2,000,000 Coin Tosses (Posted on 2003-03-03)

	Submitted by Charlie
Rating: 3.0000 (8 votes)
Solution:	(Hide)
Method 1: The distribution is the binomial distribution for 2,000,000 trials with p=1/2. We want the random variable to have the value 1,000,000. This occurs with probability C(2000000,1000000)/2^2000000, where C(n,r) is the combinations of n things taken r at a time. The numbers involved are too large to compute directly, but we can use logarithms and the definition of C(n,r). As C(n,r) = n!/(r! (n-r)!), and an approximation for ln(x!) that is excellent for large values of x is: ln(x!)=x(x+.5) – x + 1/(12x) – 1/(360x^3) + 1/(1260x^5) + ln(2*pi)/2 and since ln(a/b)=ln(a)-ln(b) and ln(2^p)=p ln(2), the natural log of our answer is ln(2,000,000!) – 2 ln(1,000,000!) – 2,000,000 ln(2) = 27017323.65031588 – 2 (12815518.38465817) – 2,000,000 (.6931471805599453) = −7.480120345950127 The natural anti-log of this is 0.000564189513564, or about 1/1772. Method 2: The binomial distribution is approximated by the normal distribution. The mean for 2,000,000 trials with p = 1/2 is 1,000,000 and the standard deviation is the square root of n*p*q or the square root of 2,000,000 * 1/2 * 1/2 . That square root is 707.1067811865476, so we want the central 1/707.1067811865476 of the bell curve, or from z-value −1/1414.213562373095 to + 1/1414.213562373095, which is 0.000564189536532, or about 1/1772.

	Subject	Author	Date
	exact solution for 400 tosses	Jer	2009-05-28 14:34:59
	Puzzle Solution	K Sengupta	2007-05-28 13:46:44
	Answer	K Sengupta	2007-05-25 11:47:39
	Another Form	Marc	2003-03-09 07:34:16
	re: Basic idea	Cory Taylor	2003-03-08 07:15:31
	re: Basic idea	Alan	2003-03-07 11:15:59
	Basic idea	Alan	2003-03-07 11:14:45
	3 answers	Cory Taylor	2003-03-04 03:53:14
	Solution	Jim Lyon	2003-03-03 07:58:47
	non-numerical solution	np_rt	2003-03-03 07:57:51