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Why doesn't a previous event affect the probability of (say) a coin showing tails?

Let's say I have a fair and unbiased coin with two sides, heads and tails.

For the first time I toss it up the probabilities of both events are equal to $\frac{1}{2}$. This much is intuitive and clear to me.

Now suppose that I toss it up $1000000000$ times and the scores are,

$501000000$ Heads

$499000000$ Tails

Now, for the $1001000000^{th}$ toss, shouldn't the probability of a tail coming up be greater than that of heads showing up?

I have seen many books which say that even for the $1001000000^{th}$ toss, the probabilities of both events are equal to $\frac{1}{2}$.

This seems wrong to me since the same books affirm that if a coin is tossed a large number of times, the quantity $\frac{heads}{tails}$ will approach $1$.

I know this is very elementary and naive, yet I had only superficially studied probability and I hope you all will bear with me.

My Objections with some of the top-voted answers

It isn't that future flips compensate for the imbalance, it is that there are so many of them it doesn't matter.

I don't get this statement. What exactly does the second sentence mean? Moreover, if what you said is true then, the following comment by a user should be wrong,

Law of large numbers

So these are contradicting each other I feel. Please bear with my lack of knowledge.

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  • $\begingroup$ Four words: law of large numbers. $\endgroup$ – Sean Roberson Jul 11 '17 at 3:36
  • $\begingroup$ You seem to think that the probability of a event should change so that the counting ratio tends to whatever it should in long run, right? I don't see any reason why this should be true, especially for $\textbf{independent}$ events but i've seen many proofs of the law of large numbers... $\endgroup$ – Nap D. Lover Jul 11 '17 at 11:02
  • $\begingroup$ And of course, no matter how many times you toss, those ratios will never be equal to exactly what they're limiting to—that's part of the definition of a limit—but they approach them arbitrarily close. $\endgroup$ – Nap D. Lover Jul 11 '17 at 11:05
  • $\begingroup$ To the edit: no they are not contradictory. If you look at my example, even if heads are ahead by $100$ after the first $100$ flips you expect the ratio $\frac {heads}{tails}$ to be close to $1$ after a million flips. The standard deviation of the number of heads in a million flips is $500$, so the error of $100$ is quite small. If that isn't small enough to ignore, think about a trillion flips. $\endgroup$ – Ross Millikan Jul 11 '17 at 13:59
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Your question mentions a 'fair and unbiased coin.' That very concept means a coin that on any given flip has a $0.5$ probability of heads and a $0.5$ of tails on any given flip, $\underline{\rm{independently}}$ of what has happened on any other flip.

To understand the statement 'It isn't that future flips compensate for the imbalance, it is that there are so many of them it doesn't matter,' let's think about an example: You want to make a $50-50$ mixture of oil and vinegar. But accidentally you start with $1$ cup of oil and $2$ cups of vinegar. Fortunately you want a large amount of the mixture, so you add $1000$ cups of oil and $1000$ cups of vinegar. You haven't fixed the imbalance of $1$ extra cup of vinegar; but you have added so much of the proper mixture, that you hardly notice the imbalance. (Your mixture went from about $66.67\%$ vinegar to about $50.02\%$ vinegar, even though there is still one full extra cup of vinegar.)

Things aren't quite as clear with the coin-flipping situation because there is no certainty that you will get exactly $1000$ heads and $1000$ tails in the next $2000$ flips; but it is spectacularly likely that the ratio of $2000$ flips will be a lot closer to $50-50$ than the first $3$ flips (that's where the law of large numbers comes in).

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The assumption for a coin is that is has no memory. That means that the chance of heads is the same on every toss. For a fair coin, that chance is $\frac 12$ regardless of the history. If you toss $100$ times and get heads every time (very unlikely, but it could happen) the most probable event after a million tosses (including the $100$ you already did) is $500050$ heads and $499950$ tails. It isn't that future flips compensate for the imbalance, it is that there are so many of them it doesn't matter. Look how close the head/tail ratio would be to $1$ at that point.

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  • $\begingroup$ Pls see edits in qn,thanks $\endgroup$ – Agile_Eagle Jul 11 '17 at 9:07
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Even in the 1000th time with a 999 tails, 0 heads, the probability of head is 1/2. But the probability that you get 1000th times tails in a row is 0.5^1000, 9.332636e-302.

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  • $\begingroup$ Exactly so how is that possible? How can a same event have different probabilities? $\endgroup$ – Agile_Eagle Jul 11 '17 at 3:29
  • $\begingroup$ also see the edit $\endgroup$ – Agile_Eagle Jul 11 '17 at 3:32
  • $\begingroup$ Pls see edits in qn,thanks $\endgroup$ – Agile_Eagle Jul 11 '17 at 9:08
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Imagine one tosses $1,000,000$ fair coins at once. Hence the outcomes of the coins are independent and then the numbers of heads and tails will be approximately (if not exactly) equal.

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  • $\begingroup$ pls see edits in qn, thanks $\endgroup$ – Agile_Eagle Jul 11 '17 at 9:07
  • $\begingroup$ I've seen your 3rd edits. You are wondering why $n$th toss still has $1/2$ probability of head or tail rather than to have more or less depending on the previous result. The thing is $n$th toss does not depend on the previous tosses. In order to get rid of this dependence, I suggested thinking of tossing the fair coins (as many as you want) at once. So there is no previous, there is only present. I hope it helps. $\endgroup$ – farruhota Jul 11 '17 at 9:34
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The ratio of heads to tails will approach $1$ as the number of times it's tossed tends towards infinity. If you've tossed the coin $1000$ times, you have hardly tossed it at all.

Another way to think about it: we agree that the first time you toss the coin, the probability of heads is $1/2$. Suppose you toss the coin $1000$ times, and you have $501$ heads and $499$ tails. You ask your friend to toss the coin for you. He does not know that you have already been tossing the coin. What is the probability that he will get heads?

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  • $\begingroup$ see edits @D_S changes to 1000000000 $\endgroup$ – Agile_Eagle Jul 11 '17 at 3:32
  • $\begingroup$ You have still hardly tossed it at all. See my edit. $\endgroup$ – D_S Jul 11 '17 at 3:41
  • $\begingroup$ I think the probability is justifiably less than 1. $\endgroup$ – Agile_Eagle Jul 11 '17 at 9:07
  • $\begingroup$ Pls see edits in qn,thanks $\endgroup$ – Agile_Eagle Jul 11 '17 at 9:07
  • $\begingroup$ You have said "For the first time I toss it up the probabilities of both events are equal to $1/2$. This much is intuitive and clear to me." Now, based on the second paragraph I wrote, you should also agree that this is true on the $1000$th time you toss the coin. $\endgroup$ – D_S Jul 11 '17 at 16:15
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The probability will always be 1/2 for each coin toss because the event physically has no tie to the previous throws. It will approach 1 because, as you approach infinity, the difference between the number of heads and tails will become smaller. The difference between heads and tails will become smaller because the probably for each are equal to each other. So, when you have 10000000000003 heads vs 10000000000017 tails, you get 0.99999999999860000000000238 or very close to 1. Just keep going infinitely and you will get closer and closer to 1.

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  • $\begingroup$ Pls see edits in qn,thanks $\endgroup$ – Agile_Eagle Jul 11 '17 at 9:07

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