The chance of picking a red sock out of a drawer of infinite socks is $1\over3$ and the chance of picking a blue sock is $2\over3$

What's the chance that if I pick $20$ socks out of these, $19$ are blue?


I tried to find the probability of $P(\text{Blue} = 19 \text{ & Red} = 1)$ and multiplying it by the number of ways this could happen.


$${P(\text{Blue} = 19 \text{ & Red} = 1)} = {2\over3}^{19} \cdot {1\over3}^1 = 0.00001504$$ Permutations: $\frac{20!}{19!} = 20$

Solution $= 20*0.00001504.$

I know this is wrong because I tried the above procedure with $P(\text{Blue} = 6 \text{ & Red} = 3)$, which intuitively should work out to 1, but did not get the result.

What am I missing here?

  • $\begingroup$ Your calculation seems right to me. Can you explain why P(blue = 6 and red = 3) should be 1? I don't see why that should happen, unless I am missing something. $\endgroup$ – Abhiram Natarajan Nov 17 '17 at 20:34

Your method is correct. Something is wrong with what you're saying here:

I tried the above procedure with P(Blue = 6 & Red = 3), which intuitively should work out to 1

You seem to be saying that if you take 9 socks then, since the probability of getting blue is 2/3, 2/3 of your 9 picks should be blue with probability 1.

This is not correct. If you pick 9 socks you can get any number of blues with non-zero probability. The expected number of blue socks will be 6. Which means if you pick 9 different socks infinitely many times, the average number of blue socks you get will be 6. You should look up expected value.

  • $\begingroup$ Thanks a lot for your answer. It appears I have mixed up some important concepts here. But am I right to say that P( blue = 6, red =3) should be higher than the probability of any deviation from that combination? i.e. (blue = 5, red = 4) $\endgroup$ – Bananafish Nov 17 '17 at 21:10
  • $\begingroup$ Yes, you're definitely right. The probability for this problem follows a binomial distribution. In your case, a trial is picking a sock, a success is picking a blue sock and the probability of success on a single trial is 2/3. Binomial distributions have a peak at their expected value and the probabilities get smaller the father you get away from it. $\endgroup$ – FullofDill Nov 17 '17 at 21:13
  • $\begingroup$ You're a gem. Thanks for being of great help to a recovering economist!... $\endgroup$ – Bananafish Nov 17 '17 at 21:20
  • $\begingroup$ No problem :). Best of luck to you. $\endgroup$ – FullofDill Nov 17 '17 at 21:27

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