MX2 Marathon (2 Viewers)

Skuxxgolfer · Jun 19, 2019

What are some hard past papers?

sharky564 · Sep 17, 2019

Skuxxgolfer said:
What are some hard past papers?

Look at BOS Trials for hard papers.

sharky564 · Sep 17, 2019

TheOnePheeph · Sep 17, 2019

sharky564 said:
$$\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.

TheOnePheeph · Sep 17, 2019

blyatman said:
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).

TheOnePheeph · Sep 17, 2019

blyatman said:
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.

TheOnePheeph · Sep 17, 2019

blyatman said:
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 · Sep 18, 2019

blyatman said:
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 said:
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:

$$\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 · Sep 18, 2019

TheOnePheeph said:
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 · Sep 18, 2019

TheOnePheeph · Sep 18, 2019

sharky564 said:
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 · Sep 18, 2019

TheOnePheeph said:
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 · Sep 18, 2019

sharky564 said:
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 · Sep 18, 2019

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 · Sep 18, 2019

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 · Sep 18, 2019

Lamberts theorem nice

DrEuler · Sep 18, 2019

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.

TheOnePheeph · Sep 18, 2019

DrEuler said:
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.

Lol I feel like such a dick arguing about maths, especially on a public forum haha. Fun nonetheless, and definitely worth it if you end up proving a question wrong.

DrEuler · Sep 18, 2019

TheOnePheeph said:
Lol I feel like such a dick arguing about maths, especially on a public forum haha. Fun nonetheless, and definitely worth it if you end up proving a question wrong.

No it's such a good thing. Debating about it really opens up your mind to others viewpoints and it can be quite beneficial for your understanding..

HeroWise · Sep 19, 2019

Do you remember what it was? @DrEuler

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