GeekDad Puzzle of the Week: When Are the Odds Even?:
Q: If we have a bag containing equal numbers of black and white marbles, and we pull out one marble, the odds of it being black are even. If we have a bag containing 120 marbles, 85 of which are black, the odds of us pulling out two marbles and them both being black is also even — (85/120)x(84/119) = 0.5 or 50%.
If the largest bag we have can hold 1,000,000 marbles, for how many sets of marbles (i.e., the 120 marbles described above are one set) can we pull two marbles and have a 50% chance of them being the same designated color? Are there any sets of marbles for which we can pull three marbles and have a 50% chance of them being the same designated color? If so, how many?
After the solution is revealed, I'll post the details of my answer.
Edit: GeekDad Puzzle Solution:In the first case you are essentially looking for integer solutions to:
a(a-1) = 2b(b-1)
There are EIGHT sets under 1 million that will result in even odds when 2 balls are drawn.
4 marbles (3 black) --> 4 x 3 = 2(3 x 2)
21 marbles (15 black) --> 21 x 20 = 2(15 x 14)
120 marbles (85 black) --> 120 x 119 = 2(85 x 84)
697 marbles (493 black) --> 697 x 696 = 2(493 x 492)
4,060 marbles (2,871 black) --> 4,060 x 4,059 = 2(2,871 x 2,870)
23,661 marbles (16,731 black) --> 23,661 x 23,660 = 2(16,731 x 16,730)
137,904 marbles (97,513 black) --> 137,904 x 137,903 = 2(97,513 x 97,512)
803,761 marbles (568,345 black) --> 803,761 x 803,760 = 2(568,345 x 568,344)
Interestingly, the next number in each sequence can be computed as follows:
a(n) = 6a(n-1) - a(n-2) - 2
So for example, the next numbers in the sequence would be:
Total balls: 6 x 803,761 - 137,904 - 2 = 4,684,660 marbles
Black balls: 6 x 568,345 - 97,513 - 2 = 3,312,555 black
Integer sequences:
A011900 and
A046090
In the second case you are looking for integer solutions to:
a(a-1)(a-2) = 2b(b-1)(b-2)
There is only ONE set under 1 million that will result in even odds when 3 balls are drawn.
6 marbles (5 black) --> 6 x 5 x 4 = 2(5 x 4 x 3)