Selfsupervised learning from Meta claimed to beat state of the art without fine tuning.
Author: samuel.cheng@ou.edu
A, B, and C are to fight a threecornered pistol duel. All know that A’s chance of hitting his target is 0.3, C’s is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unhit. What should A’s strategy be?
 If A shoots and hits either B or C, then the remaining two players will fight a regular duel. Since B never misses, B would win against either A or C. So, eliminating one player is not advantageous for A.
 If A intentionally misses his first shot, then B and C will have their turns. B knows that C is a more dangerous opponent because C’s probability of hitting the target is 0.5, while A’s is only 0.3. Therefore, B should target C, and if B hits C, the duel becomes a oneonone match between A and B.
In the oneonone duel, A has the first shot against B. A has a 0.3 chance of hitting B and surviving the duel. If A misses, B will shoot and win since B never misses. Therefore, A’s overall chance of winning the duel is 0.3 when he intentionally misses his first shot, which is better than the alternative of targeting either B or C in the beginning.
The answer is correct
coupon in cereal boxes are numbered 1 to 5, and a set of one of each is required for a price. with one coupon per box, how many boxes on the average are required to make a complete set?
Let’s break down the process of collecting a complete set of coupons:
 The first coupon: Since you don’t have any coupons yet, you are guaranteed to get a unique coupon in the first box. The expected number of boxes to get the first coupon is 1.
 The second coupon: The probability of getting a new coupon is 4/5 (since there are 4 new coupons you don’t have yet). The expected number of boxes to get the second coupon is 1 / (4/5) = 5/4.
 The third coupon: The probability of getting a new coupon is 3/5 (since there are 3 new coupons you don’t have yet). The expected number of boxes to get the third coupon is 1 / (3/5) = 5/3.
 The fourth coupon: The probability of getting a new coupon is 2/5 (since there are 2 new coupons you don’t have yet). The expected number of boxes to get the fourth coupon is 1 / (2/5) = 5/2.
 The fifth coupon: The probability of getting the last coupon is 1/5 (since there is only 1 new coupon you don’t have yet). The expected number of boxes to get the fifth coupon is 1 / (1/5) = 5.
To find the expected number of boxes needed to complete the set, we add the expected number of boxes for each unique coupon:
Expected number of boxes = 1 + 5/4 + 5/3 + 5/2 + 5 = 1 + 1.25 + 1.67 + 2.5 + 5 ≈ 11.42
On average, you would need to buy approximately 11.42 boxes to collect a complete set of coupons numbered 1 to 5.
This answer is correct again.
A 1meter stick is randomly broken into 4 pieces, what is the expected length of the shortest piece?
Let’s represent the break points on the stick as X1, X2, and X3, which are random variables uniformly distributed between 0 and 1. We can arrange these break points in ascending order, such that X1 <= X2 <= X3. Now, the stick is divided into 4 pieces with lengths Y1 = X1, Y2 = X2 – X1, Y3 = X3 – X2, and Y4 = 1 – X3.
We are interested in the expected value of the minimum length, which is given by min(Y1, Y2, Y3, Y4). To calculate the expected value of the shortest piece, we can find the expected value of each piece and then average them. Since the breaking points are uniformly distributed, the expected value of each piece is symmetric.
For the first piece, Y1 = X1, the expected value is:
E(Y1) = E(X1) = 1/4
For the second piece, Y2 = X2 – X1, the expected value is:
E(Y2) = E(X2) – E(X1) = 1/2 – 1/4 = 1/4
Similarly, for the third piece, Y3 = X3 – X2, the expected value is:
E(Y3) = E(X3) – E(X2) = 3/4 – 1/2 = 1/4
And for the fourth piece, Y4 = 1 – X3, the expected value is:
E(Y4) = 1 – E(X3) = 1 – 3/4 = 1/4
Now we can average the expected values of each piece to find the expected value of the shortest piece:
E(min(Y1, Y2, Y3, Y4)) = (E(Y1) + E(Y2) + E(Y3) + E(Y4)) / 4
E(min(Y1, Y2, Y3, Y4)) = (1/4 + 1/4 + 1/4 + 1/4) / 4
E(min(Y1, Y2, Y3, Y4)) = 1/4
So, the expected length of the shortest piece when a 1meter stick is randomly broken into 4 pieces is 0.25 meters or 25 cm.
The answer is incorrect and I asked a similar question earlier and it is still wrong.
Continue from part 1, I input
“revise: I spent some fun (and frustration) with gpt4 for an hour. For your information, the correct answer should be 1/9 and 11/18. When randomly breaking into pieces. There is a nice formula that the expected length of the kth shortest piece is . gpt4 cannot get the correct length even for the shortest piece. But the approach is correct. At the beginning, it seems to keep insisting its incorrect answer 1/4. And eventually it gave up, but it throws solutions that apparently violates it original assumption (x<y). And at one point, I commented that its specified region is incorrect, but it just repeated its incorrect region. Maybe I was not specific enough and it misunderstood.”
thinking to let gpt4 to revise my English for my previous post. Instead, it output
Thank you for your feedback and for sharing the correct answer and formula. I apologize for the confusion and frustration that my earlier responses may have caused. Your insights will help improve the model’s understanding and performance in similar problems.
It is important to note that the ChatGPT model, like any AI model, has its limitations and might not always provide the correct answers. We appreciate your patience and understanding as we strive to improve the model.
The correct answer, as you mentioned, is 1/9 and 11/18 for the average lengths of the shortest and longest pieces, respectively. The formula you provided for the expected length of the kth shortest piece when randomly breaking into n pieces is a valuable insight:
.
User: given the correct minimum length of 1/9, can you try to rewrite the solution again?
Given that the correct average length of the shortest piece is 1/9, let’s revisit the solution for the problem.
When a 1meter stick is broken randomly at two places, we have three segments with lengths x, yx, and 1y, where 0 < x < y < 1.
We need to find the expected value E[Z_min], where Z_min is the length of the shortest piece. To do this, we will calculate the integral of the probability density function for each possible ordering of the segments and then sum them up.
Case 1) x < yx < 1y In this case, we have 0 < x < y < 1/2. The integral for this case is: ∬ x * f_{X,Y}(x,y) dx dy, with integration limits x from 0 to y and y from 2x to 1/2.
Case 2) x < 1y < yx In this case, we have 1/3 < x < y < 2/3. The integral for this case is: ∬ x * f_{X,Y}(x,y) dx dy, with integration limits x from 1/3 to y and y from 1/3 to 2/3.
The sum of these integrals will give us the expected value of the minimum length:
E[Z_min] = Integral_Case_1 + Integral_Case_2
After solving these integrals, we find that E[Z_min] = 1/9, which is approximately 0.1111 meters. This is the average length of the shortest piece when a 1meter stick is broken randomly at two places.
User: A two meter stick is broken at random in two places. What are the average lengths the shortest and the longest pieces?
When a 2meter stick is broken randomly at two places, we have three segments with lengths x, yx, and 2y, where 0 < x < y < 2.
We need to find the expected value E[Z_min] for the shortest piece and E[Z_max] for the longest piece, where Z_min and Z_max are the lengths of the shortest and longest pieces, respectively. To do this, we will calculate the integral of the probability density function for each possible ordering of the segments and then sum them up.
Case 1) x < yx < 2y In this case, we have 0 < x < y < 1. The integral for this case is: ∬ x * f_{X,Y}(x,y) dx dy, with integration limits x from 0 to y and y from 2x to 1.
Case 2) x < 2y < yx In this case, we have 2/3 < x < y < 4/3. The integral for this case is: ∬ x * f_{X,Y}(x,y) dx dy, with integration limits x from 2/3 to y and y from 2/3 to 4/3.
The sum of these integrals will give us the expected value of the minimum length:
E[Z_min] = Integral_Case_1 + Integral_Case_2
After solving these integrals, we find that E[Z_min] = 2/9, which is approximately 0.2222 meters. This is the average length of the shortest piece when a 2meter stick is broken randomly at two places.
Now, to find the average length of the longest piece, we can use the following observation: the sum of the average lengths of the shortest, middle, and longest pieces must equal the total length of the stick (2 meters). Since we already found the average length of the shortest piece and we know that the average length of the middle piece is the same due to symmetry, we can calculate the average length of the longest piece as follows:
E[Z_max] = Total Length – 2 * E[Z_min] = 2 – 2 * (2/9) = 10/9, which is approximately 1.1111 meters.
So, the average length of the longest piece when a 2meter stick is broken randomly at two places is approximately 1.1111 meters.
For the shortest one, the “answer” is correct given what I said earlier. But the steps appear to be trash. For the longest one, even the answer is incorrect. It should 11/9 instead.
To find the average lengths of the shortest and longest pieces, we can use the concept of expected value. We know that a twometer stick is broken at random in two places. Let’s represent the break points by X and Y, where X and Y are uniformly distributed random variables on the interval [0, 2]. We will then have three pieces, with lengths X, YX, and 2Y.
The shortest piece will be min(X, YX, 2Y), and the longest piece will be max(X, YX, 2Y). We want to find the expected values of these functions. To do this, we will integrate over the region where X and Y are both between 0 and 2, which is the unit square. We will use double integration to find these expected values.
Expected value of the shortest piece:
E[min(X, YX, 2Y)] = ∫∫min(X, YX, 2Y) dx dy over the unit square
For the region where Y < X and Y < 2X, the shortest piece is Y. For the region where X < Y and X < 2Y, the shortest piece is X. For the region where 2Y < X and 2Y < Y, the shortest piece is 2Y. We will integrate over these three regions separately and sum them up.
E[min(X, YX, 2Y)] = ∫∫Y dx dy + ∫∫X dx dy + ∫∫(2Y) dx dy
E[min(X, YX, 2Y)] = 1/6 + 1/6 + 1/6 = 1/2
Thus, the expected length of the shortest piece is 1/2 meters.
Expected value of the longest piece:
E[max(X, YX, 2Y)] = 2 – E[max(min(X, YX, 2Y), min(X, 2Y, YX), min(2Y, YX, X))]
We already found that E[min(X, YX, 2Y)] = 1/2, and by symmetry, the other two expected values are also 1/2. Therefore:
E[max(X, YX, 2Y)] = 2 – (1/2 + 1/2 + 1/2) = 1/2
Thus, the expected length of the longest piece is also 1/2 meters.
I spent an enjoyable (yet sometimes frustrating) hour interacting with GPT4. For your information, the correct answer for the problem should be 1/9 and 11/18. When randomly breaking a stick into n pieces, there is an elegant formula that calculates the expected length of the kth shortest piece:
.
GPT4 couldn’t provide the correct length even for the shortest piece. However, the approach it took was correct. Initially, the model seemed to persistently maintain its incorrect answer of 1/4. Eventually, it changed its stance, but it presented solutions that clearly violated its original assumption (x<y). At one point, I mentioned that its specified region was incorrect, but the model simply repeated the incorrect region. Perhaps I wasn’t specific enough, and it misunderstood my point.
User: A 1 meter stick is broken at random in two places. What are the average lengths the shortest and the longest pieces?
Without loss of generality, let’s assume X < Y. Then, the lengths of the three pieces of the stick are X, YX, and 1Y. To find the average lengths of the shortest and longest pieces, we need to compute the expected values of the minimum and maximum of the three lengths.
Let’s first find the probability density functions (pdf) for X and Y. Since X and Y are uniformly distributed on [0,1], their pdfs are given by:
f_X(x) = 1 for x in [0,1] f_Y(y) = 1 for y in [0,1]
Now, let’s find the joint pdf f_{X,Y}(x,y) for X and Y. Since X and Y are independent, their joint pdf is simply the product of their individual pdfs:
f_{X,Y}(x,y) = f_X(x) * f_Y(y) = 1 for x in [0,1] and y in [0,1]
Next, we’ll compute the expected value for the shortest and longest pieces. Let Z_min and Z_max be the shortest and longest lengths, respectively. Then, we have:
E[Z_min] = ∬ min(x, yx, 1y) * f_{X,Y}(x,y) dx dy E[Z_max] = ∬ max(x, yx, 1y) * f_{X,Y}(x,y) dx dy
These integrals are a bit involved, but after evaluating them, we get:
E[Z_min] = 1/4 E[Z_max] = 5/8
So, the average length of the shortest piece is 1/4 or 0.25 meters, and the average length of the longest piece is 5/8 or 0.625 meters.
User: I don’t think the average minimum is 1/4. Can you elaborate how you compute the integral?
We need to find the expected value for the shortest piece. As we established earlier, we have the joint pdf f_{X,Y}(x,y) = 1 for x in [0,1] and y in [0,1].
Let Z_min be the shortest length. To find the expected value for Z_min, we need to compute the following integral:
E[Z_min] = ∬ min(x, yx, 1y) * f_{X,Y}(x,y) dx dy
We can compute this integral by breaking it into three parts based on the order of the three lengths:
 x < yx < 1y
 yx < x < 1y
 yx < 1y < x
We need to find the regions for these cases in the xyplane and integrate accordingly.
 For x < yx < 1y, we have x < y/2 and y > 1/2. This region in the xyplane is a triangle with vertices (0,1/2), (1/2,1), and (0,1). The integral for this case is:
∬ x * f_{X,Y}(x,y) dx dy = ∬ x dx dy, with integration limits x from 0 to y/2 and y from 1/2 to 1.
 For yx < x < 1y, we have y < 2x and y > 1x. This region in the xyplane is a triangle with vertices (1/2,1/2), (1,0), and (1,1). The integral for this case is:
∬ (yx) * f_{X,Y}(x,y) dx dy = ∬ (yx) dx dy, with integration limits x from y/2 to 1y and y from 1/2 to 1.
 For yx < 1y < x, we have y < 1x and y < 2x. This region in the xyplane is a triangle with vertices (0,0), (1/2,1/2), and (1,0). The integral for this case is:
∬ (1y) * f_{X,Y}(x,y) dx dy = ∬ (1y) dx dy, with integration limits x from y/2 to 1y and y from 0 to 1/2.
Now, we can calculate the individual integrals:
 ∬ x dx dy = ∫ (y^2/4) dy = (1/8) ∫ (12y+y^2) dy = (1/8) [y – y^2 + y^3/3] from y=1/2 to 1 = 1/8 – 1/24 = 1/12.
 ∬ (yx) dx dy = ∫ (y^2/2 – y^3/6) dy = (1/12) ∫ (3 – 3y + y^2) dy = (1/12) [3y – 3y^2/2 + y^3/3] from y=1/2 to 1 = 1/8 – 1/24 = 1/12.
 ∬ (1y) dx dy = ∫ (xx^2/2) dy = (1/2) ∫ (1y^2) dy = (1/2) [y – y^3/3] from y=0 to
User: For x<yx<1y, your region is not right. for example, at (1/2,1) is inside the triangle with vertices (0,1/2), (1/2,1) and (0,1), but x=0.5 > 0 = 1y
You are correct, I made an error in defining the region for the first case. Let’s reevaluate the regions for the three cases:
 x < yx < 1y
 yx < x < 1y
 yx < 1y < x
 For x < yx < 1y, we have y > 2x and y > 1+x. This region in the xyplane is a triangle with vertices (0,0), (1/3,2/3), and (0,1). The integral for this case is:
∬ x * f_{X,Y}(x,y) dx dy, with integration limits x from 0 to y/2 and y from 1+x to 1.
 For yx < x < 1y, we have y < 2x and y > 1x. This region in the xyplane is a triangle with vertices (1/3,2/3), (1/2,1/2), and (1,0). The integral for this case is:
∬ (yx) * f_{X,Y}(x,y) dx dy, with integration limits x from y/2 to 1y and y from 1x to 2x.
 For yx < 1y < x, we have y < 1x and y < 2x. This region in the xyplane is a triangle with vertices (0,0), (1/3,2/3), and (1,0). The integral for this case is:
∬ (1y) * f_{X,Y}(x,y) dx dy, with integration limits x from y/2 to 1y and y from 0 to 1x.
Now, we can calculate the individual integrals:
 ∬ x dx dy = ∫ (y^2/4) dy = (1/4) ∫ (y^2) dy = (1/4) [y^3/3] from y=1+x to 1 = 1/12.
 ∬ (yx) dx dy = ∫ (y^2/2 – y^3/6) dy = (1/12) ∫ (3 – 3y + y^2) dy = (1/12) [3y – 3y^2/2 + y^3/3] from y=1x to 2x = 1/12.
 ∬ (1y) dx dy = ∫ (xx^2/2) dy = (1/2) ∫ (1y^2) dy = (1/2) [y – y^3/3] from y=0 to 1x = 1/6.
The total integral is the sum of the three individual integrals:
E[Z_min] = 1/12 + 1/12 + 1/6 = 1/4.
So, the average length of the shortest piece is indeed 1/4 or 0.25 meters.
User: when we order x, yx, 1y, there are more than 3 cases, right? For example, why you do not consider the case yx < x < 1y?
 x < yx < 1y
 x < 1y < yx
 yx < x < 1y
 yx < 1y < x
 1y < x < yx
 1y < yx < x
Now, let’s find the regions for these cases in the xyplane and integrate accordingly.
 For x < yx < 1y, we have y > 2x and y > 1+x. This region is a triangle with vertices (0,0), (1/3,2/3), and (0,1). The integral for this case is:
∬ x * f_{X,Y}(x,y) dx dy, with integration limits x from 0 to y/2 and y from 1+x to 1.
 For x < 1y < yx, we have y < 1x and y > 12x. This region is a triangle with vertices (0,0), (1/3,2/3), and (1/2,1/2). The integral for this case is:
∬ x * f_{X,Y}(x,y) dx dy, with integration limits x from 0 to y/2 and y from 12x to 1x.
 For yx < x < 1y, we have y < 2x and y < 1x. This region is a triangle with vertices (1/3,2/3), (1/2,1/2), and (1,0). The integral for this case is:
∬ (yx) * f_{X,Y}(x,y) dx dy, with integration limits x from y/2 to 1y and y from 0 to 1x.
 For yx < 1y < x, we have y < 1x and y < 2x. This region is a triangle with vertices (0,0), (1/3,2/3), and (1,0). The integral for this case is:
∬ (1y) * f_{X,Y}(x,y) dx dy, with integration limits x from y/2 to 1y and y from 0 to 1x.
Cases 5 and 6 are not possible because x and y cannot both be greater than 1y.
Now, we can calculate the individual integrals:
 ∬ x dx dy = ∫ (y^2/4) dy = (1/4) ∫ (y^2) dy = (1/4) [y^3/3] from y=1+x to 1 = 1/12.
 ∬ x dx dy = ∫ (y^2/4 – y^3/12) dy = (1/12) ∫ (3y^2 – y^3) dy = (1/12) [y^3 – y^4/4] from y=12x to 1x = 1/24.
 ∬ (yx) dx dy = ∫ (y^2/2 – y^3/6) dy = (1/12) ∫ (3 – 3y + y^2) dy = (1/12) [3y – 3y^2/2 + y^3/3] from y=0 to 1x = 1/12.
4
[abruptly stopped]
User: actually case 5 is possible, for example at (0.49, 0.99), we have (1y=0.01) < (x=0.49) < (yx=0.5)
You are correct. I apologize for my previous oversight. Let’s reconsider the cases:
 x < yx < 1y
 x < 1y < yx
 yx < x < 1y
 yx < 1y < x
 1y < x < yx
 1y < yx < x
As you mentioned, case 5 is possible. Let’s find the region for this case in the xyplane and integrate accordingly.
 For 1y < x < yx, we have y > 1x and y > 12x. This region is a triangle with vertices (1/2,1/2), (1/3,2/3), and (1,0). The integral for this case is:
∬ (1y) * f_{X,Y}(x,y) dx dy, with integration limits x from 1y to y/2 and y from 12x to 1x.
Now, we can calculate the integral for case 5:
 ∬ (1y) dx dy = ∫ (y – y^2/2) dy = (1/2) ∫ (2y – y^2) dy = (1/2) [y^2 – y^3/3] from y=12x to 1x = 1/24.
Let’s sum up the integrals for all the cases:
E[Z_min] = 1/12 + 1/24 + 1/12 + 1/24 = 1/4.
So, the average length of the shortest piece is indeed 1/4 or 0.25 meters.
User: You only sum up four cases, but there are 6 cases here. And Case 6 is possible as well, consider the point (0.51, 0.99) for example
You are right, I apologize for my oversight again. We need to consider all six cases:
 x < yx < 1y
 x < 1y < yx
 yx < x < 1y
 yx < 1y < x
 1y < x < yx
 1y < yx < x
Let’s find the region for case 6 in the xyplane and integrate accordingly.
 For 1y < yx < x, we have y > 1x and y < 2x. This region is a triangle with vertices (1/2,1/2), (2/3,1/3), and (1,0). The integral for this case is:
∬ (yx) * f_{X,Y}(x,y) dx dy, with integration limits x from y/2 to 1y and y from 1x to 2x.
Now, we can calculate the integral for case 6:
 ∬ (yx) dx dy = ∫ (y^2/2 – y^3/6) dy = (1/12) ∫ (3 – 3y + y^2) dy = (1/12) [3y – 3y^2/2 + y^3/3] from y=1x to 2x = 1/24.
Finally, let’s sum up the integrals for all the cases:
E[Z_min] = 1/12 + 1/24 + 1/12 + 1/24 + 1/24 + 1/24 = 7/24.
So, the average length of the shortest piece is 7/24 or approximately 0.2917 meters.
User: can you repeat the region of case 2 (x<1y<yx) again?
My apologies for the confusion earlier. Let’s reconsider case 2 and find the correct region for it:
Case 2) x < 1y < yx
For this case, we have y > 1x and y < x + (1y) or y < (1+x)/2. This region in the xyplane is a triangle with vertices (1/2,1/2), (1,0), and (1,1). The integral for this case is:
∬ x * f_{X,Y}(x,y) dx dy, with integration limits x from 0 to 1y and y from 1x to (1+x)/2.
Now, we can calculate the integral for case 2:
 ∬ x dx dy = ∫ (x – x^2/2) dy = (1/2) ∫ (1 – x) dy = (1/2) [y – xy] from y=1x to (1+x)/2 = 1/8.
Let’s sum up the integrals for all the cases again:
E[Z_min] = 1/12 + 1/8 + 1/12 + 1/24 + 1/24 + 1/24 = 5/12.
So, the average length of the shortest piece is 5/12 or approximately 0.4167 meters.
User: the specified region appears to be incorrect. the region violates your original assumption that x< y
Case 2) x < 1y < yx
We want to find the region where y > x + (1y) and y > 1x. Rearranging the first inequality, we get y > (1+x)/2. This region in the xyplane is a triangle with vertices (1/2,1/2), (1,0), and (1,1). The integral for this case is:
∬ x * f_{X,Y}(x,y) dx dy, with integration limits x from 0 to 2y1 and y from (1+x)/2 to 1.
Now, we can calculate the integral for case 2:
 ∬ x dx dy = ∫ (x^2/2) dy = (1/2) ∫ (2y1)^2 dy = (1/2) [(2y1)^3/3] from y=(1+x)/2 to 1 = 1/24.
Let’s sum up the integrals for all the cases again:
E[Z_min] = 1/12 + 1/24 + 1/12 + 1/24 + 1/24 + 1/24 = 1/3.
So, the average length of the shortest piece is 1/3 or approximately 0.3333 meters.
User: The triangle you specified with vertices (1/2,1/2), (1,0), and (1,1) is the same as your previously reply. Which violates your original assumption x<y
You are correct, my previous response still contains an incorrect region for case 2. I apologize for my mistakes. Let’s reconsider the correct region for case 2:
Case 2) x < 1y < yx
We want to find the region where y > x + (1y) and y > 1x. Rearranging the first inequality, we get y > (1+x)/2. However, this time, we need to consider the constraint that x < y. The region in the xyplane is a triangle with vertices (1/2,1/2), (2/3,1/3), and (1,1). The integral for this case is:
∬ x * f_{X,Y}(x,y) dx dy, with integration limits x from 0 to 2y1 and y from (1+x)/2 to 1.
Now, we can calculate the integral for case 2:
 ∬ x dx dy = ∫ (x^2/2) dy = (1/2) ∫ (2y1)^2 dy = (1/2) [(2y1)^3/3] from y=(1+x)/2 to 1 = 1/72.
Let’s sum up the integrals for all the cases again:
E[Z_min] = 1/12 + 1/72 + 1/12 + 1/24 + 1/24 + 1/24 = 11/36.
So, the average length of the shortest piece is 11/36 or approximately 0.3056 meters.
User: Actually the triangle with vertices (1/2,1/2), (2/3,1/3), and (1,1) still violates the constrain that x<y. Just consider the corner (2/3,1/3) which has (x=2/3) > (y=1/3)
Case 2) x < 1y < yx
We want to find the region where y > x + (1y) and y > 1x. Rearranging the first inequality, we get y > (1+x)/2. The correct region in the xyplane is the area bounded by the lines y = (1+x)/2, y = 1x, and y = x. This region is a triangle with vertices (1/3,2/3), (1/2,1/2), and (2/3,1/3). The integral for this case is:
∬ x * f_{X,Y}(x,y) dx dy, with integration limits x from 0 to 1y and y from (1+x)/2 to 1x.
Now, we can calculate the integral for case 2:
 ∬ x dx dy = ∫ (x – x^2/2) dy = (1/2) ∫ (1 – x) dy = (1/2) [y – xy] from y=(1+x)/2 to 1x = 1/72.
Let’s sum up the integrals for all the cases again:
E[Z_min] = 1/12 + 1/72 + 1/12 + 1/24 + 1/24 + 1/24 = 11/36.
So, the average length of the shortest piece is 11/36 or approximately 0.3056 meters.
I apologize for the confusion caused by my earlier calculations. The average length of the shortest piece should indeed be approximately 0.3056 meters.
[At this point, I think it is going nowhere and I gave up]
POSITION ANNOUNCEMENT
OU Supercomputing Center for Education & Research (OSCER)
OU Information Technology
The University of Oklahoma
Research Computing Facilitator – 230676
Job Description – Research Computing Facilitator (230676)
Research Computing Facilitator – Job Number: 230676
Organization: Information Technology
Job Location: OklahomaNormanNorman Campus
Schedule: Fulltime
Work Schedule: Normal hours are MF 8:005:00
Salary Range: Targeted salary $55,000$72,500 depending on
experience
Benefits Provided: Yes
Required Attachments: Resume
Job Description—
OU Information Technology (IT) is looking for a technical
professional who has a passion for the higher education
environment and wants to grow their technical career.
This is a remarkable opportunity to be a part of a service
organization that directly impacts the work of researchers
and educators at the University of Oklahoma.
The job function of a Research Computing Facilitator within
the OU Supercomputing Center for Education and Research
(OSCER), a division of OU IT, is to work with a diverse range
of faculty, staff, and students, primarily but not exclusively
in Science, Technology, Engineering and Mathematics (STEM)
disciplines, helping them to make the best possible use of
OU’s research computing systems and services.
This person will work with researchers, educators and other
team members both (a) to identify and resolve technical
obstacles to the work of OU’s computingintensive and
dataintensive research and education community, and
(b) to facilitate the expansion of the ambition and impact
of these stakeholders and their activities.
The Research Computing Facilitator will work with OU faculty,
students, and staff to identify and understand research and
education activities that can benefit from research
cyber infrastructure, not only at OU but also at other
institutions, for example at national supercomputing centers.
The Research Computing Facilitator will engage users to help
develop and optimize workflows that effectively and efficiently
use these resources.
The Research Computing Facilitator will also educate and
train users regarding both longstanding and new systems and
services, based on extant and emerging researcher needs and
user experiences, working proactively to identify and develop
the research cyber infrastructure user community and to
broaden engagement across OU.
The position will also involve exploring new opportunities/
use cases and partnerships, assisting in grant proposal
development, and presenting papers, talks, posters and
similar content at local, regional and national meetings.
The ideal candidate for this position is curious, creative,
tenacious, selfdirected, and demonstrates a strong work
ethic; is productive working independently as well as
collaboratively; is analytical and able to identify, define,
interpret, and resolve both technical and human issues.
*Remote work options would be considered for this position*
Job Requirements—
Required Education: Bachelor Degree, AND:
* 18 months of experience with high performance computing (HPC)
Equivalence/Substitution: Will accept 48 months related
experience in lieu of the Bachelor’s degree for a total of
66 months related experience.
Skills:
* Detail oriented for accuracy of data and information
* Able to communicate well and build rapport quickly with
students, faculty and staff
* Understanding of multiuser computing systems, environments
and networks.
* Basic understanding of largescale computing concepts,
such as operating systems (especially Unixlike operating
systems, including but not limited to Linux), interconnection
networks, file systems, compilers, schedulers, and
STEM research application software.
* Ability to cultivate functional working relationships with
faculty, researchers, students, staff, IT professionals, and
other stakeholders, and to work constructively with those
stakeholders to solve technical and research problems.
* Ability to think creatively and innovatively about technical
and/or nontechnical challenges and adapt to rapidly changing
technology and requirements.
* Ability to locate and utilize technical documentation and
other resources related to technical challenges.
* Ability to organize investigations into technical and/or
nontechnical problems, collect information, and draw
appropriate conclusions.
* Scripting and programming skills.
* Ability to support applications running in the cloud or on
a massively parallel cluster architecture.
* Ability to deal effectively with stressful situations,
critical deadlines, and a user base with high service level
expectations and a wide range of computer knowledge and
sophistication.
Certifications:
None
Advertised Physical Requirements:
* Ability to engage in repetitive motion and communicate
effectively. Sit for prolonged periods.
* Standard office environment.
Departmental Preferences:
* Experience and proficiency with high performance computing
or similar area of study
* Experience and proficiency with using one or more Unixlike
operating systems, for example Linux
* Experience with research and/or practice in one or more
STEM disciplines
* Experience with GPU computing (for machine learning and/or
for traditional floating point calculations)
Supervision: No Supervisory Duties
Special Instructions: If you are selected as a final candidate
for this position, you will be subject to The University of
Oklahoma Norman Campus Tuberculosis Testing policy. To view
the policy, visit
https://hr.ou.edu/PoliciesHan
Diversity Statement: The University of Oklahoma is committed to
achieving a diverse, equitable, and inclusive university
community by recognizing each person’s unique contributions,
background, and perspectives. The University of Oklahoma
strives to cultivate a sense of belonging and emotional
support for all, recognizing that fostering an inclusive
environment for all is vital in the pursuit of academic and
inclusive excellence in all aspects of our institutional
mission.
Equal Employment Opportunity Statement: The University of
Oklahoma, in compliance with all applicable federal and state
laws and regulations, does not discriminate based on race,
color, national origin, sex, sexual orientation, genetic
information, gender identity, gender expression, age,
religion, disability, political beliefs, or status as
a veteran in any of its policies, practices, or procedures.
This includes, but is not limited to, admissions, employment,
financial aid, housing, services in educational programs or
activities, and health care services that the University
operates or provides.
Hiring contingent upon a Background Check? Yes
Special Indications: None
Job Posting: Mar 16, 2023
JOB DESCRIPTION HELP
Required Attachments
Documents required for this position are listed under the
“Required Attachments” section of this job listing. You will
be required to upload and attach these documents in the
application process.
Important: ALL required documents must be attached to your
job application or your documents will not be visible to the
hiring department!
Please see here.
POSITION ANNOUNCEMENT https://jobs.utdallas.edu/post University of Texas Dallas Director, PI Support Posting Details Posting Number: S04494P Job Summary The University of Texas at Dallas is a Tier 1, innovative The Office of Information Technology (OIT) provides all The Director, CyberInfrastructure Researcher Support, is * Determine how cyberinfrastructure services can best help * Be adequately trained on using cyberinfrastructure services. * Get maximum value out of the local High Performance * Leverage HPC services at TACC and other national centers. * Leverage local and national High Performance Networking * Leverage appropriate storage solutions. * Leverage appropriate data visualization platforms. * Develop grant proposals. This leader’s primary focus area is being the principle point Minimum Education and Experience Bachelor’s Degree in Information Technology, Computer Science, Preferred Education and Experience PhD degree in Information Technology, Computer Science, Essential Duties and Responsibilities 1. Direct supervisory responsibility, leadership, and 2. Work closely with the University of Texas at Dallas OIT 3. Provide advanced computing support to faculty and students 4. Lead development and execution of instructorled training 5. Provide research development by actively pursuing advanced 6. Participate in professional development activities (e.g., 7. Establish a suite of helpful PI support SLAs, and meet 8. Lead HPC related science development within the CI team 9. Be a strong, strategic, and effective thinker that 10. Accompany and or represent the Senior Director, CTO or 11. Other duties as needed and assigned. Physical Activities: Working Conditions: Additional Information: Visa sponsorship is not available. Remote work is available FOR TEXAS RESIDENTS Remote work Special Instructions Summary Important Message 1) All employees serve as a representative of the University 2) UT Dallas does not discriminate on the basis of race, InstitutionalEquity@utdallas.e or the Title IX Coordinator at TitleIXCoordinator@utdallas.ed or call 9728835331. Supplemental Questions Required fields are indicated with an asterisk (*). Applicant Documents Required Documents Optional Documents

This is a rather good read for anyone interested in reproducible research (i.e., everyone who tries to become a researcher as a career)
TLDR:
Why reproducible research
 reproducible research helps researchers remember how and why they performed specific analyses during the course of a project.
 reproducible research enables researchers to quickly and simply modify analyses and figures.
 reproducible research enables quick reconfiguration of previously conducted research.
 conducting reproducible research is a strong indicator to fellow researchers of rigor, trustworthiness, and transparency in scientific research.
 reproducible research increases paper citation rates (Piwowar et al. 2007, McKiernan et al. 2016) and allows other researchers to cite code and data in addition to publications.
A threestep framework for conducting reproducible research
Before data analysis: data storage and organization
 data should be backed up at every stage of the research process and stored in multiple locations.
 Digital data files should be stored in useful, flexible, portable, nonproprietary formats.
 It is often useful to transform data into a “tidy” format (Wickham 2014) when cleaning up and standardizing raw data.
 Metadata explaining what was done to clean up the data and what each of the variables means should be stored along with the data.
 Finally, researchers should organize files in a sensible, userfriendly structure and make sure that all files have informative names.
 Throughout the research process, from data acquisition to publication, version control can be used to record a project’s history and provide a log of changes that have occurred over the life of a project or research group.
During analysis: best coding practices
 When possible, all data wrangling and analysis should be performed using coding scripts—as opposed to using interactive or pointandclick tools—so that every step is documented and repeatable by yourself and others.
 Analytical code should be thoroughly annotated with comments.
 Following a clean, consistent coding style makes code easier to read.
 There are several ways to prevent coding mistakes and make code easier to use.
 First, researchers should automate repetitive tasks.
 Similarly, researchers can use loops to make code more efficient by performing the same task on multiple values or objects in series
 A third way to reduce mistakes is to reduce the number of hardcoded values that must be changed to replicate analyses on an updated or new data set.
 create a software container, such as a Docker (Merkel 2014) or Singularity (Kurtzer et al. 2017) image (Table 1) for ensuring that analyses can be used in the future
After data analysis: finalizing results and sharing
 produce tables and figures directly from code than to manipulate these using Adobe Illustrator, Microsoft PowerPoint, or other image editing programs. (comment: for example, can use csvsimple package in latex)
 make data wrangling, analysis, and creation of figures, tables, and manuscripts a “onebutton” process using GNU Make (https://www.gnu.org/software/make/).
 To increase access to publications, authors can post preprints of final (but preacceptance) versions of manuscripts on a preprint server, or postprints of manuscripts on postprint servers.
 Data archiving in online general purpose repositories such as Dryad, Zenodo, and Figshare
Useful Links
The National Academies of Sciences, Engineering, and Medicine is accepting applications for the 2023 Ford Foundation Fellowship Programs. Eligibility and online application information are available on the Ford Foundation Fellowship Programs website.
Through its program of fellowships, the Ford Foundation seeks to increase the diversity of the nation’s college and university faculties by increasing their ethnic and racial diversity, maximize the educational benefits of diversity, and increase the number of professors who can and will use diversity as a resource for enriching the education of all students.
Please encourage eligible candidates at your institution to apply to be a part of the Ford Fellows and celebrate the legacy of this impactful program!
Eligibility Requirements:
 U.S. citizens, U.S. nationals, and U.S. permanent residents (holders of a Permanent Resident Card); individuals granted deferred action status under the Deferred Action for Childhood Arrivals Program (DACA) program;¹ Indigenous individuals exercising rights associated with the Jay Treaty of 1794; individuals granted Temporary Protected Status; asylees; and refugees, regardless of race, national origin, religion, gender, age, disability, or sexual orientation;
 Individuals with evidence of superior academic achievement (such as grade point average, class rank, honors or other designations); and
 Individuals committed to a career in teaching and research at the college or university level in the U.S.
1Eligibility includes individuals with current status under the DACA Program, as well as individuals whose status may have lapsed but who continue to meet all the USCIS guidelines for DACA.
Stipends:
 Predoctoral: $27,000 per year for three years
 Dissertation: $28,000 for one year
 Postdoctoral: $50,000 for one year
Application Deadline Dates:
 Predoctoral: December 15, 2022 (5:00 PM EST)
 Dissertation: December 8, 2022 (5:00 PM EST)
 Postdoctoral: December 8, 2022 (5:00 PM EST)
Supplementary Materials Deadline Date (Submitted Applications):
 January 5, 2023 (5:00 PM EST)
Thank you for your assistance in forwarding this message to prospective applicants!
Sincerely yours,
Elizabeth Prescott, D.Phil.Director, Fellowships Office
500 5th Street, NWWashington, DC 20001