A random chord subtends an arc between 0 and 180 degrees, with equal probability anywhere between 0 and 180. The chord of the equilateral triangle subtends an arc of 120 degrees. In order for the random chord to be smaller than or equal to the side of the equilateral triangle, it must subtend less than or equal to 120 degrees. So, I think the answer would be equivalent to 120 / 180 = 2/3.

15 nickels

+ 1 quarter

= 1 dollar

15 penies

+ 1 dime

= 1 quarter

I'm thinking the answer should be the centroid or average of the starting points of all the athletes. The average is 1807.67 m, which means that the total distance covered by all athletes to arrive at the meeting point is 86881.33 m.

If I am visualizing this right the point is the mid point of AB where A and B are the two intersections.

If you use the midpoint of the two intersections, that point will always be closer to the particle on the small circle than it will be to the particle on the large circle. They will only be equal when the particles are at the intersection points.

If the two circles are of radius R1 and R2, then the point we're looking for is a distance R1 from the center of circle 2 and a distance R2 from the center of circle 1. Not sure how to prove it, but I can show it graphically.

The angular velocity of both particles with respect to the center of their circles is equal - they will both travel x degrees/unit of time. You can plot regular points around the circles starting from point A and traveling X degrees and then connect the pairs. A line of equidistant points for each pair is perpendicular to the midpoint of the connecting lines. All of those perpendicular lines will intersect at the common point I described above.

There are actually 2 common points such as I described in each circle. Which one gets used is dependent upon whether both are going in the same direction or opposite directions.

The number is 10 or 15 or 30, so the liar is S.

Even if you wanted to look at the probability of transferring the last white ball out of the remaining 9 in urn A in 4 chances, the probability would be 1/9 instead of 38%. If you get the white ball first, the probability is 1/9*1*1*1 (each 1 is the probability of getting a black ball on that draw). white ball second - 8/9 * 1/8 *1*1 = 1/9. third - 8/9 * 7/8 * 1/7 * 1 = 1/9. fourth - 8/9 * 7/8 * 6/7 *1/6 = 1/9.

As presented, the odds of choosing the last white ball in 4 tries is the same as choosing said white ball in 1 try (1/9). Better to look at it this way:

Everyone has rightly determined that there are 2 possibilities for Urn B - either 3W 3B or 4W 2B. The following equations calculate the probability of constructing either possibility.

P(3W3B) = (4/12) (3/11) (2/10) (6/9) (5/8) (4/7) = 2880/665280

P(4W2B) = (4/12) (3/11) (2/10) (1/9) (6/8) (5/7) = 720/665280

The denominator (665280) represents the total number of ways we could pull 6 balls out of Urn A. The numerators represent the number of ways each particular construction could be achieved. The relative probability of either is the ratio of each to the sum of both. That is,

P_R(3W3B) = 2880 / (2880 + 720) = 0.8

P_R(4W2B) = 720 / (2880 + 720) = 0.2

Consider the above values to be the probability of the two possible constructions of Urn B.

It could be noted that the first 3 terms of P(3W3B) and P(4W2B) are the same and could be removed to calculate the probabilities of constructing 1W2B or 0W3B, respectively. Then P_R(1W2B) and P_R(4W2B) could be calculated, and would yield the same results as P_R(3W3B) and P_R(4W2B). Just a good back-check...

If Urn B held 3B3W, then the answers to the questions are:

Q1: p(W,3B3W) = (3/6) = 0.5

Q2: p(W,3B3W) = (0/3) = 0.0

If Urn B held 4W2B, then the answers to the questions are:

Q1: p(W,4W2B) = (4/6) = 0.67

Q2: p(W,4W2B) = (1/3) = 0.33

These probabilities are of the outcome of the next pull from Urn B for each of the situations described in the OP.

To get the final answer, multiply the probability of the construction by the probability of the outcome and sum across all constructions:

Q1: p(W) = 0.5 * 0.8 + 0.67 * 0.2 = 0.533

Q2: p(W) = 0.0 * 0.8 + 0.33 * 0.2 = 0.067

E = 0

N = 4

G = 8

V = 2

I = 6

R = 1

EEN = 004 = 2*2

VIER = 2601 = 51*51

NEGEN = 40804 = 202*202

pairs of (m,n) that fit:

1,2 => 5

1,3 => 5

2,1 => 2

2,3 => 2

3,1 => 1

3,2 => 1

I think that's all

Wow - it's been a while since I've answered anything here.

The solution is rather simple.

Start with the probability that the rook is attacking the bishop. Place a rook anywhere on the chess board, and if the bishop is on the same row or column, it could be attacked. There are 7 spaces each on the row or column that the bishop could occupy and be attacked. 7 + 7 = 14 / 63 = 0.2222 is the probability that the rook is attacking the bishop.

Next calculate the probability that the bishop attacks the rook. Place the bishop anywhere on the board, and if the rook is on the same diagonal, it could be attacked. This is more difficult to calculate, because the number of spaces the bishop could attack depends upon the bishop's position. If the bishop is on the outer edge of the board, 7 spaces are available to attack. If moved into the center by one space, 9 spaces are available. Moved one space further in, the number increases to 11, and then 13 for the 4 central squares of the board. The average number of spaces a bishop could attack is then 8.75. 8.75 / 63 = 0.1389, which is the probability that the bishop is attacking the rook.

The combined probability is the sum of the two, or 0.2222 + 0.1389 = 0.3611.

Well, I think I decreased the rate correctly - that's my only question. The first new person comes along after 20 minutes, and then the rate of increase decreases to 4.95 persons/hour, so the next one will come along in 20 minutes 12 seconds. And so on.

If that's the case, then I get 149 people after 21 hours.

Because people are discrete (can't have part of a person), I assumed the rate of change was also discrete. So I might have gotten a different answer than you might have gotten with an integration using a continuously variable rate of change.

I'm guessing 5000 - double sided printing...

ELABORATE

EXQUISITE

IMMEDIATE

INDIGNANT

STAIRCASE

STRENUOUS

DETERMINE

CHAUFFEUR

FORTHWITH

REFERENCE

basically you got it right but it is a common formula to use a=v²/r where A is acceleration a vector orthogonal to the velocity. This if there is constant speed(not velocity). I am saying the angle of the road changes and this ideal speed changes to. If you can magically be going this ideal speed at every moment what is the time it takes. I simply meant you did not need to use acceleration in the direction of your velocity (though you can) you could consider the speed magically optimal at all times.

so for number 3(which i accidentally never numbered)

r=20m

theta=x/90

x is distance traveled from start of problem

so at the end of this problem which really is the middle of a turn your at

theta=(2pir/4)/90=10pi/90=pi/9

mu =.3

as for louge i meant ice luge it was the best example I had

Actually, this problem is not that bad. The easiest way I found to do it was to break the track into very short pieces, figure out the velocity at the beginning and end of each piece, take the average and figure out the drive time at the average speed for each section. You could also set up an integral function, but I'm too lazy and I like spreadsheets too much. Minimum speed at the beginning is 7.67 m/s and maximum speed at the end is 12.09 m/s. All totaled, I got 3.22 s to travel the 31.4 m quarter turn.

As an aside (and maybe even a rabbit trail), I assume that you're not actually coming off of a straight piece of track, but rather a curved piece with no banking. If you were, the theoretical maximum speed on the straight piece is infinite and the deceleration at the start of the curve would kill the driver. Assuming the infinite speed didn't get to him first...

The other problem would be a physical inability to turn the wheels instantaneously. You actually drive in a spiral when going from straight to curved sections of road to avoid a sudden change in centripital force (a=v²/r). If r goes from infinity on a straight line to, say, 20 m in the curve, you have an instantaneous change in acceleration, which is not good for your vehicle . That's why railroads layout their curves with clothoid spirals at the beginning and end, or else the trains would jump the tracks. Some highways are laid out that way as well, but most have plain old curves because drivers make the spiral themselves.

Are you sure this is the same Alex? Offering 20 times the cumulative bet is, I think, a big mistake for him. Here's why:

The ROI for any turn, as Doctor Moshe pointed out, is always the same, no matter how many numbers you bet on (assuming you place equal bets on all numbers). That ROI is 199/216. Stated another way, you expect to lose about 8 cents for every dollar you bet (1 - 199/216 = 0.08). So, if I bet a dollar each time, whether its all on one number, or fifty cents on 2, or 33 cents on 3, I expect to lose 8 cents with each roll of the dice.

If I bet on 2 numbers with each roll, my probability of getting all the way around the board in one "streak" is also calculated by Doctor Moshe. It's 0.02, which means that I should get around the board after 1/0.02 = 50 streaks. Each streak is, on average, less than 11 rolls, but let's say that they are 11 just for simplicity.

In those 50 streaks, if I keep placing one dollar bets, I expect to lose 49 * 11 * 0.08 = 43 dollars. But then I get all the way around the board, and it takes me about 11 turns to do it. I've placed 11 one-dollar bets in that time, and Alex is going to give me 20 * 11 = 220 dollars when I get around the board. So I've lost 43 dollars (or less), but I'll make 220!!

I simulated 1000 games in a spreadsheet, and the expected winnings pan out. If I start with $1000 and keep placing $0.50 on 1 and $0.50 on 2, I end up with about $1185 on average. The lowest holdings I had in any of those games was $865. So I figured I could increase my bets. I tried $5 dollars ($2.50 each on 1 & 2), and ended up with an average of $1920 at the end, with a low of $375. Increasing again to $10, I ended up with $2835 on average, but in one of the 1000 games, I lost everything before I could make it around the board. So I think a $10 bet each time, split evenly on 2 numbers, would be a pretty safe way to increase your holdings about three fold each game that Alex lets you play. You could probably do even better if you increased your bet amount as you got further around the board.

Alex should stick with the original version of the game.

A correction to my earlier post, which I can no longer edit...

a(n), b(n), c(n), and d(n) can be approximated by exponential functions of the form f(n) = Ae^Bx. So you can approximate that:

a(n) = 0.16666 e ^1.0986n

b(n) = 0.33333 e ^1.0986n

c(n) = 0.5 e ^1.0986n

d(n) = e ^1.0986n

I realized that I had messed up one of the calculations, and it looks much prettier now. I still think the answer lies with the intercepts of these functions.

And it does...

If you set up an equation:

.167 + .333Y + .5Y² = 321, then Y = 25 or -25.666 will solve.

Well, here's something interesting:

a(n), b(n), c(n), and d(n) can be approximated by exponential functions of the form f(n) = Ae^Bx. So you can approximate that:

a(n) = 0.006187 e ^1.0985n

b(n) = 0.01226 e ^1.0989n

c(n) = 0.01858 e ^1.0985n

d(n) = e ^1.0986n

I did this using regression analysis after taking the natural log of each of the numbers in the various series up to n = 35. The R² value was 0.999999 for a, b, & c, and 1.0 for d. Noticing that the exponents were all very close to 1.0986n, I decided to check what happens if I removed some of the lower values of n from the equation. As I suspected, the exponents eventually all became 1.0986n (1.0986 being the natural log of 3). So, we have 4 parallel functions when they are all graphed on a semi-logarithmic scale. I'm thinking the answer somehow has to do with the intercepts of the various equations, because the slopes are all the same. But I am not quite there yet.

How did you solve? Was it through a program or mathematical analysis?

spreadsheet I'm working on a mathematical analysis to make it more rigorous. From what I've seen (and I can tell others have seen it too), the general form of the converging function is f(n) = [a(n) + b(n)Y + c(n)Y²]/d(n), where a(n), b(n), c(n), and d(n) are distinct numerical sequences in their own right. The patterns become pretty regular for n>5, so I was thinking that if I could take limits on a(n), b(n), c(n), and d(n) as n approached infinity, I might be able to set that limit equal to 321 and directly solve for Y. Then you would obtain the solution from last time and the one I just posted, and possibly any complex solutions that might exist. But I've yet to define each of these sequences in a simple manner - just haven't spent the time on it yet. Obviously, d(n) is rather simple: d(n) = 3^n-3. And a(n) = c(n-1), so there's some simplification there. Just need to work it out...

Y could be -25.666666....

Y = 25

I think the best thing to do is first simplify to 2D to work out some relationships between the cone and sphere, and then work out the minimum ratio you seek.

Start with the picture below:

All the variable lengths can be described in terms of θ and R. So,

h = R/sin(θ)

H = R + h = R [1 + 1/sin(θ)]

B = 2H tan(θ) = 2R [1 + 1/sin(θ)] tan(θ)

Now, we can go back to 3D and plug these values into the formula for the volumes of a sphere and cone:

V_s = 4/3 πR³

V_c = H π (B/2)² / 3 = R³ [1 + 1/sin(θ)]³ [tan(θ)]² π / 3

So now everything is in terms of R and θ. R is a constant, so it doesn't really factor in, and it also makes the volume of the sphere a constant. And then, what we're really looking for is the value of θ that minimizes Vc. Well, you can take dVc/dθ, set it equal to 0, and solve for θ. Or you can do what I did and plug it into a spreadsheet. Either way, you get that θ = 19.47˚.

Taking the ratio of V_c/V_s, the R³ terms cancel out, and if θ = 19.47˚, then V_c/V_s = 2, which is what we're looking for.

Well, no, not really.

This may be oversimplifying the calculations, but I think what you're really looking for is the number of trips it takes to get a different toy when you already have 0, 1, 2, 3, ..., 9 of the 10 collectibles. So, if there are 10 toys, and I don't have any yet, then there are 10 new toys available whenever I go. So I expect to get a new toy every 10/10 = 1 trip. After that, there are 9 new toys available, so I'd expect to get a new toy every 10/9 = 1.111 trips. And so on and so forth. Adding it all up, I'd expect a complete set after 29.3 trips.

And there's always the new Batman toys coming out next month, in case I couldn't make my way to McD's every day this month...

For the 3 cylinder intersection, the shape is easiest to visualize as a cube with small pyramids on each face. The cube's dimension is sqrt(2), and the face of the cube is the base of the pyramid. Like the 2 cylinder intersection, the four connections between the base and the apex of the pyramid are not lines but quarter ellipses. This time, the radii are 1 and [2 - sqrt(2)] / 2. Another simple integration (I don't mean a real integration, but slicing it up and getting average end area volumes), and I get the volume to be 4.6863 unit³.

I'm not sure what to call the shape of the intersecting cylinders, but it's like a regular octahedron, except that instead of straight lines from the vertex to the base, they're quarter-ellipses with one radius equal to 1 unit and the other equal to sqrt(2) units. The base is square, as would be expected in a regular octahedron, with the dimension of each side equal to 2. I did a rough integration as the minor radius goes from 0 to 1 and got the volume equals 5.3333 unit³.

From time to time, I've played a dice game called Farkle. Maybe you've played it before. To start your turn, you roll 6 dice, and, if your roll scores (see below for scoring), you can do one of three things:

1. Add the score from all the scoring dice to your total for the turn and set those dice aside. Then, pick up the remaining dice and roll again.

2. If you score in more than one way (say you roll a 1 and a 5, or two 1s), you choosing only the scoring combinations you want to keep. You add the score from those dice to your score for the turn and set them aside. Then, pick up the remaining dice and roll again. [The difference between #1 and #2 is the rejection of some scoring combinations in favor of having more dice available for your next roll.]

3. Add the score from all the scoring dice to your total for the turn and stop rolling. These points become permanently attached to your score for the game.

The twist to the game is what happens if you don't score on a particular roll. This is called a Farkle, and you lose all the points you have accumulated in that turn. You still have the points from previous turns, but anything you might have gained in previous rolls that turn is lost. Here's an example of the effect of a Farkle:

Roll 1: Three 1s appear, adding 1000 to your score for the turn. You can stop rolling now and add 1000 to your score for the game. Or, you can roll again, but with only 3 dice this time.

Roll 2: Farkle - no scoring dice appear. You turn ends, and you add nothing to your score for the game, not even the 1000 points you earned on the previous roll.

If you manage to score with all 6 dice on one or multiple rolls, you get to roll all 6 again. A Farkle, however, will still cause you to lose all your points for the turn.

Scoring:

If you roll a 1, you get 100 points

If you roll a 5, you get 50 points

If you roll 3 of a kind, you get 1000, 200, 300, 400, 500, or 600 points (for rolls of three 1s, 2s, 3s, 4s, 5s, 6s respectively)

If you roll 4 of a kind, you get twice the value of 3 of a kind

If you roll 5 of a kind, you get twice the value of 4 of a kind

If you roll 6 of a kind, you get twice the value of 5 of a kind

If you roll 3 pair, you get 750

If you roll a straight (1 through 6), you get 1500

For the 3, 4, 5, 6 of a kind, 3 pair, or straight, you have to have all those dice appear on a single roll. For example, rolling a 1 on your first roll can't be followed by rolling 2, 3, 4, 5, 6 on your second roll to score a straight.

Obviously, the goal is to score as quickly as possible. Most times, the game is played to a certain limit (say first player to reach 10,000 wins). It's easy to score - but the trick is knowing when to stop.

Enough about the rules. Here's my questions:

1. Given a certain number of dice remaining to roll (1-6), what is your probable score for the next roll (assuming you keep all your scoring dice)?

2. Assuming you're required to keep all your scoring dice and keep rolling (ignore possibilities #2 & #3 above), what is the value of an average Farkle turn after the first roll? After the second? Third? Fourth?

3. When is it advantageous to apply option #2 above and not keep all your scoring dice?

4. Can you develop a rational method for stopping your turn (#3 above)?