Re: Oops I broke the laws of math.
Want to prove by induction: If there are k odd numbers in the first 2^(n-1) rows of the triangle, there are 3k odd numbers in the first 2^n rows of the triangle, having odd number in every cell of the 2^nth row.
Base case: n=1: 1st row has one odd number (1), and the first two rows contain three odd numbers (three 1's), and every cell of the 2nd row is odd.
Assume the theorem works for n >= 1. Now to prove this works for n+1.
Since every cell of the 2^nth row is odd, every cell except for the first and the last one in the (2^n+1)th row must be even.
Thus, the (2^n+2)th row must have odd numbers at 1st, 2nd, (2^n+1)th, and (2^n+2)th cell, and all remaining cells must contain even numbers,
according to the definition of pascal triangle.
Likewise, (2^n+m)th row must contain odd number at mth cell from left and from right, and even numbers inbetween, for every 1 <= m <= 2^n
This makes each half of the rows from (2^n+1)th to (2^(n+1))th contain triangle, whose borders consist of odd numbers. Thus we can apply induction hypothesis and conclude that each triangle contains the same number of odd numbers as the first 2^n rows do.
Hence the first 2^(n+1) rows contain three times the number of odd numbers in the first 2^n rows.
Also, because of induction hypothesis, each triangle's bottom must only consist of odd numbers, making the whole 2^(n+1)th row all odd.
Therefore we proved that the first 2^n rows contain three times the number of odd numbers in the first 2^(n-1) rows,
which implies that the first 2^n rows contain 3^n odd numbers.
So, the chance for hitting an odd number when throwing a dart toward the triangle of height 2^n is
(3^n) / [(2^n)(2^n+1)/2]
= 2(3^n) / (2^n)(2^n+1)
= 2(1.5^n) / (2^n+1) < 2(1.5^n) / (2^n) = 2(0.75^n)
As n goes to infinity, the chance to hit the odd number will converge to zero.
Want to prove by induction: If there are k odd numbers in the first 2^(n-1) rows of the triangle, there are 3k odd numbers in the first 2^n rows of the triangle, having odd number in every cell of the 2^nth row.
Base case: n=1: 1st row has one odd number (1), and the first two rows contain three odd numbers (three 1's), and every cell of the 2nd row is odd.
Assume the theorem works for n >= 1. Now to prove this works for n+1.
Since every cell of the 2^nth row is odd, every cell except for the first and the last one in the (2^n+1)th row must be even.
Thus, the (2^n+2)th row must have odd numbers at 1st, 2nd, (2^n+1)th, and (2^n+2)th cell, and all remaining cells must contain even numbers,
according to the definition of pascal triangle.
Likewise, (2^n+m)th row must contain odd number at mth cell from left and from right, and even numbers inbetween, for every 1 <= m <= 2^n
This makes each half of the rows from (2^n+1)th to (2^(n+1))th contain triangle, whose borders consist of odd numbers. Thus we can apply induction hypothesis and conclude that each triangle contains the same number of odd numbers as the first 2^n rows do.
Hence the first 2^(n+1) rows contain three times the number of odd numbers in the first 2^n rows.
Also, because of induction hypothesis, each triangle's bottom must only consist of odd numbers, making the whole 2^(n+1)th row all odd.
Therefore we proved that the first 2^n rows contain three times the number of odd numbers in the first 2^(n-1) rows,
which implies that the first 2^n rows contain 3^n odd numbers.
So, the chance for hitting an odd number when throwing a dart toward the triangle of height 2^n is
(3^n) / [(2^n)(2^n+1)/2]
= 2(3^n) / (2^n)(2^n+1)
= 2(1.5^n) / (2^n+1) < 2(1.5^n) / (2^n) = 2(0.75^n)
As n goes to infinity, the chance to hit the odd number will converge to zero.
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