who's up for a nice meaty probability problem?

by MaxEntropy_Man Sun Jan 20, 2013 6:53 pm

will post if there is an interest.

by bw Sun Jan 20, 2013 7:25 pm

do post.

by MaxEntropy_Man Sun Jan 20, 2013 7:29 pm

let me just post the link directly.

to those who attempt: don't post methodology for a couple of weeks until they post the solution. just post your answer so we can compare answers. i sent them my solution today. i hope tiger sons and lion daughters are enthused enough to send solutions too.

by Petrichor Sun Jan 20, 2013 9:14 pm

"0.0003215"

by Petrichor Sun Jan 20, 2013 9:16 pm

and yours?

by MaxEntropy_Man Sun Jan 20, 2013 9:41 pm

atcg wrote:and yours?

i'll wait for a few more responses. mine was higher than yours.

by Hellsangel Sun Jan 20, 2013 9:46 pm

1/216 =0.004629

by MaxEntropy_Man Sun Jan 20, 2013 9:52 pm

here is a helpful way to think about problems like this. think up and solve a related but simpler problem. in this case, a 1-d random walk played with a two-sided die with 1 & 2 marked on the two sides of the die (or alternately flip a coin). if you roll the die and get 1 you move one unit rightwards, i.e. +x, and if you roll the die and get 2, you go one unit leftwards, i.e. -x. what is the probability that you are back where you started after two rolls of the die? now generalize the result to 3-d, a six faced die, and six rolls of the die.

by MaxEntropy_Man Sun Jan 20, 2013 9:55 pm

Hellsangel wrote:1/216 =0.004629

HA - not the answer i got.

by Petrichor Sun Jan 20, 2013 9:56 pm

MaxEntropy_Man wrote:and if you roll the die and get 2, you go one unit leftwards, i.e. -x.

Should be -2 leftwards....the word used is "respectively"

Oops...you are right...withdrawing the answer.

by Guest Sun Jan 20, 2013 10:00 pm

test only https://such.forumotion.com/t10137-who-s-up-for-a-nice-meaty-probability-problem?theme_id=104384#77512

hmm

by MaxEntropy_Man Sun Jan 20, 2013 10:02 pm

atcg wrote:
MaxEntropy_Man wrote:and if you roll the die and get 2, you go one unit leftwards, i.e. -x.

Should be -2 leftwards....the word used is "respectively"

atcg: i don't think so. i think the "respectively" is just to keep track of labels. read the line that says the walker always jumps to one of its six nearest neighbors, i.e. the jump is always one lattice distance and the matchups are:

1 = +1 (x-direction), 2 = -1 (x-direction), 3 = +1 (y-direction) and so on.

at least that's how i interpreted the problem. random walk problems are fun and occur in many branches of science - statistical mechanics, transport problems like heat conduction and diffusion etc.

by Hellsangel Sun Jan 20, 2013 10:03 pm

MaxEntropy_Man wrote:
atcg wrote:
MaxEntropy_Man wrote:and if you roll the die and get 2, you go one unit leftwards, i.e. -x.

Should be -2 leftwards....the word used is "respectively"

atcg: i don't think so. i think the "respectively" is just to keep track of labels. i.e. the jump is always one lattice distance and the matchups are:

1 = +1 (x-direction), 2 = -1 (x-direction), 3 = +1 (y-direction) and so on.

at least that's how i interpreted the problem. random walk problems are fun and occur in many branches of science - statistical mechanics, transport problems like heat conduction and diffusion etc.

Your interpretation is correct. I read the problem again.

by Hellsangel Sun Jan 20, 2013 10:06 pm

The answer is that the probability every die roll be unique in the sequence of 6 throws.

6!/(6^6)?

by MaxEntropy_Man Sun Jan 20, 2013 10:10 pm

Hellsangel wrote:The answer is that the probability every die roll be unique in the sequence of 6 throws.

6!/(6^6)?

on the right track, but there are more possibilities than you have accounted for.

by Hellsangel Sun Jan 20, 2013 10:16 pm

MaxEntropy_Man wrote:
Hellsangel wrote:The answer is that the probability every die roll be unique in the sequence of 6 throws.

6!/(6^6)?

on the right track, but there are more possibilities than you have accounted for.

You are right. You may never get a move in x or y or z direction. Those possibilities need to be added.

by MaxEntropy_Man Sun Jan 20, 2013 10:29 pm

Hellsangel wrote:
MaxEntropy_Man wrote:
Hellsangel wrote:The answer is that the probability every die roll be unique in the sequence of 6 throws.

6!/(6^6)?

on the right track, but there are more possibilities than you have accounted for.

You are right. You may never get a move in x or y or z direction. Those possibilities need to be added.

no no no. that's not what i meant. every roll of the die results in one of only six possibilities, a unit move in the positive or negative x,y,or z direction. you were right about that. so the denominator of your probability expression is correct. it is your numerator that is not correct. there is more to add there.

edited to add: in retrospect i think that IS what you meant in your original reply. so yeah we are on the same page.

by Petrichor Sun Jan 20, 2013 10:34 pm

"0.00167181069958848"

by bw Sun Jan 20, 2013 10:37 pm

155/3888

by MaxEntropy_Man Sun Jan 20, 2013 10:42 pm

bw wrote:155/3888

by MaxEntropy_Man Sun Jan 20, 2013 10:49 pm

meanwhile, whenever i take an interest in football, new england loses. haven't watched one game this season before tonight. i tune in and they promptly lay an egg. calculate the odds.

by Idéfix Sun Jan 20, 2013 11:03 pm

Yay for the 49ers! Was on a plane when the game ended, and they gave us a free round of drinks to celebrate.

Interesting problem. I am tempted to solve it using Excel.

by Kris Mon Jan 21, 2013 4:38 am

panini press wrote:Yay for the 49ers! Was on a plane when the game ended, and they gave us a free round of drinks to celebrate.

Interesting problem. I am tempted to solve it using Excel.

>>> I was on my way out to check out a movie, but my younger son convinced me to watch the game. Glad I did. The niners didn't play their best game, but had some lucky breaks. Hope they are in better form on the big day.

by Kris Mon Jan 21, 2013 4:48 am

Kris wrote:
panini press wrote:Yay for the 49ers! Was on a plane when the game ended, and they gave us a free round of drinks to celebrate.

Interesting problem. I am tempted to solve it using Excel.

>>> I was on my way out to check out a movie, but my younger son convinced me to watch the game. Glad I did. The niners didn't play their best game, but had some lucky breaks. Hope they are in better form on the big day.

>>>Forgot to add: kaepernick is definitely a treat to watch

by aanjaneya Mon Jan 21, 2013 10:01 pm

I get (60 + 1080 + 720)/(6^6) = 1860/(6^6), which is approximately 4%

Without saying much, just use a symmetry argument and enumerate a small number of cases, and within each case, count the number of distinct sequences of corresponding 6-tuples.
Of course, assuming independent rolls of the standard equi-probable die.

by MaxEntropy_Man Tue Jan 22, 2013 7:23 pm

aanjaneya wrote:I get (60 + 1080 + 720)/(6^6) = 1860/(6^6), which is approximately 4%

Without saying much, just use a symmetry argument and enumerate a small number of cases, and within each case, count the number of distinct sequences of corresponding 6-tuples.
Of course, assuming independent rolls of the standard equi-probable die.

very nice. that's what i got, as did bw. aanjaneya -- are you a professional mathematician?

by Propagandhi711 Tue Jan 22, 2013 9:11 pm

Kris wrote:
Kris wrote:
panini press wrote:Yay for the 49ers! Was on a plane when the game ended, and they gave us a free round of drinks to celebrate.

Interesting problem. I am tempted to solve it using Excel.

>>> I was on my way out to check out a movie, but my younger son convinced me to watch the game. Glad I did. The niners didn't play their best game, but had some lucky breaks. Hope they are in better form on the big day.

>>>Forgot to add: kaepernick is definitely a treat to watch

overrated - they just have a great front line.

by aanjaneya Thu Jan 24, 2013 12:16 am

MaxEntropy_Man wrote:
aanjaneya wrote:I get (60 + 1080 + 720)/(6^6) = 1860/(6^6), which is approximately 4%

Without saying much, just use a symmetry argument and enumerate a small number of cases, and within each case, count the number of distinct sequences of corresponding 6-tuples.
Of course, assuming independent rolls of the standard equi-probable die.

very nice. that's what i got, as did bw. aanjaneya -- are you a professional mathematician?

Thanks for posting this. Glad that our answers agree. Yes, I work in mathematics for a living.

by bw Thu Jan 24, 2013 7:44 am

here's another question:

if the same walker takes n steps instead of 6 where n is a very large number, how far is he expected to be from the origin in the end?

by aanjaneya Thu Jan 24, 2013 1:04 pm

Intuition says it should be (0,0,0) in expectation, and intuition also says that it should not depend on whether 'n' is large or small.
i.e. one may not need a limiting condition on n \rightarrow \infty.

(Intuition due to the nice symmetry in everything; namely equi-probable faces of the standard die; for every sequence of faces of length n, there exists a neutralizing sequence of length n that also has the same probability; and so on).

In any case, here is one way to work it out.
For 'n' rolls, let the corresponding 6-dimensional vector (n_1, n_2,..., n_6) represent the number of times that the die rolls up as '1', '2', and so on till '6'.
Now (n_1,...n_6) is a random vector that follows a multinomial distribution, with the parameters 'n', and probabilities p_1, p_2,...p_6 = 1/6.
The object of interest is essentially the terminal coordinates in 3-space (i.e. (x,y,z)), which is (n_1-n_2, n_3-n_4, n_5-n_6), after 'n' rolls, where n = \sum_{i=1}^{i=6} n_i

(Due to the +1/-1 moves that specify the stochastic dynamics of this intoxicated particle).

What you are asking is the "expectation" of the above 3-tuple terminal coordinate.

So in other words, you want the expectation of the 3-dimensional random vector (n_1-n_2, n_3-n_4, n_5-n_6), where (n_1, n_2,...n_6) follows a multinomial distribution.
The expectation of each component is zero.
So intuition checks out, I think.
Do post if you agree, or see a flaw.
Thanks.

by bw Thu Jan 24, 2013 9:50 pm

actually, the distance from the origin is always positive. so i am interested in the expected value of the absolute distance. for example, if a point, constrained to lie on the x axis, is at x=1 with prob 0.5 and at x=-1 with prob 0.5, then its expected distance from the origin is 1(and not 0).

aanjaneya wrote:Intuition says it should be (0,0,0) in expectation, and intuition also says that it should not depend on whether 'n' is large or small.
i.e. one may not need a limiting condition on n \rightarrow \infty.

(Intuition due to the nice symmetry in everything; namely equi-probable faces of the standard die; for every sequence of faces of length n, there exists a neutralizing sequence of length n that also has the same probability; and so on).

In any case, here is one way to work it out.
For 'n' rolls, let the corresponding 6-dimensional vector (n_1, n_2,..., n_6) represent the number of times that the die rolls up as '1', '2', and so on till '6'.
Now (n_1,...n_6) is a random vector that follows a multinomial distribution, with the parameters 'n', and probabilities p_1, p_2,...p_6 = 1/6.
The object of interest is essentially the terminal coordinates in 3-space (i.e. (x,y,z)), which is (n_1-n_2, n_3-n_4, n_5-n_6), after 'n' rolls, where n = \sum_{i=1}^{i=6} n_i

(Due to the +1/-1 moves that specify the stochastic dynamics of this intoxicated particle).

What you are asking is the "expectation" of the above 3-tuple terminal coordinate.

So in other words, you want the expectation of the 3-dimensional random vector (n_1-n_2, n_3-n_4, n_5-n_6), where (n_1, n_2,...n_6) follows a multinomial distribution.
The expectation of each component is zero.
So intuition checks out, I think.
Do post if you agree, or see a flaw.
Thanks.

by MaxEntropy_Man Thu Jan 24, 2013 11:35 pm

as long as the jumps are uncorrelated as it is in this case, with the exception of a numerical factor, the displacement should be sqrt(N). i can compute this exactly for the simpler case of 1-d, but need to think harder about 3-d.

by aanjaneya Fri Jan 25, 2013 12:49 am

OK. The terminal coordinate is the 3-tuple, (n_1-n_2, n_3-n_4, n_5-n_6), where (n_1, n_2,...,n_6) follows a multinomial distribution.
So its expectation is indeed (0,0,0).

But bw wants the expectation of the euclidean distance of this point from the origin.
So one needs to compute the Expectation of \sqrt( (n_1-n_2)^2 + (n_3-n_4)^2 + (n_5-n_6)^2 ), with respect to the multinomial distribution of (n_1, n_2,...,n_6).

Using the multinomial distribution properties that Variance(n_i) = np(1-p), and Cov(n_i, n_j) = -np^2,
as well as the following identities, namely,
E(X^2) = Variance(X) + (E(X))^2,
and
E(XY) = Cov(X,Y) + E(X)*E(Y),
using algebra I verify that Expectation of the square of the euclidean distance of the terminal coordinate from the origin simplifies to 6np = n.

by aanjaneya Fri Jan 25, 2013 12:33 pm

one thing to note in addition:
The expectation of the square-distance in this problem is 'n'. This is indeed a very simple closed form expression.
That however does not mean that the expectation of distance is simply \sqrt(n) in closed form.
One needs the variance(distance) characterized in closed form to then further infer what is the closed form of the expectation of distance.

But what you want is indeed characterized by the expectation of \sqrt( (n_1-n_2)^2 + (n_3-n_4)^2 + (n_5-n_6)^2 ), with respect to the multinomial distribution of (n_1, n_2,...,n_6).
So you can indeed write it as a summation, since expectation for a discrete distribution may be expressed as a summation. But you want a closed-form
At first glance, it does not seem all that easy to simplify the above \sqrt to a simple closed-form (as opposed to the square of the above expression, which simplifies to 'n').

Look at the random-walk literature for further pointers. Somebody may have simplified it already for you.

by bw Fri Jan 25, 2013 8:43 pm

thanks, aanjaneya.

i wonder though if there an easier and more intuitive way of arriving at this answer that i could explain to my 7th grader (right now, he is writing a program in python and will try to plot the results).

by aanjaneya Fri Jan 25, 2013 11:21 pm

Awesome. Nothing like a computational experiment to develop and reinforce intuition in most problems.
Python is very very cool.
I will also try to think of a more intuitive explanation, though the above is the most natural one for me. i.e. to think of the quantity of interest as a function of a random variable, whose distribution may be characterized, and then work out its expectation either in closed form (preferable) or numerically.

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