I'm Mouzie and I'm TcT's friend and he's asked me to solve the following Algeebra problem.
Zee shoots a paint ball at Toby and hits him with a probability of 1/3. If not hit, Toby shoots a paint ball back at Zee and hits him with a probability of 2/3. What's the probability Zee will win if they go back and forth till somebody gets hit?
Answer: 3/7
Here's why:
Let p stand for the probability Zee survives.
Important observation: Zee's probability of survival is the same at "round 0" as it is at "round 1" (or any future round).The probability Zee survives at round 0 = prob. he hits Toby + (prob. he misses Toby)*(prob. Toby misses him)*(prob. he survives round 1)
From
Important observation above and probabilities for Zee and Toby given above
Note: If Zee hits with prob. 1/3, he misses with prob. 2/3. If Toby hits with prob. 2/3, he misses with prob. 1/3.
p=1/3+2/3*1/3*p
This is that stupid Algeebra thingee. OK. Let's hit it.
p=1/3+2/9*p cuz 2/3*1/3=2/9
p-2/9*p=1/3 cuz you can bring that 2/9*p over to the other side
7/9*p=1/3 cuz 1-2/9=9/9-2/9=7/9
p=9/7*1/3 cuz 9/7*7/9=1
p=9/21 even Zee knows how to multiply fractions
p=3/7 and then reduce them
QED (Quit Elongated Derivations)