Another nice one that people tend to refuse to accept:
You are flipping a coin. Previously, you have established that the coin has equal odds of showing heads or tails; so no funny business there. You get heads ten times in a row. What are the odds of getting heads next coinflip?
And is this equal to the odds of getting eleven times heads in a row?
odds of the next coinflip are the same as any random coinflip, 50%. Interestingly, odds of eleven consecutive coinflips with equal result is 100%*((.5)^11) = ~0.05%. The difference lies in the fact that in the first case, the "history" has already taken place; it doesn't matter what result they gave; the information is irrelevant, I might just as well have stated (heads x3-tails x3 - heads) as history. Basically, I just kept flipping the coin, until variance gave me the desired history; if I try long enough it WILL happen eventually. While in the second example, I have more uncertainty: I don't have the freedom to try, try, try again; I am calculating the odds of getting it right in one try.
Same problem works with throwing a die, or black vs. red in roulette but you have to adjust for the odds of each option (1/6 vs. 1/2 for the die, and for roulette the odds are slightly less than 50% due to the green fields.)
And for those of you who think this stuff is merely done for fun - I'm an engineer, and I have done probability calculations similar to this in that function in the past. Most difficult part is explaining any counterintuitive results to management...
Actually, this is only true if you assume that the coin MUST be a 'fair' coin. However, ten consecutive heads is some evidence that the coin isn't fair. How much - I don't know. What's the prior given the situation that the coin is actually fair vs a trick coin or something? If you flip the thing a million times and it comes up heads every single one, then sure, if you know it is a fair coin, it's still 50-50, but I would start to suspect, and I think quite rightly, that it's somehow rigged.
I had a lot of fun thinking about this just now. Assume that the coin used is (apparently) a US quarter. If there is any funny business, then the fact that we got N = 1000000 heads in a row suggests that it was a double sided coin. Suppose that there are Q quarters in circulation, and additionally D double-sided quarters (strictly, double headed). Then Bayes' theorem gives us that the probability that a randomly chosen coin is fair given that it just gave N heads in a row is (1/2)^N*Q/( (1/2)^N Q + D ).
So when is the chance of the coin being fair 50-50? Working it out, we see that it's when D/Q = (1/2)^N. In other words, assuming that there are very few fake coins, the coin is more likely than not to be fake precisely if it is more likely that a randomly chosen coin in fake than it is for a legit coin to yield N heads in a row.
So yeah, with 1 million heads in a row, it would be FAR more likely that you happened across the world's only double-headed quarter in circulation.
EDIT: In other words, if you pull a quarter out of your pocket and flip N heads in a row, and there is a least one double headed quarter for every 2^N legit quarters in circulation, then you are more likely to have happened across a fake.
Is 10 flips enough? Well, 2^10 is only 1024, and I doubt that there is a double-headed quarter for every ~1000 quarters, so probably you witnessed the rare event of flipping a 10 heads in a row rather than the even more rare event of finding a double-sided quarter as the first one you pull from your pocket.
So how many many heads would you need for a fake to be more likely? Well, I don't have stats, so I'll make a couple assumptions for an upper bound. Assume that there is at least one fake in the world. And second, let's assume that there's less than one trillion dollars worth of quarters (I suspect that there is FAR FAR less than this). Then the log base 2 of 4 trillion is about 31, so flipping 32 heads in a row would tell you that more likely than not it's a fake.
