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My friend and I are tearing each other to bits over this, hope someone can help.

Coin flip experiment:

Define a single trial as 10 coin flips of a fair coin. Perform an arbitrarily large number of trials. At some number of trials n, you notice that your distribution is extremely skewed in one direction (i.e., the "average" of your 10-flip sets is far away from 5 heads and 5 tails).

My reaction: Because you are guaranteed to hit a 5H/5T mean as n approaches infinity, the probability that the next n trials contains an equal skew in the opposite direction increases. In other words, given 2*n* trials, if the first n are skewed in one direction, than the remaining n are probably skewed in the other direction such that the overall distribution of your 2*n* trials is normal and centered around 5H/5T.

My friend's reaction: It doesn't matter if your first n trials is skewed, the next n trials should still represent an unmodified 5H/5T distribution regardless. The probability of the next n trials being skewed in the opposite direction is unchanged and low.

Who's right, and why?

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Let's say you're taking a bunch of measurements of independent process. Each measurement gives you a number: 1, 2 or 3. The average should be 2. The process is almost perfect, so you'll almost always get a 2.

You start taking measurements, and get the following results:

$$\{ 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 2, 2 \}$$

Should you now expect a series of 1s to "cancel" this out?

So far, your average is:

$$\frac{7\cdot 2 + 5\cdot 3}{12} = \frac{29}{12} \approx 2.417$$

Now, you take 100 more measurements; all are 2s.

$$\frac{7\cdot 2+5\cdot 3 + 100\cdot 2}{112} = \frac{229}{112} \approx 2.044$$

Now, another 1000 measurements, all 2s.

$$\frac{7\cdot 2+5\cdot 3 + 100\cdot 2 + 1000\cdot 2}{1112} = \frac{2229}{1112} \approx 2.004.$$

And so on.

In short, even though you got a string of 3s, you don't need a similar string of 1s to "cancel out" the effect. As your number of measurements goes to infinity, your average value will still get arbitrarily close to 2, which is what's expected.

The same can be said for flipping independent fair coins. Even given a string of heads, you don't need to get a similar string of tails to approach a 50/50 distribution as the number of trials go to infinity.

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    $\begingroup$ I like this answer because it addresses directly the confusion of the question. +1 $\endgroup$ – daniel Jul 26 '13 at 19:44
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Your friend is right. As alluded to by Daniel Fischer, the coins do not have any memory, therefore the earlier tosses are irrelevant.

If anything, if you should actually face such a situation with large $n$ you might start to doubt that the coins are fair, yet then you would rather expect the subsequent trials to be biased in the same way (not the opposite way).

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If you want to consider something really interesting, consider that as you add more flips, the probability of getting exactly half heads and tails goes down and this isn't that hard to compute mathematically as the sequence is the fraction of $2n \choose n$ divided by $(2n)^2$ for 2n flips. The denominator will grow as you increase the number of flips.

For 2 flips, this is 2/4 = 1/2.

For 4 flips, this is 3/8.

For 6 flips, this is 20/64 = 5/16.

Now, as this continues, it will keep diminishing as it is a specific outcome compared to be "near" the middle. Just something to consider as while you may get close to the middle, the exact middle keeps having lower and lower probability.

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Your friend is right.

Coin flips (at least the coin flips that most people think about) are independent of each other. The results of past flips don't effect the result of future flips. The reason that in the limit as the number of flips goes to infinity the distribution tends to 50% heads and 50% tails is that any given set of 10 flips (or any other finite number) has the same probability of having #heads/#tails = r as #heads/#tails = 1/r, so as you flip the coin many many times you will tend to have just as many sets skewed in one direction as the other.

This does not mean that consecutive, finite, sets of flips will be skewed in opposite directions because each set of 10 flips has exactly the same probability of giving any specific result. It is just that as the number of flips goes to infinity you will, statistically speaking, always end up with just as many sets of flips skewed in one direction as the other. Thus, saying that "because these first ten flips got more heads than tails, the rest of the flips we make will have more tails than heads" isn't accurate. Really what you should say is "Even though these first ten flips had more heads than tails, when we do 500 flips we're still very likely to get an almost 50 50 split because this skewed initial set will get averaged out over the many, many flips left".

It is fun to note that a set of coin flips like this is basically a random walk, and it can be shown that on average, in a set of $n$ flips the distribution will be skewed by about $\sqrt{n}$ in one direction or another. for 10 this is about 3 (we'll round to the nearest whole number) and for 500 this is about 22, however, 3/10 = .3 whereas 22/500 = .044 and as $n \to \infty$, $\sqrt{n}/n \to 0$. This is why small sets tend to be skewed more than larger sets. This is essentially because small sets of flips that make up the larger set tend to be skewed, but in opposite directions (more towards heads or more towards tails) so they have a tendency to cancel each other out.

That was kind of a long answer with a lot of words, I hope it helps.

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