Question

A somewhat easier probability problem. Adam and Eve take turns flipping coins. The first one to get heads wins.

If Eve goes first, what is the probability that she wins?

Answer 1

answer is 2/3

Answer 2

because the probability eve will win is
P(H)+P(TTH)+P(TTTTH)+P(TTTTTTH)+P(TT....H)
this is a geometric sereis .. so u can solve it and answer is 2/3

Answer 3

yes. can you find a solution without summation?

Answer 4

yes... because the sum willbe a/(1-r) where a=1/2 and r=1/2^2

Answer 5

i guess i meant a method that does not require geometric series

Answer 6

I can solve it using this recurence, \[ T=\frac12+ \frac14T \]

The chance of the Eve winning on her first turn is \( frac12 \), or she has a \( \frac14 \) chance of getting to her next turn, and like this follows , now...

Answer 7

This much simpler than solving an geometric infinite series, isn't? ;-)

Answer 8

and the reason i like this problem so much is that if you look closely, the method ffm wrote is exactly the method and reasoning one uses to derive the formula for the sum of a geometric series. the algebra is identical

Answer 9

sat, do you have more of these in your sleeves?

Answer 10

how about this one (similar reasoning) Adam and Eve roll a die. if it shows 1 or 2, Eve wins an apple from Adam and if it shows 3, 4, 5, 6 Adam wins and apple from Eve. Eve has two apples, Adam has only one. What is the probability Adam gets both Eve's apples before he loses his.

Answer 11

answer is 3/7
....

Answer 12

check the algebra i got something similar but different

Answer 13

wt did u get ? 6/7

Answer 14

no, but it is early so i could be wrong. method?

Answer 15

sat, teacher to student advice, .. I am thinking to spend some time reading this chapter : http://athenasc.com/Prob-2nd-Ch1.pdf precisely the problems is it worthy ?

Answer 16

@ffm when you wrote "easier than solving geomtric series" the hook is easier but the algebra is identical.
\[x=\frac{1}{2}+\frac{1}{4}x\]
\[x-\frac{1}{4}x=\frac{1}{2}\]

\[x(1-\frac{1}{4})=\frac{1}{2}\]
\[x=\frac{\frac{1}{2}}{1-\frac{1}{4}}\]

Answer 17

aha cute!

Answer 18

but easier I mean you don't have to go through the steps of identifying the series -> remembering the formula ( or deriving it if you can't remember ) but I can see now that thte algebra is pretty much akin.

Answer 19

look romeo and juliet!

Answer 20

not only is the algebra the same, but the reasoning is exactly the same. think about how you derive formula for solving a geometric series.

Answer 21

Yes I saw that hehe :D now please advice me what do you think about the over all chapter.

Answer 22

it looks good if you have seen the topic before, and a nice condensation. if you have not seen the topic before i would start with something more expansive

Answer 23

No, I have completed a undergrad level course on probability, but I am quite baffled to see that I can't solve quite a few problems on first reading.

Answer 24

sat, I usually derive it for n terms and then take the limit to infinity

Answer 25

then again you are mathematically mature so you can probably get a lot out of this.

Answer 26

I just find out that the book is a standard text for http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041-probabilistic-systems-analysis-and-applied-probability-spring-2006/

Answer 27

what i meant by reasoning is the same is this. usually you start with
\[x =a+ ar + ar^2 +ar^3 + ...\] and then say "multiply by r and get
\[rx =ar+ ar^2+ar^3+...\] so
\[x-rx =a\] and therefore
\[x=\frac{a}{1-r}\]and you see that this is precisely your reasoning when you wrote
\[T=\frac{1}{2}+\frac{1}{4}T\]
\[T-\frac{1}{4}T=\frac{1}{2}\] etc

Answer 28

Cute, isn't this something that we can use to prove that 0.99999999999...= 1?

Answer 29

this is the probabilistic "proof" for the algebra solution to a geometric series. of course in probabiltiy we don't have to worry about convergence because
\[p<1\] (unless of course p = 1)

Answer 30

I got it, it's the standard difference trick used in many places :)

Answer 31

sure you can write
\[\overline{.9}\] as
\[\frac{9}{10}+\frac{9}{10^2}+...\]and sum it up
\[\frac{\frac{9}{10}}{1-\frac{1}{10}}=1\]

Answer 32

Great sat, I never thought about probabilistic proof! Thanks a lot :D

Answer 33

yw
btw i got
\[\frac{4}{7}\] for the answer to second question
maybe i will post it again

Answer 34

New thread please :)