- 30 Jan '12 10:01I have a pouch with white and black marbles in it. There is a 50% chance that when I pull two marbles from the pouch, they are both white. How many of the marbles are black and how many white?

There is one solution that involves very few marbles.. but it would be nice to hear what all the possible solutions are. - 30 Jan '12 18:19

The simplest solution is*Originally posted by talzamir***I have a pouch with white and black marbles in it. There is a 50% chance that when I pull two marbles from the pouch, they are both white. How many of the marbles are black and how many white?**

There is one solution that involves very few marbles.. but it would be nice to hear what all the possible solutions are.1 black marble and 3 white marbles

but any count with the same ratio will work. - 30 Jan '12 19:13Indeed, 3 white 1 black works. 3/4 x 2/3 = 2/4 = 1/2. But that's one extreme case, the case of least marbles. From there the ratio changes. If the bag is so huge that taking one white marble out doesn't change the odds of getting another, we get

w = # of white marbles

t = total # of marbles

(w/t)^2 = 0.5

so w/t = sqrt(2)/2, a bit over 70%.

Yet that's a limit with w -> infinite.

Are there *integer* solutions for w and t to give the chance 0.5 for two white marbles other than w = 3, t = 4? - 31 Jan '12 18:42Building a bit more on this.. yes, there are other solutions. I bf'd it by starting with two white marbles and then proceeding with a simple rule;

if the chance to get two white is above 50%, add a non-white to lower it.

if the chance to get two white marbles is 50% or less, add a white to raise it.

That, far as I can tell, finds any number of solutions. Here are the first five:

3 white marbles out of 4

15 white marbles out of 21

85 white marbles out of 120

493 white marbles out of 697

2,871 white marbles out of 4,060

But whether there's a pattern to that..? Hard to tell. - 01 Feb '12 03:03 / 3 edits

If W is num of white, N is Total*Originally posted by talzamir***3 white marbles out of 4**

15 white marbles out of 21

85 white marbles out of 120

493 white marbles out of 697

2,871 white marbles out of 4,060

Chance of getting first white is

W/N

chance of getting 2nd white is

(W-1)/(N-1)

so for 50% chance

(W-1)W/((N(N-1)) = 0.5

(W-1)W = (N-1)N/2

W^2 - W = (N-1)N/2

W^2 - W + 1/4 = (N-1)N/2 + 1/4

(W - 1/2)^2 = (N-1)N/2 + 1/4

(W-1/2) = sqrt((N-1)N/2 + 1/4)

we need positive W so,

W = 1/2 + sqrt((N-1)N/2 + 1/4)

when this solution is an integer, it predicts your answers

That's all I have so far, except for the observation that if I take 2 adjacent integer solutions S and T and compute

S' = (S-1)S

T' = (T-1)T

then is seems that

T'/S' ~ 34 (although it is not necessarily an integer) - 01 Feb '12 06:05 / 1 editThat's where I got too.. to wonder if that expression is an integer. Or even rational. I solved it for N, and it looked much the same, except that the 2 is a multiplier rather than a divider, and that avoids fractions.

N^2 - N - 2W^2 + 2W = 0

N = (1 +/- sqrt(1 +8W^2 - 8W)) / 2

The dividing by 2 bit is not a problem, as the square root, if it gives a real solution, gives an odd one, so add one to that and it's even so you won't ever get half marbles.

But for which integer values of W is 8W^2 - 8W + 1 a square of some integer? - 09 Feb '12 22:48 / 4 editsSo far, I have

If you can find a triangular number T= N(N+1)/2 which also confirms to the relation T=M(M-1) where M and N are integers then

sqrt(T + 1/4)+1/2 is an integer solution for W

The set T are those triangular numbers which are twice another triangular number:

T(3){6} = 2*T(2){3} and sqrt(6 + 1/4) + 1/2 = 3

T(20){210} = 2*T(14){105} and sqrt(210 + 1/4) + 1/2 = 15

T(119){7140} = 2*T(84){3570) and sqrt(7140 + 1/4) + 1/2 = 85

t(696){242556} = 2*T(492){121278} and sqrt(252556 + 1/4) + 1/2 = 493

observing that, we can see that the Ws are N+1 for those N's where T(N) = T(M)/2 - 12 Feb '12 00:26 / 4 editsOk, here is a crazy way, owing a bit to internet research

define some "pell numbers" where pell(N) =((1+SQRT(2))^A1-(1-SQRT(2))^A1)/(2*SQRT(2))

starting N at 0 these go:

{P} = 0,1,2,5,12,29,70,169,408,985,2378,5741...

now take differences between these, so that D(N) = P(N+2)-P(N+1)

the differences are:

{D} = 1,3,7,17,41,99,239,577,1393,3363,8119,19601...

now calculate N = (D-1)/2

{N} = 1,3,8,20,49,119,288,696,1681,4059,9800....

not calculate the Nth triangular number

T(N) = N(N+1)/2

{T} = 1,6,36,210,1225,7140,41616,242556,1413721,8239770,48024900

items 1,3,7,9 etc of the above set are triangular numbers which are also squares, items 2,4,6,8 etc are triangular number which are twice another triangular number

so take items 2,4,6,8 of the above set

{T'} = 6,210,7140,242556,8239770

now, for each T calculate

W = sqrt(T + 1/4) + 1/2

this gives

{W} = 3,15,85,493,2871,16731

Now, each number in {W} is a possible number of white balls! - 12 Feb '12 18:30 / 1 editCombining the above operations all together gives the following formula for the Nth possible number of white balls:

W = SQRT((((1+root_2)^(2N+1)+(1-root_2)^(2N+1))^2+4)/32) + 0.5

this can be rearranged to:

W = SQRT(((1+root_2)^(4N+2)+(1-root_2)^(4N+2)+2)/32) + 0.5

which is about as simple as it gets, I think - 14 Feb '12 00:19 / 3 editsSigh, does nobody believe that is the right answer?

W: N: W*(W-1)/((N*(N-1))

3: 4: 0.5

15: 21: 0.5

85: 120: 0.5

493: 697: 0.5

2871: 4060: 0.5

16731: 23661: 0.5

97513: 137904: 0.5

568345: 803761: 0.5

3312555: 4684660: 0.5

19306983: 27304197: 0.5

112529341: 159140520: 0.5

655869061: 927538921: 0.5

3822685023: 5406093004: 0.5

22280241075: 31509019101: 0.5

129858761425: 183648021600: 0.5

756872327473: 1070379110497: 0.5

4411375203411: 6238626641380: 0.5

25711378892991: 36361380737781: 0.5

149856898154533: 211929657785304: 0.5