# HSC 2018-2019 MX2 Marathon (2 Viewers)

#### Skuxxgolfer

##### Member
What are some hard past papers?

#### sharky564

##### Member
What are some hard past papers?
Look at BOS Trials for hard papers.

#### sharky564

##### Member
$\bg_white \noindent Amy and Ben are flipping n fair coins and n+1 fair coins respectively. What is the probability that Ben flips more Heads than Amy?$

#### TheOnePheeph

##### Active Member
$\bg_white \noindent Amy and Ben are flipping n fair coins and n+1 fair coins respectively. What is the probability that Ben flips more Heads than Amy?$
Is the answer 1/2? You can tell by symmetry, but it can be proven using binomial probability and binomial theorem as well.

#### blyatman

##### Well-Known Member
Is the answer 1/2? You can tell by symmetry, but it can be proven using binomial probability and binomial theorem as well.
It shouldn't be exactly 0.5, since they flip a different number of coins.

#### TheOnePheeph

##### Active Member
It shouldn't be exactly 0.5, since they flip a different number of coins.
There are two cases to consider: Ben obtains more tails than Amy, or Ben obtains more heads than Amy. Since it is a fair coin, these two cases have the same probability, by symmetry. That means that the probability that Ben obtains more heads is exactly a half. This can also be proven using binomial theorem, which is attached below (This way is unnecessarily long and complex, I just wanted an extra proof to back my case up. Note I also may have made a mistake in it, I just quickly wrote it up tonight).

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#### blyatman

##### Well-Known Member
There are two cases to consider: Ben obtains more tails than Amy, or Ben obtains more heads than Amy. Since it is a fair coin, these two cases have the same probability, by symmetry. That means that the probability that Ben obtains more heads is exactly a half. This can also be proven using binomial theorem, which is attached below (This way is unnecessarily long and complex, I just wanted an extra proof to back my case up. Note I also may have made a mistake in it, I just quickly wrote it up tonight).
Hm, but if we take a more extreme example: If Amy flips 1 coin and Ben flips 100 coins, then the probability of Ben flipping more heads than Amy would be almost 100% wouldn't it? Or is this 50% result only true for n, n+1 flips? If so, then fair enough, I will take your word for it (it's too late for binomial theorem, not to mention I don't like to do math in my spare time lol).

#### TheOnePheeph

##### Active Member
Hm, but if we take a more extreme example: If Amy flips 1 coin and Ben flips 100 coins, then the probability of Ben flipping more heads than Amy would be almost 100% wouldn't it? Or is this 50% result only true for n, n+1 flips?
Yeah I believe it only works for n, (n+1). Because if you take, for example n, (n+2), then it is possible for Ben to have more tails and more heads that Amy, adding a complete new case. Since there are more cases to consider here than just having more heads or more tails, the probability will not be exactly a half. Same thing applies to n+3, n+4 etc.

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#### blyatman

##### Well-Known Member
Yeah I believe it only works for n, (n+1). Because if you take, for example n, (n+2), then it is possible for Ben to have more tails and more heads that Amy, if n is even, as he could have n/2+1 for both, or otherwise he could have the same amount of tails./heads and more heads/tails. Since there are more cases to consider here than just having more heads or more tails, the probability will not be exactly a half. Same thing applies to n+3, n+4 etc.
Yeh that's what I was originally thinking, I just extrapolated (albeit incorrectly) based on the extreme scenario. Good problem nonetheless, shows that it's counterintuitive (or intuitive?) for n, n+1.

#### TheOnePheeph

##### Active Member
Yeh that's what I was originally thinking, I just extrapolated (albeit incorrectly) based on the extreme scenario. Good problem nonetheless, shows that it's counterintuitive (or intuitive?) for n, n+1.
Yeah I didn't think it was gonna be 1/2 initially either. I ended up coming up with that evil binomial theorem solution,which resulted in 1/2, and then figured out why it worked from there. Now I have wasted my Tuesday night lol.

#### sharky564

##### Member
Yeh that's what I was originally thinking, I just extrapolated (albeit incorrectly) based on the extreme scenario. Good problem nonetheless, shows that it's counterintuitive (or intuitive?) for n, n+1.
Yeah I didn't think it was gonna be 1/2 initially either. I ended up coming up with that evil binomial theorem solution,which resulted in 1/2, and then figured out why it worked from there. Now I have wasted my Tuesday night lol.
I thought this was a very nice problem outlining the utility of symmetry:

$\bg_white \noindent After both have flipped n coins, let p_A be the probability Amy has flipped more Heads than Ben, p_B be the probability Ben has flipped more Heads than Amy, and let p_D be the probability they flipped the same number of Heads. Then, p_A = p_B by symmetry, and p_A + p_B + p_D = 1. Now, the probability Ben has flipped more Heads than Amy after his final toss is the probability of doing so when he tosses Tails, which is \frac{1}{2} p_A plus the probability he does so when he tosses Heads, which is \frac{1}{2} (p_A + p_D), so we get a total probability of \frac{1}{2} (p_A + p_A + p_D) = \frac{1}{2} (p_A + p_B + p_D) = \frac{1}{2}.$

#### sharky564

##### Member
There are two cases to consider: Ben obtains more tails than Amy, or Ben obtains more heads than Amy. Since it is a fair coin, these two cases have the same probability, by symmetry. That means that the probability that Ben obtains more heads is exactly a half. This can also be proven using binomial theorem, which is attached below (This way is unnecessarily long and complex, I just wanted an extra proof to back my case up. Note I also may have made a mistake in it, I just quickly wrote it up tonight).
Your symmetry argument unfortunately fails, as pointed out by blyatman.

I gave this prob to my year's 4u class and most agreed this answer made sense at an intuitive level, but proving it rigorously was much more of a struggle.

#### sharky564

##### Member
$\bg_white \noindent Let \mathcal{P} be a parabola with focus F. Suppose distinct points A, B, C are on this parabola. Let the tangents at A, B, C enclose triangle XYZ. Show that FXYZ is a cyclic quadrilateral.$

#### TheOnePheeph

##### Active Member
Your symmetry argument unfortunately fails, as pointed out by blyatman.

I gave this prob to my year's 4u class and most agreed this answer made sense at an intuitive level, but proving it rigorously was much more of a struggle.
Wait sorry why does this argument fail? There are only two possible cases: Ben flips more heads than Amy or Ben flips more tails than Amy, and the 2nd case obviously includes the circumstances where Ben gets the same amount of heads as Amy, as he would have to have more tails then. Then since heads and tails have the same probability, the two cases would have to be equal probability, making the probability 1/2. True straight up saying it is symmetrical is flawed, but I think this way makes sense, unless I'm missing something crucial lol.

#### sharky564

##### Member
Wait sorry why does this argument fail? There are only two possible cases: Ben flips more heads than Amy or Ben flips more tails than Amy, and the 2nd case obviously includes the circumstances where Ben gets the same amount of heads as Amy, as he would have to have more tails then. Then since heads and tails have the same probability, the two cases would have to be equal probability, making the probability 1/2. True straight up saying it is symmetrical is flawed, but I think this way makes sense, unless I'm missing something crucial lol.
No, you can have the case where they flip the same number of Heads.

#### TheOnePheeph

##### Active Member
No, you can have the case where they flip the same number of Heads.
But if they flip the same number of heads, then Ben will flip more tails, since he flips n+1 coins and Amy only flips n, therefore putting it in the 2nd case of Ben flipping more tails than Amy.

#### Carrotsticks

##### Retired
Wow! Great to see that this part of the community is still up and running. Keep it up everybody, you're all doing a great job <3

#### Skuxxgolfer

##### Member
If 10 coins are tossed 50 times, in how many cases would we expect the number of heads to exceed the number of tails?

#### HeroWise

##### Active Member
Lamberts theorem nice

#### DrEuler

##### Member
Love seeing debates about Mathematics lol, reminds me of when my schools math department and me had a huge debate in which I ended up proving a question wrong in our trial.