0 sec 0 BW (1/2) (one white ball out of 2 balls)
1 sec 0.1 B: BW→BBW (1/3)
1 sec0.2 W: BW→BWW (2/3)
2 sec 0.1.1 B: BBW→BBBW (1/4)
2 sec 0.1.2 W: BBW→BBWW (2/4)
2 sec 0.2.1 B: BWW→BBWW (2/4)
2 sec 0.2.2 W: BWW→BWWW (3/4)
3 sec 0.1.1.1 B: BBBW→BBBBW (1/5) 3 sec
3 sec 0.1.1.2 W: BBBW→BBBWW (2/5)
3 sec 0.1.2.1 B: BBWW→BBBWW (2/5)
3 sec 0.1.2.2 W: BBWW→BBWWW (3/5)
3 sec 0.2.1.1 B: BBWW→BBBWW (2/5)
3 sec 0.2.1.2 W: BBWW→BBWWW (3/5)
3 sec 0.2.2.1 B: BWWW→BBWWW (3/5)
3 sec 0.2.2.2 W: BWWW→BWWWW (4/5)
The above represents a tree of consecutive outcomes in progressive seconds. The colour-coding shows how the number of outcomes doubles every second: 2, 4, 8, 16, etc.
Write the numbers from 1 to n-1 along a row.
Write Pascal's Triangle row t (t seconds) under the numbers from 1 to n-1.
Example 1: for n=5, t=n-2=3 sec:
1 2 3 4 (1 to n-1)
1 3 3 1 (Pascal row 3) 8 outcomes.
Example 2: for n=6, t=n-2=4 sec:
1 2 3 4 5 (1 to n-1)
1 4 6 4 1 (Pascal row 4) 16 outcomes.
The Pascal numbers are tCr (also represented by C(t,r)), being the coefficients in the binomial expansion of (a+b)t). Number of outcomes is 2t (t can be replaced by n-2).
Probability of n-1 white balls is 1/2n-2, probability of 1 white ball is also 1/2n-2.
Probability of n-2 or 2 white balls is (n-2)/2n-2; probability of n-3 or 3 white balls is (n-2)(n-3)/2! = tC2.
Probability of i white balls is n-2Ci-1/2n-2, where 1≤i≤n-1, for integer i. Example: n=6, 2n-2=16:
i p i-1 n-2Ci-1
1 1/16 0 4C0=1
2 1/4 1 4C1=(n-2)/1!=4
3 3/8 2 4C2=(n-2)(n-3)/2!=4×3/2=6
4 1/4 3 4C3=(n-2)(n-3)(n-4)/3!=4×3×2/6=4
5 1/16 4 4C4=(n-2)(n-3)(n-4)(n-5)/4!=4×3×2×1/24=1