[superprismatic]

Let S be the set {(x,y,z)|x,y,z are real & x+y+z=1 & 0≤x≤1 & 0≤y≤1 & 0≤z≤1}

(this is the set of all discrete probability distributions on 3 outcomes).

Pick a random element (a,b,c) from S, i.e., pick it in such a way that each

element of S is as likely as any other of being chosen. What is the

probability that (1/6)≤a≤(2/3) and (1/6)≤b≤(2/3) and (1/6)≤c≤(2/3) ?

Edit: In that last sentence, I mean the joint probability that all three are true.

[Guest]

seems like there is a 50% chance for each individual one to be true so .5x.5x.5=.125 for all three to be true

[Guest]

bobb you overlook the fact that these are [spoiler=]not independent events.

ie if a = 2/3 we are good so far, however the probability going forward is not .5 * .5 as we have now limited the set of b and c to (1/6) for each. Why? Because a+b+c = 1

I'm missing something...but hey, it's still summer, right?

Edited August 6, 2011 by plainglazed
fixed spoiler tags

[Guest]

Just quick analysis:

Since A+B+C =1& limitations of A,B,C, 1/3≤A+B≤5/6. Probability of this is .5. With these limitations, C must be within its specified range so my answer is 50% probability.

[bonanova]

I'm replying from my iPad on which spoilers don't seem available.

So until I get to my desktop I'll just say it seems to admit use of

uniform point picking and hence a ratio of two calculable quantities.

SP, if this gives away too much, feel free to use your new powers

to Spoiler it. ;-)

[superprismatic]

I'm replying from my iPad on which spoilers don't seem available.

So until I get to my desktop I'll just say it seems to admit use of

uniform point picking and hence a ratio of two calculable quantities.

SP, if this gives away too much, feel free to use your new powers

to Spoiler it. ;-)

I was trying to think of a good hint to get people started. I don't think

I can top yours. Thanks, bonanova

[bonanova]

Back of the envelope calculation for the two-dimension case.

([16-1-2]/36) / (18/36) = 13/18.

[superprismatic]

Back of the envelope calculation for the two-dimension case.

([16-1-2]/36) / (18/36) = 13/18.

When I do the two-dimensional case, I get a probability of 1/3.

[bonanova]

When I do the two-dimensional case, I get a probability of 1/3.

Agree.

I misread the OP as {(x,y,z)|x,y,z are real & x+y+z<=1 & 0≤x≤1 & 0≤y≤1 & 0≤z≤1}

[superprismatic]

The correct [and much simpler] problem has an interesting property.

One simply slices 1/6 from each edge and 1/3 from each vertex of an equilateral triangle and finds that 1/4 of the area remains.

If the points are spread equally over the triangle [OP] then the joint probability is .25.

Interestingly, the 1/3 cuts don't have to be made.

That is, the probability that (1/6)≤a≤(2/3) and (1/6)≤b≤(2/3) and (1/6)≤c≤(2/3)

is the same as the probability that (1/6)≤a and (1/6)≤b and (1/6)≤c.

I didn't think anyone would notice that the 1/3 cuts are implied by the 1/6 ones!

[bonanova]

My post was edited and quoted at the same time - matter hits antimatter - and was anihilated.

Reconstructing ...

The correct [and much simpler] problem has an interesting property.

One simply slices 1/6 from each edge [purple shaded area]

and 1/3 from each vertex [green shaded area]

of an equilateral triangle and finds [by counting elemental

triangles] that 9/36 = 1/4 of the area remains.

If the points are spread equally over the triangle [OP] then the joint probability is .25.

Interestingly, the 1/3 [green] cuts don't have to be made.

The green area lies entirely within the purple area.

Which means, the probability that (1/6)≤a≤(2/3) and (1/6)≤b≤(2/3) and (1/6)≤c≤(2/3)

is the same as the probability that (1/6)≤a and (1/6)≤b and (1/6)≤c.