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I have following formula ("n" is number of experiments, "k" is number of successes that we expect, "p" is probability of success, "q" is probability of failure):

https://imgur.com/HXxITYZ

I used it to calculate chances of getting strictly 2 heads out of 4 tosses (order doesn't matter), 5 heads of 10 tosses, 10 heads out of 20 tosses.

2 heads out of 4 tosses have around 38% probability.

5 heads out of 10 tosses have around 25% probability.

10 heads out of 20 tosses have around 18% probability.

Looks like trend for me. And this trend is paradoxical, it means that by increasing number of tries we will decrease chance for 50/50 outcome. It seems to be at odds with what I was taught before. I was taught that as we increase number of tosses we will come closer and closer to 50/50 outcome. So if I toss a coin 10 times I can get 1 head and 9 tails, while tossing a coin 1000 times is more likely to give me result much closer to 50/50 outcome.

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  • $\begingroup$ What p and what q are you using? $\endgroup$ – Mariano Suárez-Álvarez Feb 13 '18 at 18:20
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    $\begingroup$ This is because throwing exactly 50% is not likely to happen most of the time, you can sometimes land around the 11/20 mark, or 9/20. These are still very close to 10/20, and on average you are still getting an even distribution. As you increase $n$, there are more numbers close to the 50% mark for your eventual count to finish on, so this probability is spread out between these values. $\endgroup$ – John Doe Feb 13 '18 at 18:21
  • $\begingroup$ Notice that that formula gives you the probability of getting exactly $k$ successes. As $n$ increases, it is harder and harder to have exactly $n/2$ tails. $\endgroup$ – Mariano Suárez-Álvarez Feb 13 '18 at 18:21
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    $\begingroup$ It's true that the probability of getting exactly 50-50 goes down. But the average deviation of from $.5$ gets smaller and smaller. When $n=10$ it's relatively common to see $.6$ or $.4$ (corresponding to $6/10$ or $4/10$ heads.) For $n=10000000$ it is pretty much impossible for $60\%$ or more of the flips to come up heads. $\endgroup$ – spaceisdarkgreen Feb 13 '18 at 18:33
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    $\begingroup$ Consider also that for an odd number of coin flips, the probability of exactly 50-50 is always zero $\endgroup$ – spaceisdarkgreen Feb 13 '18 at 18:36
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This is because the more you flip a coin, the less likely it will be that you get heads $exactly$ half of the time. It is true, however, that the more you flip a coin, and then compute

$$\hat{p}=\frac{\text{number of heads}}{\text{number of tosses}}$$

that $\hat{p}$ will $converge$ to $0.5$. Consider flipping a coin $10$ times and obtaining $6$ heads. Then consider flipping a coin $1000$ times and obtaining $526$ heads.

While in the second instance, we are $26$ heads away from a perfect $50/50$, proportionally, we are closer as $0.526$ is closer to $0.5$ than $0.6$.

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  • $\begingroup$ Please, add information about spreading out probability to your answer because it's incomplete without it. " As you increase n, there are more numbers close to the 50% mark for your eventual count to finish on, so this probability is spread out between these values." $\endgroup$ – user161005 Feb 13 '18 at 18:41
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    $\begingroup$ @user161005 are you talking about my comment on the question? This is not my answer.. $\endgroup$ – John Doe Feb 13 '18 at 18:42
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    $\begingroup$ Sorry, confused you with other guy. Silly me. $\endgroup$ – user161005 Feb 13 '18 at 18:52
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As it's been said, the fact that there are more and more possible values is what makes $\hat p=0.5$ less and less probable as exact value ($\hat p =\tfrac{\#heads}{\#toses}$).

Consider, for instance, the probability of $\tfrac 1 2 - \tfrac 1 {10}<\hat p_n<\tfrac 1 2 + \tfrac 1 {10}$, that is $$P\left(\tfrac 1 2 - \tfrac 1 {10}<\hat p_n<\tfrac 1 2 + \tfrac 1 {10}\right)=P\left(- \tfrac {2\sqrt n} {10}<\frac{\hat p_n-\tfrac 1 2}{\frac1{2\sqrt n}} < \tfrac {2\sqrt n} {10}\right)\approx 2\cdot\Phi\left(\tfrac {\sqrt n} {5}\right)-1$$ the approximation based on the CLT, valid for $n$ large enough.

Let's supose $n=50$. In that case we have $$P\left(\tfrac 1 2 - \tfrac 1 {10}<\hat p_{50}<\tfrac 1 2 + \tfrac 1 {10}\right)\approx 2\cdot \Phi\left(\frac{\sqrt {50}}5\right)-1\approx 0.8427.$$

Now let's do the same fot $n=100$. We get $$P\left(\tfrac 1 2 - \tfrac 1 {10}<\hat p_{100}<\tfrac 1 2 + \tfrac 1 {10}\right)\approx 2\cdot \Phi\left(\frac{\sqrt {100}}5\right)-1\approx 0.9545.$$ So the probability of $\hat p_n$ taking a value at a distance less than $0.1$ from the expected value has increased by around a 13%. But the values of $X=\#heads$ corresponding to that event went from $9$ (that is $21$ to $29$) to $19$ ($41$ to $59$), around the double. And if we kept doubling $n$ (for example), we would approximately double the number of integer values of $X$ that lead to $0.4<\hat p_n<0.6$ each time, but the total probability distributed between those values won't grow that much, fundamentally because it's very close to $1$, which is a bound, no matter $n$.

So, those probabilities which stay almost constant at some point, are distributed among more and more possible values of the measured discrete variable, and so is not surprising that the probability at each individual point decreases.

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