I am in a stats class, so I made this problem up since I finished the examples.

We have a bag of 3 white balls, 6 black balls. I choose two without replacement. What is the chance that I get the same color?

I did, and I am sorry for the lack of skills with mathype:

nCr(3,2)*(3/9) + nCr(6,2)*(6/9) / nCr(9,2)

and I got 11/36,or 30.56%. I am looking for validation

## Choose balls, odds same color?

**Moderators:** gmalivuk, Moderators General, Prelates

### Re: Choose balls, odds same color?

I'm not sure I get why you multiply terms by 3/9 or 6/9 in the numerator.

### Re: Choose balls, odds same color?

it is because i am an idiot (and combined 2 methods when I was scribbling on my notepad)

[nCr(3,2) + nCr(6,2)] / nCr(9,2) = 1/2.

[nCr(3,2) + nCr(6,2)] / nCr(9,2) = 1/2.

### Re: Choose balls, odds same color?

Yes, good. Now a followup question. Can someone find an elegant combinatorial proof that it's 1/2, which involves no computation? Specifically, I mean an involution f on the set of the 36 possible marble choices such that {x,y} are the same color if and only if f({x,y}) aren't?

I can't think of a nice one, but maybe someone else can.

I can't think of a nice one, but maybe someone else can.

### Re: Choose balls, odds same color?

Offhand, the result doesn't seem to generalize in any particularly nice way, so I doubt it's possible to find a particularly "canonical" involution. Actually, that's a somewhat interesting question in and of itself: for what positive integers [imath]x, y[/imath] is the corresponding probability still [imath]\frac{1}{2}[/imath]?

### Re: Choose balls, odds same color?

The probability that the balls are different colors is xy/((x+y) C 2) 2xy/(x+y)(x+y-1). This is clearly 1/2 iff 4xy = (x+y)(x+y-1). By algebraic manipulation, this is equivalent to (x-y)

Hence, for the probability to be 1/2, we must have x = (n

^{2}= x+y.Hence, for the probability to be 1/2, we must have x = (n

^{2}+n)/2 and y = (n^{2}-n)/2 for some integer n.- eta oin shrdlu
**Posts:**450**Joined:**Sat Jan 19, 2008 4:25 am UTC

### Re: Choose balls, odds same color?

Here's a combinatorial proof without much computation. Suppose we have a set of [imath]x={n+1\choose2}=1+\cdots+n[/imath] white balls and [imath]y={n\choose2}=1+\cdots+(n-1)[/imath] black balls, with [imath]n>1[/imath]; we wish to find a "pairing-of-pairs" on this [imath](x,y)[/imath]-ball set. Partition the white balls into [imath]n[/imath] sets, of [imath]1,2,\dots,n[/imath] balls, and similarly partition the black balls into [imath]n-1[/imath] sets, of [imath]1,2,\dots,n-1[/imath] balls. So we can label these as [imath]\{W_{11}\}\cup\{W_{21},W_{22}\}\cup\dots\cup\{W_{n1},\dots,W_{nn}\}[/imath] and [imath]\{B_{11}\}\cup\dots\cup\{B_{(n-1)1},\dots,B_{(n-1)(n-1)}\}[/imath], with the first index giving the cardinality of the subset it was chosen from and the second index counting within this subset.

Now consider a pair of balls according to which of these parts each came from. For a pair for which the white and black balls come from subsets with different numbers of balls, compute the map by exchanging the ball which came from the smaller subset for the corresponding ball of the opposite color. So a pair [imath]\{X_{ru},Y_{sv}\}[/imath] with [imath]r<s[/imath] maps to [imath]\{\bar{X}_{ru},Y_{sv}\}[/imath].

This leaves only the pairs for which the two balls come from subsets of the same cardinality. The pairs of the same color, [imath]\{W_{ru},W_{rv}\}[/imath] and [imath]\{B_{ru},B_{rv}\}[/imath] for [imath]r<n[/imath] and [imath]u<v[/imath], can be mapped to the lower and upper (strict) triangles of an [imath]r\times r[/imath] square lattice, and of course the pairs of opposite colors, [imath]\{W_{ru},B_{rv}\}[/imath], can be mapped to this whole square. So this gives us a mapping for all of the remaining pairs, except for the pairs [imath]\{W_{ru},B_{ru}\}[/imath] along the diagonals of these squares, and for the pairs [imath]\{W_{nu},W_{nv}\}[/imath] (with [imath]u<v[/imath]) among the largest white set (which has no corresponding black set).

But now we can pair up all of these remaining pairs, [imath]\{W_{ru},B_{ru}\}[/imath] with [imath]\{W_{nu},W_{n(r+1)}\}[/imath], completing the involution.

Now consider a pair of balls according to which of these parts each came from. For a pair for which the white and black balls come from subsets with different numbers of balls, compute the map by exchanging the ball which came from the smaller subset for the corresponding ball of the opposite color. So a pair [imath]\{X_{ru},Y_{sv}\}[/imath] with [imath]r<s[/imath] maps to [imath]\{\bar{X}_{ru},Y_{sv}\}[/imath].

This leaves only the pairs for which the two balls come from subsets of the same cardinality. The pairs of the same color, [imath]\{W_{ru},W_{rv}\}[/imath] and [imath]\{B_{ru},B_{rv}\}[/imath] for [imath]r<n[/imath] and [imath]u<v[/imath], can be mapped to the lower and upper (strict) triangles of an [imath]r\times r[/imath] square lattice, and of course the pairs of opposite colors, [imath]\{W_{ru},B_{rv}\}[/imath], can be mapped to this whole square. So this gives us a mapping for all of the remaining pairs, except for the pairs [imath]\{W_{ru},B_{ru}\}[/imath] along the diagonals of these squares, and for the pairs [imath]\{W_{nu},W_{nv}\}[/imath] (with [imath]u<v[/imath]) among the largest white set (which has no corresponding black set).

But now we can pair up all of these remaining pairs, [imath]\{W_{ru},B_{ru}\}[/imath] with [imath]\{W_{nu},W_{n(r+1)}\}[/imath], completing the involution.

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