Estimate how many times to flip a coin to get at least 30 heads with probability of 80%

Question

Im completely stumped by this problem. It goes as follows:

Estimate how many times a fair coin must be thrown in order to obtain at least 30 heads with a probability of 0.80.

Ive tried playing with the binomial formula to no success. As well as the negetive binomial distribution until realizing that that wouldnt be helpful.

Any help is greatly appreciated!

Answer 1

HINT: Use the normal approximation. If $X$ is the number of heads in $n$ tosses, the mean of $X$ is $\frac{n}2$, and the standard deviation is $\frac12\sqrt{n}$. Checking a normal table, I find that about $80$% of the area under the normal curve is to the right of $z=-0.84$, so you want $X=30$ to be about $0.84$ standard deviations below the mean. Use that information to set up an equation in $n$ and $\sqrt{n}$, turn it into a quadratic in $n$, and solve for $n$. Then round up.

Answer 2

Let's calculate the probabilities of at least $30$ heads if we flip the coin $n$ times $$p(n)= 2^{-n} \cdot \sum_{k \ge 30} \binom{n}{k}$$ Note that $p(n)$ is increasing with $n$, $p(n) = 0$ for $n < 30$, $p(59) = 1/2$. Here is a table of values of $p(n)$, for $50 \le n\le 99$, with $6$ decimal points: \begin{eqnarray} \left( \begin{array}{cc} 50 & 0.101319\\ 51 & 0.131219\\ 52 & 0.165875\\ 53 & 0.205051\\ 54 & 0.248309\\ 55 & 0.295027\\ 56 & 0.344440\\ 57 & 0.395683\\ 58 & 0.447842\\ 59 & 0.500000\\ 60 & 0.551289\\ 61 & 0.600924\\ 62 & 0.648232\\ 63 & 0.692672\\ 64 & 0.733846\\ 65 & 0.771490\\ 66 & 0.805474\\ 67 & 0.835784\\ 68 & 0.862505\\ 69 & 0.885800\\ 70 & 0.905892\\ 71 & 0.923044\\ 72 & 0.937541\\ 73 & 0.949678\\ 74 & 0.959747 \end{array}\right)\ \ \ \left( \begin{array}{cc} 75 & 0.968025\\ 76 & 0.974774\\ 77 & 0.980230\\ 78 & 0.984607\\ 79 & 0.988090\\ 80 & 0.990842\\ 81 & 0.993000\\ 82 & 0.994681\\ 83 & 0.995982\\ 84 & 0.996981\\ 85 & 0.997744\\ 86 & 0.998323\\ 87 & 0.998760\\ 88 & 0.999088\\ 89 & 0.999332\\ 90 & 0.999514\\ 91 & 0.999647\\ 92 & 0.999745\\ 93 & 0.999817\\ 94 & 0.999869\\ 95 & 0.999907\\ 96 & 0.999934\\ 97 & 0.999953\\ 98 & 0.999967\\ 99 & 0.999977 \end{array} \right) \end{eqnarray}

We see that we need about $66$ tosses, just $7$ over $59$, the break-even point. Well, what do you know, $89$ tosses, the break-even point to get $45$ heads, insures at least $30$ heads with probability $0.999332...$.

Here is a graphic of the $p(n)$, $30\le n \le 100$.

Let's see how approximating with a normal distributions works. The binomial random variable $X_n$ has mean $n/2$ and deviation $\sqrt{n}/2$. The standardized variable $\frac{X_n - n/2}{\sqrt{n}/2}$ is approximately normal for $n$ sufficiently large. Therefore $$p(n) \colon = p( X_n \ge 30) = p(\frac{X_n - n/2}{\sqrt{n}/2} \ge \frac{30 - n/2}{\sqrt{n}/2}) \simeq p ( X \ge \frac{60}{\sqrt{n}} - \sqrt{n})$$ for $X$ a standard normal distribution.

$$p(n) \simeq q(n) \colon = p ( X \ge \frac{60}{\sqrt{n}} - \sqrt{n}) = 1 - \text{Cdf}_{N(0,1)} ( \frac{60}{\sqrt{n}} - \sqrt{n})$$

Note that $q(n)$ is strictly increasing with $n$. One notes that $q(66)=0.769910..$ and $q(67) = 0.803776...$. Hence, with the normal approximation, we need about $67$ tosses. Note that in general $q(n)$ approximates $p(n)$ from below, so any number obtained with the normal approximation will be OK.