HSC 2015 MX2 Permutations & Combinations Marathon (archive) (4 Viewers)

RealiseNothing · Aug 14, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
You're not including 0 L's 1L's etc...

You are though.

Truncating after the 8th W gives a bijection between the ways Aristotle can win and the arrangements of 8 W's 5 L's.

RealiseNothing · Aug 14, 2015

Re: 2015 permutation X2 marathon

RealiseNothing said:
Either I'm really derping with this question but surely the total outcomes isn't $2^{13}$ ?

Like in reality the sequences WWWWWWWWWWWWW and WWWWWWWWLLLLL should be equivalent, but you are counting them as different by doing this.

RealiseNothing · Aug 14, 2015

Re: 2015 permutation X2 marathon

I think the answer should be $\frac{3}{7}$

RealiseNothing · Aug 14, 2015

Re: 2015 permutation X2 marathon

Arrange 8 W's and 5 L's for Aristotle

$\frac{13!}{8!~5!}=1287$

Arrange 6 W's 7 L's for Socrates

$\frac{13!}{6!~7!}=1716$

Answer is

$\frac{1287}{1287+1716} = \frac{3}{7}$

faisalabdul16 · Aug 14, 2015

Re: 2015 permutation X2 marathon

RealiseNothing said:
Either I'm really derping with this question but surely the total outcomes isn't $2^{13}$ ?

It is. 13 games (extreme case) if played, for each game there is 2 outcomes. So in total, there is 2^13 outcomes

RealiseNothing · Aug 14, 2015

Re: 2015 permutation X2 marathon

faisalabdul16 said:
It is. 13 games (extreme case) if played, for each game there is 2 outcomes. So in total, there is 2^13 outcomes

Should we not be truncating after the 8th win though?

RealiseNothing · Aug 14, 2015

Re: 2015 permutation X2 marathon

Like I'm pretty confident it's 3/7

(if it isn't im going to look like the biggest dickhead)

InteGrand · Aug 14, 2015

Re: 2015 permutation X2 marathon

Haven't thought about the Q much, just decided to verify the answer by the 'boring way'. Considering cases with k losses (0 ≤ k ≤ 5) and summing probabilities over all these cases, it is also 595/2048.

Drsoccerball · Aug 14, 2015

Re: 2015 permutation X2 marathon

InteGrand said:
Haven't thought about the Q much, just decided to verify the answer by the 'boring way'. Considering cases with k losses (0 ≤ k ≤ 5) and summing probabilities over all these cases, it is also 595/2048.

I tried that and i didnt get that answer

InteGrand · Aug 14, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
I tried that and i didnt get that answer

Did you place the condition that the last game played must be a W?

RealiseNothing · Aug 14, 2015

Re: 2015 permutation X2 marathon

I think I have to fix up my bijection actually.

Drsoccerball · Aug 14, 2015

Re: 2015 permutation X2 marathon

InteGrand said:
Did you place the condition that the last game played must be a W?

yes

RealiseNothing · Aug 14, 2015

Re: 2015 permutation X2 marathon

RealiseNothing said:
I think I have to fix up my bijection actually.

Nevermind it's fine.

I don't know anymore lol

RealiseNothing · Aug 14, 2015

Re: 2015 permutation X2 marathon

The problem is I can't find anything wrong with my method which is doing my head in.

We have a bijection between arranging W's and L's and the ways Aristotle can win, but yet something isn't adding up.

InteGrand · Aug 14, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
yes

$With $k$ losses ($k\in \{0,1,2,3,4,5\}$), we have 8 $W$'s and $k$ $L$'s ($W$ representing a win for Aristotle, $L$ a loss), so a total of $8+k$ games. If we restrict the last game to be a $W$, then for the first $7+k$ games, we need to order 7 $W$'s and $k$ $L$'s, which is done in $\binom{7+k}{7}$ ways. So the probability of Aristotle winning in $7+k$ games is $P_k (A \text{ wins})=\binom{7+k}{7}\cdot\left(\frac{1}{2} \right)^{8+k}$. Hence the total probability $\rho$ is $\rho = \sum _{k=0}^5 \binom{7+k}{7}\cdot\left(\frac{1}{2} \right)^{8+k} = \frac{595}{2048}$.$

InteGrand · Aug 14, 2015

Re: 2015 permutation X2 marathon

RealiseNothing said:
The problem is I can't find anything wrong with my method which is doing my head in.

We have a bijection between arranging W's and L's and the ways Aristotle can win, but yet something isn't adding up.

$How did you arrive at your probability? You got the number of ways Aristotle can win right (another way to calculate this is $\sum _{k=0} ^5 \binom{7+k}{7}=1287$), but maybe screwed up the probability part?$

Drsoccerball · Aug 14, 2015

Re: 2015 permutation X2 marathon

InteGrand said:
$With $k$ losses ($k\in \{0,1,2,3,4,5\}$), we have 8 $W$'s and $k$ $L$'s ($W$ representing a win for Aristotle, $L$ a loss), so a total of $8+k$ games. If we restrict the last game to be a $W$, then for the first $7+k$ games, we need to order 7 $W$'s and $k$ $L$'s, which is done in $\binom{7+k}{7}$ ways. So the probability of Aristotle winning in $7+k$ games is $P_k (A \text{ wins})=\binom{7+k}{7}\cdot\left(\frac{1}{2} \right)^{8+k}$. Hence the total probability $\rho$ is $\rho = \sum _{k=0}^5 \binom{7+k}{7}\cdot\left(\frac{1}{2} \right)^{8+k} = \frac{595}{2048}$.$

why is it 7 and not 8 ?
EDIT: fixed the last?

InteGrand · Aug 14, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
why is it 7 and not 8 ?

$There's a total of $8$ wins, but the last game is fixed as a win, so in terms of \emph{ordering} games, we only need to consider ordering the first 7 $W$'s in the first $7+k$ games ($0\leq k\leq 5$, $k$ integer, equal to the number of losses).$

RealiseNothing · Aug 14, 2015

Re: 2015 permutation X2 marathon

InteGrand said:
$How did you arrive at your probability? You got the number of ways Aristotle can win right (another way to calculate this is $\sum _{k=0} ^5 \binom{7+k}{7}=1287$), but maybe screwed up the probability part?$

Yep I got A winning in 1287 ways.

Probably did something really stupid with the probability.

InteGrand · Aug 14, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
why is it 7 and not 8 ?
EDIT: fixed the last?

The last game must be a W for Aristotle to win the competition.

HSC 2015 MX2 Permutations & Combinations Marathon (archive) (4 Viewers)

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