Dumbest thoughts of the day - Resolved

I'm sure there's some more elegant solution to this problem but... I don't remember much from stochastic processes. I'll offer one bit of the solution.

Let X_1, X_2, X_3, ... be the objective "intelligence rating" for the first, second, third, and so on thought. Then {X} is an i.i.d random variable. For simplicity, thoughts are on a continuous scale, the average thought will be given the value 0, and the distribution of thoughts is symmetrical. Thus {X} is a continuous random variable symmetrically centered at 0.

Furthermore, let X_n+1 = X_n when X_n min{X_1, X_2, ... , X_n-1}, and thus X_n+1 is the thought "That previous thought was the dumbest thought I've had today".

We can define S_k to be the time of the k-th occurrence where X_n+1 = X_n given X_n min{X_1, X_2, ... X_n-1}. What we're seeking then is, for a fixed N thoughts in a day, max { k : S_k = N }.

To make things even simpler, let I_k be the interval time between S_k-1 and S_k.

After X_1 occurs, we are looking for X_n X_1. Since X is i.i.d., P(X_n X_1) is a constant for all n. Let that probability equal p and q = 1-p.

Then Es_1 = Ei_1 = Sum p^n * q * (n+2) from n=0 to n=infinity. This geometric series can be simplified into:

Es_1 = 2 - p/(p-1)

Since Ex_1 = 0, p = .5. Then Es_1 = 3.

Es_2 = Es_1 + Ei_2.

... now to keep going you would a) have to know the distribution and b) find what E[X_S_1 | X_S_1 X_1] and repeat the process till you find max{ k : Es_k = N}. Its a brute force method but I think this would work.

I disagree with above solution, since you require knowing the distribution, but as I said above, the distribution is actually irrelevant to the answer. The answer should not vary based on distribution, hence you can't need to know it. Thank you for the effort though.

I am not sure if treating this as a stochastic process is the way to go though. I have been trying to find a decent combinatoric solution, which I think is feasible if you just arrange all N thoughts in order of intelligence, and then determine in how many ways they can fall out in that order.

However, to be fair, I don't know the answer or the solution. I did run a few simulations though, and they have been consistent with my combinatorial approach. (The simulations involved uniform distribution).

Are you assuming your thoughts are on a discrete scale?? And also how are you using combinatorics??? Because you don't know how many ties there are in intelligence ratings which is what thwarts a ranking methodology (e.g. if the 4th dumbest thought happens after the 5th dumbest thought and they are equally low in intelligence rating, then strictly speaking, it would be incorrect to say "That was the dumbest thing I have thought so far today"

I was just thinking, if you use a uniform distribution, then the problem becomes really easy. Let X be distributed uniformly from [-1, 1] and let N be say 50

Then Es_1 = 3 as shown before

Given the uniform distribution, E[X_S_1 | X_S_1 X_1] = -0.5

Thus now p=0.75 and Ei_2 = 2 + .75 / .25 = 5 and therefore Es_2 = 3 + 5 = 8

We can then continue with E[X_S_2 | X_S_1 X_S_1] = -0.75 and p = 0.875

TPride, thanks for taking the time to think about this. I am not using discrete scaling, as mentioned in original post, so no two will ever be equal to each other. However, I still disagree with you based on the reasons I stated above - the distribution does not affect the answer, hence anything that uses the distribution must be flawed. I tested this via simulation, and the answer I achieved via combinatorics was confirmed for both uniform and normal distribution. I think that is sufficient evidence to conclude the distribution does not matter.

My reasoning is as follows:

Start with a small n, 1 through 4 for example. Sort the thoughts in order of stupidity (dumbest is #1). Then list all the possible orders in which these thoughts can fall out - there should be a total of n! of these possibilities. My assumption that appears to be correct but that I can't prove is that each permutation is just as likely. This seems intuitive for small samples, and I think should scale up. Let me know your thoughts about this assumption.

So if you have 1 thought, it must come first and obviously the expectation is 1.

If you have 2 thoughts and sort them in order of stupidity as described above, then there are 2! = 2 possibilities, each equally likely. Either the dumbest one comes first and other one second, or reverse. So expectation is 1/2 * 1 + 1/2 * 2 = 1.5

If you have 3 thooughts, the following are the 3!=6 possibilities for how they can fall out:

Keep in mind that the list below is sorted by stupidity, dumbest first, and the number denotes the order in which you had the thought. 1 2 3 (dumbest thought came first, so you could say "dumbest thing I thought today" only that one time.) 1 3 2 (dumbest thought came first, so you could say "dumbest thing I thought today" only that one time.) 2 1 3 (here, the dumbest thought came out of your head 2nd, and the 2nd dumbest came out first, so 2) 2 3 1 (here, the dumbest thought came out of your head 3rd, and the 2nd dumbest came out first, so 2) 3 1 2 (here, the dumbest thought came out 3rd, and the 2nd dumbest 1st, so 2) 3 2 1 (here, your thoughts got dumber each time, so 3) So expectation is 1/6(21 + 32 + 13) = 11/6 = 1.833. This has been confirmed to be correct via simulation for normal and uniform thoughts.

If you have 4 thoughts, there are 4! = 24 possibilities. I will not list them out, thought I did on paper. The expectation turns out to be 1/24(61+112+63+1*4) = 50/24 = 2.083. This has been confirmed to be correct via simulation for normal and uniform thoughts.

My problem is I am not sure how to scale this up to n, but I do believe the approach is correct. I have been looking for a pattern that I can express combinatorically for the coefficients within the expectation - so the 6, 11, 6 and 1 in the n=4 case. However, I am looking into determining the exact coefficients for n=5 through 8 or 9 via computer to have more sample data.

Admittedly, I do not know stochastic calc so I do not exactly follow your steps, but could you find your answer for n=3 or n=4 using your method (assuming uniform or normal, since that seems to be a part of it) and see if it matches the figures above?

You are close so here is a hint. Given how complicated you are trying to make this, I'm surprised you haven't commented on the obvious pattern.

You commented that for n = 1, 2, 3, 4 you had answers of E(n) = 1, 1.5, 1.833, 2.083.

Let's take a look at delta_E(n): 1, 0.5, 0.33, 0.25.

Do we notice any patterns here?

The answer to your question is the summation for n from 1 to infinity of 1/n (if we assume you can have infinity thoughts in one day). If I recall these are known as Taylor series, and if I recall correctly from my calculus days, this one does not converge. So if you are having problems answering this brain teaser, fear not! There is no answer (the result is infinite).

-BigBaby

P.S. I haven't done Taylor series since early 2006 during my senior year of high school, so if that's all right it will be a significant boost to my already considerable opinion of my own faculty with numbers. I think my hedge fund is going to love me.

BigBaby needs to never comment again so he can retain the highest ratio of insight:posts in the history of WSO.

I'm dumber having read this thread.

Too much time on your hands. I thought this was the investment banking message board. You're about 14 math courses too high.

Wasn't the answer obvious...