HSC 2016 MX2 Combinatorics Marathon (archive) (3 Viewers)

[leehuan]

Re: HSC 2016 MX2 Combinatorics Marathon

Biggest assumption that the HSC makes is that dice have 6 sides

 

[seanieg89]

Re: HSC 2016 MX2 Combinatorics Marathon

A-5 is definitely a straight in the vast majority of poker variants.

 

[braintic]

Re: HSC 2016 MX2 Combinatorics Marathon

Another poker question:

In how many ways can you be dealt a card-high hand (ie. not even a pair ..... a nothing hand)? (assuming of course that order is not important)
Do this without counting all the other hands and subtracting from the total.

 

[braintic]

Re: HSC 2016 MX2 Combinatorics Marathon

Open to all. And the method IS related to a past HSC question.

 

[Paradoxica]

Re: HSC 2016 MX2 Combinatorics Marathon

Open to all. And the method IS related to a past HSC question.

My answer to a) is 5/9

My answer to b) is 1+√(⅔)

I will post my solutions iff my answers are verified to be correct.

 

[seanieg89]

Re: HSC 2016 MX2 Combinatorics Marathon

My answer to a) is 5/9

My answer to b) is 1+√(⅔)

I will post my solutions iff my answers are verified to be correct.

Those numbers look good to me .

 

[Paradoxica]

Re: HSC 2016 MX2 Combinatorics Marathon

 

[InteGrand]

Re: HSC 2016 MX2 Combinatorics Marathon

I think braintic is asking for part (b) what length x (or L) of segment we should ask for in order for the probability that X ≤ L is 1/3, where X := |A – B|, where A and B are the r.v.'s corresponding to the positions of the two cuts.

(In other words, the inverse CDF of X evaluated at 1/3.)

So unless I've misinterpreted something, I think we'd need x to be the solution with the negative root, i.e. 1 – √(2/3). With this x, the probability that X ≤ x will be 1/3. Can't use 1 + √(2/3), because the probability that X ≤ 1 + √(2/3) is trivially 1, because the difference |A – B| (which is X, the length of the inner segment) is always going to be between 0 and 1.

In general, I believe that Pr(X ≤ x) will be 1 – (1 – x)², for 0 ≤ x ≤ 1 (with Pr(X ≤ x) being 0 if x < 0 and 1 if x > 1). So we set 1 – (1 – x)² equal to 1/3 for braintic's part (b), and note that 1 – x is non-negative when deciding on which root to take.

 

[Paradoxica]

Re: HSC 2016 MX2 Combinatorics Marathon

I think braintic is asking for part (b) what length x (or L) of segment we should ask for in order for the probability that X ≤ L is 1/3, where X := |A – B|, where A and B are the r.v.'s corresponding to the positions of the two cuts.

(In other words, the inverse CDF of X evaluated at 1/3.)

So unless I've misinterpreted something, I think we'd need x to be the solution with the negative root, i.e. 1 – √(2/3). With this x, the probability that X ≤ x will be 1/3. Can't use 1 + √(2/3), because the probability that X ≤ 1 + √(2/3) is trivially 1, because the difference |A – B| (which is X, the length of the inner segment) is always going to be between 0 and 1.

In general, I believe that Pr(X ≤ x) will be 1 – (1 – x)², for 0 ≤ x ≤ 1 (with Pr(X ≤ x) being 0 if x < 0 and 1 if x > 1). So we set 1 – (1 – x)² equal to 1/3 for braintic's part (b), and note that 1 – x is non-negative when deciding on which root to take.

Oh wait, I see what you're getting at. No, when I said let the interval length be L, I was referring to the entire interval, not it's segment.

 

[seanieg89]

Re: HSC 2016 MX2 Combinatorics Marathon

I think braintic is asking for part (b) what length x (or L) of segment we should ask for in order for the probability that X ≤ L is 1/3, where X := |A – B|, where A and B are the r.v.'s corresponding to the positions of the two cuts.

(In other words, the inverse CDF of X evaluated at 1/3.)

So unless I've misinterpreted something, I think we'd need x to be the solution with the negative root, i.e. 1 – √(2/3). With this x, the probability that X ≤ x will be 1/3. Can't use 1 + √(2/3), because the probability that X ≤ 1 + √(2/3) is trivially 1, because the difference |A – B| (which is X, the length of the inner segment) is always going to be between 0 and 1.

In general, I believe that Pr(X ≤ x) will be 1 – (1 – x)², for 0 ≤ x ≤ 1 (with Pr(X ≤ x) being 0 if x < 0 and 1 if x > 1). So we set 1 – (1 – x)² equal to 1/3 for braintic's part (b), and note that 1 – x is non-negative when deciding on which root to take.

I interpreted it as asking what length should we specify (instead of 1) for the whole interval, in order to have the probability of the middle interval having length at most 1/3 equal to 1/3?

Paradoxica's is a correct answer to this question.

(It's almost equivalent to your interpretation, which is probably what braintic meant.)

 

[Paradoxica]

Re: HSC 2016 MX2 Combinatorics Marathon

I interpreted it as asking what length should we specify (instead of 1) for the whole interval, in order to have the probability of the middle interval having length at most 1/3 equal to 1/3?

Paradoxica is a correct answer to this question.

"a" correct answer? (that implies more than one, methinks)

Or maybe I'm just reading too far in.

 

[seanieg89]

Re: HSC 2016 MX2 Combinatorics Marathon

Nope, just meant that others might argue differently to reach to same conclusion. That is the unique length.

 

[InteGrand]

Re: HSC 2016 MX2 Combinatorics Marathon

I interpreted it as asking what length should we specify (instead of 1) for the whole interval, in order to have the probability of the middle interval having length at most 1/3 equal to 1/3?

Paradoxica is a correct answer to this question.

Ah I see. I (mis)interpreted it as asking for what middle length (out of the unit interval) should we ask for in order for that probability (Pr(|A – B| ≤ x) to be 1/3, and didn't read Paradoxica's solution too closely, and thought he was doing the same problem I was thinking of.

The reason I thought this was that I misinterpreted braintic's "the answer to part (a)" as meaning "Pr(|A - B| ≤ x)" (where A and B are uniform on the unit interval, i.e. just the same as the initial r.v.'s), and the "length" in question referring to the x. So I misinterpreted it as meaning "If I want 'the probability that |A - B| is no more than x' to be 1/3, what x should I choose?".

 

[Paradoxica]

Re: HSC 2016 MX2 Combinatorics Marathon

Ah I see. I (mis)interpreted it as asking for what middle length (out of the unit interval) should we ask for in order for that probability (Pr(|A – B| ≤ x) to be 1/3, and didn't read Paradoxica's solution too closely, and thought he was doing the same problem I was thinking of.

The reason I thought this was that I misinterpreted braintic's "the answer to part (a)" as meaning "Pr(|A - B| ≤ x)" (where A and B are uniform on the unit interval, i.e. just the same as the initial r.v.'s), and the "length" in question referring to the x.

Fair enough. I was briefly confused when I first read the question, because I was unsure exactly what it was asking.

Guess the question needs a bit of rewording I suppose. Ah well.

 

[seanieg89]

Re: HSC 2016 MX2 Combinatorics Marathon

Ah I see. I (mis)interpreted it as asking for what middle length (out of the unit interval) should we ask for in order for that probability (Pr(|A – B| ≤ x) to be 1/3, and didn't read Paradoxica's solution too closely, and thought he was doing the same problem I was thinking of.

The reason I thought this was that I misinterpreted braintic's "the answer to part (a)" as meaning "Pr(|A - B| ≤ x)" (where A and B are uniform on the unit interval, i.e. just the same as the initial r.v.'s), and the "length" in question referring to the x.

That could well have been his intended question. This is why having rigorous notation for probabilistic statements and their proofs is important .

 

[InteGrand]

Re: HSC 2016 MX2 Combinatorics Marathon

That could well have been his intended question. This is why having rigorous notation for probabilistic statements and their proofs is important .

Yeah . So for clarification purposes, what was the Q. you and Paradoxica were doing again? Basically, this?:

Let A, B be (independent and) uniformly distributed on [0, L] (L > 0, real). Suppose Pr(|A – B| ≤ 1/3) = 1/3. Find L.

??

 

[Paradoxica]

Re: HSC 2016 MX2 Combinatorics Marathon

Yeah . So for clarification purposes, what was the Q. you and Paradoxica were doing again? Basically, this?:

Let A, B be (independent and) uniformly distributed on [0, L] (L > 0, real). Suppose Pr(|A – B| ≤ 1/3) = 1/3. Find L.

??

Yes, that is indeed what we interpreted it as.

For completeness, could you (unless you already have) give your solution as well?

 

[InteGrand]

Re: HSC 2016 MX2 Combinatorics Marathon

Yes, that is indeed what we interpreted it as.

For completeness, could you (unless you already have) give your solution as well?

Let x be in [0,1]. Consider the unit square in the a-b plane (the region where A,B's values are taken). Basically find the area on the unit square satisfying |a – b| ≤ x. To do this, sketch the lines b = a+x and b = a-x, and you'll see they intersect the square in some places and cut off these triangles. The desired region is the square minus these triangles. The two triangles have combined area (1 - x)². (This is found by finding the coordinates of the intersection points of the lines with the square and then using simple "base times height".)

So the desired region has area 1 – (1 - x)² (subtracting the triangles' area from the unit square). Due to the uniform distributions, this is exactly the probability that |A – B| ≤ x.

So Pr(X ≤ x) = 1 – (1 - x)², for 0 ≤ x ≤ 1. So set this equal to 1/3 for the desired x, and remember than 1 - x ≥ 0 when deciding on the root.

I think it's similar method to what you used (still haven't carefully read over that, but basically it's based on finding the area of the desired region I think).

 

[Paradoxica]

Re: HSC 2016 MX2 Combinatorics Marathon

Let x be in [0,1]. Consider the unit square in the a-b plane (the region where A,B's values are taken). Basically find the area on the unit square satisfying |a – b| ≤ . To do this, sketch the lines b = a+x and b = a-x, and you'll see they intersect the square in some places and cut off these triangles. The desired region is the square minus these triangles. The two triangles have combined area (1 - x)². (This is found by finding the coordinates of the intersection points of the lines with the square and then using simple "base times height".)

So the desired region has area 1 – (1 - x)² (subtracting the triangles' area from the unit square). Due to the uniform distributions, this is exactly the probability that |A – B| ≤ x.

So Pr(X ≤ x) = 1 – (1 - x)², for 0 ≤ x ≤ 1. So set this equal to 1/3 for the desired x, and remember than 1 - x ≥ 0 when deciding on the root.

Interesting how the two interpretations give complementary answers which arise from the same equation...

 

[braintic]

Re: HSC 2016 MX2 Combinatorics Marathon

It hadn't occurred to me that my question was ambiguous. My intention was that the original interval remains length 1, and the length of the middle segment would change to get a probability of 1/3.