Both William and Tell hit their target half the time. They decide to fight a duel in which they exchange shots until one is hit. What is the chance that the one that shoots first will win?

supermanaccz
Nov 3, 2017

#2**+2 **

Both William and Tell hit their target half the time. They decide to fight a duel in which they exchange shots until one is hit. What is the chance that the one that shoots first will win?

I like this question.

I will assume Tell goes first becasue William gets to go first all the rest of the time so it is Tell's turn to go first this time.

The prob that the Tell wins on his first shot is 0.5

The prob that the William misses on his 1st shot and Tell hits on his second shot is 0.5*0.5*0.5

The prob that the William misses on his 1st and 2nd shots and Tell hits on his 3rd shot is 0.5*0.5*0.5*0.5*0.5

etc

The prob that Tell wins is the limiting sum of a GP with a=0.5 and r=0.5^2

So

\(\begin{align} P(Tell\; wins) &= 0.5+0.5^3+0.5^5+.....\\ &=\frac{0.5}{1-0.5^2}\\ &=\frac{1}{2}*\frac{4}{3}\\ &=\frac{2}{3} \end{align}\)

Melody
Nov 3, 2017

#2**+2 **

Best Answer

Both William and Tell hit their target half the time. They decide to fight a duel in which they exchange shots until one is hit. What is the chance that the one that shoots first will win?

I like this question.

I will assume Tell goes first becasue William gets to go first all the rest of the time so it is Tell's turn to go first this time.

The prob that the Tell wins on his first shot is 0.5

The prob that the William misses on his 1st shot and Tell hits on his second shot is 0.5*0.5*0.5

The prob that the William misses on his 1st and 2nd shots and Tell hits on his 3rd shot is 0.5*0.5*0.5*0.5*0.5

etc

The prob that Tell wins is the limiting sum of a GP with a=0.5 and r=0.5^2

So

\(\begin{align} P(Tell\; wins) &= 0.5+0.5^3+0.5^5+.....\\ &=\frac{0.5}{1-0.5^2}\\ &=\frac{1}{2}*\frac{4}{3}\\ &=\frac{2}{3} \end{align}\)

Melody
Nov 3, 2017