Winning at darts

Printable View

January 31st 2010, 05:06 PM
Anonymous1

Winning at darts

Two boys take turns throwing darts at a target. Al throws first and hits with probability $\frac{1}{5}.$ Bob throws second and hits with probability b. What value of b makes each boy have probability $\frac{1}{2}$ of winning?

What I've got:

$P(Al Wins) = \frac{1}{5} \sum_{i=0}^{\infty} (\frac{4}{5} (1-b))^i = \frac{1+4b}{5}$

$P(Bob Wins) = b \sum_{i=0}^{\infty} (\frac{4}{5})^{i+1} (1-b)^i = \frac{4b}{1+4b}.$

Then I set them equal and got $16b^2 -12b + 1 = 0$
Which makes no sense. What am I doing wrong?
February 1st 2010, 01:30 AM
mr fantastic

Quote:

Originally Posted by Anonymous1

Two boys take turns throwing darts at a target. Al throws first and hits with probability $\frac{1}{5}.$ Bob throws second and hits with probability b. What value of b makes each boy have probability $\frac{1}{2}$ of winning?

What I've got:

$P(Al Wins) = \frac{1}{5} \sum_{i=0}^{\infty} (\frac{4}{5} (1-b))^i = \frac{1+4b}{5}$ Mr F says: No. This is equal to ${\color{red} \frac{1}{1 + 4b}}$ . Recheck your calculation.

$P(Bob Wins) = b \sum_{i=0}^{\infty} (\frac{4}{5})^{i+1} (1-b)^i = \frac{4b}{1+4b}.$

Then I set them equal and got $16b^2 -12b + 1 = 0$
Which makes no sense. What am I doing wrong?

..
February 1st 2010, 10:38 AM
Soroban

Hello, Anonymous1!

Mr f is right . . . Your algebra is off.

Quote:

Two boys take turns throwing darts at a target.
Al throws first and hits with probability $\tfrac{1}{5}.$
Bob throws second and hits with probability $b.$
What value of $b$ makes each boy have probability $\tfrac{1}{2}$ of winning?

What I've got:

$P(\text{Al wins}) \;=\; \underbrace{\frac{1}{5} \sum_{n=0}^{\infty} \bigg[\tfrac{4}{5} (1-b)\bigg]^n}_{\text{This is correct}} \;\;=\;\; \underbrace{\frac{1+4b}{5}}_{\text{This is not}}$

The infinite series has: first term $a = 1\:\text{ and }\:\text{common ratio }r = \tfrac{4}{5}(1-b)$

. . Its sum is: . $\frac{a}{1-r} \;=\;\frac{1}{1 - \frac{4}{5}(1-b)} \;=\;\frac{1}{1-\frac{4}{5} + \frac{4}{5}b} \;=\; \frac{1}{\frac{1}{5} + \frac{4}{5}b} \;=\;\frac{5}{1 + 4b}$

. . Hence: . $P(\text{Al wins}) \:=\;\frac{1}{5}\cdot\frac{5}{1+4b} \;=\;\frac{1}{1+4b}$

Then: . $P(\text{Al wins}) \:=\:\frac{1}{2} \quad\Rightarrow\quad \frac{1}{1+4b} \:=\:\frac{1}{2}$

. . From which we get: . $b \,=\,\frac{1}{4}$