Thread: How Good Are You At Reasoning?

Are you gonna mark my paper Xei?

I have no idea about your one Dianeva.
I'm gonna just go out on a limb and say 366.
If there were 365 and each birthday has an equal chance, then none would share a birthday.
If you add one more person, their birthday would fall on the same day as someone elses.
But that's not a 50% chance of at least two people sharing the same birthday, that's a 100% chance.
So I dunno lol

182.5?

Oh I see you said "at least" a 50% chance. So I guess I may not be wrong. Probably am anyway lol

182.5 (well 183 whole people) would be the answer to how many people would be needed for a 50% chance that at least one of their birthdays lies on a particular day (ex: a 50% chance that at least one person's birthday is June 10th). But looking for any shared birthday is a different question and the answer is much smaller.

Stubert needs to change his avatar because I keep thinking at first glance that he's tommo lol.

Originally Posted by PhilosopherStoned

Actually, I got both of those correct.

Looking back at mine, I see that I had misread his question as he implies that all cards with even numbers are red on the other side. So the result is of my version #3 as wrong. As for your answer to #6 - it's not necessarily correct


As for those of you lacking an adequate explanation as to how one is to arrive at the answer belonging to question #3. A friend of mine explained it as such: "Suppose ten thousand people take the test. 1% (100 people) have cancer and of those 80 will test positive. Of the 10,000 people, 9900 do not have cancer. 9.6% of those will test positive, which is 950 people. A total of 1030 people test positive, of which 80 have cancer. So if you receive a positive result, the probability you have cancer is 80/1030 = 0.078, abut one chance in 13."

Originally Posted by Dianeva

182.5 (well 183 whole people) would be the answer to how many people would be needed for a 50% chance that at least one of their birthdays lies on a particular day (ex: a 50% chance that at least one person's birthday is June 10th). But looking for any shared birthday is a different question and the answer is much smaller.

Actually the correct answer for that problem 253.

For n people the chances of having at least one birthday lying on a given day is

1 - (364/365)^n

Or (364/365)^n for it not being on a given day.

Chances of it being on a given day for:
n=1, 0.27%
n=10, 2.71%
n=20, 5.34%
n=50, 12.82%
n=100, 23.99%
n=200, 42.23%
n=250, 49.93%
n=300, 56.09%
n=400, 66.63%
n=500, 74.63%
n=750, 87.22%
n=1000, 93.57%
n=1500, 98.37%
n=2000, 99.59%
n=4400, ~100%





For the original birthday problem, 1 - 365!/((365-n)!*365^n) is the chance of having a shared birthday in a group of n people.

Thanks. I saw your edits lol. Before seeing your final answer I did it myself too and got 253 as well.

Final answer obtained from what you had but solving for n, then replacing p with 0.5 and calculating.

p = 1 - (364/365)^n
n = log[base 364/365] (1 - p)

To be fair, I hadn't actually worked it out before. I was copying that answer from some website. I thought those things never lied!
Do the newer graphing calculators let you solve logarithms the 'normal' way with a custom base? I'm sick of having to solve for log[base a]b with log b / log a.

Originally Posted by Dianeva

182.5 (well 183 whole people) would be the answer to how many people would be needed for a 50% chance that at least one of their birthdays lies on a particular day (ex: a 50% chance that at least one person's birthday is June 10th). But looking for any shared birthday is a different question and the answer is much smaller.

Stubert needs to change his avatar because I keep thinking at first glance that he's tommo lol.

How can it be smaller?

Hmmmm....

Gonna think type here....
If there's only 2 days and I need to find the amount of people needed to have a 50% chance of them having a birthday on the same day, the answer would be 2 people.

6 days, there is a 1/6 chance that 1 person will have a birthday on any given day. So.... fuck I just had this.... So to have a 3/6 chance of two people having the same birthday, so.... fuck.... This isn't gonna work, I don't know maths at all lol

Ok, ignore that I think.... Unless I was on the right track, then tell me lol

I think 1/365 chance of any given birthday. But you need it at a 182.5/365 chance.

So you need 91.25 people.

10 possible days. 1/10 chance for any given day. Need a 50% chance of two people on any given same day. If you have two people only, given the 1/10 chance, it raises it to a 1/20 chance. To get it down to a 1/10 chance again you would need ?2 people?
If I fucked that up I'll fuck up the rest of this... but here goes.

365 days. 1/365 chance for any given day. Need 50% chance of two people on any given same day. If there are two people, the chance is 1/730. If there are two people the chance is 1/365. If there are 3 people, the chance is 1/182.5. If there are 4 people, the chance is 1/91.25. If there are 4 people, the chance is 1/45.625. If there are 5 people, the chance is 1/22.8125. If there are 6 people, the chance is 1/11.40625. If there are 7 people, the chance is 1/5.703125. If there are 8 people, the chance is 1/2.8515625.

I'm just gonna say 9 people.

If I were bad at reasoning it would be likely that I'd come up with a bad reason as to why I was good at reasoning (due to my lack of reasoning skills leading me to falsely conclude I was good at it.) Since I haven't done that I must be good at reasoning.

Okay, here's the answers. They've pretty much all been done now, though there are simpler options:

Spoiler for Answers:

1. Coin n gets flipped for every number that divides n except 1, so coins with an odd number of divisors will be heads, and an even number will give tails.

For each number p that divides n, you get a second number q such that p*q = n, for instance, for n = 12

1*12
2*6
3*4,

so the only way a number can have an odd number of divisors is if for one of the divisors the second number is the same, e.g. for 36

1*36
2*18
3*12
4*9
6*6.

In other words, p = q, i.e. n = p*q = p^2, i.e. n is a square number.


2. A line can intersect another straight line at most once. Each time your cut hits another cut, or hits the edge of the pizza, you add another piece. So you get 1 piece for 0 cuts, 1 + 1 = 2 pieces for 1 cut, 2 + 2 = 4 pieces for 2 cuts, 4 + 3 = 7 pieces for 3 cuts, ... , 22 pieces for 6 cuts.

Technically all we have proved is that 22 is the maximum; it could be that it's impossible to intersect each line with a previous cut. But we can draw a diagram to show that it is possible:



And thus the maximum number of pieces for 6 cuts is 22.

It seems 'obvious' that you can always cut the previous lines no matter how many there are, but can you prove it..?


4. Simple math; total people claiming to be tree-huggers is 80 + 950 = 1030, total people claiming to be tree-huggers who are tree-huggers is 80, thus the fraction is 80/1030.

This was not a hard question, so why did I include it? Because it's actually exactly the same as question 3. The probability that somebody is a tree hugger is 100/10,000 = 1%, the probability that you find out they are when they are is 80/100 = 80%, and the probability that you think falsely that they are when they aren't is 950/9900 = 9.6%.


3. Like all of these questions, you don't actually need to know any maths. You just need to know that the probability of something being A when there are n things which are A and m things in total is n/m, which is obvious.

Draw a rectangle representing 10,000 average people. Split it into two groups, those with cancer and those without, and work out the number of people in each group. Then further split each group according to positive and negative test results, so you have four groups, (those with cancer who test positive, those with cancer who test negative, those without cancer who test positive, those without cancer who test negative). Work out the numbers in each and you can then easily work out the ratio of the number people who have cancer and positive tests to the total number of people who have positive tests.

The thing about this question is that if you are clear headed and define exactly what the question means, you easily see that it is the same question as 3, which is just simple arithmetic. Amazingly only 15% of doctors were able to do the question, but far more are able to do it when you change it into a problem about numbers of people rather than probabilities.


5. We obviously need to check that the card with an even number is red on the other side. We do not need to check the red card; if it was odd on the other side then that wouldn't contradict even cards being red (and for the same reason you don't need to turn over the card with a 1). We do however need to check the blue card, because if it were even on the other side then it would contradict even cards being red.

Again hardly anybody gets this, yet they do get the next question, which is exactly the same but with different words.


6. Obviously you need to check that the 16 year old isn't drinking alcohol and that the person drinking beer isn't under 18.


7. If my claim were false, we could have a group of n people where no two of them had the same number of friends, i.e. n people all with a different number of friends.

There are only n options (0 friends, 1 friend, ... , n - 1 friends), thus we must have one person with 0 friends, another person with 1 friend, ... , one person with n - 1 friends, i.e. friends with everybody.

However, if we have somebody who is friends with everybody, we can't have somebody who is friends with nobody, as they would have to be friends with at least this person.

Originally Posted by tommo

I'm just gonna say 9 people.

Given n people, you need to work out the probability that none of them shares a birthday, i.e. n people have n different birthdays, i.e. person 2 doesn't have person 1's birthday, person 3 doesn't have person 2 or person 1's birthday, etc.

As for the other questions you asked for, 1 is wrong because the assertion that all coins get flipped to tails isn't right (just consider coin 4), 2 is correct and for the right reason, 3 is definitely wrong because it would mean testing positive for cancer doesn't make it more likely that you have cancer, 4 I'm guessing was an arithmetic slip, 5 is wrong because a red card having an odd number on the back doesn't contradict my claim that all cards with even numbers on the back are red, for 6 you need to check the drink of the 16 year old, the reason for 7 is wrong (why don't you have to check it for groups of size 3 or more; my claim is about groups of any size?). For more detail check the spoiler.

So apparently the answer to "How good are you at reasoning" is "pretty damn good, but can make silly errors". Which is nothing new to me.

Your answer to #7 is definitely more concise. Why I didn't think to phrase it like that, I don't know :p

I guess it's something you pick up in a maths degree; how to hone in on the crux of your argument and make it as simple and lucid as possible. It always makes the problem look easy, but most of the time it's infuriatingly hard to actually do. You have to learn quite quickly that your 'obvious' argument that you can't quite put into words is in fact probably wrong.

Question 7 is in fact a question from one of the official example sheets for the graph theory course here at Cambridge (well, the exact question was 'show that all graphs of size 2 or greater have two or more vertices of the same order'), so well done to those who got it.

If you want to have a go at honing your logic, here's quite a similar question from a different course:


8. On Hawaii, everybody has mobile phones, but naturally there are some people who don't know the phone numbers of other people (and if Alohi has Laka's number, Laka doesn't necessarily have Alohi's number).

We say that Alohi can communicate with Laka if Alohi knows the number of somebody who can then phone somebody else and so on until finally somebody phones Laka.

It can be shown that this splits the island into a distinct number of communicating groups, in which everybody can communicate with everybody else in the group.

Show that there is at least one communicating group which can't contact anybody outside the group.

[Part 2: show this isn't true for an infinite island].

Originally Posted by Xei

I guess it's something you pick up in a maths degree; how to hone in on the crux of your argument and make it as simple and lucid as possible

Originally Posted by Xei

5. We obviously need to check that the card with an even number is red on the other side. We do not need to check the red card; if it was odd on the other side then that wouldn't contradict even cards being red (and for the same reason you don't need to turn over the card with a 1). We do however need to check the blue card, because if it were even on the other side then it would contradict even cards being red.

I disagree with this. That's what I first thought of. But, you said 5. I have four cards with numbers on one side and colours on the other. I claim that cards with even numbers are red on the other side. I put the cards in front of you.

1, 2, red, blue.

If red were odd, it would contradict the claim. If 1 was red it would also contradict the claim. If 2 was blue, it would also contradict the claim. If blue was 2 it would also contradict the claim.

It is not the same as the drinking question like you said. The drinking question only goes one way. The 21 year old or whatever will not be illegal no matter what. A red card can have an odd number though. Which would contradict your claim.

Originally Posted by Xei

Given n people, you need to work out the probability that none of them shares a birthday, i.e. n people have n different birthdays, i.e. person 2 doesn't have person 1's birthday, person 3 doesn't have person 2 or person 1's birthday, etc.

Again.

1 Knew the first one was wrong lol
2 is correct and for the right reason - Pwned
3 is definitely wrong - Knew that too lol
4 I'm guessing was an arithmetic slip - Nope, although I guess you could call it that. I just haven't done maths in a long time and I'm not good at it.
5 is wrong because a red card having an odd number on the back doesn't contradict my claim that all cards with even numbers on the back are red - As stated, I disagree with this.
6 you need to check the drink of the 16 year old - I said that hapharzadly, but fair enough, was just mucking around.

the reason for 7 is wrong (why don't you have to check it for groups of size 3 or more; my claim is about groups of any size?). For more detail check the spoiler.
You said "7. In any group of (two or more) people, there are always at least two people with the same number of friends in that group. True or false?"
So I just went with 2 people. Not thinking you meant it would have to apply to groups with over 2 people as well.

Originally Posted by Xei

7. If my claim were false, we could have a group of n people where no two of them had the same number of friends, i.e. n people all with a different number of friends.

There are only n options (0 friends, 1 friend, ... , n - 1 friends), thus we must have one person with 0 friends, another person with 1 friend, ... , one person with n - 1 friends, i.e. friends with everybody.

However, if we have somebody who is friends with everybody, we can't have somebody who is friends with nobody, as they would have to be friends with at least this person.

I have no idea what you mean here.

Originally Posted by tommo

I disagree with this. That's what I first thought of. But, you said 5. I have four cards with numbers on one side and colours on the other. I claim that cards with even numbers are red on the other side. I put the cards in front of you.

1, 2, red, blue.

If red were odd, it would contradict the claim. If 1 was red it would also contradict the claim. If 2 was blue, it would also contradict the claim. If blue was 2 it would also contradict the claim.

It is not the same as the drinking question like you said. The drinking question only goes one way. The 21 year old or whatever will not be illegal no matter what. A red card can have an odd number though. Which would contradict your claim.

It's subtle but 'the cards with even numbers are red' is not the same thing as 'the cards with even numbers are the red cards'. The latter means the evens and only the evens are red and thus a red card must be even, the former does not. There could be three red cards in that deck, but it would still be true that 'the even cards are red' (but not true that the even cards are the red cards).

I have no idea what you mean here.

Not sure how I can simplify it. The claim is that in any group of people, some people have the same number of friends. If we think the claim is not true, we must be claiming that we can find a group where everybody has a different number of friends. We show that that is impossible. It's a proof by contradiction. It's the same method you would use to prove the claim that everybody on Earth but you is in your room.

... just means fill in the blanks in the obvious way for the rest of the cases.

Originally Posted by Xei

It's subtle but 'the cards with even numbers are red' is not the same thing as 'the cards with even numbers are the red cards'. The latter means the evens and only the evens are red and thus a red card must be even, the former does not. There could be three red cards in that deck, but it would still be true that 'the even cards are red' (but not true that the even cards are the red cards).

Ok I'm honestly not trolling here or anything. But I simply don't agree.
Even if you are correct, it is incredibly vague and therefore not a good question.

I claim that cards with even numbers are red on the other side. I put the cards in front of you.

1, 2, red, blue.
red, blue 3 4

You said the cards with even numbers are red on the other side.

In this case, I am showing a possible result from flipping them all over, with the bottom line being the opposite side.

1 is red on the other side, proving your claim false - Because the red card has an odd number
2 is blue, proving your claim false - Because the even number is blue
red is 3, proving your claim false - Because the red card has an odd number
blue is 4, proving your claim false - Because the even number is blue

"We do not need to check the red card; if it was odd on the other side then that wouldn't contradict even cards being red"
See, the way you worded the question makes it seem like red cards also have to be even numbers. In fact, it's the only correct way to interpret your question.

I never said that there were two reds and two blues. I also never said that there were cards with a 3 and a 4 on the back.

If this had been the question you would be correct, but it wasn't, and it isn't ambiguous. I made no statements at all about how many of each there were, and it isn't implicit.

I never said there were two of each either....
Never mind.... It's obvious you don't understand what I'm saying. I'll leave it at that.

Well no I don't really understand, because what you're saying is logically wrong and thus impossible to understand. 'Even cards are red' doesn't mean 'red cards are even'. 'English people are British' doesn't mean 'British people are English'. This just follows from the basic meaning of the words and I still don't think you've explained it.

red=even even=red.

If the question said "all even cards are red, but not all red cards are even" then I'd say your answer was correct, and it was what I was thinking in the first place. I just changed my mind because it seemed too obvious.

Well, 'all even cards are red, but not all red cards are even' is equivalent to 'even cards are red'. It's just straight from the definition of how qualification works. It's clear when you consider more intuitive adjectives in exactly the same grammatical structure. For example, 'bats are mammals' means 'all bats are mammals, but not all mammals are necessarily bats'.

Originally Posted by tommo

But I simply don't agree.

Do you often find yourself disagreeing with logic?

Nope. Just Chuck Testa.

Originally Posted by Xei

1. Coin n gets flipped for every number that divides n except 1, so coins with an odd number of divisors will be heads, and an even number will give tails.

For each number p that divides n, you get a second number q such that p*q = n, for instance, for n = 12

1*12
2*6
3*4,

so the only way a number can have an odd number of divisors is if for one of the divisors the second number is the same, e.g. for 36

1*36
2*18
3*12
4*9
6*6.

In other words, p = q, i.e. n = p*q = p^2, i.e. n is a square number.

This is sufficient to claim that the coins that are heads up will be squares. As I pointed out earlier, it's not sufficient to claim that the set of heads-up coins is equivalent to the sets of squares, i.e. that "the sqaures" is the answer. For that, you need to show that every square will have an odd number of divisors...

I don't see how it isn't clear from the construction that a square will have an odd number of divisors? It will have an even number of non-square factors and one square factor. i.e. means equivalence and you can follow the implication backward from the last line if you really want.

I guess it depends on how rigourous you want to be. Normally when a proposition is of the "if and only if" style, one expects to see both sides at least acknowledged.

As for 8:

The problem isn't clearly stated. It could occur that everybody is in the same group (everyone has everyone else;s phone number), in which case we'd have to admit the empty group, hence making the problem trivial. It's also not specified that these are maximal groups in which case the assertion isn't correct either. Simply take the group that can't contact anyone and split it in two.

So lets assume:

1) each group has some maximality condition associated with it. I picked: If person p is not in group G then either p cannot contact anyone in G or no one in G can contact p.
2) there are at least two such groups.



Take a finite group. Pick another group that someone in this group can contact. Continue this way with the second group so that someone in the second group can contact the third group and so on. There are three possibilities.

1) we will eventually pick a group in which there are no external contacts
2) we will loop back around to a group we have already picked


Suppose the second occurs. Trim the chain from the beginning to first occurance of the group that was picked twice. Then any person in any of the groups can contact any person in any of the other groups that occur by following the chain to the group that contains the other person. But that person can then contact the first person in the same way. This contradicts the assumption of maximality and leaves only the first option which was to be proved.

For the second part, we construct an example. Let person n be able to contact person n+1 and n+2 if n is even and person n-1 if n is odd.

Note that it doesn't matter how many people we have, just how many groups we have.