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Author Topic: Maths thread.  (Read 307148 times)

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scott_pilgrim

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Re: Maths thread.
« Reply #425 on: May 27, 2015, 04:37:34 pm »
+1

I don't think it's a game theory problem, because a candy jar game like that assumes that people aren't perfectly rational (or else they'd just be able to count the number exactly).  If you're just using the overestimate vs. underestimate as a tiebreaker, I think it would have basically no impact on how you play, especially if you're talking about a number around 10,000.  I guess it also depends on how many people are playing (you would prefer to overestimate with more players since a tie is more likely), but you'd need a lot of players I think before you cared about that.

You could do it Price Is Right-style and say whoever guesses closest without going over (or under, since you want people to make bigger guesses) wins.

A fun related game theory problem: You have n people bidding on a $100 bill.  They bid secretly, can't communicate with each other, and can only bid in integer numbers of dollars.  Whoever bids the highest amount without tying with someone else wins.  How much do you bid?  (I remember hearing a problem like this but I have no idea how to solve it...)
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Kirian

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Re: Maths thread.
« Reply #426 on: May 27, 2015, 05:12:34 pm »
+3

Random game theory question I was wondering (I don't know any game theory, so your explanations won't make sense, I just want an answer (discuss your explanations with each other of course if you want)):

Think of the candy jar game.  There is a certain amount of candy in a jar and several people guess how much there is.  Whoever guesses the closest wins.  Now, here's the catch:  If two people guess the same amount above and below (say the answer is 207 and two people guessed 205 and 209), then the person with the higher number wins.  How will this affect your strategy?

And, a hard mode I guess: I was thinking of this because of a candy jar I'm making with nerds.  I haven't finished counting, but there will be at least 10,000 nerds in it.  It looks like there's less nerds than there are for some reason, and when I had 7000 in there my brother thought there were 2000.  Would knowing that everybody will guess lower than the actual amount affect the answer to the first question?

Yet your jar still has fewer nerds than this thread.
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sudgy

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Re: Maths thread.
« Reply #427 on: May 27, 2015, 05:17:49 pm »
0

Okay, think of it this way:  There are p players, whose guesses average the actual number n with a standard deviation σ (this is how your answer is found too, since you can't count the actual number).  With that tiebreaker rule, how different would you want your guess to be from the normal way of guessing?

Hard mode would be where the average guess is lower than n.
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eHalcyon

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Re: Maths thread.
« Reply #428 on: May 27, 2015, 05:30:00 pm »
+2

And, a hard mode I guess: I was thinking of this because of a candy jar I'm making with nerds.  I haven't finished counting, but there will be at least 10,000 nerds in it.  It looks like there's less nerds than there are for some reason, and when I had 7000 in there my brother thought there were 2000.  Would knowing that everybody will guess lower than the actual amount affect the answer to the first question?

10,000 seems like a lot of people to be making just one candy jar.  Is it a really big jar?  That's a pretty impressive turnout though.  Be careful with fire codes, especially if your brother is so bad at counting that he loses 5000 RSVPs.  Make sure your location can safely hold such a mass of people.

Edit: didn't see there was a second page where Kirian sort of made this joke already...
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eHalcyon

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Re: Maths thread.
« Reply #429 on: May 27, 2015, 05:56:14 pm »
+1

A fun related game theory problem: You have n people bidding on a $100 bill.  They bid secretly, can't communicate with each other, and can only bid in integer numbers of dollars.  Whoever bids the highest amount without tying with someone else wins.  How much do you bid?  (I remember hearing a problem like this but I have no idea how to solve it...)

Reminds me of the Dollar Auction, which is hilarious.
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liopoil

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Re: Maths thread.
« Reply #430 on: May 27, 2015, 09:32:01 pm »
0

A fun related game theory problem: You have n people bidding on a $100 bill.  They bid secretly, can't communicate with each other, and can only bid in integer numbers of dollars.  Whoever bids the highest amount without tying with someone else wins.  How much do you bid?  (I remember hearing a problem like this but I have no idea how to solve it...)
This is super-dependent on n. For n = 2, both bid 99. For other n...
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eHalcyon

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Re: Maths thread.
« Reply #431 on: May 27, 2015, 09:40:23 pm »
0

A fun related game theory problem: You have n people bidding on a $100 bill.  They bid secretly, can't communicate with each other, and can only bid in integer numbers of dollars.  Whoever bids the highest amount without tying with someone else wins.  How much do you bid?  (I remember hearing a problem like this but I have no idea how to solve it...)
This is super-dependent on n. For n = 2, both bid 99. For other n...

n = 3.  If the other two bid 99, I should just bid 98.  But if somebody else bids 98, then 99 is the place to be again...
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Kirian

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Re: Maths thread.
« Reply #432 on: May 27, 2015, 09:53:16 pm »
0

A fun related game theory problem: You have n people bidding on a $100 bill.  They bid secretly, can't communicate with each other, and can only bid in integer numbers of dollars.  Whoever bids the highest amount without tying with someone else wins.  How much do you bid?  (I remember hearing a problem like this but I have no idea how to solve it...)
This is super-dependent on n. For n = 2, both bid 99. For other n...

n = 3.  If the other two bid 99, I should just bid 98.  But if somebody else bids 98, then 99 is the place to be again...

I feel like this must be related at least somewhat to this:

http://en.wikipedia.org/wiki/Unexpected_hanging_paradox
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Re: Maths thread.
« Reply #433 on: May 27, 2015, 09:54:17 pm »
0

A fun related game theory problem: You have n people bidding on a $100 bill.  They bid secretly, can't communicate with each other, and can only bid in integer numbers of dollars.  Whoever bids the highest amount without tying with someone else wins.  How much do you bid?  (I remember hearing a problem like this but I have no idea how to solve it...)
This is super-dependent on n. For n = 2, both bid 99. For other n...

n = 3.  If the other two bid 99, I should just bid 98.  But if somebody else bids 98, then 99 is the place to be again...
If the two others bid 99, you should bid 1... or 0... or -1 dollars as that too is an integer
For odd numbered n, I would just bid a negative number high in magnitude and hope everyone else cancels out, which, assuming that they are "perfectly rational" people, that should happen probably
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liopoil

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Re: Maths thread.
« Reply #434 on: May 27, 2015, 09:54:51 pm »
0

A fun related game theory problem: You have n people bidding on a $100 bill.  They bid secretly, can't communicate with each other, and can only bid in integer numbers of dollars.  Whoever bids the highest amount without tying with someone else wins.  How much do you bid?  (I remember hearing a problem like this but I have no idea how to solve it...)
This is super-dependent on n. For n = 2, both bid 99. For other n...

n = 3.  If the other two bid 99, I should just bid 98.  But if somebody else bids 98, then 99 is the place to be again...
No, if the other two people bid the same amount, I want to bid 0 dollars. Or, since he didn't specify positive integer, a  negative number with very large absolute value.

PPE: ninja'd :(
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liopoil

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Re: Maths thread.
« Reply #435 on: May 27, 2015, 09:56:07 pm »
+3

For odd numbered n, I would just bid a negative number high in magnitude and hope everyone else cancels out, which, assuming that they are "perfectly rational" people, that should happen probably
This is flawed. Since this game is symmetric, all players ought to adopt the same strategy, and for n > 2 I am almost certain that it is a mixed strategy.
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Ghacob

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Re: Maths thread.
« Reply #436 on: May 27, 2015, 10:11:57 pm »
0

For odd numbered n, I would just bid a negative number high in magnitude and hope everyone else cancels out, which, assuming that they are "perfectly rational" people, that should happen probably
This is flawed. Since this game is symmetric, all players ought to adopt the same strategy, and for n > 2 I am almost certain that it is a mixed strategy.

The flaw comes with how victory isn't well defined here. If the goal is to maximize profit*P(victory), as might seem reasonable, the winning strategy is the bet as large (in magnitude) a negative number as you can fit on the page (unfortunately negative infinity is not an integer)

A more reasonable goal(which I'll be thinking in terms of from now on) might be to maximize P(victory) where victory will net you a positive profit (someone betting $100 dollars or more would surely increase your chances of winning, but seems a bit... hmm...

I feel like there's some way you could improve that so that we could remove the cap but regardless

I feel like the correct answer relies on a normal curve centered at 100-n/2, although using n or n/4 seem not unreasonable
..
Can anyone else tell that I'm not doing real math, just relying on intuition?
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GendoIkari

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Re: Maths thread.
« Reply #437 on: May 27, 2015, 10:28:22 pm »
0


A fun related game theory problem: You have n people bidding on a $100 bill.  They bid secretly, can't communicate with each other, and can only bid in integer numbers of dollars.  Whoever bids the highest amount without tying with someone else wins.  How much do you bid?  (I remember hearing a problem like this but I have no idea how to solve it...)

This is the same game that I tried to start in the forum games section, unsuccessfully: http://forum.dominionstrategy.com/index.php?topic=6372.0
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Titandrake

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Re: Maths thread.
« Reply #438 on: May 27, 2015, 10:32:50 pm »
+2

Under the following assumptions + simplifications:
  • The goal of each player is to maximize their expected profit.
  • The only valid bids are $0 and $99
  • For the optimal strategy, no one person can deviate and perform better.
  • The players are greedy - if there exist multiple strategies that satisfy the deviation policy, then each player chooses the strategy that maximizes the probability of a winner.

Then for n = 3 the optimal strategy ends up being bidding $0 10/11 of the time and $99 1/11 of the time.

Suppose two players bet $0 with probability p and $99 with probability 1-p. The third bets with probabilities q, 1-q respectively. Then, the expected profit is

100 * q * (1-p)^2 + (1-q) * p^2

Now, we want to maximize this profit with respect to q. Note that this is linear in q - it equals

(99p^2 - 200p + 100) q + (constant with respect to q)

For p < 10/11, this coefficient is positive, and q = 1 outperforms p
For 1 >= p > 10/11, this coefficent is negative, and q = 0 outperforms p

So, the only value of p for which no deviation can improve expected profit is p = 10/11.

Now, in principle you can generalize this to the game where you can make any bet from $0 to $99, if you add more variables, but at that point this approach quickly becomes infeasible. I think even with a computer it's not solvable exactly, since I think you'd end up with non-linear constraints, but I could be wrong.
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liopoil

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Re: Maths thread.
« Reply #439 on: May 27, 2015, 10:36:00 pm »
+1

Yeah, I don't know about that. In any game, the goal is to maximize expected payoff, not 'P(victory)', because not all victories as you call them are equal. Your payoff is 0 if you don't get the $100, or if you bet $100. If you do get the $100, your payoff is 100 - your bid. The game where we are trying to name the greatest uniquely named integer less than or equal to x is an entirely different game.

PPE: Titandrake figured something out! Yeah, with 100 pure strategies I don't see this having a nice answer.
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GendoIkari

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Re: Maths thread.
« Reply #440 on: May 27, 2015, 10:37:43 pm »
0

Under the following assumptions + simplifications:
  • The goal of each player is to maximize their expected profit.
  • The only valid bids are $0 and $99
  • For the optimal strategy, no one person can deviate and perform better.
  • The players are greedy - if there exist multiple strategies that satisfy the deviation policy, then each player chooses the strategy that maximizes the probability of a winner.

Then for n = 3 the optimal strategy ends up being bidding $0 10/11 of the time and $99 1/11 of the time.

Suppose two players bet $0 with probability p and $99 with probability 1-p. The third bets with probabilities q, 1-q respectively. Then, the expected profit is

100 * q * (1-p)^2 + (1-q) * p^2

Now, we want to maximize this profit with respect to q. Note that this is linear in q - it equals

(99p^2 - 200p + 100) q + (constant with respect to q)

For p < 10/11, this coefficient is positive, and q = 1 outperforms p
For 1 >= p > 10/11, this coefficent is negative, and q = 0 outperforms p

So, the only value of p for which no deviation can improve expected profit is p = 10/11.

Now, in principle you can generalize this to the game where you can make any bet from $0 to $99, if you add more variables, but at that point this approach quickly becomes infeasible. I think even with a computer it's not solvable exactly, since I think you'd end up with non-linear constraints, but I could be wrong.

I've never thought of this as a solvable game with an optimal solution. Logically, if there were an optimal solution, then all players could follow that optimal solution, which would mean that all players bid the same amount, leaving no winner. Therefore there cannot exist an optimal solution. It's a mix of luck and mind games.
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liopoil

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Re: Maths thread.
« Reply #441 on: May 27, 2015, 10:39:38 pm »
+1


A fun related game theory problem: You have n people bidding on a $100 bill.  They bid secretly, can't communicate with each other, and can only bid in integer numbers of dollars.  Whoever bids the highest amount without tying with someone else wins.  How much do you bid?  (I remember hearing a problem like this but I have no idea how to solve it...)

This is the same game that I tried to start in the forum games section, unsuccessfully: http://forum.dominionstrategy.com/index.php?topic=6372.0
Well, this is Gherald's game. Hey, I see I signed up for that, over two years ago, wow! Have I won yet?

I've never thought of this as a solvable game with an optimal solution. Logically, if there were an optimal solution, then all players could follow that optimal solution, which would mean that all players bid the same amount, leaving no winner. Therefore there cannot exist an optimal solution. It's a mix of luck and mind games.
No, all games are solvable. The solution is not to bid the same thing every time, it is to bid different amounts with different probabilities, just like in rock-paper-scissors the solution is to throw each with 1/3 probability.
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GendoIkari

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Re: Maths thread.
« Reply #442 on: May 27, 2015, 10:41:39 pm »
0

Yeah, I don't know about that. In any game, the goal is to maximize expected payoff, not 'P(victory)', because not all victories as you call them are equal. Your payoff is 0 if you don't get the $100, or if you bet $100. If you do get the $100, your payoff is 100 - your bid. The game where we are trying to name the greatest uniquely named integer less than or equal to x is an entirely different game.

PPE: Titandrake figured something out! Yeah, with 100 pure strategies I don't see this having a nice answer.

Hmm, ok the idea of wanting to not only win, but win by the most, does factor into your decision. But it's still the same game with the same rules. Just reverse the numbers in my game:

"You have to name an integer that is less than 100. Whoever names the largest unique integer wins." You can say "less than 100" in the rules, because bidding 100 or more is like automatically losing; even if you technically would "win" the game in that case, you wouldn't want to. So the only difference is that with my game, there's no incentive given to winning with a larger number instead of winning with a smaller number. But you can easily decide that you want to win with a larger number instead of just winning period.
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GendoIkari

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Re: Maths thread.
« Reply #443 on: May 27, 2015, 10:45:01 pm »
0


A fun related game theory problem: You have n people bidding on a $100 bill.  They bid secretly, can't communicate with each other, and can only bid in integer numbers of dollars.  Whoever bids the highest amount without tying with someone else wins.  How much do you bid?  (I remember hearing a problem like this but I have no idea how to solve it...)

This is the same game that I tried to start in the forum games section, unsuccessfully: http://forum.dominionstrategy.com/index.php?topic=6372.0
Well, this is Gherald's game. Hey, I see I signed up for that, over two years ago, wow! Have I won yet?

I've never thought of this as a solvable game with an optimal solution. Logically, if there were an optimal solution, then all players could follow that optimal solution, which would mean that all players bid the same amount, leaving no winner. Therefore there cannot exist an optimal solution. It's a mix of luck and mind games.
No, all games are solvable. The solution is not to bid the same thing every time, it is to bid different amounts with different probabilities, just like in rock-paper-scissors the solution is to throw each with 1/3 probability.

Not sure what you mean by all games are solvable. Many games, including this one, have a large portion of mind-games built in. There isn't simply a "single best possible move". And the idea of not doing the same thing every time only matters when playing multiple times in a row. For a single game being played, what you should bid is going to be based on what you know about the players you are playing against, and will still rely on a lot of luck. There's no logical/mathematical way to make the "best" move.

Rock-Paper-Scissors is a great example. Doing each thing at complete random might be the best strategy when playing against a computer who knows human habits. But when playing against a human, the best strategy is to use what you know about human habits, like a computer can do. Which is why a trained computer will beat a human most of the time.
« Last Edit: May 27, 2015, 10:46:19 pm by GendoIkari »
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liopoil

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Re: Maths thread.
« Reply #444 on: May 27, 2015, 10:52:33 pm »
+2

The single best move is to generate a random number between 0 to 1, then bid accordingly based on some rules you make so that you make each bid with the desired probability.

I mean that any game with well-defined payoffs has a Nash equilibrium, that is, there exists a set of mixed strategies for each player such that it is impossible for any player to improve their expected payoff by deviating from their strategy. A mixed strategy is a strategy in which you choose each of the 'pure' strategies, e.g., bidding a certain value, with some probability. Dominion has a Nash equilibrium, an optimal way to play, but it is so disgusting that trying to find it is not practical.

There is still plenty of luck involved, but there is a specific strategy that maximizes your odds, and if you play any other strategy against this strategy your odds will be slightly worse.
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Re: Maths thread.
« Reply #445 on: May 27, 2015, 10:56:57 pm »
+1

Rock-Paper-Scissors is a great example. Doing each thing at complete random might be the best strategy when playing against a computer who knows human habits. But when playing against a human, the best strategy is to use what you know about human habits, like a computer can do. Which is why a trained computer will beat a human most of the time.
Doing each thing at random is the best strategy when playing a player who will play perfectly well. If you are willing to assume that your opponent plays poorly, sure you can take advantage of that, but in doing so you are playing imperfectly and would lose to another player who plays better.

Okay, that is not quite true for rock-paper-scissors, because playing rock every time does as well as anything against pure random, but the point is that any other strategy besides pure random can be exploited. This includes the computer who is trained to look for human patterns; a computer who looked for patterns of a computer looking for human patterns would win.
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Re: Maths thread.
« Reply #446 on: May 27, 2015, 10:59:50 pm »
+2

Oops, yes, bid 1, not 98...

And to the rest of this conversation,  :o

Really though, I haven't done game theory stuff in a while; I forgot about the possibility of mixed strategies.

@Gendo, we're talking Game Theory, so we assume perfectly rational players.  These players are basically computers, capable of perfectly random selection and with no human habits to exploit.
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Re: Maths thread.
« Reply #447 on: May 28, 2015, 08:45:21 am »
0


A fun related game theory problem: You have n people bidding on a $100 bill.  They bid secretly, can't communicate with each other, and can only bid in integer numbers of dollars.  Whoever bids the highest amount without tying with someone else wins.  How much do you bid?  (I remember hearing a problem like this but I have no idea how to solve it...)

This is the same game that I tried to start in the forum games section, unsuccessfully: http://forum.dominionstrategy.com/index.php?topic=6372.0

Using this form of the problem, which I like a whole lot better, lead to a few insights not available in the profit oriented version
In this where the only goal is optimizing chance of victory:
It's visibly clear that you wouldn't want to choose more than n and I feel like it's not smart to choose more than n/2 either

So I see the optimal strategy for a sufficently large n (with a smaller n, it might be worth hoping everyone else cancels out) as randomly choosing a number between 1 and n/2, although I'm not sure if you should but some sort of curve on that probability


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Re: Maths thread.
« Reply #448 on: May 28, 2015, 09:07:59 am »
0

Oops, yes, bid 1, not 98...

And to the rest of this conversation,  :o

Really though, I haven't done game theory stuff in a while; I forgot about the possibility of mixed strategies.

@Gendo, we're talking Game Theory, so we assume perfectly rational players.  These players are basically computers, capable of perfectly random selection and with no human habits to exploit.
Oops, yes, bid 1, not 98...

And to the rest of this conversation,  :o

Really though, I haven't done game theory stuff in a while; I forgot about the possibility of mixed strategies.

@Gendo, we're talking Game Theory, so we assume perfectly rational players.  These players are basically computers, capable of perfectly random selection and with no human habits to exploit.
Disclaimer: just woke up.

I don't think it is so easy to assume perfectly rational. If there is a Nash equilibrium, sure you can do. But with the 100$, n player who bids highest without tieing wins, I'm not so sure there is one.
Nash equilibrium basically means that if all player behave accordingly to, it does not make sense for a single player to behave differently.
But this one is more prisoners dilemma like (even more extreme as there is no incentive to cooperate at all)as a single player you always have the incentive to bid 100, if you assume that the Nash is not leading to someone else saying 100 with prob >1/2.  ...    ...
Ok forget what I said, this probably us the Nash eq:
Say a random number so that the Total prob that two people say 100 is 1/2, 99 is 1/4, 98 is 1/8 etc.  If you deviate from this, you are hurting yourself as you are more likely to collide.

Edit: ok maybe you also don't want to get to 1/2, but maximize the prob of "you say 100, but n-1 copies of you don't say 100" the other prob you get by recursion anyway

Edit2: also I have ignored that you have to bid, this biases to smaller numbers as higher numbers lower your profit
« Last Edit: May 28, 2015, 10:13:08 am by DStu »
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Re: Maths thread.
« Reply #449 on: May 28, 2015, 11:29:50 am »
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Then nobody says anything about my more defined problem...
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm
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