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

[InteGrand]

Re: 2015 permutation X2 marathon

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

1/2 ? It doesn't matter if he switches.

 

[braintic]

Re: 2015 permutation X2 marathon

Something that is interesting about the Monty Hall problem (assuming the host always reveals a goat, so the probabilities are 1/3 and 2/3):

If someone walks onto the set after the host has opened a door (so he doesn't know the player's original choice) and bets on the location of the car, it will be a 50-50 bet for him.
So each person will have a different probability for each door, yet BOTH sets of probabilities will be borne out by repeated trials.

 

[InteGrand]

Re: 2015 permutation X2 marathon

Something that is interesting about the Monty Hall problem (assuming the host always reveals a goat, so the probabilities are 1/3 and 2/3):

If someone walks onto the set after the host has opened a door (so he doesn't know the player's original choice) and bets on the location of the car, it will be a 50-50 bet for him.
So each person will have a different probability for each door, yet BOTH sets of probabilities will be borne out by repeated trials.

Yep. The knowledge of the observer is everything.

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

Are these all similar to the "not looking" pick from a bag full of marbles ?

 

[braintic]

Re: 2015 permutation X2 marathon

Are these all similar to the "not looking" pick from a bag full of marbles ?

Aren't all picks "not looking"? It is hard to pick randomly if you look.

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

Aren't all picks "not looking"? It is hard to pick randomly if you look.

No like the resulting marbles. If I don't look at put it back vs if i look then put it back.

 

[Silly Sausage]

Re: 2015 permutation X2 marathon

You can think of it as this way: If the player picks the door with the goat initially and switches, his chances of winning the goat is 100%. Since the probabilities of which of the two doors with the car behind it do NOT change and that the player does know which door the car is behind, the probability of winning the car is . If he doesn't switch his chances of winning the car is obviously .

 

[InteGrand]

Re: 2015 permutation X2 marathon

You can think of it as this way: If the player picks the door with the goat initially and switches, his chances of winning the goat is 100%. Since the probabilities of which of the two doors with the car behind it do NOT change and that the player does know which door the car is behind, the probability of winning the car is . If he doesn't switch his chances of winning the car is obviously .

This was how I used to think of it (and still do mostly), but the problem with it is that I don't think it really explains the case where Monty picks the doors randomly and happens to show a goat.

 

[rand_althor]

Re: 2015 permutation X2 marathon

Fred lives on one of 10 islands sitting in a vast lake. One day a package drops uniformly at random on one of the ten islands. Fred can't swim and has no boat, but luckily there is a teleporter on each island. Each teleporter teleports to only one of the other ten teleporters (note that it may teleport to itself), and no two teleporters teleport to the same teleporter. If the configuration of the teleporters is chosen uniformly at random from all configurations that satisfy these constraints, compute the probability that Fred can get to the package using the teleporters.
Answer: 11/20

 

[deboiz]

Re: 2015 permutation X2 marathon

Wait so whats the answer? I reckon its still better to switch

 

[deboiz]

Re: 2015 permutation X2 marathon

Fred lives on one of 10 islands sitting in a vast lake. One day a package drops uniformly at random on one of the ten islands. Fred can't swim and has no boat, but luckily there is a teleporter on each island. Each teleporter teleports to only one of the other ten teleporters (note that it may teleport to itself), and no two teleporters teleport to the same teleporter. If the configuration of the teleporters is chosen uniformly at random from all configurations that satisfy these constraints, compute the probability that Fred can get to the package using the teleporters.
Answer: 11/20

OK well i got 55% is that right hahaha (assuming that not using a teleporter counts as 'using the teleporters')

working: 0.1 + 0.9/(9*10) * (1 + 2 +...+9)

basically i counted permutations l0l

 

[rand_althor]

Re: 2015 permutation X2 marathon

OK well i got 55% is that right hahaha (assuming that not using a teleporter counts as 'using the teleporters')

working: 0.1 + 0.9/(9*10) * (1 + 2 +...+9)

basically i counted permutations l0l

Yep that's the correct answer. I'm guessing the 0.1 is if it lands on the island he is on. Can you explain the second part?

 

[deboiz]

Re: 2015 permutation X2 marathon

SWEET yeah 0.1 is if he lands on the island he's on.

Ok, my explaining skills are actually pretty shit, but here goes.

I numbered the islands from 0 to 9
and then i made a 10 digit number, where first digit is the island that the teleporter on 0 is directed to, 2nd digit is the island that island 1's teleporter goes to

eg. 9876543210

teleporter on 0 is directed to 9

So then, I let the dude start on island 0, and the randomly dropped gift go on island 1 (by symmetry it doesn't matter)

I counted the permutations that were valid, and divided by total permutations

now let's consider a 'chain', which is a cycle of islands that the guy can cycle through. if the situation was 1023456789, then the guy would continually cycle between 0 and 1, so this is a cycle of length 2, and 1203456789 would be a cycle length 3 (cycling through 0, 1, 2)

so consider a cycle of length k, and count the valid permutations provided 0 and 1 are linked in a chain

We pick the k-2 remaining members of the chain
8C(k-2)

We permute them, however noting that a chain 123 is equivalent as chain 231 (sorta a circular thing) BUT different to 213
k! / k

now we permute the remaining 10-k items
(10-k)!

and divide by total perms
10!

so we got

8C(k-2)*(k-1)!(10-k)! / 10!

simplifying to (k-1)/90

so then sum up k= 2, 3, 4, 5, ... 10

and you get (1 + 2 +...+9)

NOW PUTTING THE PIECES TOGETHER

you got 0.1 chance for landing on the island he's on

0.9 for not landing on the island he is on

within the 0.9 you got the probabilty of (1+2+..+9) / 90

hence the answer 0.1 + 0.9/(9*10) * (1 + 2 +...+9) = 0.55

 

[kawaiipotato]

Re: 2015 permutation X2 marathon


Answer: 11/50

 

[RealiseNothing]

Re: 2015 permutation X2 marathon

Two people play a game (call them A and B).

1. Numbers between 1 and 30 inclusive are written on balls and put in a bag.

2. Person A chooses a number, then person B chooses a number.

3. A ball is then drawn.

The closest person to the number drawn wins an amount equal to the number drawn in dollars. Eg. If A chooses 10, B chooses 20, and 16 is drawn, then B wins $16.

Suppose A and B both play optimally, what numbers do A and B choose?

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

Two people play a game (call them A and B).

1. Numbers between 1 and 30 inclusive are written on balls and put in a bag.

2. Person A chooses a number, then person B chooses a number.

3. A ball is then drawn.

The closest person to the number drawn wins an amount equal to the number drawn in dollars. Eg. If A chooses 10, B chooses 20, and 16 is drawn, then B wins $16.

Suppose A and B both play optimally, what numbers do A and B choose?

LOL 30 so they both get $30

 

[InteGrand]

Re: 2015 permutation X2 marathon

LOL 30 so they both get $30

Don't think they're allowed to choose the same number. (And the money the winner gets is the number drawn from the bag.)

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

Don't think they're allowed to choose the same number. (And the money the winner gets is the number drawn from the bag.)

Well if they play wouldn't one pick 1 and the other picks 30?

 

[InteGrand]

Re: 2015 permutation X2 marathon

Well if they play wouldn't one pick 1 and the other picks 30?

No, doubt it.