BrainDen.com - Brain Teasers  # markweitzman

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1. First the problem should be clarified to explicitly make clear what three points at random on a circle mean - otherwise we start to get into trouble with Bertrand's paradox - https://en.wikipedia.org/wiki/Bertrand_paradox_(probability). But assuming the simple definition that the points are distributed uniformly according to equal arc length (or equal angles) in a range from zero to 2*pi, I will give an analytic solution, that is probably way harder than necessary for this problem but very general and useful if the probability distribution is different from uniform. So starting from the point (1,0) let x,y,z be the value of the cut points arc length measured counterclockwise and for simplicity for now use a range of 0-1 instead of 0-2*Pi (afterwards we can scale up the answer by multiplying by 2pi). So essentially we have chosen three points in the 0-1 interval uniformly, this can be visualized as choosing three points from a cube of side 1 uniformly. There are six possible orderings for the values of the three points, so the cube is divided into six regions, but by simple symmetry arguments one can show that the value of integrating the arc length containing the point (1,0) on the circle is the same for all regions of the cube. So assume x<y<z this corresponds to the region of the cube 0<x<1, x<y<1, y<z<1. The arc length of the point containing (1,0) on the circle is given by 1+x-z (easy to see this as total length of circle 1 minus middle interval z-x. So integrating 1+ x - z over the region of the cube given by the prior intervals gives the average arc length corresponding to the x<y<z ordering. The evaluation of the multiple integral is routine, and the result is 1/12. Multiplying by 6 to account for all the orderings, and by 2*pi to rescale as indicated above leads to an answer of pi.
2. Here is a png image of the Markov chain for my post above.
3. I have found two ways to solve this problem analytically - the first not so rigorous and the second more rigorous. First method: Let A be the event student A rolls a twelve which has a probability of 1/36. Then the expected number of rolls for A to roll a 12 is 36 which can easily be calculated from the following equation: (1) E(A trials) = 1/36 + 35/36*[ E(A trials)+1] solving this equation leads to E(A trials)=36. in general if the probability per trial is p the expected number of trials till success is 1/p. For B the calculation is a little more complicated: (2) E(B trials) = 1/6*[1/6 * 2 + 5/6 * (E(B trials) + 2)] + 5/6* [E(B trials)+1] solving this equation leads to E(B trials) = 42. Now comes the non-rigorous part - effectively the expected value of B rolling consecutive 7's being equal to 42 rolls corresponds to a 1/42 chance on each roll even though this does not correspond to the actual B events (for example B can't win on the first roll). So using these relative probabilities in a standard way: P(A) = 1/36 P(B) = 1/42 P(A)/[P(A)+P(B)] = 7/13 probability A wins before B wins. This is the correct answer 0.538461 as the simulations confirm. Now for the more rigorous way. Essentially this a Markov chain analysis (For reference see P. 364 Introduction to Probability - Bertsekas, Tsitsiklis). There are 4 states in the chain - two are absorbing - one where A wins I will call a12, the other where B wins I will call a77. The other two states are transient states, the start state which I will denote by as, and the state where the previous roll was a 7, I will denote by a17. Each variable (as, a17, a77, a12) represents the probability of ending in the final state where A wins (i.e. a roll of 12 before two consecutive 7's) given that they are starting in the current state represented by the variable. The following equations apply: a12 = 1 a77 =0 a17 = 29/36 * as + 6/36 * a77 + 1/36 * a12 as = 6/36 * a17 + 29/36 * as + 1/36 * a12 Solving these equations leads to as = 7/13.
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