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One is blindfolded to choose 11 socks randomly, find the probability of choosing 11 reds in a row

resolution12meh

New member
Jul 6, 2019
2
I realise this is probably quite easy and basic but I just cant get comfortable with calculators to work it our with any certainty.

There are 100 socks in a drawer. 84 are Red. 16 are White. If you are blindfolded & have to choose 11 socks randomly, what is the odds or what it more likely:

1) Picking 11 reds in a row.

OR

2) Picking 10 whites and 1 red


I know the answer is 2, but just need the correct calculation, math and percentage behind each option.

Thanks in advance.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Why are you not able to do this yourself? As your title says, this is pretty basic!

There are 100 socks, 84 red. The probability the first sock chosen is red is 84/100. There are then 99 socks, 83 red. The probability the second sock chosen is red is 83/99.
There are then 98 socks, 82 red. The probability the third sock chosen is red is 82/98.

Get the idea? The probability that 11 red socks in a row are red is [tex]\frac{84(83)(82)\cdot\cdot\cdot (75)(74)}{100(99)(98)\cdot\cdot\cdot (91)(90)}[/tex].
We can write [tex]84(83)(82)\cdot\cdot\cdot(75)(74)[/tex] as [tex]\frac{84(83)(82)\cdot\cdot\cdot (3)(2)(1)}{73(72)\cdot\cdot\cdot(3)(2)(1)}= \frac{84!}{73!}[/tex] and write [tex]100(99)(98)\cdot\cdot\cdot (91)(90)[/tex] as [tex]\frac{100!}{89!}[/tex] so that answer can be written in the shorter form [tex]\frac{84!}{73!}\frac{89!}{100!}[/tex].


10 white and one red sock can be done similarly. There are 100 socks and 16 white socks. The probability the first sock chosen is white is 16/100. There are then 99 socks, 15 of them white. The probability the second sock chosen is white is 15/99. Continuing in that way, the probability the first 10 socks chosen are white is [tex]\frac{16}{100}\frac{15}{99}\cdot\cdot\cdot\frac{7}{91}= \frac{16!}{6!}\frac{100!}{90!}[/tex]. There are then 89 socks left, 84 of them red. The probability the last sock chosen is red is [tex]\frac{84}{89}[/tex]. The probability the first 10 socks chosen are white and the last sock chosen is red is
[tex]\frac{16!}{6!}\frac{100!}{90!}\frac{84}{89}[/tex].


That is the probability the first ten socks chosen are white and the last one red. A little thought should show that the probability "the first sock is red and the last 10 white" is exactly the same just with the fraction [tex]\frac{84}{89}[/tex] at the front instead of at the back. And, in fact, the red sock being chosen at any place among the 10 white is exactly the same calculation, just with the fraction [tex]\frac{84}{89}[/tex] in a different position among the white sock fractions.

So to find the probability of "10 white one red" in any order, we need to multiply that fraction by 10:
[tex]10\frac{16!}{6!}\frac{100!}{90!}\frac{84}{89}[/tex].

I will leave it to you to actually multiply those (do a lot of cancelling first!).
 

resolution12meh

New member
Jul 6, 2019
2
Thank you for the answer and its really appreciated. Unfortunately, as good with numbers financially as i am, the math behind your answer is complicated to me and im not 100% sure on working it out unfortunately. :-(

I would be super appreciative if someone could state in simple terms the 1 in x chance or the %age chance for each probability.

Thank you in advance anyone.
 

Gaffer

New member
Nov 22, 2019
1
HallsofIvy's explanation is correct but there are some mistakes in formulas. So, for the first case (picking 11 reds in a row) the probability is:

${P}_{1} = \frac{84!}{73!}\frac{89!}{100!}$

For the second case (picking 10 white and one red sock) using the same reasoning we get the probability of choosing 10 whites in a row:

$\frac{16!}{6!}\frac{90!}{100!}$

and because the probability the 11-th sock then chosen is red is 84/90 and also because it can be chosen before or after or in between of 10 red socks this probability must be multiplied by 11. So, as a result we have the following probability for the second case:

${P}_{2} = \frac{16!}{6!}\frac{90!}{100!}\frac{84}{90} \cdot 11$

Using a simple online factorial calculator we can get:

${P}_{1} = 0.131$, ${P}_{2} = 4.75e-9$

So the odds in favor of the first case are:

${P}_{1} / {P}_{2} = 27612527 / 1$

that is absolutely logical because there are much more red socks than white ones.