mpauls Posted June 13, 2009 Share Posted June 13, 2009 I'm sure this will interest at least a few people. I think its a good exercise though it's not related to anything specifically important, but if you don't think about this sort of thing it may provide a catalyst to get you started. Let's see who will get it first: If we have 6 players competing in a round robin game of dice. Each player will play a total of 5 games, that is, one game between each player. (To organize this conceptually it may be useful to think about this as five rounds of three games between two teams.) The top 3 players get a prize (assume its an equal prize). (i) We have a three sided die, equally weighted, and marked: ‘A,’B,’C (ii) There is only one roll per game and it makes no difference who rolls the die (iii) ‘A is the most favorable outcome, ‘B the worst. ‘C gives both players the same result. Payoffs : If the roller throws ‘A he gets 3 points and his opponent gets 0 points, if the roller throws ‘C he gets 0 points and his opponent gets 3 points, if the roller throws ‘B both players get 1 point. We don’t care about the the order of the combinations. After the 15 total games are played (5 total games per player) there are a total of 21 different possible combinations of ‘A, ‘B, ‘C, (without respect to the order in which the combinations occurred) however in terms of the payoffs there are only 15 different combinations (b/c some of the different combinations give the same numerical ‘payoff’ result.) Here is some info to help get you started: Total possible results and their payoffs are in a list below. Also consider for example, if one player tallies 5 ‘A rolls then no other player can tally 5 ‘A rolls, rather each other player must have at least 1 ‘C. So there is a bit of conditional probability going on here so think Bayesian. The question is to figure out which three winning combinations occur together and the probability of these results occurring. Results Payoffs (‘A,‘B,‘C) (5,0,0) 15 (4,0,1) 12 (3,0,2) 9 (2,0,3) 6 (1,0,4) 3 (0,0,5) 0 (4,1,0) 13 (3,1,1) 10 (2,1,2) 7 (1,1,3) 4 (0,1,4) 1 (3,2,0) 11 (2,2,1) 8 (1,2,2) 5 (0,2,3) 2 (2,3,0) 9 (1,3,1) 6 (0,3,2) 3 (1,4,0) 7 (0,4,1) 4 (0,5,0) 5 Link to comment Share on other sites More sharing options...

Partner24 Posted June 13, 2009 Share Posted June 13, 2009 I like this kind of mental exercice. Thank you for posting it. I'll begin to plunge and answer your question very partially (I may be wrong) I have one chance out of three to win each time I play (1 chance to win, 1 chance to breakeven, 1 chance to lose). The odds that I will win every 5 games (5.0.0) are 0,4%, quite small. If I have to bet 100$ for each tournament and the person who get (5,0,0) win a 30000$ sum from the house, the odds are slightly in my favor and the house will goes bankrupt over time. I'll check this out for answering your question fully. Cheers! Link to comment Share on other sites More sharing options...

mpauls Posted June 17, 2009 Author Share Posted June 17, 2009 The attachment might help. Link to comment Share on other sites More sharing options...

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