Actually, it is a step beyond the birthday problem. We went through the birthday problem and moved it to cards to make the maths easier in another thread. But this goes a step further by limiting the number of matches.

This is quite easy, I reckon. I'll come back with the answer in a few mins.

Given k decks of j cards, we want the chances of n pairs coming up.

The number of ways of having a unordered pile of k cards (i.e. one from each deck) so that there are n pairs in the pile is:

jCn x (j-n)C(k-2n). jCn is counting the number of options for n pairs cards that are the same, then (j-n)C(k-2n) picks the remaining distinct cards.

To find the number of different ordered lines of k cards, we must find the number of permutations of k cards with n pairs in them
This is k!/2^n, k! being the number of permutations when all cards are different, and then divide by 2 for every pair.

So the number of different ways we can draw exactly n pairs is:

jCn x (j-n)C(k-2n) x k!/2^n.

So to find the probability, we just divide by the total number of possibilities: j x k.
leaving us:
P(n pairs from k decks of j)=jCn x (j-n)C(k-2n) x k!/(j x k x2^n)

Which can be simplified a bit, of course.

For 1 pair and 20 decks of 50, I got 1.697E+30

which is clearly dodgy, so I'm gunna have to check my spreadsheet.

Because it should be j^k as the total number of possibilities, silly me.

I wish I was as good as math... I am technically two years ahead ( algebra 2 freshman year) but i am not good at math by any means I guess I'm just good at passing classes...

As good as that at math is what I meant

I now get 0.08445968. Hopefully there aren't any lingering errors in that.

ok, for 20 decks of 52 cards, each with 4 cards of the same number... Usually a pair in a card game is like "a pair of 3's" so suit doesn't matter.

thus you have the equivalent system of 20 decks of 13 cards, and it is not possible for you to draw only one pair.

In the system decsribed above - 20 deck, one random card drawn from each deck, asusme you randomly draw an Ace from the first deck, 2, 3, 4, 5... from the next 11 decks. You draw a king from the 13th deck, and from the 14th deck you are guarenteed to make a pair. - I might point out that if you didn't want to use the normal definition of 'pair' you could probably have avoided tlakign about cards at all..

So let's assume you require that we match both number AND suit.
your chances of drawing a card from the 1st deck are 100%
you then have 19 chances of making an exact match - each chance is 1/52 - that is 19/52 chance of making a exact pair (at least)

Now given that you draw a pair, the specific card doesn't matter, and the order that you draw the cards in doesn't matter. Thus Assuming you draw a pair from any two decks - What is the chance that every other drawn card will be unique.

you've got 18 cards to draw, and each must not match either each other or the 1 pair you have drawn - is this an easier problem?

51/52 chance of the first card not matching the one pair you have.
50/52 of the next card still not making a pair...


so we have (19/52) * (51*50*49*48*47*46*45*44*43*42*41*40*39*38*37*36*35*34) / 52^18 - 0.008446 which i think doesn't take into account the number of combinations in which this can occur... em...

though that it matches ghostmaker's answer except by a factor of 10 suggests to me that i've got it right.

yeah, my 19/52, and 51*50*...*34 / 52^18 don't care what orders they happen in; but i'm not sure i fully understand the combination thing - would appreciate if anyone can explain it to me.

- not combinaiton, arrangments...

the alternative route to an answer is work out the probability of at least 1 match, then subtract the probability of more than 1 match.

at least one would be: draw card - second card matching is 1/52
assuming you you haven't gotten a match you add the chance the third card will match either: 2/52 by the probability that you haven't gotten a match (which is 1 - 1/52) = 2/52 * 51/52
assuming you still don't have a match, we add the chance of a match on the third card 3/52*(1 - 51/52 - 2/52*51/52 )... and so on. (this seems longer and harder to do - and is therefore wrong :P )

Couldn't you just do...

1-((51/52) x (50/52) x (49/52) x (48/52) etc, etc, to (33/52)) ?

No, that'd give you the chance that at least one pair was drawn.
If you want a single pair, uh...give me some time. I doubt I'll do this. But meanwhile, he's another one I'd like an answer to.

A test has 20 questions, all of them multiple choice. Each question has 5 answers; A, B, C, D, and E. I would like to know the probability of getting exactly 12 of them E's.

P(12 Es)=(0.2^12)*(0.8^8)*20C12=8.66*10^-5

Ghost, does that not sound very small?

12/20 is pretty high. so to satisfy my reasoning skills, the average would be 4/20.
once you start deviating from this average you very quickly move to unlikely results. (that would be the normal of the distribution) there is obviosuly 1 possible test with all answers being E, however there are ~95,360,000,000,000 possible different solutions, so 1 is very small and unlikely.

How do we get a feel for the likelyhood of 12 rather than 20? any suggestions?

Then again the odds are very small too.

For a specific orientation of the 12 Es and 8 Not Es, the odds are clearly (0.2^12)*(0.8^8). Then you just multiply by the number of orientations, giving what I have above.

Look up a binomial distribution table to check that one.

i'm still confused by when it is necceasry to multiply by the orientations or not :(

When you care about the order they come in, you don't, when you don't care, you do. :)

but i'm fairly sure there is a way to count when you don't care about the order which includes that multiplication as an addition... i think...