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The following is a question which I personally find badly constructed, and would like to gather some opinion on:

Suppose you select a coin from a pile that is thought to have probability of landing heads $p\sim\mathcal{N}(\frac12, \frac1{100})$. The probability of landing heads three times in a row lies within $(\frac1{N+1},\frac1N)$, where $N\in\mathbb{Z}^+$. Find $N$.

This makes absolutely zero sense to me. First of all, now $p$ has a range spanning $\mathbb{R}$ which is unacceptable for a probability. Furthermore, even if you just assumed that the probability matched the mean exactly, that would be $N=8$. Where would the $N+1=9$ come into play? Am I missing something completely, or is this a badly constructed question?

Ice Tea
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    It is not a perfect question and the probability $p$ is outside $[0,1]$ is less than $0.000001$. Perhaps you could treat this as a censored normal distribution if it makes the question easier, but I would not bother. The probability of heads falling three times in a row is $\int p^3 ,f(p),dp$, which is not exactly $\frac18$. The question is really asking you to calculate this or at least the integer part of its reciprocal – Henry Oct 16 '21 at 00:36
  • Have you thought about the Central Limit Theorem? You've omitted a word from the question ... "Suppose you select a coin from ..." The Central Limit Theorem applies to averages and to sums (since an average is based on a summation). The concept for theorem is that as you average more and more, the interval around the mean narrows. – Gwendolyn Anderson Oct 16 '21 at 02:21
  • @GwendolynAnderson CLT doesn't apply here. The canonical example for CLT is flipping the coin $n$ times for a large number of trials and the number of heads converges in distribution to $\mathcal{N}(np,np(1-p))$, which is not the case here. – Ice Tea Oct 16 '21 at 07:19
  • @Henry that makes a lot more sense! How would you rephrase the question? – Ice Tea Oct 16 '21 at 07:20
  • Oh "a pile." I see. No, you don't need a large number of trials, three trials is fine. The sample deviation will be $\frac{\sigma}{\sqrt(n)}$. You are dealing with a normal distribution so you don't need a large sample for the sum or average to also be normally distributed by the CLT. So you can form a confidence interval around the mean of $\frac12$ using the sample variance. The theory rests on the CLT. The mean of three tosses is $\frac12 \times \frac12 \times \frac12$ and the results vary around that by the sample deviation of how the mean of the coins drawn vary around $\frac12$. – Gwendolyn Anderson Oct 18 '21 at 02:28
  • I see that the sum of Normals is not going to work for solving this problem. I was expecting the answer to be more straightforward but it does seem that the integration is the only way to solve this for $P(HHH)$. – Gwendolyn Anderson Oct 20 '21 at 01:10

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Essentially the question is asking you to find $\mathbb E[p^3]$, which is not exactly $\frac{1}{8}$.

You do not need to do this precisely, as it wants the answer as the integer part of its reciprocal. This leaves room for some interpretation of the "normal distribution" for $p$. As stated, the probability $p$ is in $[0,1]$ would be $\Phi(5)-\Phi(-5) \approx 0.9999994266$, where $\Phi()$ is the cumulative distribution function of a standard normal distribution.

So it will not be too dramatic if we ignore this point and just find $\int\limits_{-\infty}^\infty p^3 \,f(p)\,dp$. If you do the calculus, this turns out to be $\dfrac{7}{50}=0.14$.

Alternatively, we could condition on $p$ being in $[0,1]$ and find $\frac{\int\limits_{0}^1 p^3 \,f(p)\,dp}{\int\limits_{0}^1 f(p)\,dp}=\dfrac{7}{50} - \dfrac{3}{20\sqrt{2\pi}\exp(25/2)(\Phi(5)-\Phi(-5))} \approx 0.139999777$.

Both of these are about $\dfrac{1}{7.143}$ so between $\dfrac{1}{7}$ and $\dfrac{1}{7+1}$.

Henry
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  • In R, something like 1/mean(rnorm(10^6,1/2,sqrt(1/100))^3) will give about 7.14 or 7.15, which is good enough for these purposes, so you do not need to do the calculus. – Henry Oct 16 '21 at 09:51
  • Do you have a conceptual explanation to offer as to why the chance of tossing three heads would be greater for these biased coins than for fair coins, when the distribution is even about the mean of $\frac12$? How do you think of this intuitively, outside of the calculations? Which outcomes will appear less than for fair coins? – Gwendolyn Anderson Oct 20 '21 at 01:14
  • @GwendolynAnderson Suppose you had a coin with probability $0.9$ of heads; then three heads has probability $0.729$. Meanwhile for a coin with probability $0.1$ of heads, three heads has probability $0.001$. The average of these is $0.365$, well above $0.125$. Something similar happens with all symmetric cases because the high probability cases do not fall as much and the low probability cases do not have far to fall. The original question here is symmetric so it is not a surprise it sees a similar bias, albeit smaller as it is closer to $\frac12 \to \frac18$ – Henry Oct 20 '21 at 01:21
  • Since it's symmetric it seems that the chance of two-of-a-kind would have to be less for the biased coins than for fair coins since it is the same thing as one-of-a-kind, interchanging H and T as identical distributions. I'll mull over the basis of this theory a bit, and its basis, since I do know how distributions bounded below by zero will have upward biases. Thanks for your insights. – Gwendolyn Anderson Oct 20 '21 at 01:57
  • I might be being stupid but is there a 'nice' way of doing the calculus for integrating the normal PDF multiplied by $p^n$? – DPI Nov 01 '23 at 12:30
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    @David If $X\sim N(\mu, \sigma^2)$ then the raw third moment is $E[X^3] = \mu^3+3\mu \sigma^2$ (see https://math.stackexchange.com/a/4030443/6460 for others - you can use integration by parts), which here is $(\frac1{2})^3+3\times \frac12\times\frac1{100} =0.14$. – Henry Nov 01 '23 at 12:43
  • @Henry thanks, I thought you were doing the integral analytically as if it was some well known thing haha – DPI Nov 01 '23 at 12:55
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    @David - do it once in your life for $E[X^0]$ (hard) to get the coefficient in the normal density, then again for $E[X^1]$ (by substitution) and $E[X^2]$ (by parts) to demonstrate the mean and variance, then once more for a recurrence for $E[X^n]$ (by parts) and then never do it again but instead look it up. – Henry Nov 01 '23 at 13:02
  • @Henry I seem to remember doing it for $E[X^0]$ in an undergrad class (not actually related to prob/stats, but the integral came up). Sadly that was 8-10 years ago aha! – DPI Nov 01 '23 at 16:11