Classic Type Game Theory Problem

[pzhon]

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"Uniform distribution on [0,1]" is standard terminology for Lebesgue measure. And if m is Lebesgue measure, then m([0,1])=1 and m({x})=0 for all x. The statement of the problem makes perfect sense and the existence of the uniform distribution on [a,b] is proven in any first-year graduate real analysis course, and some undergraduate ones.

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You're joking, right?

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No. It's standard terminology.

Mathworld: uniform distribution

[jason_t]

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"Uniform distribution on [0,1]" is standard terminology for Lebesgue measure. And if m is Lebesgue measure, then m([0,1])=1 and m({x})=0 for all x. The statement of the problem makes perfect sense and the existence of the uniform distribution on [a,b] is proven in any first-year graduate real analysis course, and some undergraduate ones.

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You're joking, right?

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No. It's standard terminology.

Mathworld: uniform distribution

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When I said "You're joking, right?" I was questioning his assertion that the existence of the Lebesgue measure implies that the problem makes sense. It doesn't.

I already clarified exactly what I meant when I used the phrase "uniform distribution." We need a probability measure on [0,1] that assigns nonzero equal mass to all the singletons. The Lebesgue measure almost satisfies this, but it assigns zero mass to each singleton.

I'm not a probabilist. I wanted to state that it's impossible to select real numbers uniformly at random. I used the phrase "uniform distribution." If that's not the right phrase it doesn't matter because I clarified exactly what I meant.

[jason1990]

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Suppose m is a measure on [0,1] such that m([0,1]) = 1 and m({x}) = m({y}) for any x, y in [0,1]. That's what we mean by a uniform distribution.

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No it's not. That would allow too many measures. For example, any probability measure on [0,1] which is absolutely continuous with respect to Lebesgue measure has this property.

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No. What I mean by "uniform distribution" is a measure that assigns nonzero equal measure to each of the singletons in the space. This is what we need so that we can pick real numbers uniformly at random. If you want a continuous uniform distribution that's a different issue.

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Okay, so that's what *you* mean. But that's not what anyone else means. Everyone else understands it to mean Lebesgue measure. And when it means Lebesgue measure, the problem makes perfect sense.

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A uniform distribution is one which is invariant under translations, rotations, and reflections.

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The notions of translations, rotations and reflections don't even make sense on a general measure space.

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No one's talking about a general measure space.

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If m({x}) > 0 for some x in [0,1] choose n so large that n * m({x}) > 1. Then choose n distinct x_1, ... x_n in [0,1]. We'd have

1 = m([0,1]) >= m({x_1, ..., x_n}) = m({x_1}) + ... + m({x_n}) = n * m({x_1}) > 1

so that 1 > 1. This is a contradiction. Hence m({x}) = 0 for any x in [0,1]. Oops.

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What does "oops" mean here? Do you think you have arrived at a contradiction to the existence of m?

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We want the event {x} to have nonzero probability. That's the oops. We concluded that there is no measure assigning nonzero equal weights to all the points.

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No. *You* want the event {x} to have nonzero probability.

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"Uniform distribution on [0,1]" is standard terminology for Lebesgue measure. And if m is Lebesgue measure, then m([0,1])=1 and m({x})=0 for all x. The statement of the problem makes perfect sense and the existence of the uniform distribution on [a,b] is proven in any first-year graduate real analysis course, and some undergraduate ones.

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You're joking, right? I just proved for you that you can't assign nonzero equal mass to each of the points. You can't pick real numbers uniformly at random.

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I'm not joking. The uniform distribution on [a,b] is a constant times Lebesgue measure. Its existence is proved in first-year courses.

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Are you, by chance, an algebraist?

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No. Don't insult me. I'm a complex analyst. I study very serious problems which to a certain extent involve about trying to find very special measures on certain compact subsets of the complex place. The notions of Hausdorff measure, doubling measures, Calderon-Zygmund theory of singular integrals, capacity, etc. all play very major roles in what I study. Capacity as you might know arises by assigning a very special probability measure to the boundary of a compact set. cf. the Painleve problem, the Denjoy conjecture, the Vitushkin conjecture and Tolsa's theorem.

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I thought I was insulting the algebraists. [img]/images/graemlins/wink.gif[/img]

I'm a probabilist. And in probability, when someone uses the term "uniform distribution on [0,1]", they mean Lebesgue measure. You obviously think the term means something else, but you're wrong. Honestly, I'm stunned. Clearly you know enough so that this is a semantic issue and not a mathematical one, but how can your use of terminology be so off base? If you called a Markov process a martingale, it might be understandable since you're not a probabilist. But the term you're misusing is so basic and taught in every undergraduate probability and statistics course there is. Did someone hack your account? Is this jason_t's younger brother?

[David Sklansky]

Let me explain something to everyone. Bill Chen had no problem with the problem. Thus there isn't one.

[jason_t]

A few things quickly (I'm working on mathematics tonight [img]/images/graemlins/laugh.gif[/img]); I'll write more later.

What does it mean to "[deal a real number]"? I would assume it means choose one uniformly and at random. That, in my mind, means you have a probability measure and each real number is assigned the same nonzero mass. Where am I thinking about this incorrectly?


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I'm a probabilist. And in probability, when someone uses the term "uniform distribution on [0,1]", they mean Lebesgue measure. You obviously think the term means something else, but you're wrong. Honestly, I'm stunned. Clearly you know enough so that this is a semantic issue and not a mathematical one, but how can your use of terminology be so off base? If you called a Markov process a martingale, it might be understandable since you're not a probabilist. But the term you're misusing is so basic and taught in every undergraduate probability and statistics course there is. Did someone hack your account? Is this jason_t's younger brother?

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I've never taken probability. Undergraduate or otherwise. All of my knowledge of it is informal. I never have a need for anything formal in my work. I don't think I've made any mathematical errors in my posts, just semantical ones.

[jason1990]

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What does it mean to "[deal a real number]"? I would assume it means choose one uniformly and at random. That, in my mind, means you have a probability measure and each real number is assigned the same nonzero mass.

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But you proved that it cannot mean this. So what can it mean? We have to have a probability measure. And we certainly want that measure to have the property that if X is the chosen number, then P(a<X<b) depends only on b-a.

Here's my attempt at a rigorous formulation of the problem.
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Let X, Y, U, V be independent, uniformly distributed random variables, where "uniform" is meant in the standard probabilistic sense. Let

h:[0,1]x[0,1] --> {0,1} and
g:{0,1}x[0,1] --> {0,1}

be Borel-measurable. Define

S_A = h(X,Y) and S_B = g(h(X,Y),Y).

Define

C_A = U*S_A + X*(1 - S_A) and
C_B = V*S_B + Y*(1 - S_B).

Define

E = E(h,g) = 50P(C_A > C_B) - 50P(C_A < C_B).

Find max_h(min_g(E)) and min_g(max_h(E)) and find functions h and g such that E(h,g) attains these values.
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Interpretations:
X,Y - original cards dealt to A and B, respectively
U,V - new cards dealt, if they choose to switch
h,g - strategies
S_* - 1 if * switches, 0 otherwise
C_* - *'s card after deciding whether to switch
E - A's expectation
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P.S. You can take X,Y,U,V as the four coordinate projections of [0,1]^4 onto [0,1]. Then C_A and C_B are functions from [0,1]^4 to R and, for example, P(C_A > C_B) is just

m({x:C_A(x) > C_B(x)}),

where m is Lebesgue measure on R^4.

[Darryl_P]

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I also estimate the overall edge to player A as 8%, though I didn't proofread that part. That is lower than I expected.


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I've already done the overall EV calculation and it's:

EV(A) = 0.5062
EV(B) = 0.4938

and I'm pretty sure it's right. In case it is, can I be mentioned in a footnote in your published work as the first person to correctly determine the EV? Thanks.

[felson]

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I've never taken probability. Undergraduate or otherwise.

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In all seriousness, I think the misunderstanding lies here. You don't know what a continuous-valued random variable is, or what a probability density function is (and why it is different from a probability mass function). It may be non-intuitive, but a continuous-valued RV may take on a particular value, even though the probability of that event is zero (although, I believe, the pdf must be nonzero).

[jason_t]

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I've never taken probability. Undergraduate or otherwise.

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In all seriousness, I think the misunderstanding lies here. You don't know what a continuous-valued random variable is, or what a probability density function is (and why it is different from a probability mass function). It may be non-intuitive, but a continuous-valued RV may take on a particular value, even though the probability of that event is zero (although, I believe, the pdf must be nonzero).

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I know what a random variable is. (It's just a measurable function, right? Maybe we want {x: f(x) = +-oo} to have zero measure too.)

[felson]

I'm not familiar enough with measure theory to define all the terms for you as rigorously as you would like. But wikipedia should provide the intuition. You'll have to consult a textbook or someone else for the formalism.