1. ## Tossing coin problem

Miranda and Rob play a game by each tossing a fair coin. The game consists of tossing the two coins together, until for the first time either two heads appear when Miranda wins, or two tails appear when Rob wins.

1) Show that the probability that Miranda wins at or before the nth toss is $\frac{1}{2}-\frac{1}{2^{n+1}}$

2) Show that the probability that the game is decided at or before the nth toss is $1-\frac{1}{2^n}$

Sorry guys, made a bit of a mistake in the question. Should have looked at it better

2. Do you have any other information?

3. Originally Posted by BertieWheen
Do you have any other information?
No that's all the information given in the question.

4. 1. A little weird, will have to think about it for a minute.

2. The question doesn't say Miranda or Rob has to throw the winning toss, so on either persons turn there are only four possible outcomes: {HH, TT, TH, HT}. Thus on any one turn, there is a 0.5 chance the game will end in a win for any one of the players, which makes sense as there is a 0.5 chance the game will end on the "first" toss:

$1-\frac{1}{2^{1}} = 0.5$

This is a sort of "cummulative" (well it is) distribution for "the probability that that game will end in 'n' number of turns"; which is what the question is asking. For example, what is the probability that the game will end at or before the 2nd turn - 0.75.

5. Originally Posted by ANDS!
1. A little weird, will have to think about it for a minute.

2. The question doesn't say Miranda or Rob has to throw the winning toss, so on either persons turn there are only four possible outcomes: {HH, TT, TH, HT}. Thus on any one turn, there is a 0.5 chance the game will end in a win for any one of the players, which makes sense as there is a 0.5 chance the game will end on the "first" toss:

$1-\frac{1}{2^{1}} = 0.5$

This is a sort of "cummulative" (well it is) distribution for "the probability that that game will end in 'n' number of turns"; which is what the question is asking. For example, what is the probability that the game will end at or before the 2nd turn - 0.75.
Yeah the 1st one put me off so that's why I asked

And for the second one, I think you're saying the formula is

$1-\frac{1}{2^{x}}$ where $x$ is the turn number (e.g 1st turn, 2nd turn etc)

Is that right? I'm still getting to grips with statistics.

PS. "cumulative" only has one "m".

EDIT: Well it is in England anyway, I don't know where you're from and whether it's spelt the same there (e.g "Colour" in England is "Color" in America)

EDIT 2: Ahh the formula's in the question :P Didn't see it there

6. I don't see how the formula in part 1 can be correct, since for n = 1, it gives 3/8. The probability that Miranda wins at or before the 1st toss is 1/4, not 3/8.

As for part 2, you can try this explanation on for size.

Like ANDS! said, the probability of a game ending for any round is 1/2. So, let p(n) denote the probability that the game ends by the nth round. Then

p(1) = 1/2

p(2) = 1/2 + (1/2)(1/2)

p(3) = 1/2 + (1/2)(1/2) + (1/2)(1/2)(1/2)

...

where the numbers come from: the probability that the game has ended on a previous round plus the probability that the game has not ended on a previous round times the probability that the game ends on this round given that the game has not ended on a previous round.

7. Good explanation undefined The colours give a good explanation visually, however I'm unsure how acevipa could write that down... I have a feeling the solution lays with powers of 0.5...

New explanation

$p(1) = 0.5$
$p(2) = 0.5 + 0.5^2$
$p(3) = 0.5 + 0.5^2 + 0.5^3$
$p(4) = 0.5 + 0.5^2 + 0.5^3 + 0.5^4$
$p(5) = 0.5 + 0.5^2 + 0.5^3 + 0.5^4 + 0.5^5$
$p(6) = 0.5 + 0.5^2 + 0.5^3 + 0.5^4 + 0.5^5 + 0.5^6$
$p(7) = 0.5 + 0.5^2 + 0.5^3 + 0.5^4 + 0.5^5 + 0.5^6 + 0.5^7$
...

8. Originally Posted by BertieWheen
Good explanation undefined The colours give a good explanation visually, however I'm unsure how acevipa could write that down...
A tree could be easily written on paper. On the computer it takes a bit longer...

An upward branch means the game continues, and a downward branch means the game ends. This tree is for n = 3. We add the probabilities for leaves circled in red, 1/2 + 1/4 + 1/8 = 7/8.

9. Sorry guys, made a bit of a mistake with the first part of the question. It makes a lot more sense now.

10. Originally Posted by acevipa
Sorry guys, made a bit of a mistake with the first part of the question. It makes a lot more sense now.
Using a tree works for part 1 as well. It yields:

$m_1=\frac{1}{4}$

$m_2=\frac{1}{4}+\left(\frac{1}{2}\right)\left(\fra c{1}{4}\right)$

$m_3=\frac{1}{4}+\left(\frac{1}{2}\right)\left(\fra c{1}{4}\right)+\left(\frac{1}{2}\right)^2\left(\fr ac{1}{4}\right)$

$m_4=\frac{1}{4}+\left(\frac{1}{2}\right)\left(\fra c{1}{4}\right)+\left(\frac{1}{2}\right)^2\left(\fr ac{1}{4}\right)+\left(\frac{1}{2}\right)^3\left(\f rac{1}{4}\right)$

...

$m_n=\frac{1}{4}\sum_{i=0}^{n-1}\left(\frac{1}{2}\right)^i$

Keep in mind that we could use formal notation instead of a tree (you get a bunch of things that look like $P(A\cap B) = P(A|B)P(B)$), it's just more of a bother in my opinion, unless we are doing something that needs to be that rigorous.

11. Originally Posted by undefined
Using a tree works for part 1 as well. It yields:

$m_1=\frac{1}{4}$

$m_2=\frac{1}{4}+\left(\frac{1}{2}\right)\left(\fra c{1}{4}\right)$

$m_3=\frac{1}{4}+\left(\frac{1}{2}\right)\left(\fra c{1}{4}\right)+\left(\frac{1}{2}\right)^2\left(\fr ac{1}{4}\right)$

$m_4=\frac{1}{4}+\left(\frac{1}{2}\right)\left(\fra c{1}{4}\right)+\left(\frac{1}{2}\right)^2\left(\fr ac{1}{4}\right)+\left(\frac{1}{2}\right)^3\left(\f rac{1}{4}\right)$

...

$m_n=\frac{1}{4}\sum_{i=0}^{n-1}\left(\frac{1}{2}\right)^i$

Keep in mind that we could use formal notation instead of a tree (you get a bunch of things that look like $P(A\cap B) = P(A|B)P(B)$), it's just more of a bother in my opinion, unless we are doing something that needs to be that rigorous.
That's strangely beautiful.
I love patterns ^_^

acevipa, is there anything else you need help with for this question?

12. Originally Posted by undefined
Using a tree works for part 1 as well. It yields:

$m_1=\frac{1}{4}$

$m_2=\frac{1}{4}+\left(\frac{1}{2}\right)\left(\fra c{1}{4}\right)$

$m_3=\frac{1}{4}+\left(\frac{1}{2}\right)\left(\fra c{1}{4}\right)+\left(\frac{1}{2}\right)^2\left(\fr ac{1}{4}\right)$

$m_4=\frac{1}{4}+\left(\frac{1}{2}\right)\left(\fra c{1}{4}\right)+\left(\frac{1}{2}\right)^2\left(\fr ac{1}{4}\right)+\left(\frac{1}{2}\right)^3\left(\f rac{1}{4}\right)$

...

$m_n=\frac{1}{4}\sum_{i=0}^{n-1}\left(\frac{1}{2}\right)^i$

Keep in mind that we could use formal notation instead of a tree (you get a bunch of things that look like $P(A\cap B) = P(A|B)P(B)$), it's just more of a bother in my opinion, unless we are doing something that needs to be that rigorous.
How does the formula relate to part 1

13. Originally Posted by BertieWheen
That's strangely beautiful.
I love patterns ^_^

acevipa, is there anything else you need help with for this question?
So would it be like this:

$
m_n=\frac{1}{4}+\left(\frac{1}{2}\right)\left(\fra c{1}{4}\right)+\left(\frac{1}{2}\right)^2\left(\fr ac{1}{4}\right)+\left(\frac{1}{2}\right)^3\left(\f rac{1}{4}\right)+...+\left(\frac{1}{4}\right)\left (\frac{1}{2}\right)^{n-1}
$

$=\frac{1}{4}\left[1+\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right )^2+...+\left(\frac{1}{2}\right)^{n-1}\right]$

$=\left(\frac{1}{2}\right)^2\left[\frac{1-\left(\frac{1}{2}\right)^{n}}{\frac{1}{2}}\right]$

$=\left(\frac{1}{2}\right) \left(1-\frac{1}{2^n}\right)$

$=\frac{1}{2}-\frac{1}{2^{n+1}}$

14. Originally Posted by acevipa
How does the formula relate to part 1
$\sum_{i=0}^{n-1}\left(\frac{1}{2}\right)^i$ is an geometric series. The general formula is

$\sum_{k=0}^nar^k=a\frac{1-r^{n+1}}{1-r}$.

Apply this and you will get the expression given in the problem.

What I gave for part 2 above is also in terms of a geometric series, by the way, I just didn't write using the summation symbol.

15. Thanks everyone for all your help