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Probability of getting high heads total in long coin-toss sequence

If I flip a coin 300 times, what is the probability of getting head at least 200 times?

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  • 3
    $\begingroup$ Innuendo.${{}}$ $\endgroup$ – Git Gud May 28 '15 at 15:15
  • $\begingroup$ Sorry, trying not to burst out laughing right now. Anyways, what have you tried so far? $\endgroup$ – Brenton May 28 '15 at 15:15
  • $\begingroup$ Do you know about binomial distribution and its approximation with a normal distribution? $\endgroup$ – ajotatxe May 28 '15 at 15:16
  • $\begingroup$ See math.stackexchange.com/questions/346005/… . But you may use the normal approximation for calculating the probability (see Central limit theorem). $\endgroup$ – Stephan Kulla May 28 '15 at 15:18
  • $\begingroup$ I am new in probability theory, can anyone suggest me what are the steps of learning probability theory with proper source or link? $\endgroup$ – M S Hossain May 28 '15 at 15:20
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Note that the coin we assume is a fair coin: $p=\frac{1}{2}$

The number of times we flip the coin is: $n=300$

The number of times we are interested in achieving is $k=200$

Thus, the average is $\mu = np = 150$

The standard deviation is $\sigma = \sqrt{np(1-p)} = \sqrt{75} = 5\sqrt{3}\approx 8.66$

So, letting $X$ be binomially distributed $X\sim B(n,p)$ and $Y$ be normally distributed $Y\sim N(\mu,\sigma)$, and $Z$ be the standard normal distribution $Z\sim N(0,1)$ we use the normal approximation to the binomial:

$$Pr(X\geq 200) \approxeq Pr(Y\geq 199.5) = Pr(Z\geq \frac{199.5 - 150}{5\sqrt{3}})\approx Pr(Z\geq 5.7158)$$

I.e. it is the probability of appearing 5.7158 standard deviations above the mean.

This is a very unlikely event. Most tables don't even go that high. Using a calculator such as this one you find the probability is around $5.47\times 10^{-9}$

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