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ThudnBlunder
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It's a Truel World
« on: Dec 19th, 2008, 10:38am » |
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I have just noticed that there are now lots of new interactive demos at the Mathematica site. I've been playing around with the Truel demo. In a simplified truel one player is assumed to be a sure shot, in which case the worst player (who shoots first) should deliberately miss until one of the others (who shoot at each other, the sure shot firing last) is killed. But when we merely assume
1 > a > b > c > 0, I don't think this is necessarily true. Interestingly, in the demo mentioned above it is possible to choose numbers so that each player has an equal chance (1/3) of surviving. So for what probabilities a,b,c can this happen?
Let R(X : Y) = Ratio of X winning : Y winning a duel, with X shooting first
and
let X hit the target with probability x
and
let Y hit the target with probability y
Then R(X : Y) = x/(1 - x)y
Now assume 3 duellists A,B,C have respective probabilities a,b,c where 1  a  b c 0
and
let F = R(C : B)
let G = R(C : A)
let H = R(B : A)
Then we get
a = (FH - G)/GH
b = (FH - G)/FH
c = (FH - G)/(FH - G + H)
Let A(s) = probability of A surviving the truel
Let B(s) = probability of B surviving the truel
Let C(s) = probability of C surviving the truel
Assuming C shoots first and adopts the strategy of deliberately missing, we get
A(s) = 1/[(G + 1)(H + 1)]
B(s) = H/[(F + 1)(H + 1)]
C(s) = 1 - A(s) - B(s)
and when A(s) = B(s) = C(s) = 1/3 we get
(G + 1)(H + 1) = 3
(F + 1)(H + 1) = 3H
From this we get
F = 1 - G
H = (2 - G)/(1 + G)
and finally
a = 2(1 - 2G)/[G(2 - G)]
b = 2(1 - 2G)/[(1 - G)(2 - G)]
c = 2(1 - 2G)/(4 - 5G)
provided
3 -  7  G  1/2 (because 1 a 0)
For example, choosing G = 2/5 we get A(s) = B(s) = C(s) = 1/3 when
a = 5/8
b = 5/12
c = 1/5
F = 3/5
H = 8/7
On the demo this corresponds (with a die size of 120) to (A,B,C) = (24,50,75)
Note that the demo has a < b < c, not a > b > c
Considering a simplified truel with a = 1, b = 1/2, c = 1/3
F = (1/3)/[(2/3)(1/2)]
= 1
G = (1/3)/[(2/3)(1)]
= 1/2
H = (1/2)/[(1/2)(1)]
= 1
Therefore
A(s) = 1/[(3/2)(2)]
= 1/3
B(s) = 1/[(2)(2)]
= 1/4
C(s) = 1 - 1/3 - 1/4
= 5/12
This is confirmed by the demo using 12 die size and (A,B,C) = (4,6,12).
Unfortunately, I have just realized that to run interactively the Truel demo requires Mathematica7, no less. 
So I guess you'll just to have to take my word for it. 
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| « Last Edit: Mar 14th, 2009, 12:44pm by ThudnBlunder » |
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Re: It's a Truel World
« Reply #1 on: Jan 12th, 2009, 10:31pm » |
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By popular demand  , here is some more analysis and results:
Let A(s), B(s), C(s) be the respective probabilities of A, B, C surviving the truel when C, the worst shot, chooses to delibarately miss.
Let A(s)*, B(s)*, C(s)* be the respective probabilities of A, B, C surviving the truel when C chooses to shoot at A, the best shot.
Then we get
a2(1 - b)(1 - c)
A(s) = ------------------------
(a + b - ab)(a + c - ac)
b2(1 - c)
B(s) = ------------------------
(a + b - ab))(b + c - bc)
C(s) = 1 - A(s) - B(s)
a2(1 - b)(1 - c)2
A(s)* = -------------------------------------
(a + c - ac)[b + (1 - b)(a + c - ac)]
b[c + b(1 - c)2]
B(s)* = -------------------------------------
(b + c - bc)[a + (1 - a)(b + c - bc)]
C(s)* = 1 - A(s)* - B(s)*
C should deliberately miss when C(s) > C(s)* and shoot at A when C(s) < C(s)*
That is, as
A(s)* + B(s)* <> A(s) + B(s)
After some tedious algebra this simplifies to a[a(1 - b)2 - b]
Miss if c > ----------------------- ........................(1)
a2(1 - b)2 + b2(1 - a)
In particular, it is evident from this inequality that C should always choose to miss whenever a < b/(1 - b)2
In other words, always choose to miss when H > 1 - b (where H = the ratio of probability of B killing A to A killing B.)
For the simplified truel that has a = 1, this means whenever b > (3 - 5)/2 = 1 - phi = 0.382...
And when a = 1, equation (1) becomes
Miss if c > 1 - b/(1 - b)2
eg. letting b = 1/3, we choose to miss if c > 1/4 [When c = 1/4, C(s) = C(s)* = 1/3]
So assume b = 1/3, c = 1/5 < 1/4
Then
A(s) = 100/525
B(s) = 280/525
C(s) = 145/525
and
A(s)* = 224/525
B(s)* = 155/525
C(s)* = 146/525
And C(s)* > C(s), as expected. (However, Wolfram's aforementioned truel demo baldly claims that missing is the optimum strategy.)
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| « Last Edit: Jan 19th, 2009, 1:06am by ThudnBlunder » |
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Re: It's a Truel World
« Reply #2 on: Jan 30th, 2009, 2:12pm » |
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I wrote to the author of the Truel Demo informing him of the above analysis and giving a link. He replied, "I'm not sure I understand your analysis. It is always advantageous for the weakest player to wait until they have the first shot on one opponent." I think this can be read as, " I do not understand your analysis. This is because I have not studied it, as I believe a priori that your conclusion is wrong." This in turn may be interpreted as, "Every example of a specific truel I have seen involved a sure shot (a = 1), a good shot (9/10 b 1/2), and a poor shot (c 1/2). In each case the poor shot does best by shooting into the air until the other two have sorted out their differences. Therefore your conclusion is wrong and I do not understand your analysis." However, when a sure shot and two very poor shots are involved it is not wrong. I recalculated, assuming a = 1 from the outset.
Then we get:
A(s) = (1 - b)(1 - c)
B(s) = (1 - c)b2/(b + c - bc)
A(s)* = (1 - b)(1 - c)2
B(s)* = b[c + b(1 - c)2]/(b + c - bc)
Calculation of C(s) - C(s)* still looks quite messy but fortunately many terms cancel and we are left with
C(s) - C(s)* = c2[c(1 - b)2 - (1 - b)2 - b]/(b + c - bc)
C should shoot into the air when C(s) - C(s)* > 0
and
C(s) - C(s)* > 0 when c(1 - b)2 - (1 - b)2 - b > 0
That is, when
c > (b2 - 3b + 1)/(b2 - 2b + 1)
And b2 - 3b + 1 < 0 when 1 > b > (3 - 5)/2 = 1 - phi = 0.382..., as before.
CONCLUSION:
C should shoot into the air when c > 1 - b/(1 - b)2
In particular, C should always shoot into the air when b > 1 - phi = 0.382...
Otherwise C should shoot at A, the sure shot.
MORAL: Even truellists need to keep one eye on the Golden Ratio, as there is a phine line between life and death. LOL
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| « Last Edit: Feb 1st, 2009, 6:02am by ThudnBlunder » |
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towr
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Re: It's a Truel World
« Reply #3 on: Jan 30th, 2009, 2:33pm » |
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Even if he didn't believe/understand your analysis; at the very least he could have done a quick simulation to find out that it is indeed not always advantageous for the weakest player to wait.
One counter-example is enough to disprove a proposition.
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Eigenray
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Re: It's a Truel World
« Reply #4 on: Jan 30th, 2009, 6:31pm » |
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I'm not sure I understand your analysis either, so here's mine. The truelists shoot in the order A,B,C,A,..., with probabilities a<b<c.
Let [p,q] = p/(p+q-pq) denote the probability that P wins starting a duel against Q. Let p' denote 1-p.
Let's assume for now that c=1, that B aims at C, and C aims at B.
If A deliberately misses, then the probability of A winning is simply
Pm(A) = b [a,b] + b' [a,1].
If A shoots at C, the probability is
Ps(A) = a [b,a]' + a' Pm(A)
= a [b,a]' + a' b [a,b] + a' b' [a,1].
Now
Pm(A) - Ps(A) = [ a(1-b)2 - b2 + 3b - 1 ] a2/(a+b-ab),
which is positive exactly when
a > ( b2 - 3b + 1) / (1-b)2.
If b > (3-5)/2 ~ .382, then A should always miss.
In the other direction, under the assumption a < b, if b <  , then A should shoot at C, where ~ .318 satisfies = (2-3+1)/(1-)2.
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| « Last Edit: Jan 30th, 2009, 6:51pm by Eigenray » |
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ThudnBlunder
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on Jan 30th, 2009, 6:31pm, Eigenray wrote:
In the other direction, under the assumption a < b, if b < , then A should shoot at C, where ~ .318 satisfies = (2-3+1)/(1-)2. |
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Oops, I forgot to determine that point. Yep, in these here parts there's many a slip 'twixt holster and hand.
on Jan 30th, 2009, 6:31pm, Eigenray wrote:| I'm not sure I understand your analysis either, so here's mine. |
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But what is there not to understand? High-school maths, clearly explained at every step? Anyway, because when all is said and done at the end of the day it goes without saying that in a very real sense a picture is worth a thousand words, below is a diagram which achieves the best of both worlds by taking into account all sides of the argument. LOL The shaded area represents when A should shoot at C. My answer agrees with yours, except I did not include the point of intersection of the two functions.
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| « Last Edit: Mar 14th, 2009, 12:50pm by ThudnBlunder » |
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Re: It's a Truel World
« Reply #6 on: Feb 3rd, 2009, 11:18am » |
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Obviously no longer suspected of being congenetically related to the Usenet crank James Harris, I received a friendly reply from the author of the Truel Demo (Ed Pegg, Jr., a noted metagrobologist), agreeing with the above analysis. (Or perhaps it was Eigenray's imprimatur that did it.  ) He says he will change the demo, but when shooting in the air is not the best option for the poorest shot I can't find any rational solutions to the equal-opportunity-of-survival scenario that is mentioned (and partly analysed in my first post).
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| « Last Edit: Feb 22nd, 2009, 5:14pm by ThudnBlunder » |
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Re: It's a Truel World
« Reply #7 on: Feb 3rd, 2009, 1:28pm » |
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on Jan 31st, 2009, 4:59am, ThudanBlunder wrote:
But what is there not to understand? High-school maths, clearly explained at every step? |
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Well I didn't really try too hard to understand it; I just found it easier to do my own analysis. For example, you take a > b > c instead of a < b < c. What order do they shoot in, and who targets whom? Then there are all these unexplained "we get"s. But I agree with your results.
Suppose that A misses and B and C target each other, and that a < b < c are rational. Solving B(s) = C(s) gives a as a rational expression of b,c. Now solving A(s) = B(s) for c shows that b2 - 16b + 16 is a square, so we can set b = 8 + t + 12/t for some rational t. Solving now for a and c, and requiring 0 < a < b < c < 1, this gives the parameterization
a = (t+2)/(t+1)
b = (t+2)(t+6)/t
c = -(t+2)(t+6)/(2t+6)
7 - 5 < t < -2
Now, with these probabilities, the condition for A to miss is
A(s) - A*(s) = t/3 (10 + 7t + t2) / (t2+10t+36) > 0
which is true for all t in (-5+7, -2).
Of course by allowing t to be real we get a parametrization of all real solutions as well, so we can conclude that under the given assumptions, if A(s) = B(s) = C(s) = 1/3, then it is preferable for A to miss.
For an integer parametrization we can take
a = m/(m+n)
b = m(4n-m)/[n(2n+m)]
c = m(4n-m)/[2n(n-m)]
where 0 < m/n < 3-7.
Assuming gcd(m,n)=1 the least common denominator is at least (m+n)n(2n+m)(n-m)/36, which is  (n4). The smallest couple solutions are
{21, 44, 77}/84, {24, 50, 75}/120, {35, 72, 90}/280, {132, 273, 364}/858, {280, 585, 936}/1260, {220, 456, 627}/1320, ...
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| « Last Edit: Feb 3rd, 2009, 2:21pm by Eigenray » |
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ThudnBlunder
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Re: It's a Truel World
« Reply #8 on: Feb 3rd, 2009, 2:01pm » |
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on Feb 3rd, 2009, 1:28pm, Eigenray wrote:| What order do they shoot in, and who targets whom? Then there are all these unexplained "we get"s. |
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I thought I had covered the (well-known) assumptions in my first post. Maybe not. Never mind.
Funny, my English teacher never liked 'get' either. 
Your analysis looks interesting.
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| « Last Edit: Feb 4th, 2009, 12:30pm by ThudnBlunder » |
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Re: It's a Truel World
« Reply #9 on: Feb 4th, 2009, 12:55pm » |
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Eigenray, your mathematical dexterity constantly amazes me.......
on Feb 3rd, 2009, 1:28pm, Eigenray wrote:
Of course by allowing t to be real we get a parametrization of all real solutions as well, so we can conclude that under the given assumptions, if A(s) = B(s) = C(s) = 1/3, then it is preferable for A to miss.
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.......but are you concluding that when A(s) = B(s) = C(s) = 1/3, it is always better to miss, even allowing irrational a,b,c?
You see, I have an unposted proof in my margin here that, when it is better to shoot at C,
there exist irrational a,b,c such that A(s) = B(s) = C(s) = 1/3
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| « Last Edit: Feb 5th, 2009, 9:32am by ThudnBlunder » |
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Re: It's a Truel World
« Reply #10 on: Feb 4th, 2009, 3:10pm » |
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on Feb 4th, 2009, 12:55pm, ThudanBlunder wrote:| But are you concluding that when A(s) = B(s) = C(s) = 1/3, it is always better to miss, even allowing irrational a,b,c? |
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Yes, assuming a < b < c. The rational points are dense on the curve A(s) = B(s) = C(s) so restricting to them doesn't change much.
It follows from the parametrization you gave as well:
Quote:c = 2(1 - 2G)/[G(2 - G)]
b = 2(1 - 2G)/[(1 - G)(2 - G)]
a = 2(1 - 2G)/(4 - 5G)
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which gives A*-A = (1-G)(4+G)(1-2G) / [3(2-G)(G2-3G+5)] > 0 for G in (3-7, 1/2).
Quote:You see, I have an unposted proof in my margin here that, when it is better to shoot at C,
there exist irrational a,b,c such that A(s) = B(s) = C(s) = 1/3
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I'm not sure what you mean by that. "When it is better to shoot at C" depends on the values of a,b,c.
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ThudnBlunder
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Re: It's a Truel World
« Reply #11 on: Feb 4th, 2009, 3:50pm » |
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on Feb 4th, 2009, 3:10pm, Eigenray wrote:
I'm not sure what you mean by that. "When it is better to shoot at C" depends on the values of a,b,c. |
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Well, there is no point in shooting at somebody who is shooting into the air. (Unless he is also whistling Dixie out of tune, say. In which case, be certain to hire a sure shot.) So it must mean shooting at the sure shot.
In other words, when (using your notation)
c[c(1 - b)2 - b]
a < ---------------------
c2(1 - b)2 + b2(1 - c)
as established in an earlier post.
And when c = 1 this is represented by the blue shaded area in the previous diagram.
Will get back to you.
Now, in which book did I write in the margin........
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| « Last Edit: Feb 9th, 2009, 8:24am by ThudnBlunder » |
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Re: It's a Truel World
« Reply #12 on: Feb 5th, 2009, 4:21pm » |
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As a counterexample (that even towr himself would be proud of) is worth a million 'we get's, consider
a = 0.017122...
b = 0.020405...
c = 0.042448....
where the original radicals have been converted to decimals.
And observe that
i) a < b < c
ii) A(s)* = B(s)* = C(s)* = 1/3
iii) a < 1 - bc/{[c(1 - b)]2 + (1 - c)b2} as in my previous post
iv) (10 - 2)/3 = 0.387426... < G = a/c(1 - a) = 0.410384... < 2 - 1 = 0.414214...
This last constraint replaces 3 - 7 G < 1/2 when helping B (shall we call it) is preferable.
It is, in any case, stricter than 3 - 7 G < 1/2 and entirely enclosed by it.
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| « Last Edit: Feb 9th, 2009, 8:17am by ThudnBlunder » |
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Re: It's a Truel World
« Reply #13 on: Feb 5th, 2009, 4:40pm » |
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on Feb 5th, 2009, 4:21pm, ThudanBlunder wrote:| ii) A(s)* = B(s)* = C(s)* = 1/3 |
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But your question was about A(s) = B(s) = C(s) = 1/3, not A*(s).
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Re: It's a Truel World
« Reply #14 on: Feb 5th, 2009, 5:19pm » |
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on Feb 5th, 2009, 4:40pm, Eigenray wrote:
But your question was about A(s) = B(s) = C(s) = 1/3, not A*(s). |
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That's true. Careless of me. I thought my intended meaning was clear.
That is, (again, using your notation)
when A(s) > A(s)* we seek equal opportunity of survival with A(s) = B(s) = C(s) = 1/3
and
when A(s) < A(s)* we seek equal opportunity of survival with A(s)* = B(s)* = C(s)* = 1/3
I don't think there are any rational solutions for the second part. If so, rewriting part of the demo will be a problem.
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| « Last Edit: Mar 14th, 2009, 12:54pm by ThudnBlunder » |
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Eigenray
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Okay....
Solving A* = B* gives
c = a ( a2(2b-1) - b2(a-1)2 ) / [ (1-a)(2a2 + ab - 4a2b - b2 + ab2) ]
Solving B* = C* gives
a4 - b*(2a3 + 4a4) + b2(12a3-5a2) + b3*(8a4-17a3+9a2) + b4*(a4 - 5a3 + 9a2 - 7a + 2) = 0.
Using Mathematica's root numbering, let b(a,k) = Root[(...)/.b->#&, k]. Define
L = 0.1067220725
H = 0.8932779253
M = 0.2631209943
Then:
b(a,1) and b(a,2) are always real, but only in [0,1] for L < a < H. However, b(a,1) < a on this interval, while we only have a < b < c < 1 with b = b(a,2) for L < a < M.
b(a,3) and b(a,4) are real for a < L or a > H. But b(a,3) < a on both, and b(a,4) < a for a > H. For a < L, we have a < b < c < 1 with b = b(a,4).
Therefore let b(a) be b(a,4) for a < L, and b(a,2) for a > L. Then 0 < a < b < c < 1 for 0 < a < M.
Now it turns out that A* - A > 0 only for a < 0.1188376889. So:
The solutions to A* = B* = C* = 1/3, a < b < c, are parametrized by 0 < a < M = 0.2631209943. Under this condition, A should shoot and C when a < 0.1188376889, and miss otherwise.
Attached are the plots of a,b,c in (red, green, blue) along the two curves: on the left is A(s) = B(s) = C(s) = 1/3, and on the right is A*(s) = B*(s) = C*(s) = 1/3.
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Re: It's a Truel World
« Reply #16 on: Feb 5th, 2009, 6:10pm » |
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on Feb 5th, 2009, 4:21pm, ThudanBlunder wrote:As a counterexample (that even towr himself would be proud of) is worth a thousand 'get's, consider
a = 0.017122...
b = 0.020405...
c = 0.042448....
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Are you sure about this? For these values I get
(A*(s), B*(s), C*(s)) = (0.369689, 0.259028, 0.371283)
and my formulas are the same as the ones in your second post (switching a and c).
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| « Last Edit: Feb 5th, 2009, 6:12pm by Eigenray » |
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Re: It's a Truel World
« Reply #17 on: Feb 5th, 2009, 6:36pm » |
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on Feb 5th, 2009, 6:10pm, Eigenray wrote:
Are you sure about this? For these values I get
(A*(s), B*(s), C*(s)) = (0.369689, 0.259028, 0.371283)
and my formulas are the same as the ones in your second post (switching a and c).
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As those equations looked a bit messy to deal with, I parameterized them (using my notation) into
A(s)* = 1/(G + 1)(FH + H + 1)
B(s)* = [H(F2 + F + 1) - FG]/(F + 1)(FH + H + 1)
and they both came out to exactly 1/3 on my humble Maplesoft calculator. 
And the condition to help B became
F < G(GH + G + 1)/H(GH + G + H)
which also checked out.
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| « Last Edit: Feb 6th, 2009, 4:22am by ThudnBlunder » |
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Eigenray
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Re: It's a Truel World
« Reply #18 on: Feb 5th, 2009, 6:46pm » |
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on Feb 5th, 2009, 6:36pm, ThudanBlunder wrote:
As those formulas looked a bit messy I parameterized them (using my notation) into
A(s)* = 1/(G + 1)(FH + H + 1)
B(s)* = [H(F2 + F + 1) - FG]/(F + 1)(FH + H + 1)
and they both came out to exactly 1/3 on my calculator. :-/ |
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That's what I get as well but, using your notation,
(c,b,a) = (0.017122, 0.020405, 0.042448),
(F,G,H) = (0.853726, 0.410391, 0.490719),
A* = 0.371283, B* = 0.259028.
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ThudnBlunder
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Re: It's a Truel World
« Reply #19 on: Feb 5th, 2009, 8:22pm » |
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on Feb 5th, 2009, 6:46pm, Eigenray wrote:
That's what I get as well but, using your notation...
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I will have to have another look. (But from tomorrow I may for a few weeks be confined to lurking.)
I do know that I started out with F = 3/5
and when we consider
A(s) = B(s) = C(s) = 1/3
we get then
G = 2/5
H = 8/7
subject to 3 - 7  G < 1/2
(as in my first post)
And when we consider
A(s)* = B(s)* = C(s)* = 1/3
we ge have
G = (8593 - 73)/48 = 0.410384...
H = (158593 - 455)/1328 = 0.704425
subject to (10 - 2)/3 G < 2 - 1
So perhaps our G matches, allowing rounding error. But not our F or H.
However, unlike yours, my F is right by definition!
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| « Last Edit: Mar 14th, 2009, 1:29pm by ThudnBlunder » |
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Eigenray
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I see, it looks like your b is just off; it should be 0.0290331 or so. But that doesn't really matter.
Using F,G,H really simplifies things! I found the following rational point:
a = 1780225612071980440487233690 / 42514698180175799467200302197
b = 890112806035990220243616845 / 12666774444361618608640463004
c = 326459223625633611840607911167657350 / 3023918895788355583589056125975619389
(a,b,c) = (0.0418732, 0.0702715, 0.107959)
Needless to say, it was not by brute force. There is an elliptic curve involved.
Solving A*=1/3 for F,
F = (2-G-H-GH)/((1+G)H),
and then solving B*=1/3 for H shows that
D2 = G4 + 12G - 4
is a perfect square. Setting
X = 2(D+G2), Y = 12 + 4G(D+G2)
puts us in normal form
E : Y2 = X3 + 16X + 144.
This curve has rank 2, generated by P=(-4,4) and Q=(0,12), and no torsion. So every rational point on E is of the form nP + mQ for some integers m,n.
For each point on E, we get G, and two choices for H, giving two possibilities for (a,b,c); i.e., there are two rational functions from E to triples (a,b,c) with A*=B*=C*=1/3. It turns out they only differ by precomposition with the involution R <-> -Q-R of E, so we can fix either of them WLOG.
Now, not all points satisfy 0 < a < b < c < 1, and A* > A. The "smallest" one that does is
-3P + 5Q = ( -52236003526944484 / 33628523310064225,
66246444689296793086565564 / 6166826011846075282843375 ),
which corresponds to the solution listed above.
In order of increasing maximum denominator of (a,b,c), the "valid" points are
-3P+5Q, 2P+7Q, -5P-13Q, 12P+8Q, 5P-12Q, ...
These points correspond to solutions having maximum denominators with log10 rounding to
36, 43, 120, 137, 142, ...
so you would need very large dice. The base-10 logs of the least common denominators round to
84, 101, 292, 331, 349, ...
So, if the players are to be equally likely to survive when A shoots at C, then they need a die with more faces than there are particles in the universe!
Mathematica notebook attached. Magma commands:
Code:
E:=EllipticCurve([16,144]);
TorsionSubgroup(E);
RankBounds(E);
Generators(E);
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Code:
Abelian Group of order 1
2 2
[ (-4 : 4 : 1), (0 : -12 : 1) ]
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| « Last Edit: Feb 6th, 2009, 1:12am by Eigenray » |
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ThudnBlunder
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Re: It's a Truel World
« Reply #21 on: Feb 6th, 2009, 1:01am » |
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Great stuff!!
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| « Last Edit: Oct 4th, 2009, 10:16am by ThudnBlunder » |
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