Duel (Posted on 2003-03-06)

	Submitted by np_rt
Rating: 3.3333 (6 votes)
Solution:	(Hide)
The first thing to do is to calculate the probability of winning if you shoot first. Let P(n)=probability of hitting your rival on your nth turn if you shoot first. P(1)=1/2 P(2)=(1/2)*(1/2)*(1/2) The first 1/2 is you missing on your first try, the second is your rival missing, the third is you hitting him. P(3)=(1/2)*(1/2)*(1/2)*(1/2)*(1/2) ...etc In general, P(n)=(1/2)*((1/4)^(n-1)) The probability of you winning if you shoot first is the sum of all P(n)'s, which forms an infinite geometric series, assuming you guys can shoot forever, with the first term a=1/2 and the ratio r=1/4. Using the formula for the sum of this infinite geometric series (which converges), S=a/(1-r), you'd end up with a probability of 2/3. And consequently, the probability of winning for the person shooting second is 1/3. Next there are two cases to consider, one where the coin lands heads and other tails. P(winning)=P(heads)*P(going 1st)+P(tails)*P(going 2nd) =(1/3)*(2/3)+(2/3)*(1/3) =4/9 Now, if you were to choose tails, your probability of winning would be 5/9.

	Subject	Author	Date
	Puzzle Solution	K Sengupta	2008-05-14 15:53:26
	Answer	K Sengupta	2008-05-13 05:53:46
	another method	Cory Taylor	2003-03-07 04:49:25
	solution	Charlie	2003-03-07 04:08:18