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HSC 2015 MX2 Permutations & Combinations Marathon (archive) (1 Viewer)

Thread starter Drsoccerball
Start date Aug 8, 2015

Status: Not open for further replies.

D

Drsoccerball

Well-Known Member

#281

Re: 2015 permutation X2 marathon

Probability that all go through one gate(each):
n/n^n x (n-1)/n^n x...x1/n^n

B

braintic

Well-Known Member

#282

Re: 2015 permutation X2 marathon

At Wimbledon, 128 players compete in a knockout tournament.
In this imaginary Wimbledon, the players are seeded 1 to 128, but the seeds are not used to determine the draw - the draw is done randomly.
Assume that every game is won by the higher seed.

(a) What is the probability that the number 2 seed finishes runner-up?

(b) What is the probability that the number 3 seed at least reaches the semi-finals?

Librah

Not_the_pad

#283

Re: 2015 permutation X2 marathon

I has 4 pairs of different coloured jelly beans, if i place these 8 jellybeans in a row, how many different arrangements are possible if no two of the same coloured jellybeans are next to one another.

D

Drsoccerball

Well-Known Member

#284

Re: 2015 permutation X2 marathon

braintic said:
At Wimbledon, 128 players compete in a knockout tournament.
In this imaginary Wimbledon, the players are seeded 1 to 128, but the seeds are not used to determine the draw - the draw is done randomly.
Assume that every game is won by the higher seed.

(a) What is the probability that the number 2 seed finishes runner-up?

(b) What is the probability that the number 3 seed at least reaches the semi-finals?

a) (1/2)^6
b) (1/2)^5 + (1/2)^6 +(1/2)^7
? whats the answer

G

glittergal96

Active Member

#285

Re: 2015 permutation X2 marathon

braintic said:
At Wimbledon, 128 players compete in a knockout tournament.
In this imaginary Wimbledon, the players are seeded 1 to 128, but the seeds are not used to determine the draw - the draw is done randomly.
Assume that every game is won by the higher seed.

(a) What is the probability that the number 2 seed finishes runner-up?

(b) What is the probability that the number 3 seed at least reaches the semi-finals?

a) Seed #2 will reach the finals and hence end up runner up precisely when he is on the opposite side of the draw to player 1. This happens with probability 64/127.

b) Seed #3 will reach the semis precisely when he is the top seeded player in his quarter of the draw. Ie the top two players must lie in the remaining three quarters of the draw. This happens with probability (96/127).(95/126).

G

glittergal96

Active Member

#286

Re: 2015 permutation X2 marathon

Use common sense to test your attempts before asking for answer.

This model isnt HUGELY inaccurate apart from when seeds are close.

In your experience, does the 2nd best player in the world only reach the finals once every 64 grand slams?

RecklessRick

Active Member

#287

Re: 2015 permutation X2 marathon

glittergal96 said:
In your experience, does the 2nd best player in the world only reach the finals once every 64 grand slams?

Grand slam draws are not random, they are determined based on seeds.

G

glittergal96

Active Member

#288

Re: 2015 permutation X2 marathon

RecklessRick said:
Grand slam draws are not random, they are determined based on seeds.

Am well aware of this. So this fact is to the benefit of seed #2 player trying to reach the finals in real life vs this model.

On the flipside, the fact that he cannot possibly lose to any of the players other than seed #1 in this model is a big point in favour of how frequently he will reach the finals in this model vs real life.

These two facts cancel each other out to some extent.

In any case, alarm bells should be ringing in your head at the number 1/64.

B

braintic

Well-Known Member

#289

Re: 2015 permutation X2 marathon

$$A game of 'Speed Blue Ball' is played between three teams, with only one winner.$

$$A knockout tournament is played between $3^n$ teams, in a balanced draw of $n$ rounds.$

$Assume all teams are equally likely to win a match against all opponents$

$$Prove by induction that the probability that two particular teams meet in the tournament is $3^{1-n}$

This is similar to a past HSC question.
(Could we allow 48 hours for 2015ers to answer before jumping in. If they get it wrong or have no idea, prompt them with hints.)

D

Drsoccerball

Well-Known Member

#290

Re: 2015 permutation X2 marathon

braintic said:
$$A game of 'Speed Blue Ball' is played between three teams, with only one winner.$

$$A knockout tournament is played between $3^n$ teams, in a balanced draw of $n$ rounds.$

$Assume all teams are equally likely to win a match against all opponents$

$$Prove by induction that the probability that two particular teams meet in the tournament is $3^{1-n}$

This is similar to a past HSC question.
(Could we allow 48 hours for 2015ers to answer before jumping in. If they get it wrong or have no idea, prompt them with hints.)

$1) $ Prove for n=1 $$

$3^0=1 $ Therefore if three teams then they have to play each other, hence n=1 is true $$

$2) $ Assume $ n=k $ is true for k integers $ >0$

$3^k $ teams, probability $ = 3^{1-k}$

$3) $ Prove $ n=k+1$

$3^{k+1}$ teams $= 3^{-k} $ probability $$

$$ From assumption: if there is k+1 teams $ 3^{k+1} $ probability= 3^{1-k} \cdot ( $ another win $)$

$$ Since the probability of another win $ = \frac{1}{3} $ The probability of $ 3^{k+1} $ teams $ = 3^{-k}$

$\therefore n=k+ 1 $ is true $$

$Insert induction conclusion.$

B

braintic

Well-Known Member

#291

Re: 2015 permutation X2 marathon

Drsoccerball said:
$1) $ Prove for n=1 $$

$3^0=1 $ Therefore if three teams then they have to play each other, hence n=1 is true $$

$2) $ Assume $ n=k $ is true for k integers $ >0$

$3^k $ teams, probability $ = 3^{1-k}$

$3) $ Prove $ n=k+1$

$3^{k+1}$ teams $= 3^{-k} $ probability $$

$$ From assumption: if there is k+1 teams $ 3^{k+1} $ probability= 3^{1-k} \cdot ( $ another win $)$

$$ Since the probability of another win $ = \frac{1}{3} $ The probability of $ 3^{k+1} $ teams $ = 3^{-k}$

$\therefore n=k+ 1 $ is true $$

$Insert induction conclusion.$

I'm afraid the inductive step needs a lot more work than that.
There is more than just adding 'another win'.
You have to consider the chance of them meeting in the first round, or meeting in subsequent rounds. We are considering the probability two teams meet - it is not obvious that adding an extra round multiplies the probability by 1/3, even though it turns out that way in the end.

Last edited: Sep 29, 2015

Zlatman

Member

#292

Re: 2015 permutation X2 marathon

braintic said:
$$A game of 'Speed Blue Ball' is played between three teams, with only one winner.$

$$A knockout tournament is played between $3^n$ teams, in a balanced draw of $n$ rounds.$

$Assume all teams are equally likely to win a match against all opponents$

$$Prove by induction that the probability that two particular teams meet in the tournament is $3^{1-n}$

This is similar to a past HSC question.
(Could we allow 48 hours for 2015ers to answer before jumping in. If they get it wrong or have no idea, prompt them with hints.)

I hope this image is clear enough:
http://imgur.com/dhLssUm

EDIT: Latex added

$$Prove true for n=1:$

$\begin{align*} 1 \: round &= 3 \: teams \\ &= 1 \: game \\ P &= 1 \\ &= 3^0 \\ &= 3^{1-1} \end{align*}$

$$Therefore, this statement is true for $n=1$.$

$Assume true for $n=k$:$

$$i.e. $ P(k\:rounds) = 3^{1-k}$

$Prove true for $n=k+1$:$

$$i.e. $ P(k+1 \: rounds) = 3^{-k}$

$\begin{align*}P(meeting \: in \: k+1 \: rounds) &= P(meeting \: round \: one) + P(meeting \: in \: following \: k \: rounds) \\ &= \frac{2}{3^{k+1}} + P(both \: teams \: winning) * P(meeting \: in \: k \: rounds) \\ &= \frac{2}{3^{k+1}} + \frac{1}{3} * \frac{1}{3} * 3^{1-k} (by \: the \: assumption) \\ &= \frac{2}{3^{k+1}} + \frac{1}{3^{k+1}} \\ &= \frac{3}{3^{k+1}} \\ &= \frac{1}{3^{k}} \\ &= 3^{-k} \end{align*}$

$Therefore, by mathematical induction, the probability two particular teams will meet in the tournament is $ 3^{1-n} $.$

Last edited: Sep 29, 2015

B

braintic

Well-Known Member

#293

Re: 2015 permutation X2 marathon

Zlatman said:
I hope this image is clear enough:
http://imgur.com/dhLssUm

EDIT: Latex added

$$Prove true for n=1:$

$\begin{align*} 1 \: round &= 3 \: teams \\ &= 1 \: game \\ P &= 1 \\ &= 3^0 \\ &= 3^{1-1} \end{align*}$

$$Therefore, this statement is true for $n=1$.$

$Assume true for $n=k$:$

$$i.e. $ P(k\:rounds) = 3^{1-k}$

$Prove true for $n=k+1$:$

$$i.e. $ P(k+1 \: rounds) = 3^{-k}$

$\begin{align*}P(meeting \: in \: k+1 \: rounds) &= P(meeting \: round \: one) + P(meeting \: in \: following \: k \: rounds) \\ &= \frac{2}{3^{k+1}} + P(both \: teams \: winning) * P(meeting \: in \: k \: rounds) \\ &= \frac{2}{3^{k+1}} + \frac{1}{3} * \frac{1}{3} * 3^{1-k} (by \: the \: assumption) \\ &= \frac{2}{3^{k+1}} + \frac{1}{3^{k+1}} \\ &= \frac{3}{3^{k+1}} \\ &= \frac{1}{3^{k}} \\ &= 3^{-k} \end{align*}$

$Therefore, by mathematical induction, the probability two particular teams will meet in the tournament is $ 3^{1-n} $.$

On the right track. But the probability that they meet in round one is actually:

$\frac{2}{3^{k+1}-1}$

This mistake simplifies the algebra drastically.

Zlatman

Member

#294

Re: 2015 permutation X2 marathon

braintic said:
On the right track. But the probability that they meet in round one is actually:

$\frac{2}{3^{k+1}-1}$

This mistake simplifies the algebra drastically.

Oh, right! Yeah, something seemed off in that first round, I couldn't figure it out. Thanks, I'll try to fix that!

Zlatman

Member

#295

Re: 2015 permutation X2 marathon

$$Prove true for n=1:$

$\begin{align*} 1 \: round &= 3 \: teams \\ &= 1 \: game \\ P &= 1 \\ &= 3^0 \\ &= 3^{1-1} \end{align*}$

$$Therefore, this statement is true for $n=1$.$

$Assume true for $n=k$:$

$$i.e. $ P(k\:rounds) = 3^{1-k}$

$Prove true for $n=k+1$:$

$$i.e. $ P(k+1 \: rounds) = 3^{-k}$

$\begin{align*}P(meeting \: in \: k+1 \: rounds) &= P(meeting \: round \: one) + P(meeting \: in \: following \: k \: rounds) \\ &= \frac{2}{3^{k+1} - 1} + P(both \: teams \: winning) * P(meeting \: in \: k \: rounds) \\ &= \frac{2}{3^{k+1} - 1} + \frac{3^{k+1} - 3}{3^{k+1} - 1} * \frac{1}{3} * \frac{1}{3} * 3^{1-k} (by \: the \: assumption) \\ &= \frac{2}{3^{k+1} - 1} + \frac{1 - 3^{-k}}{3^{k+1} - 1} \\ &= \frac{3 - 3^{-k}}{3^{k+1} - 1} \\ &= \frac{3^{k+1} - 1}{3^{k+1} - 1} * 3^{-k} \\ &= 3^{-k} \end{align*}$

$Therefore, by mathematical induction, the probability two particular teams will meet in the tournament is $ 3^{1-n} $.$

B

braintic

Well-Known Member

#296

Re: 2015 permutation X2 marathon

Zlatman said:
$$Prove true for n=1:$

$\begin{align*} 1 \: round &= 3 \: teams \\ &= 1 \: game \\ P &= 1 \\ &= 3^0 \\ &= 3^{1-1} \end{align*}$

$$Therefore, this statement is true for $n=1$.$

$Assume true for $n=k$:$

$$i.e. $ P(k\:rounds) = 3^{1-k}$

$Prove true for $n=k+1$:$

$$i.e. $ P(k+1 \: rounds) = 3^{-k}$

$\begin{align*}P(meeting \: in \: k+1 \: rounds) &= P(meeting \: round \: one) + P(meeting \: in \: following \: k \: rounds) \\ &= \frac{2}{3^{k+1} - 1} + P(both \: teams \: winning) * P(meeting \: in \: k \: rounds) \\ &= \frac{2}{3^{k+1} - 1} + \frac{3^{k+1} - 3}{3^{k+1} - 1} * \frac{1}{3} * \frac{1}{3} * 3^{1-k} (by \: the \: assumption) \\ &= \frac{2}{3^{k+1} - 1} + \frac{1 - 3^{-k}}{3^{k+1} - 1} \\ &= \frac{3 - 3^{-k}}{3^{k+1} - 1} \\ &= \frac{3^{k+1} - 1}{3^{k+1} - 1} * 3^{-k} \\ &= 3^{-k} \end{align*}$

$Therefore, by mathematical induction, the probability two particular teams will meet in the tournament is $ 3^{1-n} $.$

Perfect!

B

braintic

Well-Known Member

#297

Re: 2015 permutation X2 marathon

Arthur, Bernard and Charles fight a three way pistol duel.

Arthur hits his target 3 times out of 10, Bernard never misses his target, and Charles hits his target 50% of the time.
(All combatants know all of these statistics.)

They take turns to fire cyclically in the order Arthur, Bernard, Charlie (starting with Arthur), each turn involving shooting once at the shooter's choice of target. Assume that someone dies when hit, at which point the remaining two take turns to fire (beginning with whoever didn't make the fatal shot).

Given that Bernard makes the most sensible choices for his survival, what should Arthur's strategy be to maximise his chance of survival?

(An 'obvious' answer is most likely wrong)

Last edited: Sep 29, 2015

D

Drsoccerball

Well-Known Member

#298

Re: 2015 permutation X2 marathon

braintic said:
Arthur, Bernard and Charles fight a three way pistol duel.

Arthur hits his target 3 times out of 10, Bernard never misses his target, and Charles hits his target 50% of the time.
(All combatants know all of these statistics.)

They take turns to fire cyclically in the order Arthur, Bernard, Charlie (starting with Arthur), each turn involving shooting once at the shooter's choice of target. Assume that someone dies when hit, at which point the remaining two take turns to fire (beginning with whoever didn't make the fatal shot).

Given that Bernard makes the most sensible choices for his survival, what should Arthur's strategy be to maximise his chance of survival?

(An 'obvious' answer is most likely wrong)

Probability that A kills B on the first go and eventually kills C = $\frac{9}{120}$

Probability that A goes for B misses but kills him later: $\frac{7}{10} \cdot \frac{3}{10} = \frac{21}{100}$

$\therefore $ Total probability to survive if he goes for B first $ : \frac{363}{1300}$

$$ This is then compared to if he goes for C first then he misses and then kills B $:$

$\frac{70}{100} \cdot \frac{30}{100}$

$\therefore $ he would have a Higher probability of surviving if he shoots B first $$

B

braintic

Well-Known Member

#299

Re: 2015 permutation X2 marathon

Drsoccerball said:
Probability that A kills B on the first go and eventually kills C = $\frac{9}{120}$

Probability that A goes for B misses but kills him later: $\frac{7}{10} \cdot \frac{3}{10} = \frac{21}{100}$

$\therefore $ Total probability to survive if he goes for B first $ : \frac{363}{1300}$

$$ This is then compared to if he goes for C first then he misses and then kills B $:$

$\frac{70}{100} \cdot \frac{30}{100}$

$\therefore $ he would have a Higher probability of surviving if he shoots B first $$

You haven't considered a third option! (And no - it has nothing to do with shooting himself)

D

Drsoccerball

Well-Known Member

#300

Re: 2015 permutation X2 marathon

braintic said:
You haven't considered a third option! (And no - it has nothing to do with shooting himself)

I dont see the last case

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