help with probability

[sao123]

Its been years since I did this in college, but now I have a simple problem to solve and im drawing a blank.

If I was making a MTG deck of 60 cards.
In this deck I have 4 identical cards, and 56 other cards I dont care about. During the course of a card game, I draw 25 cards. whats the probability I drew at least one of those 4?
I dont care what position I draw it in, just that I draw it.

I thought it would be
(4 choose 1) / (60 choose 32) + (4 choose 2) / (60 choose 32) + (4 choose 3) / (60 choose 32) + (4 choose 4) / (60 choose 32) =
but that gives me 1.44x 10-16 which is completely retarded.

how do i do this again?

 

[Freshgeardude]

p(4,1)/p(60,1) which is 4/60 which is 1/15


i believe

 

[sao123]

p(4,1)/p(60,1) which is 4/60 which is 1/15


i believe

thats if I draw 1 card, not 25.

 

[esun]

You have sort of the right idea but your denominator is off. This is a counting problem really.

So you want to count the number of hands of 25 cards that have at least one of 4 specific cards from the deck out of all possible hands of 25 cards.

You've broken it down into hands of 25 cards with exactly 1 of those cards, 2 of those cards, 3 of those cards, or 4 of those cards. Let's look at the case of having just one of those cards first.

So you have a hand containing 1 card from a set of four and 24 cards from a set of 56. How many such hands are there? Well that 1 card could be any of the four, so that's 4 hands right there (more generally, 4 choose 1). Then those 24 cards can be chosen any way (order doesn't matter) from those 56. So we have (56 choose 24) of those. That gives us (4 choose 1) * (56 choose 24) hands with 1 card from the set of four.

Generalizing, we have (4 choose 2) * (56 choose 23), (4 choose 3) * (56 choose 22), and (4 choose 4) * (56 choose 21) hands having 2, 3, or 4 of those cards we're looking for. So we sum all of those up first to get the total number of hands with at least one of those cards.

Now for the total number of hands, we simply have (60 choose 25), so we divide by that. This gives 0.892624607.

Note that this is related to the hypergeometric distribution: http://en.wikipedia.org/wiki/Hypergeometric_distribution

Here we have N = 60 (total number of cards), n = 25 (number of cards drawn), m = 4 (number of identical cards), and k = 1, 2, 3, 4 (probability of getting exactly 1, 2, 3, or 4 successes).

So you do P(X=1) + P(X=2) + P(X=3) + P(X=4) using the formula given and you should get the same result.

 

[Carom]

I agree. I had done it in Excel via
=HYPGEOMDIST(4,25,4,60)
=HYPGEOMDIST(3,25,4,60)
=HYPGEOMDIST(2,25,4,60)
=HYPGEOMDIST(1,25,4,60)
Sum= .8926 and was wondering that the probability was that high. It is indeed.

 

[sao123]

much thanks.

 

[KoolAidKid]

An easier way:

1 - probability of drawing none
= 1 - (56/60)*(55/59)*...*(32/36)
= 1 - (35*34*33*32)/(60*59*58*57)
= .8926

 

[Pheran]

Stats applied to Magic - this thread is so geeky it makes me smile.

Also, thumbs up to KoolAidKid for pointing out the better solution. Lots of stats problems are much simpler to solve if you calculate the opposite probability and subtract it from 1.

 

[gorcorps]

Its been years since I did this in college, but now I have a simple problem to solve and im drawing a blank.

If I was making a MTG deck of 60 cards.
In this deck I have 4 identical cards, and 56 other cards I dont care about. During the course of a card game, I draw 25 cards. whats the probability I drew at least one of those 4?
I dont care what position I draw it in, just that I draw it.

I thought it would be
(4 choose 1) / (60 choose 32) + (4 choose 2) / (60 choose 32) + (4 choose 3) / (60 choose 32) + (4 choose 4) / (60 choose 32) =
but that gives me 1.44x 10-16 which is completely retarded.

how do i do this again?

P(you dying a virgin) = 1.0

 

[sao123]

P(you dying a virgin) = 0.0

I have a wife and child.

 

[GodlessAstronomer]

Haha pwnt.

 

[classy]

I have a wife and child.

Look at little man, hehehehehehe, cute kid, congrats

 

[GodlessAstronomer]

That kid can raise one eyebrow. That's awesome, not many people can do that well.

 

[lxskllr]

Look at little man, hehehehehehe, cute kid, congrats

He looks old in that photo. He looks like he's sitting there silently judging :^D

 

[GodlessAstronomer]

Babby picture fixed.

 

[lxskllr]

He needs poker cards in his hands, a glass of whisky, and a cigarette burning in an ashtray :^D

 

[bommy261]

lol