# Probability of exact match of peak of expected distribution for 1000 fair coin flips.

In random events such a fair coin tosses, it is easy to predict over a long run of trials (flips) how many heads and how many tails should come up. However, very rarely does that number match the expected # exactly. For example, if you flip a fair coin $1000$ times you would expect about $500$ heads and about $500$ tails but you would be much more likely not to get exactly $500$ of either. So my question is what is the probability of getting exactly $500$ heads and $500$ tails out of $1000$ fair coin flips? It happens sometimes but not often so there should be a probability of it happening. If it is not the most probable outcome, then what is? $499$ + $501$, $498$ + $502$...?

I guess another way of asking this question is what % of the area under the bell curve for the graph of the possible outcomes for 1000 fair coin flips is exactly 500 heads? Is it something like a 5% chance for example?

Another example (for reference only) is random digits like in pi. If you take say $1$ million digits after the decimal point, you would expect $100,000$ of each of the digits $0$ thru $9$ but none of them in reality have exactly $100,000$ (although digit $8$ is very close with $99,985$ occurrences). It also seems that the larger the # of trials (coinflips or digits of pi for example), the harder it is to match the normal distribution exactly.

• What do you understand by "expected normal distribution"? – Henry Oct 2 '14 at 18:34
• I don't know if that is the right term but I mean the amount that probability would tell us should happen. For example with coin flips, they say 50% heads and 50% tails but for 1000 flips you rarely see 500 of each so if my question/title is not clear, how can I word it better? – David Oct 2 '14 at 18:35
• @David First example: You can approximate the binomial distribution by the normal distribution. The greater the sample size is, the better is the appoximation. See here:en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem – callculus Oct 2 '14 at 18:48
• To also address the particular case you asked, the number $X$ of heads has a binomial distribution with parameters $n=1000$ and success probability $p=1/2$, so that $$P(X=500)={1000\choose500}(1/2)^{1000}\approx 0.02522502.$$ – binkyhorse Oct 2 '14 at 18:52
• Well that explains why it is so rare. About 2.5%. – David Oct 2 '14 at 18:57

If you are considering a small enough number $N$ of flips, you can compute the probability directly of getting $H$ heads out of $N$ flips using the binomial distribution. In general, if the probability of getting heads for one flip is $p$ then the mostly likely number of heads will be either the integer above or below $Np$.
If you don't want to calculate with the binomial distribution exactly, and instead want to use a normal distribution, then still the most likely value will be $Np$ and so if you use the value of the normal distrubtion density at that value $Np$ rounded up or down to the nearest integer and multiply the density by 1, i.e. leave it unchanged (because #heads are in increments of 1), you will get a good estimate for the probability of getting $Np$ heads either rounded up or rounded down to the nearest integer. All other numbers of heads that are more than 1 away from $Np$ will have lower probability. However it will be very unlikely that you get exactly $Np$ heads rounded up or rounded down if $Np$ is large. If you want a confidence interval (i.e. I'm $95\%$ sure the number of heads will be within $K$ of $Np$) then you can use the standard deviation of the normal distribution approximation and just take plus or minus 2 or 3 standard deviations. You can look up the normal distribution to figure out the exact width of the confidence interval and the confidence amount when you take plus or minus 2 standard deviations or 3 standard deviations.