# Even odds

This evening, I found a nice probabilistic puzzle on http://www.futilitycloset.com/“A bag contains 16 billiard balls, some white and some black. You draw two balls at the same time. It is equally likely that the two will be the same color as different colors. What is the proportion of colors within the bag?”
To be honest, I did not understood the answer on the blog, but if we write it down, we want to solve

Let us count: if  is the total number of balls, and if  is the number of white balls then

I.e. we want to solve a polynomial equation (of order 2) in , or to be more precise, in

If  is equal to 16, then  is either 6 or 10. It can be visualized below

```> balls=function(n=16){
+ NB=rep(NA,n)
+ for(k in 2:(n-2)){
+ NB[k]=(k*(k-1)+(n-k)*(n-k-1))
+ }
+ k=which(NB==n*(n-1)/2)
+ if(length(k)>0){
+ plot(1:n,NB,type="b")
+ abline(h=n*(n-1)/2,col="red")
+ points((1:n)[k],NB[k],pch=19,col="red")}
+ return((1:n)[k])}
> balls()
[1]  6 10```

But more generally, we can seek other ‘s and other pairs of solutions of such a problem. I am not good in arithmetic, so let us run some codes. And what we get is quite nice: if  admits a pair of solutions, then  is the squared of another integer, say . Further, the difference between  and  is precisely . And  will be one of the answers when the total number of balls will be . Thus, recursively, it is extremely simple to get all possible answers. Below, we have  and the difference between  and ,

```> for(s in 4:1000){
+ b=balls(s)
+ if(length(b)>0) print(c(s,b,diff(b)))
+ }
[1] 9 3 6 3
[1] 16  6 10  4
[1] 25 10 15  5
[1] 36 15 21  6
[1] 49 21 28  7
[1] 64 28 36  8
[1] 81 36 45  9
[1] 100  45  55  10
[1] 121  55  66  11
[1] 144  66  78  12
[1] 169  78  91  13
[1] 196  91 105  14
[1] 225 105 120  15
[1] 256 120 136  16
[1] 289 136 153  17
[1] 324 153 171  18
[1] 361 171 190  19
[1] 400 190 210  20
[1] 441 210 231  21
[1] 484 231 253  22
[1] 529 253 276  23
[1] 576 276 300  24
[1] 625 300 325  25
[1] 676 325 351  26
[1] 729 351 378  27
[1] 784 378 406  28
[1] 841 406 435  29
[1] 900 435 465  30
[1] 961 465 496  31```

Thus, given , consider an urn with  balls. We draw two balls at the same time. It is equally likely that the two will be the same color as different colors. Then the number of colors within the bag are respectively

Finally, observe that the ‘s are well known, from Pascal’s triangle,

also known as triangular numbers,

Maths can be magic, sometimes…

OpenEdition suggests that you cite this post as follows:
Arthur Charpentier (January 22, 2012). Even odds. Freakonometrics. Retrieved September 16, 2024 from https://doi.org/10.58079/oujz

## 3 thoughts on “Even odds”

1. you’re right, thanks for pointing out the typo. I used the second one, so I wrote that one….

2. Antonio says:

or n(n-1)/2

3. Antonio says:

Maybe it is k(k-1)+(n-k)(n-k-1)=2k(n-k) in place of k(k-1)+(n-k)(n-k-1)=n(n-1)?

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