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Witherweaver

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« Reply #250 on: February 11, 2015, 10:33:56 pm »
+1

Any time you have terms like

sqrt{a^2+b^2x^2}
sqrt{a^2-b^2x^2}
sqrt{a^2x^2-b^2}

draw a relevant triangle and make the same kind of argument.  The "correct" substitution to make will simply fall out of labeling the sides/angles.

And note that {a^2+b^2}^{n/2} for any integer n is just a nice monomial of those expressions, so you can expect an integrand that looks like powers of trig functions.
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DStu

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« Reply #251 on: February 12, 2015, 03:32:36 am »
+1

Given two randomly selected points in a unit square, what is the probability that the distance between them is greater than 1/2?
(2015 AMC A, modified)

Which probability distribution, and which distance?
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Titandrake

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« Reply #252 on: February 12, 2015, 05:05:58 am »
+3

Given two randomly selected points in a unit square, what is the probability that the distance between them is greater than 1/2?
(2015 AMC A, modified)

Which probability distribution, and which distance?

It's a high school math competition, come on

(Uniform and Euclidean, although you're free to try other parameters if you want to.)
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DStu

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« Reply #253 on: February 12, 2015, 05:15:14 am »
0

Given two randomly selected points in a unit square, what is the probability that the distance between them is greater than 1/2?
(2015 AMC A, modified)

Which probability distribution, and which distance?

It's a high school math competition, come on

(Uniform and Euclidean, although you're free to try other parameters if you want to.)

If I wouldn't have asked this I would have to think about it, and I don't have time for this...
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Ozle

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« Reply #254 on: February 12, 2015, 08:43:45 am »
+3

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Witherweaver

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« Reply #255 on: February 12, 2015, 09:20:29 am »
+2

6 Moats?
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Polk5440

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« Reply #256 on: February 12, 2015, 09:28:21 am »
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This is an integral we would ask when I was a student instructor for calculus when we wanted to identify who the A students really were. Those that can make progress on this integral have really internalized integration methods and trig identities (you have to want to see a trig identity instead of what you have; a tall order for B students) or can reason at a high level of insight/intuition.
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Watno

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« Reply #257 on: February 12, 2015, 09:32:15 am »
+1

This is an integral we would ask when I was a student instructor for calculus when we wanted to identify who the A students really were. Those that can make progress on this integral have really internalized integration methods and trig identities (you have to want to see a trig identity instead of what you have; a tall order for B students) or can reason at a high level of insight/intuition.
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Witherweaver

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« Reply #258 on: February 12, 2015, 09:33:31 am »
0

This is an integral we would ask when I was a student instructor for calculus when we wanted to identify who the A students really were. Those that can make progress on this integral have really internalized integration methods and trig identities (you have to want to see a trig identity instead of what you have; a tall order for B students) or can reason at a high level of insight/intuition.

Which one?  The \sqrt{1-x^2} one?
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sudgy

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« Reply #259 on: February 12, 2015, 11:22:31 am »
0

Probability has to be from 0 to 1...
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Kirian

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« Reply #260 on: February 12, 2015, 12:01:18 pm »
0

Probability has to be from 0 to 1...

This is do likely it happens more than 100% of the time.
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Ozle

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« Reply #261 on: February 12, 2015, 01:49:10 pm »
+1

Probability has to be from 0 to 1...

6 1's then. Sorry should have been slightly more specific.

Whats next?
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Polk5440

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« Reply #262 on: February 12, 2015, 08:56:27 pm »
0

This is an integral we would ask when I was a student instructor for calculus when we wanted to identify who the A students really were. Those that can make progress on this integral have really internalized integration methods and trig identities (you have to want to see a trig identity instead of what you have; a tall order for B students) or can reason at a high level of insight/intuition.

Which one?  The \sqrt{1-x^2} one?

Yes.
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silverspawn

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« Reply #263 on: February 12, 2015, 09:24:51 pm »
0

This is an integral we would ask when I was a student instructor for calculus when we wanted to identify who the A students really were. Those that can make progress on this integral have really internalized integration methods and trig identities (you have to want to see a trig identity instead of what you have; a tall order for B students) or can reason at a high level of insight/intuition.

Which one?  The \sqrt{1-x^2} one?

Yes.

yea I am good aren't I? ^_^
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liopoil

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« Reply #264 on: February 16, 2015, 03:48:34 pm »
+2

Here's a game theory problem I made and solved yesterday. It's relatively easy compared to the other problems posted here.

We're playing a game of Rock, Paper, Scissors, with a twist; I am never allowed to throw paper. Throwing scissors is not recommended for you. In the case where we both throw the same thing, there are no payoffs and we simply throw again.  If there are some whole number n throws in a row where we both throw the same thing, I win the throw. The winner of a throw has a payoff of 1 and the loser a payoff of -1. What is the optimal strategy for each player and what is your expected payoff for this game, in terms of n?
« Last Edit: February 16, 2015, 04:09:09 pm by liopoil »
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XerxesPraelor

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« Reply #265 on: February 16, 2015, 06:04:55 pm »
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Well, it's going to be Rock most of the time, Paper every once in a while, I think.
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liopoil

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« Reply #266 on: February 16, 2015, 06:22:34 pm »
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Well, it's going to be Rock most of the time, Paper every once in a while, I think.
Correct! Now, just how often?
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Kirian

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« Reply #267 on: February 16, 2015, 06:22:39 pm »
+4

Here's a game theory problem I made and solved yesterday. It's relatively easy compared to the other problems posted here.

We're playing a game of Rock, Paper, Scissors, with a twist; I am never allowed to throw paper. Throwing scissors is not recommended for you. In the case where we both throw the same thing, there are no payoffs and we simply throw again.  If there are some whole number n throws in a row where we both throw the same thing, I win the throw. The winner of a throw has a payoff of 1 and the loser a payoff of -1. What is the optimal strategy for each player and what is your expected payoff for this game, in terms of n?

The best strategy is                                                                         Moat                                                                                                                                            .

I'm not sure about the expected payoff, but I'm certain it's close to                 42
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scott_pilgrim

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« Reply #268 on: February 16, 2015, 06:43:39 pm »
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I'm not sure I understand the game.  Do I get -1 for an entire sequence of consecutive drawn throws?  Or is it -1 for each consecutive drawn throw after the first?  And shouldn't the expected payoff be independent of n, since we would need to be finding the optimal strategy for both players anyway?
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liopoil

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« Reply #269 on: February 16, 2015, 06:47:49 pm »
0

I'm not sure I understand the game.  Do I get -1 for an entire sequence of consecutive drawn throws?  Or is it -1 for each consecutive drawn throw after the first?  And shouldn't the expected payoff be independent of n, since we would need to be finding the optimal strategy for both players anyway?
You get -1 for the last throw only. All drawn throws except the last have no payoffs at all. You could say that when there's a draw, decrease n by one, then play again, and if n = 0, I win.

Neither the expected payoff nor the optimal strategy is independent of n.
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scott_pilgrim

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« Reply #270 on: February 16, 2015, 08:02:09 pm »
0

Okay, the reason I'm confused is because, the way I'm reading it, n is dependent on how you play, not vice versa.  So I could (maybe) answer the questions you asked, and then give you an expected value of n, but you're saying the answer should be dependent on n.

I think I get it now though.  The game is, we play "weird rock paper scissors" n times.  I thought the game was, we play it an arbitrary number of times, and then n is "the number of consecutive drawn throws".

Is that right?  Honestly the problem really reads like my original interpretation...
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Witherweaver

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« Reply #271 on: February 16, 2015, 08:18:29 pm »
0

Okay, the reason I'm confused is because, the way I'm reading it, n is dependent on how you play, not vice versa.  So I could (maybe) answer the questions you asked, and then give you an expected value of n, but you're saying the answer should be dependent on n.

I think I get it now though.  The game is, we play "weird rock paper scissors" n times.  I thought the game was, we play it an arbitrary number of times, and then n is "the number of consecutive drawn throws".

Is that right?  Honestly the problem really reads like my original interpretation...

I think fix an n ahead of time.  Say, n=2.  Then the game is:

1) Play Rock, Paper, Scissors, Player A cannot throw Paper.
2) On a tie, replay
3) Player A win condition: Tie twice in a row
4) Player B win condition: Win any single throw
5) Play until a winner

Winner gets +1, Loser gets -1.
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Titandrake

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« Reply #272 on: February 16, 2015, 08:24:58 pm »
+2

Finding an optimal strategy is easier if you frame it this way.

1) Let's play Rock, Paper, Scissors, n.
2) Both players make a throw, player A cannot throw paper.
3) If player B wins, player B gets +1, player A gets -1, done
4) If player A wins, player A gets +1, player B gets -1, done
5) If a draw happens, and n = 1, player A gets +1, player B gets -1, done
6) Otherwise, play a game of Rock, Paper, Scissors, n-1, and use the payout from that instead.

From here you can try some inductive/recursive definition of the winning strategy based on n. Don't have the time to actually work it out, but there's the framework.
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liopoil

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« Reply #273 on: February 16, 2015, 09:22:09 pm »
+1

Okay, the reason I'm confused is because, the way I'm reading it, n is dependent on how you play, not vice versa.  So I could (maybe) answer the questions you asked, and then give you an expected value of n, but you're saying the answer should be dependent on n.

I think I get it now though.  The game is, we play "weird rock paper scissors" n times.  I thought the game was, we play it an arbitrary number of times, and then n is "the number of consecutive drawn throws".

Is that right?  Honestly the problem really reads like my original interpretation...
We only play one game. If there's a tie, that game didn't count and we play the game again, except n is decreased by one. Witherweaver and Titandrake have it right. I didn't make it clear that n is defined beforehand. n = 0 is a game, in that game I always win immediately. n = 1 is a different game, where I win ties. n = 2 is a game where if we tie, we instead play a game where I win ties, and so on. How best to play depends on which game we are playing, that is, the value of n. So we can define our strategy and expected payoff in terms of n.
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scott_pilgrim

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« Reply #274 on: February 17, 2015, 02:02:05 am »
0

Okay, so I think both players throw rock with probability n/(n+1), and the expected payout is (n-1)/(n+1).

Here's a proof by induction:

For n=1, the problem is trivial.  I would make a payoff matrix but I'm too lazy to figure out how tables work.  It's -1 for rock/rock, +1 for rock/paper, +1 for rock/scissors, and -1 for paper/scissors.  So each player should choose between their two options with probability 0.5, and the expected payout is 0.

Now suppose the solution holds for n.  We argue it holds for n+1.  The payoff matrix for n+1 will look the same as it did for n=1 for rock/paper, rock/scissors, and paper/scissors, but the rock/rock element will have the expected payout for n, which is (n-1)/(n+1).  So if p is player 1's probability of choosing rock and q is player 2's probability of choosing rock, then player 2 wants to maximize pq(n-1)/(n+1)+p(1-q)+(1-p)q-(1-p)(1-q).  So take the derivative with respect to q and set it equal to 0, to get p(n-1)/(n+1)-p+1-p+1-p=0, so p(n-1)/(n+1)-3p+2=0, so p=(n+1)/(n+2), as desired.  Since the equation is symmetric for p and q, we can make the same argument for q.

So now the expected payout (for n+1) will be ((n+1)/(n+2))^2*(n-1)/(n+1)+2((n+1)/(n+2))(1-(n+1)/(n+2))-(1-(n+1)/(n+2))^2 which is awful to simplify by hand.  But wolfram says it's n/(n+2) so it's right.

Cool problem!
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