I’ve been trying to figure out what a general rule would be for the following type of scenario:
Say there are 10 cards face down, and 3 of them are spades. You randomly pick two. What are the chances of picking two spades?
I’d be interesting in knowing if there’s an equation for this sort of thing. Also, could you explain the logic or rationale behind the equation?
Thanks…
There are 3 ways to pick two cards such that you have two spades. There are 10 choose 2 ways to pick 2 cards. 10 choose two is defined as (I could be wrong, but here is what I remember) 10!/(8!2!).
So your total probability for choosing two cards from 10 where exactly 3 are spades such that you get 2 spades is 3/(10!/(8!2!))
This comes out to 3/45. About 7%. Someone please correct my math if necessary, but the idea is sound. You take the number of ways to achieve the outcome, divide by the total number of outcomes possible.
TheNerd has the correct answer – 3/45. This is according to the equation, which follows essentially from the longer, more straightforward process :
While it isn’t made explicitly clear, I’m assuming we’re picking the cards sequentially (i.e. like picking them up), or picking two separate cards at the same time (which would have the same result as picking sequentially). So it goes like this :
The odds of picking a spade first are 3 spades / 10 cards = 3/10. After the first spade is taken, there are 9 cards, 2 of which are spades. So the odds of getting a second spade are 2/9. Both these events must occur, so you multiply the probabilities together and get 3/10*2/9 = 3/45, which is the answer from the formula.
Deriving the formula is a bit lengthy, but perhaps you can see how it proceeds. (translation : I’m too lazy to do it right now.)
The point about picking the cards sequentially is the idea of “sampling without replacement” as opposed to “sampling with replacement”, which would be the case if you, say, picked a card, then put it back, re-shuffled and picked another card. The answer to this problem is left as an exercise for the poster.
Your math is perfect, Nerd. Another way of solving it is observing there is a three in ten chance that the first card you pick is a spade, and then a two in nine chance that the second will also be a spade.
3/10 * 2/9 = 6/90 = 3/45
Let me give a crack and coming up with the “rationale” beihnd the (10!/8!2!) comes from in a more general way.
What we are trying to ge tis the number of ways to draw n (2) cards from M (10) total.
Now suppose you needed to draw just one card. Well the number of posible draws there is easy…10. Now suppose we need to draw two, as in the example. Well there are 10 possibilities for the first card and nine for the second. So 10 times 9 (10!/8!) possibilites…not quite. Because that double counts picking a&b as well as b&a. So we need to divide by 2 (2!) in order to get to the 45 number we have in the answer.
Now let’s look at three cards. Well following our example above that is 10x9x8 (10!/7!). But agin we have counted omse more than once. a&b&c, a&c&b, b&a&c, b&c&a, c&a&b, c&b&a. So in this case there are six (3!), or generally we the first pick can be any one of 3, the second any one of two, and the third just one…3x2x1. So we divde by 6 to get the number in this case being 120.
So the first half is easy to see M!/(M-N)! because we are really just trying to keep one more possibility as we increase N. The other half 1/N! is to eliminate the multiple counting of the same combinations.
If the order in which the cards was selected mattered then you wouldn’t divide by N!.
Wow! Thanks for the clear, detailed replies, guys!
I had thought about the solution that Greg Charlespresents (which reduces the math nicely) but was confusing myself earlier, thinking that I couldn’t make the assumption that the first card would be a spade when drawing the second. Of course, since the question presumes that the first card drawn is a spade (otherwise you can’t get two spades), one can safely make that assumption.
Tretiak, thanks for going through the factorial math.
The simpler solution (3/10 * 2/9) involves a way of thinking about the problem in two steps, (probability for first card) times (probability for second). That’s what I was stuck on. It gets messy when it’s not clear how the first action may affect subsequent actions.
Now I see there is another heuristic that is more generally applied, (total number of matching cases) / (total number of cases).
To add a little something:
You’re finding the probability of two events. When the events are connected by an “or” (as in, draw a spade or a heart) you add the event probabilties. When the events are connected with an “and”, (as in your problem) you multiply. In your problem there’s a twist - the probability of the second event is dependednt upon the outcome of the first, so you have to take that into consideration.
A math-intensive reference for this stuff: http://www.math.uah.edu/stat/prob/prob5.html
(talking now to the general audence) This site has a graduate-level run down of the infamous Monty Hall Problem