I know this has been asked before, but I think not in this exact way, so here goes:

Suppose you're going to bet on the flip of a coin. Your bet is always "HEADS", but the amount of your bet may vary, from 1 dollar to 100 dollars. You have deep pockets, so you can bet on thousands of flips.

The first 99 flips were heads. I know that the probability of the coin coming up heads is 0.5, but the probability of the flipper flipping 100 in a row is 1/2^100.

Assuming you can make dozens, or hundreds of bets, shouldn't you be betting more on heads at this point? Considering that head/tails average will approach 0.5 for large n, wouldn't it be reasonable to start betting 2 dollars instead of 1 dollar on heads? You're not assuming heads is "due", but at some point, with a fair coin, the probability of H/T can be expected to average out.

To make this more obvious, suppose the first 99,999 flips have been tails, and the coin is fair. The probability of 100,000 tails is the same as 99,999 tails + 1 heads: 1/2^100000, but if the law of large numbers is correct there must be a time at which it's smarter to bet 2 dollars on heads than 1 dollar on heads. If the wager should be different, is there a way to tell how much larger or smaller the wager should be?

I know that the law of large numbers speaks more to the dilution of a run in the overall account, but it would seem that when you're generating a large data set, that you should start regressing to the mean at some point. I've read a little about Bayesian analysis, and I naively assumed that the conditional probability of heads after a run of 99,999 tails should be higher than 0.5. I'm sure I'm wrong, but I need it explained, I guess.

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    $\begingroup$ A Bayesian outlook would compute a revised estimate for the probability of head, and after $99,999$ tails would sensibly revise the estimate of the probability of tail upwards. $\endgroup$ Commented Jun 24, 2014 at 4:05

2 Answers 2


If the coin is fair then it doesn't matter which side you bet on, the odds are the same. But if you witnessed 99 flips and all were heads you might question your assumption that the coin is fair and act accordingly.


Here are a couple of comments which may help.

First, a probabilistic view. Let $E$ be the event that the coin turns up heads $100$ times in a row. Then it is true, as you state, that $$P(E)=\frac{1}{2^{100}}\ ,$$ which is very small. The thing is, however, that if you have already seen $99$ heads turn up, this probability is irrelevant. What you should be considering is $$P(E\,|\,\hbox{the first $99$ tosses are heads})\ ,$$ and this is just $\frac{1}{2}$.

Second, a "behavioural" view. Suppose that you have just seen heads turn up $99$ times and are considering whether to adjust your bet accordingly. Would it make any difference to your decision if you then learned that the previous $99$ tosses, before you arrived, had all been tails?

Third, although it is true, as you say, that the expected ratio of heads to total tosses is $\frac{1}{2}$, it is not true that the expected absolute difference between the number of heads and tails is $0$. On the contrary, it is roughly $\sqrt n$, where $n$ is the number of tosses. If you get $99$ heads in a row, there is no reason to expect a future swing towards tails. Indeed, precisely because you expect heads and tails equally often, you would then expect that after $1000099$ tosses, the proportion of heads would be about $\frac{500099}{500000}$, which is very close to $\frac{1}{2}$.

Hope this helps.

  • $\begingroup$ Haha I'd like to make a quibble. The difference between the number of heads and tails is itself a random variable, and we shouldn't "expect" (in the English sense) it to be $\sqrt{n}$. Moreover the expectation of the difference is not $\sqrt{n}$ either but some constant multiple of it. After all the heads and tails switch leading infinitely often with probability 1, and at each switch the difference is 0. Thus after 1000099 tosses the proportion of heads would not be about $\frac{500099}{500000}$ because the 6 significant figure precision is incorrect. $\endgroup$
    – user21820
    Commented Jun 24, 2014 at 4:52

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