Past HSC Q8. (1 Viewer)

[KFunk]

Hey guys, this is a question I just did from a past HSC paper. I'm looking at the last part of the question ( the 2^1-n bit) in particular but I've written up the rest of it as a lead in. I got the answer by extrapolating the pattern, factorising and summing but the way the answer looks I swear there must be some kind of simple elegant solution to it . Anyhow, if you have the time, give it a go at it and let me know if you can find a simple way to do the last part. Here it is:


"Eight players enter a knock-out singles tennis tournament in which each of the first four round winners plays one second round game to decide who enters the final.

Assuming that all players are equally likely to win a game, show that the probability that two particular entrants play each other in the tournament is ¹/₄

Also show that if sixteen persons enter the tournament, then the probability that two players meet is ¹/₈

Prove that for a similar knock-out tournament for 2ⁿ players, that the probability that two players meet is 2^1-n"

 

[damo676767]

i got it out

number of matches over number of pairs

=(2^n -1 ) / (2^n)C2

=2(2^n -1)(2^n -2)!/2^n!

=2(2^n -1)(2^n -2)!/ 2^n (2^n -1) (2^n - 2)!

=2/2^n

=2^(1-n)

 

[Stefano]

i got it out

number of matches over number of pairs

=(2^n -1 ) / (2^n)C2

=2(2^n -1)(2^n -2)!/2^n!

=2(2^n -1)(2^n -2)!/ 2^n (2^n -1) (2^n - 2)!

=2/2^n

=2^(1-n)

Good work. Care to explain your reasoning?

 

[KFunk]

Interesting, I'd like to know your reasoning as well. For example, how does ^(2ⁿ)C₂ represent the 'number of pairs'?

 

[haboozin]

isn't that from like 1982 hsc... wow so old more than 20 yrs

didnt they make 4unit easier after that year.

 

[Stefano]

isn't that from like 1982 hsc... wow so old more than 20 yrs

LOL - you've done all past papers back to 1982...? wow. . . . .

 

[Sepulchres]

If you use that formula for the first one, you get 1/7. Coincidentally, you also get this if you find the probability of playing a particular player in the FIRST round. ie. 7 other people, P(playng 1 particular)= 1/7.

So yea, reasoning please.

 

[damo676767]

Interesting, I'd like to know your reasoning as well. For example, how does ^(2ⁿ)C₂ represent the 'number of pairs'?

ok, i assume every body gets my working, its just my reasoning

first of all there are 2^n people, and everybody bar 1 gets knocked out. There for there will be 2^n - 1 people knocked out, THat meens there are 2^n - 1 matches.

as for total number of pairs

^XC_Y


meens how many ways can X things be aragned into groups of Y
(of course it meens many other things too)


x = 2^n as there are that many people

y = 2 as they are pared

If you use that formula for the first one, you get 1/7. Coincidentally, you also get this if you find the probability of playing a particular player in the FIRST round. ie. 7 other people, P(playng 1 particular)= 1/7.

So yea, reasoning please.

the formula i ended up with is

p=2^(1-n)

in the first circum stance there are 8 people

8=2^n
n=3

p=2^(1-3)
= 1/8




any other questions

 

[Ogden_Nash]

That's a really nice method damo but KFunk, there's nothing wrong with yours. In fact, judging by the first two parts to the question, it's probably the way they wanted you to approach it. It's not that messy either, considering it's a Q8.

 

[KFunk]

any other questions

Just one .

as for total number of pairs

^XC_Y


meens how many ways can X things be aragned into groups of Y
(of course it meens many other things too)

^XC_Y is the number of size-y subsets of a size-x set (more or less) so then ^2ⁿC₂ represents the number of unique pairings that can be made from 2ⁿ people. What I still don't understand is why dividing the number of individual matches by the number of unique pairings yields the probability of two particular people versing each other in a match.

 

[Sepulchres]

EDIT: dont worry. I was mistaken

 

[damo676767]

Just one .



^XC_Y is the number of size-y subsets of a size-x set (more or less) so then ^2ⁿC₂ represents the number of unique pairings that can be made from 2ⁿ people. What I still don't understand is why dividing the number of individual matches by the number of unique pairings yields the probability of two particular people versing each other in a match.

ok lets simplify it, i am rolling a die

there is 1 way for me to get a 5

there are 6 posible way

there for the probability of me getting a 5 is 1/6

it is the same thing

the number of matches = the number of people that are paired against each other

which is the amount of possible ways that it can come true


and the amount of possible pares is ^{(2^n)}C₂


do you understand that?

 

[KFunk]

the number of matches = the number of people that are paired against each other

which is the amount of possible ways that it can come true


and the amount of possible pares is ^{(2^n)}C₂


do you understand that?

I'm afraid not. I don't think my brain is made to understand your logic. Part of it is that I'm not sure what you're refering to with all thes things. The number of people paired against each other is 2^n, but becomes 2^(n-1) in round two and so on... Saying that the number of matches (2^n - 1) is equal to something which is variable depending on what you refer to (and never seems to equal it) doesn't make sense to me.

 

[Sepulchres]

I think I'm slowly getting it...

I'm afraid not. I don't think my brain is made to understand your logic. Part of it is that I'm not sure what you're refering to with all thes things. The number of people paired against each other is 2^n, but becomes 2^(n-1) in round two and so on... Saying that the number of matches (2^n - 1) is equal to something which is variable depending on what you refer to (and never seems to equal it) doesn't make sense to me.

[(2^n) -1] is the total number of matches played in a tournament. Ie. In the first tournament with 8 people, 4 are played in the first, 2 in the second plus the final =7 games in total.

2^nC2 is the total number of unique pairs possible.

So, the number of pairings which _actually_ occur in the tournament over the total _possible_ number of pairs in the tournament is how he worked it out.

However, I think that P(win)= 1/2 has a significance in this question and would be needed to work it out otherwise, esp as it can be written at 2^-1?

I dont know, this is quite confusing.

 

[damo676767]

I'm afraid not. I don't think my brain is made to understand your logic.

ok then
back to basics


the way that probibility works is the amount of ways that it will be true over the amount of ways it can happen

eg
flipping a coin getting heads
1(only 1 head)/2(only 2 things that can happen, heads or tails)

rolling a die getting less than 3
2(can only be 1 or 2)/6(it could be 6 posible numbers, 1-6)
1/3


so if we are both playing in the tennis tournament with 2ⁿ people

the answer is

the amount of chances we have of being pared together / the total amount of pairs



the amount of chances we have of being parded together is the number of matches.

so how many matches are there going to be

well everybody bar 1 losses a match, when you lose yo are out. there are 2ⁿ people therfore there are 2ⁿ - 1 matches

there fore the amount of chances we have of being pared together = 2ⁿ - 1

but this could be and of the pairs, so to make sure it is us we must devide it by the total amount of pairs it could be


the total amount of pairs is ^2ⁿC₂



is that clearer or did i just say the same thing

 

[Sepulchres]

Yea, I get it now. Good stuff.

 

[KFunk]

the total amount of pairs is ^2ⁿC₂

This is my main hang up
^2ⁿC₂ = 2^n-1(2ⁿ - 1)

In the first round you have 2^n-1 pairs, then 2^n-2 in the next and so on until you have 1 pair in the last round. If you sum these you get (2ⁿ - 1), i.e. the total number of pairs formed during the tournament is the same as the number of individual matches (which you can deduce logically as well). But this number ≠ ^2ⁿC₂. What are you representing with ^2ⁿC₂ and how did you arrive at it?

 

[Sepulchres]

This is my main hang up
^2ⁿC₂ = 2^n-1(2ⁿ - 1)

In the first round you have 2^n-1 pairs, then 2^n-2 in the next and so on until you have 1 pair in the last round. If you sum these you get (2ⁿ - 1), i.e. the total number of pairs formed during the tournament is the same as the number of individual matches (which you can deduce logically as well). But this number ≠ ^2ⁿC₂. What are you representing with ^2ⁿC₂ and how did you arrive at it?

Yea, its cause the number of pairs that ARE actually formed during the tournament is [(2^n) -1]. Conversely, (2^n)C2 corresponds to the number of POSSIBLE pairs, not how many ARE actually formed during the tournament. Knaw mean.

 

[damo676767]

This is my main hang up
^2ⁿC₂ = 2^n-1(2ⁿ - 1)

In the first round you have 2^n-1 pairs, then 2^n-2 in the next and so on until you have 1 pair in the last round. If you sum these you get (2ⁿ - 1), i.e. the total number of pairs formed during the tournament is the same as the number of individual matches (which you can deduce logically as well). But this number ≠ ^2ⁿC₂. What are you representing with ^2ⁿC₂ and how did you arrive at it?

i dont think you understand

2ⁿ - 1 is the amount of pairs that actually happen

^2ⁿC₂ is the total amount of ways that they can be paired up. but not all of them will play each other.

 

[KFunk]

It's clicking into place, thanks for bearing with me .