suppose you have a perfectly round disk. You put three legs randomly on this disk to form a table. Supposing the legs are perfectly perpendicular to the disk and are attached to the disk firmly, what is chance that the table will not fall when you flip the disk or in other word when you put the table to stand on its legs? 28 AnswersShow More Responses I got 1/4th, can you guys post how you got your answers? 1/2, if three legs cover > 180 degrees, the table can stand. heres what I did. I cut the circle into quarters and then said that the table will not stand if all 3 legs are in the same half of the circle. there is (1/2)^3 odds of putting all 3 legs in the same half, and 4 different halfs (top half, bottom half, left half, and right half) for this to happen. --> 1 - 4 * (1/2)^3 = 1/2 anyone see a problem with this? I don't really understand why you said there are 4 different halfs. You could have have cut the circle up into eights and then there are 8 halfs. You could have cut the circle up into 10000ths and then argued that there are 10000 halfs. The way I think you are supposed to do it is this. Place the first leg, it doesn't matter where you place it. Now, what is the chance that the 2nd and third leg are on the same half as the first leg. So, it's 2C2 * (1/2)^2 = 1/4. Chance of this not happening = 3/4. What am I missing? I think its 1/2 also. Same reasoning as the comment above, but I think the 1st AND 2nd leg can be anywhere, because there is no established "half" until there are 2 legs on the table. The only place the 2nd leg cant be is exactly 180 degrees away from the 1st (the probability of which is about 0, because 179.9 degrees in either direction is already fine). So the probabilty of all 3 legs being on the same half is 1 x 1 x 1/2 Why not think the opposite from the beginning? What is the probability that it could not stand? It means that the legs are in the same half circle. 3*(1/2)*(1/2) = 3/4 Then the answer is 1/4 Brett, why are you multiplying by 3...the odds that the three are on the same side is as follows 1 peg can go anywhere. 50% chance that the next peg is on the same side and another 50% chance that the 3rd peg is on the same side therefore shouldn't it be (1/2)*(1/2) = 1/4 chance that the table falls (aka all pegs on the same side) and there fore probability that the table stands is 1-1/4 = 3/4 so isn't 3/4 your final answer n/360, where n is the angle of the 2 legs Ok, here is mine, the first one is always not a problem, so let's jump to the 2nd stand In order to drop, the second must lie inside the half circle with the 1st one so, however, if you think of that in the clockwise and anti clockwise location, you will immediately discover that where ever it is, the 2nd is some how within the 180degree ranges. So, the problem leads to the third stand. So, if you can see the big picture, it is not hard to realized that there is only one safe zone for the 3rd stand, which is the opposite region of the one between 1st and 2nd (say 2nd located at n), so, the probability of it being dropped will be n/360 if using only integers => 1- n/360 being the right answer (to me). And if you sum up all the probabilities and combination, you will soon see that 1/2 is the right answer. Hope it can help! Show More Responses It's 1/2. 1/pi * integral from 0 to pi (x/pi) Fix a leg. Join this point to the opposite side of the disk to split the circle into half. If both legs lie in the same half it will topple. Effectively what is the probability of two coins the same in two tosses. We have HH,HT,TH,TT so 1/2. It will stand 1/2 of the time. The answer is 1/4. The answer is not 1/2. It's not 1/2, and it's not 1/4. Salalah had it right. Draw a circle. Draw the epicenter. Draw your first random leg (any point). Draw a line from it, through the epicenter, and to the circumference (your line will be longer than the radius). Draw a second leg (any point). Draw a line from it, through the epicenter, to the circumference. Your third leg needs to fall in the resulting pie piece. The probability of it falling there will be the the angle from point 1 to the epicenter to point 2, divided by 360 degrees. I believe it is the same with cutting the circle 3 times and probability that all of the pieces are smaller than 0.5* circumference because in my opinion, table will stand if the center of mass is inside of the triangle made by three cut. If one of the pieces is bigger than the 0.5, then the triangle could not capture the center so the answer is 1/4. 1/4. here is how to get the answer mathematically: angles between every two legs: x,y,z then x+y+z=2pi and restrictions are 0 1. Let the center be O 2. The 1st leg is at A --> AO cuts the circle at A1 and A2, with A1 being on the other side of A w.r.t O 3. The 2nd leg is at B --> BO cuts the circle at B1 and B2, with B1 being on the other side of B w.r.t O 4. The 3rd leg is at C. In order that the table doesn't fall, C has to be in the pie OA1B1 5. Let the angle AOB be phi 6. The probability that the table doesn't fall is phi / 2pi 7. Since A and B are uniformly distributed within the circle, the expected value of phi is \int_{0}^{pi} phi * dphi = pi^2 / 2 8. --> the probability that the table doesn't fall is : pi^2 / 2 / 2pi = pi / 4 Am I wrong any where ? Show More Responses point 7 is not clear. The sector/pie in some of the above solutions is right but there is a bit more to it. The point A can be anywhere on a radius. Placement of the first peg will establish a radius. Also it will not matter where on this radius the first peg is. It could be close to centre or near the circumference. The second peg can lie anywhere in a tiny sector projecting angle of dtheta say. So with the first point A being anywhere on its radius and B anywhere in the sector, the third leg must fall in the sector of angle theta on other side of the circle . so the probability for B is 1/2*(r^2)*dtheta/pi*r^2 and for C is (1/pi)*sin*cos so take the integral of this will give 1/(2*pi^2) The answer is 1/4. Place one leg on the unit circle and define this point as phi = 0. Then place the next leg on the unit circle. Note the position of placement is uniformly distributed from 0 to 2pi. Now consider if the second leg is placed at a position x (x radian away from the first leg), draw a diagram and notice that for stability, draw two straight lines from each leg and going through the centre. The third leg must be placed in the region opposite of the other 2 legs, so that the three legs make a triangle which contains the centre of mass of the table. So if xpi, stability probability = (2pi - x)/2pi We need to sum over all possible values of x (0 to 2pi) and not forget normalisation (in this case 1/2pi) So probability of stability = (1/2pi)*[( integral of x/2pi from 0 to pi) + (integral of (2pi-x)/2pi) from pi to 2pi)] = 1/4 Suppose 1st leg placed at A and 2nd at B. Draw a diameter through A. If the 3rd leg is placed on same side of diameter as B (prob = 0.5) it falls. Else it does not fall (prob = 1-0.5=0.5). Answer: 0.5. |