I hate her. She’s such a racist.
No, the point is that ANY two people have the same birthday. If you have 367 people, it is 100% certain that you will find at least two people sharing a birthday. They can’t all have been born on different dates.
Anyway, my favourite probability paradox:
Suppose you take three poker chips, and mark them las follows. On the first chip, you draw an** X** on each side. On the second, you draw an O on each side. On the third you draw an X on one sidre and an O on the other.
X-X
X-O
O-O
Now you play a game with your friend. You put the chips in a bag, your opponent pulsl one out without looking and put it down on the table. You then try to guess what is on the face down side. You and your oponent both stake a dollar. Guess right, you win both dollars, guess wrong and your opponent takes both dollars.
So, your opponent pulls a chip out, and an X is face up. What should you guess for the face down side? Should you guss X? should you guess O? Or maybe it doesn’t matter?
I’d guess X. The probability of the chip being chip #1 is 2/3 vs. being chip #2 is 1/3. This is just a variation of the Monty Hall Problem.
I agree with nivlac: there are three different Xs, of which two have an X on the other side.
Small nitpick: It should be m!/((m - n)! m[sup]n[/sup]) if n is sample size and m is number of outcomes.
I believe you’re correct. MikeS said he had switched the roles of m and n.
I’ve been lurking on this thread and the XX XO OO chip thing is driving me nuts. I don’t see why it wouldn’t be an equal likelihood of getting X or O. After all, once you pick out the chip and see there’s an X, it can be one of two possible chips, either XX or XO. You won’t know until you turn the chip over, so the probability is 1/2 XX and 1/2 XO… right?
The two Xs on either side of the XX chip are distinct, even though they’re indistinguishable. Imagine that one’s painted red, one’s painted blue, and the X on the XO chip is painted yellow–is it clear then that the answer is 2/3?
If you pick out a chip, a third of the time it’s going to be the XX chip. A third of the time it’s going to be the XO chip–but half of that time, it’s going to come up O.
So, the XX chip comes up X twice as often as the XO chip comes up X.
I see that half the time the XO chip will come up O, but once it’s come up X , I don’t get how that changes the probability of the situation in question.
I guess I’m just feeling dense today. I sort of see what you’re saying, that there are three different outcomes. Using the different color thing ultrafilter mentioned (let’s call the different Xs : X1 and X2 and X3, and assume one coin is X1/X2, and the other is X3/O)
X1/X2 (I see X1 face up)
X3/O (I see X3 face up)
X2/X1 (I see X2 face up)
and two of those options mean that I have the XX coin rather than the XO coin, so 2/3 probability.
But it still doesn’t freakin’ make sense 
There are two coins with Xs on them, and I know I’ve picked one of them, meaning I could have one of two possible coins. Considering that the exact same probability holds true for O (If you see an O, that means it’s more likely that you’ve gotten the OO coin than the OX one) it seems as if we’re reaching the conclusion that we’re much more likely to pick either the OO or the XX coin out of a bag, when of course we should have 1/3 chance of picking each. After all, if you did the same experiment and got an O, according to the above logic, you should have twice as much chance of having an OO in your hand than an OX… but the OX coin, in other words, seems much less likely to be picked! How can that be true if we’re picking randomly?
head hurts
In my post, I said each coin has a 1/3 chance of being picked. So, it’s not true.
It’s just that half the time, the XO coin comes up O. So, the XX coin comes up X twice as often–that’s what “half” means. 
I see your problem. Take two aspirin (if you’re not allergic to NSAIDs) and listen to some old Clapton
There are three X’s. Instead of them being identical, suppose we label them 1,2,3.
chips are :
X1-X2
X3-O
O-O
So, if you have an X upwards it could be
X1 up, X2 underneath.
X2 up, X1 underneath.
X3 up, O underneath.
So, 2 of the three options gives you an X underneath.
Another way of looking at it: You have 2 chances out of three of picking a double chip. Always bet that you got the double.
BTW, don’t feel so bad about not getting it right away. The problem is meant to confuse the uninitiated. That’s why it’s an interesting problem. As I mentioned before, if you want more pain/fun, google the Monty Hall Problem.
A few days ago I wrote a 2nd year engineering statistics final, which I probably failed. This thread is reminding me of why I hate stats so much. Ugggh :smack:
There are three coins. Of them, two are the same on both sides. So you probably got one that was the same on both sides.
If I recall my statistics correctly it’s ( 364/365 + 363 / 365 + 362 / 365 … ) and so on until you get the percentage you want (say, 50%).
It’s an example of the fact that human beings generally aren’t very good with thinking intuitively about probabilities. Steven Pinker devotes a section of his wonderful book How the Mind Works to this. He says:
One example is the so-called gambler’s fallacy. If a roulette wheel comes up red five times in a row, lots of people think it’s due to come up black. This is wrong; the wheel has no memory, and each spin is independent of what happened before. As a rule of thumb, using past history to predict future events is a good survival tool; lots of things in nature work this way. But when something has been specifically designed to behave randomly, the rule breaks down.
Seeing as the first two terms add up to almost 2, I think your recall’s a bit off.
Well, this has been a fun thread. Thanks for the replies… I think I’m getting it. (Maybe.)