probability of randomly choosing 2 correct items out of 4 in a 10 item sample size

Question

I am attempting to see what the probability of randomly selecting 2 out of 4 top S&P sectors out of 10 possible sectors to choose from when you have 4 guesses to get the 2. I have a method I have used to try and predict the top 2 sectors that is able to accurately select at least 1 of the top 4 92% of the time, and select 2 of the top 4 a little over half the time when I have 4 guesses. I am looking to see how these percentages compare to the probability of selecting at least 2 of the top 4 sectors out of 10 with 4 guesses by simple random selection. Hopefully this comparison will show me the effectiveness of my process.

Answer 1

This is a textbook example of the hypergeometric distribution, one of the very first distributions taught in any introductory course in probability or statistics. With $N$ total balls, $K$ of which red and the remaining $N-K$ of which are white, if we were to pull $n$ of those balls without replacement the probability that exactly $k$ of them are red and the remaining $n-k$ of them are white will be given by the random variable $X$ counting the number of red balls pulled where

$$\Pr(X=k) = \dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$

Here in your specific problem, that is $N=10,~ K=4,~ n=4$

The table of results, both for the probability to get exactly $k$ and to get at least $k$ will be:

$$\begin{array}{|l|r|r|r|r|}\hline k&\Pr(X=k)&\Pr(X=k)\approx\square&\Pr(X\geq k)&\Pr(X\geq k)\approx \square\\ \hline 0&\dfrac{\binom{6}{4}\binom{4}{0}}{\binom{10}{4}}=\dfrac{1}{14}&0.0714&1&1\\ 1&\dfrac{\binom{6}{3}\binom{4}{1}}{\binom{10}{4}}=\dfrac{8}{21}&0.381&\dfrac{13}{14}&0.9286\\ 2&\dfrac{3}{7}&0.4286&\dfrac{23}{42}&0.5476\\ 3&\dfrac{4}{35}&0.1143&\dfrac{5}{42}&0.119\\4&\dfrac{1}{210}&0.0048&\dfrac{1}{210}&0.0048\\\hline \end{array}$$

You say your "method" gets at least one with probability around 92% and at least two a bit over half the time... it sounds like your "method" is not any better than guessing purely randomly.

Answer 2

There are $\left(\begin{array}{c}10\\4\end{array} \right) = 210$ ways to choose 4 items out of 10. You are interested only in those where you pick at least 2 (i.e. 2, 3 or 4) out of the fixed set of 4.

There are $\left(\begin{array}{c}4\\m\end{array} \right) \times \left(\begin{array}{c}10-4\\4-m\end{array} \right)$ to pick exactly $m$ desired items out of 4.

For $m=2,3,4$ we get

$\left(\begin{array}{c}4\\2\end{array} \right) \times \left(\begin{array}{c}6\\2\end{array} \right) = 90$ $\;\;\;\;$ $\left(\begin{array}{c}4\\3\end{array} \right) \times \left(\begin{array}{c}6\\1\end{array} \right) = 24$ $\;\;\;\;$ $\left(\begin{array}{c}4\\4\end{array} \right) \times \left(\begin{array}{c}6\\0\end{array} \right) = 1$

Thus you probability is: $$p = \frac{90 + 24 + 1}{210} \approx 0.54762$$