Probability and the game of RISK
Date: 04/21/2000 at 19:48:46 From: Josh Peretti Subject: Probability and the game of RISK. In the classic game of Risk the attacker attacks with 3 dice and the defender defends with 2 dice. What is the probability that the defender will lose 2 battalions? What is the probability that they will each lose 1 battalion? What is the probability that the attacker will lose 2 battalions? It's a very well-known game, but just in case you don't know, here are the relevant rules: The attacker first rolls his 3 dice and discards the lowest number. The defender then rolls his 2 dice. At this point the attacker matches his highest number with the highest number of the defender and his lowest number with the defender's lowest number. If the attacker's highest number is greater than the defender's highest number, the attacker wins. If it is equal or lesser, the attacker loses. The same is done for the pair of lowest numbers. I've been discussing this with a group of junior-high math teachers, and we have no idea how to figure it out. Thanks. Josh Peretti
Date: 04/22/2000 at 17:49:15 From: Doctor Mitteldorf Subject: Re: Probability and the game of RISK. Dear Josh - One way to go about this is the brute force technique. I encourage you to actually do this, because it is straightforward and transparent, and it will help you understand where any more abstract approach is coming from. The brute force technique would be to model the situation with a computer. You'll have 5 different dice, each with values from 1 to 6, so you can program with 5 nested loops for a1:=1 to 6 do for a2:= 1 to 6 do for a3:=1 to 6 do for b1:=1 to 6 do for b2:= 1 to 6 do Inside all these loops, you'll have some if...then statements about what happens. First, compare to find the higher and lower of the two b's; then find the highest and middle of the two a's. Compare a_high and b_high, a_mid and b_low, and tally up the results. This tally will be run for each of the 7776 (6*6*6*6*6) possible outcomes. The program will count how many of these 7776 possibilities result in victory by the attacker, etc. The program will take half an hour of thinking in which you will come to understand the situation you're modeling much more intimately. It will run in less than the blink of an eye. -------------------- A more abstract approach would be to come up with probability distributions for the attacker's dice and the defender's dice. Let's start with the defender's high die. There are 6*6 = 36 possible outcomes to consider. In 11 of them, the highest die is a 6, so the probability of the higher of the defender's dice being a 6 is 11/36. How did I count this? The first die can be 6 and the second die can be anything (that's 6 possibilities) or the first die can be 1 ... 5 and the second die can be a 6 (5 more possibilities.) Similarly, you can tabulate the probabilities for the defender's highest die: Higher cases die out of 36 6 11 5 9 4 7 3 5 2 3 1 1 The probabilities for the lower of the 2 dice are just reversed; the probability of a 1 is 11/36, of a 2 is 9/36, etc. Can you see why? I'll leave it to you to do the similar but more complicated analysis for the attacker. Start with the number of cases out of 6*6*6 = 216 in which the highest of the three dice is a 6. There are 91 of them, hence the probability for the highest die being a 6 is 91/216. Continue to find the complete probability table for the highest die. Then move to the middle die, and compute its probability table. Of course, you'll discard the lowest die. Now you have a probability table for the highest of three and for the higher of 2. From that, you can pair each of the 36 pairings, add them up, and have the result for the higher die; similarly for the lower die. It's a good deal of tabulation and calculation no matter how you do it, but it's a shortcut compared to what the computer does, which is to take each of the 7776 cases separately. The results I get are as follows: Attacker wins: 2890/7776 (0.372) Defender wins: 2275/7776 (0.293) A and D Split: 2611/7776 (0.336) Thanks for the review of the Risk rules. The last time I played Risk was about 1965, I think. I hope you'll continue and complete the calculation, both as a computer program and as a hand calculation. Maybe the two will even come out the same. In any case, I hope you'll write back and either report your completed results, or tell me exactly what you did and how far you got with it before you got stuck. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
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