In a system where 2 events are equally likely, is there a limit to the # of times 1 outcome can appear in seq?

Question

So, let's take a "fair" quarter. We would expect heads and tails to both come up 1/2 the time. Basic math says the chance of getting 10 heads in a row is (1/2)^10. I know this. BUT, are there any theories that say no, if p(A) = 0.5, in a fair system you will NEVER see more than X in a row, pragmatically?

Answer 1

How long does it take you to flip a coin?
If the odds of getting x heads in a row are so low that it most likely will take you longer than the age of the universe to get that many heads in a row, then it is fair to say that it will never happen.

And as for hcbiochem's answer, if I have gotten 20,000 heads in a row and I now flip the coin one more time, the odds of getting a head are still 50%.

Answer 2

It is never impossible to get x number of heads in a row. Each new coin toss has the same odds, not depending at all on the outcome of the previous toss. So it is always mathematically possible to get 10 heads in a row, or even 100. But not very likely!

Answer 3

there is no absolute limit. As you flip the coin more and more, the chances of getting the same outcome every time decreases. It will never be zero, but it will approach zero. You are essentially cutting your chances in half every time you do it again. You can cut a number in half infinitely many times and it will never reach zero. It will be very close to zero, but it will never reach it.

Answer 4

No there isn't any proof that I'm aware of. As you say, the probability of an additional heads just decreases with each flip of the coin.

Answer 5

No there is no limit but the probability (at the outset, i.e before any events have occurred) of getting a sequence of x events rapidly diminishes as x increases, never however, reaching zero.

Bramble.