HSC 2016 MX2 Combinatorics Marathon (archive) (2 Viewers)

[InteGrand]

Re: HSC 2016 MX2 Combinatorics Marathon

Quasi-journal-like paper on the variants

Blog post on the variants

Another blog post

An officially published paper???

Ah yeah Monty Small is another. Knew there was another variant rhyming with 'Hall' that I'd forgotten.

 

[braintic]

Re: HSC 2016 MX2 Combinatorics Marathon

A reminder that there are still two questions unanswered:







(2) Three door Monty Hall problem - but this time when the host has two doors to choose from, he is biased towards one selection. Let's say he picks the leftmost of the two remaining doors with probability p. Furthermore, you know which door he favours.

This time your probabilities of winning by switching are affected by which door the host opens.
Find formulae in terms of p for the probability he wins based on which door the host opens.
Explain what happens in the case when p = 0 or 1.
And find the expected number of wins per game by switching (in the variable p case).

 

[Carrotsticks]

Re: HSC 2016 MX2 Combinatorics Marathon

2n soldier applicants, choose a team of n, then choose k reserves from n leftovers. Or choose k reserves first, then n team members from remainder of 2n-k.

 

[seanieg89]

Re: HSC 2016 MX2 Combinatorics Marathon

Spoiler for second Monty Hall variant: (highlight to read)

I think this is just an exercise in conditional probability. We let A(k) denote the event that the k-th door is correct and B(k) denote the event that Monty opens the k-th door. (Where we let door 1 be the door that we opened.)

Then:

P(B(2)|A(1))=p
P(B(3)|A(1))=1-p
P(B(2)|A(3))=1
P(B(3)|A(2))=1

Then

P(B(2))=(1+p)/3
P(B(3))=(2-p)/3

and

P(A(2)|B(3))=P(B(3)|A(2))P(A(2))/P(B(3))=1/(2-p)

and similarly

P(A(3)|B(2))=1/(1+p).

I gunned for these conditional probabilities from the start because they are precisely what we want. They tell us the expected value of switching given that Monty has opened the 3rd door (resp 2nd door). Note that they both reduce to the familiar 2/3 if p=1/2.

Now I think the expected value of the strategy of switching regardless (which is still optimal because both of the probabilities computed above are at least 1/2) is still exactly 2/3. This is because the switching strategy is still successful if and only if your original choice is incorrect.

So as it happens, the extra information you have (knowledge of p) does not change how you play this variant of Monty Hall. However, if there were more interactions (eg you could "double down" or something) after Monty opens a door, this could be different, especially if p greatly differs from 1/2.

Eg in the extreme case of p=1 (essentially same as p=0 by symmetry of doors), Monty will always open door 2 if he has the choice.
So if he does open door 2, this is not telling us much at all, in fact if we compute the probabilities above we obtain P(A(3)|B(2))=1/2. Switching is exactly as good as staying.

If he opens door 3 though, this is telling us a great deal, as this means he did not have a choice! I.e. there is a 100% chance that door 2 contains the car, and at this point I would offer Monty a sizeable sidebet .

 

[Paradoxica]

Re: HSC 2016 MX2 Combinatorics Marathon

How many ways can a 2*n grid of squares be tiled by 2*1 dominoes?

Consider analysing small values of n before making your conclusion.

 

[KingOfActing]

Re: HSC 2016 MX2 Combinatorics Marathon

How many ways can a 2*n grid of squares be tiled by 2*1 dominoes?

Consider analysing small values of n before making your conclusion.

Fibonacci (+ a shifted index)?

Starting with n = 1, the sequence is 1,2,3,5,8,13,...

The argument is to denote a_n the amount of tilings possible with a 2xn grid. Obviously a_1 = 1, a_2 = 2

Then consider n > 2

The grid will either end with one vertically (with a 2x(n-1) grid left), or by two horizontally (with a 2x(n-2) grid left). Hence the amount of tilings follows

 

[porcupinetree]

Re: HSC 2016 MX2 Combinatorics Marathon

 

[Paradoxica]

Re: HSC 2016 MX2 Combinatorics Marathon

Wow

Stop with the eggplants already.

 

[braintic]

Re: HSC 2016 MX2 Combinatorics Marathon

2n soldier applicants, choose a team of n, then choose k reserves from n leftovers. Or choose k reserves first, then n team members from remainder of 2n-k.

It took me a while to figure out you had cross multiplied.

I was thinking more of a probability story:

LHS is the probability that if you choose k from the 2n, they will all come from a specified subset of n.
RHS is the probability that if you choose a group of n from the 2n, they will not contain k specified people.

 

[braintic]

Re: HSC 2016 MX2 Combinatorics Marathon

In a particular maths test sat by a class of 31 students, no two students get the same mark.
11 of these students are randomly chosen.
The medians of the class and the chosen group are calculated.
What is the probability that the two medians are equal?

 

[braintic]

Re: HSC 2016 MX2 Combinatorics Marathon

In a particular maths test sat by a class of 31 students, no two students get the same mark.
11 of these students are randomly chosen.
The medians of the class and the chosen group are calculated.
What is the probability that the two medians are equal?

No takers? It really is quite easy once you look at it the right way. Definitely a 3U question.

 

[Ambility]

Re: HSC 2016 MX2 Combinatorics Marathon

In a particular maths test sat by a class of 31 students, no two students get the same mark.
11 of these students are randomly chosen.
The medians of the class and the chosen group are calculated.
What is the probability that the two medians are equal?

 

[math man]

Re: HSC 2016 MX2 Combinatorics Marathon

Don't know if this question has been asked, but...
NEW
Q. In a standard deck of cards (52 cards) there are 4 suits and the cards are numbered ace, 2-10 jack, queen and king.
Find the probability of getting a straight?
(A straight is any five cards in a row, 10, jack, queen, King, ace counts as straight)

 

[wu345]

Re: HSC 2016 MX2 Combinatorics Marathon

Don't know if this question has been asked, but...
NEW
Q. In a standard deck of cards (52 cards) there are 4 suits and the cards are numbered ace, 2-10 jack, queen and king.
Find the probability of getting a straight?
(A straight is any five cards in a row, 10, jack, queen, King, ace counts as straight)

Number of lowest cards of straight (i.e 2-10)=9*4 (assuming ace-2-3-4-5 doesn't count?)
Therefore number of straights=9*4*(4^4)
P (straights)=9*4^5/52C5
=192/54145

 

[leehuan]

Re: HSC 2016 MX2 Combinatorics Marathon

Number of lowest cards of straight (i.e 2-10)=9*4 (assuming ace-2-3-4-5 doesn't count?)
Therefore number of straights=9*4*(4^4)
P (straights)=9*4^5/52C5
=192/54145

A2345 doesn't count but technically in poker a straight needs to eliminate the straight flush

 

[math man]

Re: HSC 2016 MX2 Combinatorics Marathon

Number of lowest cards of straight (i.e 2-10)=9*4 (assuming ace-2-3-4-5 doesn't count?)
Therefore number of straights=9*4*(4^4)
P (straights)=9*4^5/52C5
=192/54145

Ace-5 does count. And straight flush is not a straight

 

[leehuan]

Re: HSC 2016 MX2 Combinatorics Marathon

Yeah so it's P(5 in a row) - P(S.F.)

Need to make it clear that A-5 counts though because conventionally in poker it does not.

 

[math man]

Re: HSC 2016 MX2 Combinatorics Marathon

Yeah so it's P(5 in a row) - P(S.F.)

Need to make it clear that A-5 counts though because conventionally in poker it does not.

What poker do you play? All poker and Texas hold 'em I've played count it.
Also that's why these types of questions would never be asked in HSC , too many rules and not everyone knows them

 

[leehuan]

Re: HSC 2016 MX2 Combinatorics Marathon

Either YPP rigged it or I have never seen A2345 count when I play Texas Hold'em

If that's so I might need to watch a poker video or two. I legit have never seen it count yet.
Of course, I don't gamble with real money though.

I know that the latter statement is true though. But given the nature of BoS I don't really mind the question being posted.

 

[math man]

Re: HSC 2016 MX2 Combinatorics Marathon

Either YPP rigged it or I have never seen A2345 count when I play Texas Hold'em

If that's so I might need to watch a poker video or two. I legit have never seen it count yet.
Of course, I don't gamble with real money though.

I know that the latter statement is true though. But given the nature of BoS I don't really mind the question being posted.

The HSC has to be careful with questions they give cause a lot of foreigners come to nsw and they don't know all these concepts of rules