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March 26[edit]

Question that is harder than it looks[edit]

Each episode of the TV quiz show Pointless features four couples, one of which wins the show. Non-winning couples get asked back for one more appearance (two shows in total). Winning couples do not get asked back. Assuming that all four couples in each show have an equal chance of winning, what is the long-term average number of returning couples present in each show? This is not homework. It is a question that I devised myself out of interest. I have arrived at an answer but I am not certain it is correct. I would like to check it against someone else's answer. 109.153.236.169 (talk) 02:47, 26 March 2015 (UTC)[reply]

I get

1{\tfrac {5}{7}}

. -- BenRG (talk) 06:47, 26 March 2015 (UTC)[reply]

I get that too. Abecedare (talk) 07:42, 26 March 2015 (UTC)[reply]

Likewise. It's a Markov chain problem, with the transition matrix

{\begin{bmatrix}0&0&0&1\\0&0&3/4&1/4\\0&1/2&1/2&0\\1/4&3/4&0&0\end{bmatrix}}

. -- Meni Rosenfeld (talk) 11:12, 26 March 2015 (UTC)[reply]

Also, for a general number n of couples, I get

{\frac {n(n-1)}{2n-1}}

. -- Meni Rosenfeld (talk) 11:19, 26 March 2015 (UTC)[reply]

(OP) Thanks for the replies everyone. I'm wondering now about the necessity for the condition "all four couples in each show have an equal chance of winning". I believe that the answer would be different if the likelihood of winning was somehow influenced by whether a couple was a returning couple or a new couple. However, suppose that some couples are inherently better than others at the game, and that couples retain the same ability for each show on which they appear. Does this make a difference to the 12/7 answer? At first I thought not, but now I am having doubts. 217.44.208.212 (talk) 13:28, 26 March 2015 (UTC)[reply]

What matters is the probability that a returning couple wins vs. a new couple. If there is no "experience" factor, so an individual couple has the same chance if winning whether they are new or returning, then the probability of a returning couple winning is the same as if all couples have the same chance of winning and you get the same answer. But if returning couples have learned from experience and have a better chance of winning than they had as new couples, or if their morale is so broken by there loss that they have a lower chance of winning on the second try, that will change the answer. --RDBury (talk) 13:47, 26 March 2015 (UTC)[reply]

Thanks, I understand all the part about the "experience" factor. However, it seems to me that "an individual couple has the same chance of winning whether they are new or returning" is not possible unless all couples have 1/4 chance of winning (or the numbers are artificially fixed in a way that does not make sense in reality). This is because they will be with different contestants in each case. If the contestants' chances add up to 1 in the first game, then, when couples come back and are mixed up with a different group, the chances won't generally add to 1. The way I tried to think of it was to assign a "strength" number to each couple, which they retain throughout. The chance that a couple will win is their "strength" divided by the total of all couples in the game. For example, if couples' strengths are 1, 3, 2 and 2, then the chance that the first couple wins is 1/(1+ 3 + 2 + 2). In this model, does the 12/7 result still hold? 217.44.208.212 (talk) 14:27, 26 March 2015 (UTC)[reply]

I don't think what RDBury said is true. Returning couples are likely to be less skilled, and this affects the result. It's not trivial to model symbolically, but I ran a simulation on a model that assumes there are "winners" and "losers", in equal amounts. A winner will always win against a loser (losers have a strength of 1, winners have an arbitrarily high strength). I got a ratio of ~1.658 - with a small, but statistically significant, difference from 12/7.

You're right, I missed the "weeding out" factor. --RDBury (talk) 16:56, 26 March 2015 (UTC)[reply]

If couples are inherently similar, but affected by experience, this is easier to model. -- Meni Rosenfeld (talk) 14:39, 26 March 2015 (UTC)[reply]

Here's an approximate graph of the value, as a function of the percentage of losers: http://i.imgur.com/BAeD0fI.png (Edit: see below for an accurate graph) -- Meni Rosenfeld (talk) 15:14, 26 March 2015 (UTC)[reply]

Thanks, I think your comment "Returning couples are likely to be less skilled" is very useful, and would explain why the 12/7 number is no longer correct if varying skill levels are allowed. 217.44.208.212 (talk) 15:00, 26 March 2015 (UTC)[reply]

Also, to further clarify why this results in a lower value: Returners are less skilled, thus less likely to win, thus it is more likely that a newcomer will win and be denied the opportunity to return. Thus, this result in an overall lower level of returners. In the winner/loser model, the value is minimal with roughly 68.8% losers, and a value of ~1.6438. -- Meni Rosenfeld (talk) 15:14, 26 March 2015 (UTC)[reply]

Ok, I have symbolic results now. With an equal number of winners and losers, the value is 735718463 / 443497368, which is 1.65890... . The minimal value of 1.643630087... is obtained for 69.08278956...% losers. The closed-form formula for a general proportion p of losers is:

4 (12960 - 1440 p - 3632 p^2 - 13328 p^3 + 30922 p^4 - 10636 p^5 + 797 p^6 - 14780 p^7 + 21384 p^8 - 12394 p^9 + 4417 p^10 - 1120 p^11 - 1199 p^12 + 1302 p^13 - 338 p^14 + 96 p^15 - 63 p^16 + 12 p^17) /
(30240 - 2400 p - 6960 p^2 - 28496 p^3 + 68202 p^4 - 27554 p^5 + 1419 p^6 - 29175 p^7 + 48060 p^8 - 31547 p^9 + 10440 p^10 - 2597 p^11 - 126 p^12 + 1113 p^13 - 471 p^14 + 160 p^15 - 84 p^16 + 16 p^17)

An accurate graph can be found at http://i.imgur.com/vri1YLF.png.

Mathematica code:

 n = 4;
 
 States = {{0, 0, 0, n}}; Clear[RevState];
 
 RevState[x_] := (AppendTo[States, x]; RevState[x] = Length[States])
 
 TransitionMatrix[p_] := (
   Clear[trans];
   trans[x_, y_] := 0;
   i = 1;
   (*Run over source states*)While[i <= Length[States],
     {RL, RW, NL, NW} = States[[i]];
     y = Table[0, {4}];
     (*Choose whether newcomer or returner wins*)Do[
      If[RW + NW > 0,
       y[[1]] = NL;
       y[[2]] = NW - j;
       q = If[j == 0, RW/(RW + NW), NW/(RW + NW)],
       y[[1]] = NL - j;
       y[[2]] = 0;
       q = If[j == 0, RL/(RL + NL), NL/(RL + NL)]
       ];
      new = n - y[[1]] - y[[2]];
      (*Choose how many winners for next round*)Do[y[[3]] = k;
       y[[4]] = new - k;
       If[q > 0, 
        trans[i, RevState[y]] = 
         q Binomial[new, k] If[k == 0, 1, p^k] If[new == k, 
           1, (1 - p)^(new - k)]], {k, 0, new}], {j, 0, 1}];
     i++
     ]
    Table[trans[i, j], {i, 1, Length[States]}, {j, 1, Length[States]}]
   )
 
 Val[p_] := Module[{x},
   x = Eigenvectors[Transpose[TransitionMatrix[p]]][[1]];
   x = x/Total[x];
   Sum[x[[i]] States[[i, j]], {j, 1, 2}, {i, 1, Length[States]}]]

-- Meni Rosenfeld (talk) 18:04, 26 March 2015 (UTC)[reply]

Wow! 217.44.208.212 (talk) 21:51, 26 March 2015 (UTC)[reply]

Decoupled mathematical models[edit]

I am trying to find a good source that describes and defines decoupled mathematical models. I'm thinking less of situations where there is truely no interaction between variables, as described here. I'm thinking more of situations where decoupling is used as a choice to simplify a problem when modelling. Such a choice will lead to two or more models where the output from one model is used to define the input of the other, but there is no feedback between the two (except in special cases where iteration is used). I was hoping there would be something on this on Wikipedia, but I can't find it. Any ideas on the best place to look? A Google search turns up lots of things, some irrelevant and many just being examples of decoupling, rather than a good general description.

Yaris678 (talk) 12:12, 26 March 2015 (UTC)[reply]

Before I get too deep in a possibly irrelevant answer, is KAM theory at all relevant? There one starts with an integrable system (so "decoupled" into harmonic oscillators), and then perturbs away from the integrability. Sławomir Biały (talk) 12:40, 26 March 2015 (UTC)[reply]

It sounds like you're talking about operator splitting (wut, no article? try a book's introduction), although maybe you had another kind of modeling in mind. --Tardis (talk) 14:01, 26 March 2015 (UTC)[reply]

Hi guys. Thanks for the suggestions, but I don't think what I am talking about is those things. It is more like submodelling, as described here. In that example, in reality, the motion in the stress-concentration area will feed back and affect motion far from this area, but the detailed model of the stress concentration area just takes its boundary conditions from the coarser model, and doesn't feed back. Yaris678 (talk) 14:25, 26 March 2015 (UTC)[reply]

I think Central_limit_theorem#CLT_under_weak_dependence might spirit you're looking for. But in that case, the theorem just holds without the assumption of strict independence. Probably more appropriate is anything that's called "separation of time scales" or similar. See e.g. this paper [1] about Michaelis–Menten_kinetics and how similar modeling ideas can be applied elsewhere. The idea is not the the fast and slow time scales are truly independent/decoupled, just mostly independent, and that's often good enough. Searching google scholar for thinks like /model [simplify/simple] decouple (assumption) [area of application]/ [2] should also help you find relevant materials. SemanticMantis (talk) 15:31, 26 March 2015 (UTC)[reply]

Also, "mean field assumption" or "mean field approximation" will get you models where a sort of decoupling assumption is made, even though it is often known to not be literally true. This type of thing comes up a lot in my field (ecological modeling, e.g. the pair approximation [3]), but it's a rather different sort of modeling than the examples you've given. But if you want refs in ecological modeling, I can give you plenty of those. SemanticMantis (talk) 15:37, 26 March 2015 (UTC)[reply]

Hi SemanticMantis, You are in the right area and I think it has helped me clarify in my mind what I am after. Simplifying a model by approximating a system by two or more decoupled systems is exactly what I am talking about. I don't know if I will find source that describes this exactly. I'm looking for quite a general source. I suppose if there is an application it is in modelling mechanical systems. Yaris678 (talk) 16:59, 26 March 2015 (UTC)[reply]

I'm glad I got at least the right area, but I'm not aware of a good general reference on the topic. Sadly, mathematical modeling is divided into fairly distinct silos - so model techniques in physics, engineering, biology, chemistry etc. are often uninformed by what the others are doing. There's also the fact that advances in pure and applied math are fairly decoupled :) Further, approaches are fairly distinct in their methods: discrete/continuous, analytic/simulation, etc. Anyway, another thing you might be interested in is multiscale modeling, and this book [4]. Moment_closure is another common way that an assumption of decoupling can help provide tractable and informative models.

The general idea of decoupling (or approximating weak dependence by indenpendence) assumptions should be at least mentioned in any good book on mathematical modeling, but types of decoupling assumptions will vary (by field of application and mathematical methods).

At the risk of stating the obvious, if you are at a university, you could also contact your local reference librarian, or perhaps a professor of mechanical engineering or applied math. Hope that helps, SemanticMantis (talk) 18:36, 26 March 2015 (UTC)[reply]

I'm not in a university, unfortunately.

Multiscale modelling is a great example of the sort of thing I am talking about. I can imagine that there will be a small-scale model, which informs a coarse-scale model on certain physical properties and then the coarse-scale model informs a different small-scale model with boundary conditions. So information goes up and down the scale but never feeds back, keeping the models decoupled.

Yaris678 (talk) 10:42, 27 March 2015 (UTC)[reply]

Ok, well that's about all I can come up with right now. FYI, many universities will grant library cards to members of the public, you might want to look into that if there is a university near where you live. SemanticMantis (talk) 14:15, 27 March 2015 (UTC)[reply]

Laplace transform understanding[edit]

I am trying to understand the meaning of the Laplace domain, as it seems to be weirder than the Fourier domain. I asked a question on that page: Talk:Laplace_transform#Intuition_and_inverse_transforming_the_Dirac_delta feel free to answer it there as I think my confusion points to a shortcoming in that page. —Ben FrantzDale (talk) 18:46, 26 March 2015 (UTC)[reply]

I think what's going on is just what you said: the Laplace transform is massively redundant, so you can't expect an arbitrary C → C function to be the Laplace transform of something. In particular, per Inverse Laplace transform#Mellin's inverse formula, for any sufficiently large γ the values on the line Re(s) = γ are sufficient to reconstruct the original function (and hence the whole Laplace transform).

If you consider a Laplace transform restricted to Re(s) = γ, call it

F_{\gamma }(s)=\int _{0}^{\infty }e^{-(\gamma +is)t}f(t)\,dt

, then Mellin's inverse formula becomes

f(t)={\frac {1}{2\pi i}}\int _{-\infty }^{\infty }e^{(\gamma +is)t}F_{\gamma }(s)\,ds

. This looks a lot like the Fourier transform and its inverse. If you like, you could think of the Laplace transform as a bunch of side-by-side Fourier-like transforms of the function (one of which (F₀ in my notation) actually is the Fourier transform). If there are no singularities, any one of them is sufficient to recover the original function. -- BenRG (talk) 01:44, 27 March 2015 (UTC)[reply]

Thanks for the reply. That makes sense. The fact that the Fourier transform appears along the complex axis is also mentioned here, which I just watched for the first time. Very interesting! —Ben FrantzDale (talk) 02:50, 27 March 2015 (UTC)[reply]