Probabilities in a Dice Game Using Markov AnalysisDate: 03/22/2004 at 21:36:05 From: John Subject: probability Anne, Bob and Carmel are going to take turns rolling a fair die, in the order Anne, Bob, Carmel, Anne, Bob Carmel, etc. The first person to roll a 6 will win $100. a) Find the probability that Anne will win $100 if the first four numbers rolled are 3, 2, 5, 5. b) Find the probability that Anne will win $100 if a draw is to be declared with nobody winning $100 in the event of a 6 not being rolled within the first 8 rolls in total. c) Suppose that a draw is to be declared if ever two 1's are rolled in a row (by two different players). Also suppose that the game is in fact over and did not end in a draw. Find the probability that Anne is the player who won $100. I was able to do part (a) by directly summing an infinite series and I checked this by performing a first step analysis. However, I am unsure how to solve the rest of the problems. Date: 03/25/2004 at 19:57:37 From: Doctor Mitteldorf Subject: Re: probability Dear John, The Markov method uses "state vectors" and matrices, and is probably more abstract than you're comfortable with, unless you're an extraordinary 17-year-old. Here's one way to solve a problem like this without having to use an infinite series. The probability of A winning on her first roll is (1/6). The probability that no one prevails in round 1, so we have to go to round 2, is (5/6)^3. But now A is in the same situation she was in at the beginning of round 1. We can use this fact to make an equation. Let's say A's total probability of winning is p. Then one part of p is the probability she'll win in the first round, (1/6). The second part of p is just p times the probability that she'll make it to round 2. So our equation is p = 1/6 + (5/6)^3 * p We can solve this for p, to find p = 36/91 = 0.3956. I have been mulling your problem (c) with the two-1's rule for much of the day, and I don't have a handy explanation. The best I can do is a Markov analysis, or a computer simulation. Computer simulations are my specialty--they're fun and they're easy for problems like this, and they give you an instant answer. If you have a little knowledge of BASIC or C++, perhaps you'd like to try it. The answer I get is that Ann wins 0.35357 of the time. Incidentally, the same computer simulation has convinced me that the "draw" situation with two-1's in a row occurs exactly 0.125 or 1/8 of the time. Here's the reasoning and the algebra that supports that prediction. Think of the "system" as having two states. As long as the game isn't over, then either a 1 has just been rolled, or anything else has just been rolled (2,3,4,5). Let u be the probability of two 1's eventually preceding a 6 if we're in state 1, and let v be the probability of a draw eventually starting in state 2. We'll write two equations connecting u and v that come from the probabilities for the next roll of the die, starting from state 1 or state 2. Starting in state 1 (a 1 has just been rolled), there's a 1/6 chance of a 1, ending in a draw there's a 1/6 chance of a 6, ending in a win there's a 4/6 chance of anything else -> state 2 From this, we write the equation: u = 1/6 (1) + 1/6 (0) + 4/6 (v) (our first equation) Starting in state 2 (a 2,3,4 or 5 has just been rolled, or just begun), there's a 1/6 chance of a 1, -> state 1 there's a 1/6 chance of a 6, ending in a win there's a 4/6 chance of anything else -> state 2 From this, we write the equation: v = 1/6 (u) + 1/6 (0) + 4/6 (v) (our second equation) Solve the two equations together, to find that v = 1/8, agreeing with my numerical experiment. It also tells us that u = 1/4. This method is in the spirit of Markov analysis, because we're thinking in terms of "state space" and "transition probabilities". In a full Markov analysis, you'd make a matrix of transition probabilities from each possible state to every other possible state. This suggests how we might go about solving the full problem, the probability of Ann winning. It also suggests why it might be messy and tedious! In the full problem, there are 6 states in the system: Either a 1 has just been rolled, or not. And Ann, Bob, or Carmel might be next to roll. Thus 2 * 3 = 6 states. Using the same kind of analysis that I just illustrated in the simple problem, you could define 6 probabilities u,v,w,x,y and z for Ann winning starting in each of the 6 states. Then you could write down 6 equations that describe the probabilities for going from each of those 6 states to each of the other states. Then you'd have 6 equations in 6 unknowns, and you could solve the whole mess. Of course, it won't be quite so bad, because the 3 players rotate in fixed order, so, for example, the equation for u will just have a w and an x in it, but no v, y and z. Still it might take some thinking to write down the equations, and a few pages of algebra to solve them. Let me know if you take it on... - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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