HSC 2013 MX2 Marathon (archive) (31 Viewers)

[Sy123]

Re: HSC 2013 4U Marathon

Yeah that makes more sense.
So then would it be......

a)
P($20) = 10C0 (18/37)^10(19/37) + 10C1(18/37)^11(19/37) + 10C2(18/37)^12(19/37)^2 + 10C3(18/37)^13(19/37)^3 +10C4(18/37)^14(19/37)^4 + 10C5(18/37)^15(19/37)^5 + 10C6(18/37)^16(19/37)^6 +10C7(18/37)^17(19/37)^7 + 10C8(18/37)^18(19/37)^8 + 10C9(18/37)^19(19/37)^9

P($0) = 1- P($20)


b) If he makes a single bet with his money, he will have a higher probability of winning.
However, he has a higher probability of losing as well.
If he only wants to make a few dollars, it is better for him to bet $1 at a time.
If he only wants double-or-nothing then he must do it all in one bet.

?????? For some reason I am thinking that I missed something again .

But for example, if we denote W as win, and L as Loss, he can go:

WLWLWLWLWLWLWLWLWLWLWLWLWLWLWLWLWLWLWL WWWWW WWWWW and then win.
And this can go on infinitely many times, so infinite series are involved. I was experimenting with $2, whether he wins up to 4.

When he has $2, (also I made probability p so its easier), I seperated each decision into two groups, so each game he plays will have a structure like:

(WL) , (WL) , (LW) , (WL) , .... (WL) , (WW).

It doesn't matter what order the L and W's are after each partition, so the probability is



So yeah this might help some other people in trying to find the answer to the problem.

 

[Sy123]

Re: HSC 2013 4U Marathon

A man walks into a casino with $10 and the only bet he knows how to make is on a colour in roulette (the probability of winning this bet is 18/37, and you win as much as you bet).

Rather than spending all his money at once with one bet, he decides to "play it safe" and bet $1 at a time until he either reaches $20 or goes broke.

a) What is the probability that he will walk away with $20 and what is the probability that he will go broke?

b) Is this better or worse than he would do if he just made a single bet with all of his money?

Ok I may have figured it out, first lets let the probability of winning be p for simplicity sake.

We break the games into W and L, where W and L denotes win and loss respectively. We can create a string of W's and L's to represent the game, each game we can pair a W and an L.

At the end of every game, there must be 10 more W's than L's, and for each L there must be a W pair there (if he wins), we are not concerned about order here since its all multiplication.

So, it should be:



Where p = 18/37



Right? maybe? Why isn't winning and losing the game complementary? :/

b) Depends whether for that specific value of p, whether P(win) > p.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

P($0) = 1- P($20)

I'm not entirely sure that this is true. There would have to be some probability that he never hits $20 or broke, so that would have to be taken into account. However, for this to happen he would have to go on infinitely without reaching $20 or $0. So the probability of this would be extremely low, perhaps that low that it's negligible or even 0% chance of happening? Just something I'm considering.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Right? maybe? Why isn't winning and losing the game complementary? :/

I don't think it has to be. Re:above post.

Although it very well could be that they are complementary, haven't made up my mind yet. Will give it a better go later after I finish trying to make this conics question.

 

[anomalousdecay]

Re: HSC 2013 4U Marathon

Yeah, I think we need to finish series in 2-unit before I can do that .

 

[anomalousdecay]

Re: HSC 2013 4U Marathon

Dying for that conics question.............

 

[qrpw]

Re: HSC 2013 4U Marathon

A man walks into a casino with $10 and the only bet he knows how to make is on a colour in roulette (the probability of winning this bet is 18/37, and you win as much as you bet).

Rather than spending all his money at once with one bet, he decides to "play it safe" and bet $1 at a time until he either reaches $20 or goes broke.

a) What is the probability that he will walk away with $20 and what is the probability that he will go broke?

b) Is this better or worse than he would do if he just made a single bet with all of his money?

We can consider part (a) as a random walk on the number line from 0 to 20, starting at 10, with probability 18/37 of moving to the right by one, and 19/37 to the left. Let p(x) be the probability that the walker will reach 20 before reaching 0, when he is at x.

Therefore we can say:






Now we need to find a function that satisfies these conditions.


Where a, b are constants.

Substituting in our first two conditions, and solving for a and b,






Lastly, substitute x=10 to find the probability of walking to 20 before walking to 0. Skipping the algebra, we'll arrive at



(b)Which is worse than if he made a single bet.

I'm not entirely sure that this is true. There would have to be some probability that he never hits $20 or broke, so that would have to be taken into account. However, for this to happen he would have to go on infinitely without reaching $20 or $0. So the probability of this would be extremely low, perhaps that low that it's negligible or even 0% chance of happening? Just something I'm considering.

Random walks on Z2 are always recurrent so there is no probability of wandering around infinitely.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Consider an ellipse with equation:



A parabola is constructed such that it's focus and directix is the same as the ellipse's positive focus and directrix.

It is observed that when the equation of the parabola and ellipse are solved simultaneously, the resulting polynomial has no real roots. Prove this observation.

 

[Sy123]

Re: HSC 2013 4U Marathon

Consider an ellipse with equation:



A parabola is constructed such that it's focus and directix is the same as the ellipse's positive focus and directrix.

It is observed that when the equation of the parabola and ellipse are solved simultaneously, the resulting polynomial has no real roots. Prove this observation.

Have you found a neat geometric way of doing it?

I can do it quite easily with manually equating the equations and showing that it has no real roots. But I was wondering if you found a nicer way of doing this.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Have you found a neat geometric way of doing it?

I can do it quite easily with manually equating the equations and showing that it has no real roots. But I was wondering if you found a nicer way of doing this.

I have, and it is very elegant imo, but I'll wait to see if anyone else finds one first. I'll pm you my solution however.

 

[Sy123]

Re: HSC 2013 4U Marathon

bump.

Difficulty: 1.5/5.

The condition: exists if the polynomials are all unique, since



Then there exists some non-negative integer k such that:

o=l+k

Then

And then the polynomials are not unique (q and z are not unique), therefore in order for there to exist unique polynomials q and z, o < l

==========

Is that it?

 

[Carrotsticks]

Re: HSC 2013 4U Marathon

Consider an ellipse with equation:



A parabola is constructed such that it's focus and directix is the same as the ellipse's positive focus and directrix.

It is observed that when the equation of the parabola and ellipse are solved simultaneously, the resulting polynomial has no real roots. Prove this observation.

This is not possible because the parabola's focus and directrix are equidistant from its vertex.

However, the ellipse's foci and directrix are not equidistant.

Suppose they are, so a/e = ae, which implies that e=1.

However, if e=1, the conic section is a parabola, which contradicts it being an ellipse in the first place.

Never mind! For a moment I thought that it also had to share the same point of intersection as the ellipse.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

This is not possible because the parabola's focus and directrix are equidistant from its vertex.

However, the ellipse's foci and directrix are not equidistant.

Suppose they are, so a/e = ae, which implies that e=1.

However, if e=1, the conic section is a parabola, which contradicts it being an ellipse in the first place.

Quoting for fail

 

[Carrotsticks]

Re: HSC 2013 4U Marathon

Quoting for fail

We all have derp moments =)

Anyway here's my revised answer, which funnily enough is pretty similar to my original response.

 

[Carrotsticks]

Re: HSC 2013 4U Marathon

And by e > 1, I actually meant that 0 < e < 1.

=)

EDIT: And by "ellipse" in the second paragraph, I actually mean "parabola".

Clearly not having a good night.

 

[hayabusaboston]

Re: HSC 2013 4U Marathon

We all have derp moments =)

Anyway here's my revised answer, which funnily enough is pretty similar to my original response.

Well Realise, it looks like Carrot's got it hahaa.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

And by e > 1, I actually meant that 0 < e < 1.

=)

=) =) =) =) =) =) =) =)

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Sorry in advanced to the people sitting the BOS trials this year.

 

[Trebla]

Re: HSC 2013 4U Marathon

Sorry in advanced to the people sitting the BOS trials this year.

Editing time!

 

[Carrotsticks]

Re: HSC 2013 4U Marathon

Editing time!

lol oh you