5%
 May 5th, 2017, 09:18 AM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,364 Thanks: 1270 So you are saying there are 10 unique cards out of a possible 500 that you are interested in, packs are 8 cards that are chosen randomly and independently, and you want to know how many packs you need to buy to have a probability of P, somewhere close to 1, to get these 10 cards. Is that correct?
 May 6th, 2017, 12:27 AM #4 Newbie   Joined: May 2017 From: Mandurah, Western Australia Posts: 3 Thanks: 0 Not quite. I just meant to use the numbers 500 and 10 as examples to say that there are a finite number of unique cards, and the cards that I want are a subset of these. I.e. There isn't an infinite amount of different cards. If I bought cards in packs of 8, roughly how many packs would I have had to buy to end up with 10 cards that each, independently, had a 1.5% odds of being pulled, per card. EDIT: In retrospect, there was no reason for me not to actually give the real example: https://steam.shadowverse.com/drawrates?pack_id=10001 If I have 10 legendaries, irrespective of what they were, roughly how many packs would I have had to buy to have a ~50% and ~90% confidence. Sorry if this was unclear. Last edited by Thenakedgun; May 6th, 2017 at 12:41 AM. Reason: Updating w/ Link
 May 6th, 2017, 08:44 AM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,364 Thanks: 1270 I'll take a look at that link time permitting but right off there's a problem. $\dfrac{1}{0.015} = 66.\bar{6} \not \in \mathbb{N}$ so the probabilities must be non-uniform.
 May 6th, 2017, 06:16 PM #6 Senior Member   Joined: Jun 2014 From: USA Posts: 493 Thanks: 36 The proper way to address this question is to consider a binomial distribution. The probability mass function follows from a binomial distribution: Let $p$ = your chances of success when drawing a card Let $n$ = the number of independent cards drawn When a random variable $X$ follows a binomial distribution with parameters $n \in \mathbb{N}$ and $p \in [ \,0,1] \,$ such as this, then the probability of getting $k$ successes out of $n$ cards is given by the probability mass function: $$Pr( \,X = k) \, = \binom{n}{k}p^k( \,1 - p) \,^{n-k} = \frac{n!}{k!( \,n-k) \,!}p^k( \,1 - p) \,^{n-k}$$ The cumulative distribution function is...: $$Pr( \,X \leq k) \, = \sum_{i = 0}^{z}\binom{n}{i}p^i( \,1 - p) \,^{n-i}$$ ...where $z$ is the greatest integer less than or equal to $k$. The cumulative distribution function can also be expressed as an integral: $$Pr( \,X \leq k) \, = ( \,n - k) \,\binom{n}{k} \int_0^{1-p}t^{n-k-1}( \,1 - t) \,^kdt$$ https://en.wikipedia.org/wiki/Binomial_distribution edit: PS - When $n$ is large enough, an approximation of can be made using the normal distribution. Last edited by AplanisTophet; May 6th, 2017 at 06:28 PM.
May 6th, 2017, 06:26 PM   #7
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Quote:
 Originally Posted by AplanisTophet The proper way to address this question is to consider a binomial distribution. The probability mass function follows from a binomial distribution:
he's got M different cards though, and 10 of those that he's interested in.

Even if you repartition the cards into the 10 he's interested in and the rest he's not, you still have to distinguish the success of getting each of the 10 cards. It's quite likely you'll get 2 or more of one of the 10 before getting all 10 of them.

A believe a multinomial distribution is the way to go.

May 6th, 2017, 06:32 PM   #8
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Emphasis mine:

Quote:
 Originally Posted by Thenakedgun The odds of me pulling one of the cards I want from this set is a constant 1.5% per card. If I have 10 copies of the card that I wanted presently, how many packs would I have had to open? I realise that there won't be a set number due to the nature of the question, but let's make it: "The odds that I'd have 10 copies after x packs is 50%" and "The odds that I'd have 10 after y packs is 90%"
Quote:
 Originally Posted by romsek he's got M different cards though, and 10 of those that he's interested in. Even if you repartition the cards into the 10 he's interested in and the rest he's not, you still have to distinguish the success of getting each of the 10 cards. It's quite likely you'll get 2 or more of one of the 10 before getting all 10 of them. A believe a multinomial distribution is the way to go.
He wants 10 of the same type of card I believe (a legendary card)... He's happy with 10 of the same legendary card or 10 different legendary cards, but he's got a 1.5% success rate per card.

Last edited by AplanisTophet; May 6th, 2017 at 06:34 PM.

May 6th, 2017, 06:40 PM   #9
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Quote:
 Originally Posted by AplanisTophet Emphasis mine: He wants 10 of the same type of card I believe (a legendary card)... He's happy with 10 of the same legendary card or 10 different legendary cards, but he's got a 1.5% success rate per card.
ahh, if that's the case then your analysis is correct.

but he does use the phrase "10 of these 500". That does make me think he wants 10 different types of cards.

Perhaps OP can shed some light. I'm not familiar enough with MTG or whatever this is to know if wanting 10 of the same card makes sense.

May 6th, 2017, 07:07 PM   #10
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Quote:
 Originally Posted by romsek ahh, if that's the case then your analysis is correct. but he does use the phrase "10 of these 500". That does make me think he wants 10 different types of cards. Perhaps OP can shed some light. I'm not familiar enough with MTG or whatever this is to know if wanting 10 of the same card makes sense.

Yeah, I caught this too (emphasis mine):

Quote:
 Originally Posted by Thenakedgun If I have 10 legendaries, irrespective of what they were, roughly how many packs would I have had to buy to have a ~50% and ~90% confidence. Sorry if this was unclear.

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