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When a fair die is flipped $72,000$ times, with what probability can we assert that the number $5$ occurs between $11,500$ and $12,500$ times?

Work: Would it be: $$\sum_{k=11501}^{12499} \binom{72000}{k} \left(\frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{72000-k}?$$

gebruiker
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Aymy
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  • That would be the exact, but difficult way. Have you heard about the central limit theorem? – Arthur Mar 21 '16 at 15:53
  • No I haven't I will look it up – Aymy Mar 21 '16 at 15:53
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    It will be binomially distributed. The normal approximation to the binomial will be very close to accurate. Note: $(72000/k)$ is not $\binom{72000}{k}$ the first is a division, the second is the binomial coefficient. Very different things. – JMoravitz Mar 21 '16 at 15:53
  • Second thing to note about typesetting, $\sum\limits_{k=11501}^{12499}$ is what you want, not what you wrote. Rightclick someone else's math and select "Show math as > TeX commands" to see how to type it the same way. Third thing to note, if your exponent is longer than one character, it needs to be enclosed in braces. $(5/6)^{72000-k}$ – JMoravitz Mar 21 '16 at 15:54
  • The central limit theorem says that since you're throwing so many dice, the number of fives is roughly normal distributed. – Arthur Mar 21 '16 at 15:55

1 Answers1

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The exact probability of lying strictly between $11500$ and $12500$ will be as you mean to write:

$$\sum\limits_{k=11501}^{12499} \binom{72000}{k}(1/6)^k(5/6)^{72000-k}$$

This answer is not very useful to us though because it is incredibly tedious to compute any further, even with the help of a computer.

or between $11500$ and $12500$ inclusive changing the limits on the summation to instead be $\sum\limits_{11500}^{12500}$, the difference being whether we interpret the statement "between" as $11500<k<12500$ or $11500\leq k\leq 12500$.

Let us instead approximate the scenario using a normal distribution. To do so, we need to figure out two things about our actual distribution (on top of the fact that it is binomially distributed).

The mean (average, or expected value), and the standard deviation. We know from earlier work that the mean is $\mu = np$ for binomial distributions and that the standard deviation is $\sigma = \sqrt{npq}$

$\mu = np = 72000\cdot \frac{1}{6}=12000$

$\sigma = \sqrt{npq} = \sqrt{72000\cdot \frac{1}{6}\cdot \frac{5}{6}}=10000$

So, we expect the actual results to be well described by a normal distribution with mean $12000$ and standard deviation $10000$.

Letting $X$ be our original binomially distributed random variable counting the number of successes on the die roll, $Y$ be our related normal distribution that well approximates it, and $Z$ be the standard normal distribution, we have:

$Pr(11501\leq X\leq 12499) \approx Pr(11500.5\leq Y\leq 12499.5)= Pr(\frac{11500.5-12000}{10000}\leq Z\leq \frac{12499.5-12000}{10000})$

$=Pr(-0.04995\leq Z\leq 0.04995)=Pr(Z\leq 0.04995) - Pr(Z\leq -0.04995)$

These values can be looked up on a table like any of these.

$\approx 0.5199 - 0.4801 = 0.03984$

Note: we had to shift by $.5$ when converting the probability from $X$ to $Y$ because when looking at the picture for the histogram, the original question asks us to add the areas of the rectangles. The rectangles themselves are width $1$ and are centered on the integers. Had we used the same bounds for $Y$ as we did for $X$, we would have ignored the left half of the rectangle on the left, as well as the right half of the rectangle on the right. While it may not be noticeable in a problem with such large numbers here, in a smaller example it could significantly change the outcome.

Also note: the normal approximation to the binomial will only give good results in the case that both $np$ and $nq$ are at least five. If either is too low, then the approximation will give a significant estimation to achieving a negative value, which is of course impossible for a binomial distribution to achieve, thus skewing all results.

JMoravitz
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