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Author Topic: Maths thread.  (Read 133125 times)

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Iridium

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Re: Maths thread.
« Reply #1050 on: December 04, 2019, 11:09:09 am »
+1

Have any of you guys taken the Putnam? I'm taking it this Saturday and want to know what you guys think about it.
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teamlyle

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Re: Maths thread.
« Reply #1051 on: December 04, 2019, 01:51:57 pm »
+1

I'm taking it too!
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Re: Maths thread.
« Reply #1052 on: December 04, 2019, 03:25:23 pm »
+1

Yea it's my last year for it.
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Titandrake

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Re: Maths thread.
« Reply #1053 on: December 04, 2019, 03:29:58 pm »
+1

The Putnam is something where you generally don't need that much college-level math to understand the proofs of all the questions, but figuring the proofs out is a different story.

Tip-wise, scoring on the Putnam is very bimodal, it's basically 10/9/8 if you get it or 0/1 if you don't get it. If you want full credit, you'll need to be careful to describe all the edge cases and how your proof handles them, and given how bimodal it is, you really want to get the full 10 points if you know how the proof works.
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blueblimp

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Re: Maths thread.
« Reply #1054 on: December 05, 2019, 03:38:32 am »
0

I wrote the Putnam the maximum 4 times. It was fun. I enjoyed it more than the other math contests I've tried. I'd suggest practicing specifically past Putnam questions to get used to its particular style if you haven't, though it's a little late to do that now.

Quote
Tip-wise, scoring on the Putnam is very bimodal, it's basically 10/9/8 if you get it or 0/1 if you don't get it.

I've been told this too, but somehow, my first time writing, I managed to get a 6 (I think it was).
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Iridium

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Re: Maths thread.
« Reply #1055 on: December 08, 2019, 02:26:47 pm »
0

 This is my first time taking it, and I'm pretty sure I got almost a 10 this time. I solved B1, but probably with a couple logical holes here and there. How'd you guys find it?
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heron

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Re: Maths thread.
« Reply #1056 on: December 08, 2019, 04:30:44 pm »
0

I found it about the same as usual; I was very disappointed after the test when I realized I misread B5. For some reason I thought it said f(n) = F_{2n + 1} instead of f(2n + 1) = F_{2n + 1}. I had solved a problem that was basically the same as the actual problem before, but the technique could not solve the problem they way I had read it. If I had read more carefully I would almost surely have got it...

I think B6 was unusually easy this year; it was definitely the problem that took me the least time to solve.
A1 and B1 were perhaps more difficult/time consuming than usual.
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Iridium

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Re: Maths thread.
« Reply #1057 on: December 08, 2019, 06:23:13 pm »
0

I have a lot of catching up to do then lol, I had no idea what I was doing for all of them except B1. How did you do B6?
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Re: Maths thread.
« Reply #1058 on: December 10, 2019, 02:06:23 pm »
0

What does "margin of error" in a poll correspond to? Is an x% margin of error "assuming we've sampled from a perfect binomial distribution, for each candidate, with 95% confidence their true support lies in a x% relative radius around their reported support?"
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pacovf

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Re: Maths thread.
« Reply #1059 on: December 10, 2019, 03:02:00 pm »
0

Thatís the least I would do, if I thought the only source of uncertainty was the statistics of polling a subset of the population. If they have information on the bias introduced by their sampling procedure, that could be folded into their margin of error too. People also tend to give different answers to a poll than when voting (for example), which could also be folded in if the pollsters have some way to account for that.

Alternatively, they could use their polling history to determine their previous prediction error.
« Last Edit: December 11, 2019, 12:18:42 am by pacovf »
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silverspawn

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Re: Maths thread.
« Reply #1060 on: December 10, 2019, 03:06:00 pm »
0

Wait, but if the pollster says that their own poll has a certain margin of error, I was assuming that has a commonly agreed upon meaning.
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pacovf

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Re: Maths thread.
« Reply #1061 on: December 10, 2019, 07:28:25 pm »
0

Sorry, I should have started my post by saying that I do not work in this domain. A quick check seems to point that the margin of error is defined as half the width of the confidence interval of the measure, accounting only for statistics and any intentional deviation from random sampling done by the pollsters.
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shraeye

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Re: Maths thread.
« Reply #1062 on: December 12, 2019, 01:11:11 pm »
+2

Yeah, the standard idea on Margin of Error is that it is there to account for the fact that the poll is based on a sample of voters.

In theory, if every single person were polled and responded, then the margin of error would be 0.  We know the EXACT proportion who said Yes (for example).

Whenever a smaller sample is used to represent the entire population, there is a chance that things went wrong.  Maybe this may be some bias the pollster brings (even if it's unintentional), what if the pollster feels more comfortable talking to smiling middle-aged couples, and less comfortable talking to tattooed hipsters, and what if the tattooed hipsters leaned No more often and the smilers leaned Yes?

But that is the sort of bias that good pollsters usually try very hard to eliminate.  The bigger issue is just white-noise randomness.  If there were 20 voters (10 yes, 10 no), and I asked 6 of them their opinion, it turns out that only 37% of the time will I hear 3 Yes and 3 No.  In general, the bigger the sample, the more likely it is that my results either A) is exactly the true proportion or B) is close to the true proportion.

Specifically the Margin of Error (MoE) is a way to represent a Confidence Interval (CI), from Result-MoE up to Result+MoE.
Confidence Intervals are built based on percentages: such as, "here is a 95% confidence interval."
What a 95% confidence interval means is this:  "if our poll's proportion of Yes (53.6%, for example) was also the actual proportion of Yes in the entire group of voters, AND we re-ran the poll and built a new CI of the same size; well then 95% of those intervals would also include the value 53.6%"

Which is a weird definition.  But anyway, the main point is that polls are not 100% accurate representations of reality.  Smaller MoE means more people were polled (or results were more extreme, but that usually has a smaller impact overall)
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Re: Maths thread.
« Reply #1063 on: December 12, 2019, 02:24:10 pm »
0

Is it relative percentage or percentage points? If the MoE is 10% and we measure p=0,2. does that mean CI = [0.1, 0.3] or CI = [0.18, 0.22]?
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shraeye

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Re: Maths thread.
« Reply #1064 on: December 12, 2019, 03:46:47 pm »
0

It's not normally relative, so it would be .1 to .3 in that example
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pacovf

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Re: Maths thread.
« Reply #1065 on: December 13, 2019, 04:53:32 am »
+3

What a 95% confidence interval means is this:  "if our poll's proportion of Yes (53.6%, for example) was also the actual proportion of Yes in the entire group of voters, AND we re-ran the poll and built a new CI of the same size; well then 95% of those intervals would also include the value 53.6%"

Which is a weird definition.

So I constantly forget and then relearn the rigorous definition of a confidence interval, but I think this one is equivalent to the one that says: "All values outside the 95% CI have less than a 5% chance to give a poll result this 'extreme' (as in, far from the real value) or more." Which is a bit more intuitive for me.

In other words, the CI is not so much keeping the "possible" real values, as it is rejecting all values for which the poll results would be very unlikely.

...Hopefully I'm not just saying something that is obvious for everyone.
« Last Edit: December 13, 2019, 04:57:59 am by pacovf »
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Re: Maths thread.
« Reply #1066 on: December 21, 2019, 10:39:58 am »
0

I'm not sure if it's true that they're equivalent, but if it is then your reformulation is a definite improvement.
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gamesou

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Re: Maths thread.
« Reply #1067 on: January 19, 2020, 02:04:15 am »
+1

Here is a little riddle of my own

A cooperative game is played between n players (n being at least 3) with 2n+2 cards: n+1 cards are white and n+1 cards are black. 2 cards are dealt to each player, who places them on their forehead (ŗ la Hanabi) so that everyone but them can look at the cards. The 2 undealt cards remain secret.

Players play cyclically ; at each turn a player may either pass or guess their card colors (2 black / 2 white / 1 of each). If wrong, the team lost. If true, that player stops playing. The game continues, and the winning condition is that everybody correctly guessed their card colors.

You goal is to devise a strategy that always succeds. Obviously no communication is allowed after the deal.
« Last Edit: January 19, 2020, 04:11:32 am by gamesou »
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hhelibebcnofnena

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Re: Maths thread.
« Reply #1068 on: January 20, 2020, 10:17:37 pm »
+3

I can solve the game in 5 rounds. It might be possible in fewer, but I don't think so. Also, I'm assuming they're aloud to communicate this game-plan beforehand?

1st round:
Everyone passes unless the two hidden cards are same-color (even if they can figure out their cards through some other method). If that is the case, anyone who also has the same two colors will see that only one color is missing and will guess correctly. Everyone then knows the missing two cards, and can proceed to guess correctly.

2nd round:
Everyone passes unless they see someone with two whites and someone else with two blacks. If that is sighted, the two hidden cards must be white/black. Everyone uses that knowledge (communicated through the fact of the person guessing) to proceed to a correct guess.

3rd round:
Everyone passes unless they see someone with two whites (There can only be one such person at this point). If they see someone with two whites, they guess white/black, because that has to be correct after the previous rounds. The one person who doesn't see anyone with two whites knows that they are that person.

4th round:
Same as 3rd round but with double black. These two rounds are interchangeable, but must remain separate, and the order must be decided beforehand.

5th round:
The only remaining possibility is that everyone has white/black. Everyone guesses that.
« Last Edit: January 20, 2020, 10:18:55 pm by hhelibebcnofnena »
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gamesou

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Re: Maths thread.
« Reply #1069 on: January 21, 2020, 02:45:57 am »
0

I can solve the game in 5 rounds. It might be possible in fewer, but I don't think so. Also, I'm assuming they're aloud to communicate this game-plan beforehand?

Well done! This is better than the solution I had in mind :-)
(yes, communication before game is allowed).
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bitwise

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Re: Maths thread.
« Reply #1070 on: January 21, 2020, 06:58:32 pm »
0

I can solve the game in 5 rounds. It might be possible in fewer, but I don't think so. Also, I'm assuming they're aloud to communicate this game-plan beforehand?

1st round:
Everyone passes unless the two hidden cards are same-color (even if they can figure out their cards through some other method). If that is the case, anyone who also has the same two colors will see that only one color is missing and will guess correctly. Everyone then knows the missing two cards, and can proceed to guess correctly.

2nd round:
Everyone passes unless they see someone with two whites and someone else with two blacks. If that is sighted, the two hidden cards must be white/black. Everyone uses that knowledge (communicated through the fact of the person guessing) to proceed to a correct guess.

3rd round:
Everyone passes unless they see someone with two whites (There can only be one such person at this point). If they see someone with two whites, they guess white/black, because that has to be correct after the previous rounds. The one person who doesn't see anyone with two whites knows that they are that person.

4th round:
Same as 3rd round but with double black. These two rounds are interchangeable, but must remain separate, and the order must be decided beforehand.

5th round:
The only remaining possibility is that everyone has white/black. Everyone guesses that.

I think you can combine the 3rd and 4th round.
Combined 3rd+4th round:
Everyone now knows that there is now at most one person with two cards of the same color. If that color is black, the person two before the person with 2 black cards will guess, and otherwise, the person one before the person with 2 white cards will guess.
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MiX

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Re: Maths thread.
« Reply #1071 on: January 21, 2020, 07:02:10 pm »
0

I think you can combine the 3rd and 4th round.
Combined 3rd+4th round:
Everyone now knows that there is now at most one person with two cards of the same color. If that color is black, the person two before the person with 2 black cards will guess, and otherwise, the person one before the person with 2 white cards will guess.

But turn order's random: the all-black or all-white player could be the first in the round.
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bitwise

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Re: Maths thread.
« Reply #1072 on: January 21, 2020, 07:05:55 pm »
0

I think you can combine the 3rd and 4th round.
Combined 3rd+4th round:
Everyone now knows that there is now at most one person with two cards of the same color. If that color is black, the person two before the person with 2 black cards will guess, and otherwise, the person one before the person with 2 white cards will guess.

But turn order's random: the all-black or all-white player could be the first in the round.
It should be okay if you consider the last player to be "before" the first player (as would happen in the cyclic order). Once the order is determined, every person has a unique person before them and a different unique person 2 before them.
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MiX

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Re: Maths thread.
« Reply #1073 on: January 21, 2020, 07:13:12 pm »
0

I think you can combine the 3rd and 4th round.
Combined 3rd+4th round:
Everyone now knows that there is now at most one person with two cards of the same color. If that color is black, the person two before the person with 2 black cards will guess, and otherwise, the person one before the person with 2 white cards will guess.

But turn order's random: the all-black or all-white player could be the first in the round.
It should be okay if you consider the last player to be "before" the first player (as would happen in the cyclic order). Once the order is determined, every person has a unique person before them and a different unique person 2 before them.
But you wouldn't know what you are if you're second-to-last on the second round, right? So how could you guess to signal then? Unless you do know? If so I don't think it's obvious.
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bitwise

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Re: Maths thread.
« Reply #1074 on: January 21, 2020, 07:18:07 pm »
0

I think you can combine the 3rd and 4th round.
Combined 3rd+4th round:
Everyone now knows that there is now at most one person with two cards of the same color. If that color is black, the person two before the person with 2 black cards will guess, and otherwise, the person one before the person with 2 white cards will guess.

But turn order's random: the all-black or all-white player could be the first in the round.
It should be okay if you consider the last player to be "before" the first player (as would happen in the cyclic order). Once the order is determined, every person has a unique person before them and a different unique person 2 before them.
But you wouldn't know what you are if you're second-to-last on the second round, right? So how could you guess to signal then? Unless you do know? If so I don't think it's obvious.
The person would be guessing/signaling in the third round, not the second round. It's not that the person must signal one/two turns before the third round of the all-black/all-white person, it's that the person one/two spaces before the all-black/all-white person must signal in the third round.
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