Probabilty question I'm stuck on

Jan 2010
15
0
Just doing some practice probability questions and came across this which confused me:

Two darts players​
A and B throw alternately at a board and the
1st to score a bull wins the game. The outcomes of different throws are independent and on each throw
A has probability pA of scoring a bull while B has probability pB of scoring a bull. If A has the 1st throw, calculate the probability that A wins the game.


I'm not sure what type of question this is (though i would guess at geometric). I'm not sure how to work this out as there is no information about the number of throws and I'm also unsure about how to bring independence into this.
Any advice would be welcome
 
Dec 2009
3,120
1,342
Just doing some practice probability questions and came across this which confused me:

Two darts players​
A and B throw alternately at a board and the
1st to score a bull wins the game. The outcomes of different throws are independent and on each throw
A has probability pA of scoring a bull while B has probability pB of scoring a bull. If A has the 1st throw, calculate the probability that A wins the game.


I'm not sure what type of question this is (though i would guess at geometric). I'm not sure how to work this out as there is no information about the number of throws and I'm also unsure about how to bring independence into this.
Any advice would be welcome
Yes, schteve,

it's a geometric series for which you must evaluate \(\displaystyle S_{\infty}\)

A wins if he scores on the 1st throw,
or misses on the 1st, B misses, A scores on the 2nd,
or A misses, B misses, A misses, B misses, A scores on the 3rd,
and so on...........

The probability is

\(\displaystyle P_A+(1-P_A)(1-P_B)P_A+(1-P_A)(1-P_B)(1-P_A)(1-P_B)P_A+...\)

\(\displaystyle =P_A+(1-P_A)(1-P_B)P_A+[(1-P_A)(1-P_B)]^2P_A+[(1-P_A)(1-P_B)]^3P_A+....\)

a geometric series with first term \(\displaystyle P_A\) and multiplier \(\displaystyle [(1-P_A)(1-P_B)]\)

If you let \(\displaystyle Q_A=1-P_A,\ Q_B=1-P_B\)

it is neater to express.