My Statistics Are Beyond Rusty . . .

[Yellow Mage]

Say I have a deck of cards. A really, really large deck of cards. Specifically, 100 million is the deck count. It consists of 1 million instances each of 100 different cards. They are, by the hand of God Himself, shuffled randomly, and you start to draw without looking. What is the minimum you would want to draw to have, within 99.9% certainty, at least 1 instance of each of the 100 different cards?

. . .

My neighbor is a biologist and she wants a very rough estimate on some super complicated situation involving antibodies of some sort. She came to me for assistance, but, well, see the title of the thread . . .

More importantly, though, how would I even go about solving this problem?

[TheGrandMom]

Re: My Statistics Are Beyond Rusty . . .

Jeez...God must have fricken huge ass hands!!

[Aeni]

Re: My Statistics Are Beyond Rusty . . .

Say I have a deck of cards. A really, really large deck of cards. Specifically, 100 million is the deck count. It consists of 1 million instances each of 100 different cards. They are, by the hand of God Himself, shuffled randomly, and you start to draw without looking. What is the minimum you would want to draw to have, within 99.9% certainty, at least 1 instance of each of the 100 different cards?

. . .

My neighbor is a biologist and she wants a very rough estimate on some super complicated situation involving antibodies of some sort. She came to me for assistance, but, well, see the title of the thread . . .

More importantly, though, how would I even go about solving this problem?

Isn't this a standard deviation problem?

Standard deviation - Wikipedia, the free encyclopedia

You may have to reapply it in another manner for your particular problem but you can get the gist of it from reading that article. Incidentally, if what you said is true, your neighbor is working on a problem that I think was recently published in Nature where a team was doing something similar and that particular result came up (the statistical numeration) But I think she should really seek someone from the math department at a local university for help.

[Feba]

Re: My Statistics Are Beyond Rusty . . .

What is the minimum you would want to draw to have, within 99.9% certainty, at least 1 instance of each of the 100 different cards?

The correct answer is to draw one hundred cards and let the power of bravado do the rest.

EDIT: Or use this

[ItazuraNhomango]

Re: My Statistics Are Beyond Rusty . . .

More importantly, though, how would I even go about solving this problem?

Can't help you too much, but the type of math you're looking for may be 'Probability' instead of 'Statistics'.

You may also want to rephrase the question as "What is the probability of having N different types of cards when drawing M number of cards when the chance of obtaining any particular type card is exactly 1%?"

[Icemage]

Re: My Statistics Are Beyond Rusty . . .

Your neighbor is a biologist and doesn't know how to do a probability analysis or where to get one? Oy vey.

With a deck of N = 100 million total cards of 100 types, each with 1M instances, to be absolutely certain that you have one of every card, you need to draw 99,000,001 cards.

99.9% certainty is a much trickier proposition; there's probably a way to represent it, but you'd probably have to ask a math professor, and it would take quite a bit of time.

Off the cuff (note: I'm NOT a mathemetician, and this is simply an approximation), with a pool of 1M of each card, for small numbers, each card you draw effectively does not deplete the available pool of cards to draw from, and so your chance of drawing any one specific card is roughly at all times 1 in 100, at least until you've drawn at least a hundred thousand cards or so - and in the vast majority of cases, it will remain roughly 1 in 100 for almost all cases. It's probably likely that you will get a full set of 100 quite often within the first hundred thousand draws.

More interestingly, you can probably simplify the problem by classifying the cards into "cards I have already drawn" and "cards I haven't drawn" for each draw. That still leaves it as an incredibly complex problem, but it reduces the order of complexity significantly.

You're going to have to be more specific about the exact numbers you're dealing with if you want someone to actually take a stab at solving this problem because it's a LOT of freaking work to arrive at it due to the scope of the numbers.


Icemage

[Mog]

Re: My Statistics Are Beyond Rusty . . .

I'd calculate the probability of that happening for you, but I don't think my calculator could handle 99999900!. Those are huge numbers you're dealing with.

[KingOfZeal]

Re: My Statistics Are Beyond Rusty . . .

Going off what Icemage started, couldn't you just say:

There needs to be a .1% chance that you do NOT have one of every card, or (10M * .001) 10,000 cards. So couldn't you just reduce the total from absolute certainty (the 99,000,001 you mentioned) by that much, giving 98,990,001 cards? That should give you a 99.9% chance of having at least one of every card. That is, there is a 1 in 10,000 chance that you drew every card EXCEPT one of them.

[Icemage]

Re: My Statistics Are Beyond Rusty . . .

Going off what Icemage started, couldn't you just say:

There needs to be a .1% chance that you do NOT have one of every card, or (10M * .001) 10,000 cards. So couldn't you just reduce the total from absolute certainty (the 99,000,001 you mentioned) by that much, giving 98,990,001 cards? That should give you a 99.9% chance of having at least one of every card. That is, there is a 1 in 10,000 chance that you drew every card EXCEPT one of them.

That's not how probability works.

For a simplification of the problem, you can look at it this way:

Each time you draw a card, one of two things happens:

(a) You drew a card you've never drawn before.

(b) You draw a card you already have a copy of.

The interesting thing about this, is that in case A, there are always 1,000,000 of that card in the deck when you draw the first one. and all of them shift into the "already drawn" category once one is drawn.

So on any given draw number N, there are 100M - N + 1 cards to be drawn. And of those, there are only multiples of 1M undrawn cards (i.e. after the first draw, there are exactly 99,999,999 cards left in the deck, with 99,000,000 of them undrawn. After the second draw, there are 99,999,998 cards left, with either 99,000,000 undrawn or 98,000,000 undrawn, etc.).

Still horrendously complex, but "manageable".


Icemage