>That is the theoretical probability that A wins the series, given that A >has won the first game.
I guess I can mention now, without the risk of spoiling anything, that the essence of this problem can be traced al the way back to the origins of the theory of probability (see http://www.businessweek.com/chapter/chap4.htm and http://people.brandeis.edu/~cecchett/Textbook%20inserts/The%20Story%20of%20Blaise%20Pascal.htm). ________________________________________ From: programming-boun...@jsoftware.com [programming-boun...@jsoftware.com] on behalf of Kip Murray [k...@math.uh.edu] Sent: Monday, February 13, 2012 2:02 PM To: Programming forum Subject: Re: [Jprogramming] Challenge 5 Super Bowl Supposition Here is a partial solution to Linda's Challenge 5. --Kip Murray Let us call the teams A and B. A has won the first game of a seven game series, and there are up to six games to go. I deal with the question of whether A wins the series, ignoring the matter of how many games are required. Each game is played until there is a winner -- there are no ties. Let us imagine the teams play all six remaining games even if the series is decided before the sixth remaining game. Because the teams are equally matched, the six remaining games are equivalent to six tosses of a fair coin with sides 0 and 1, where 1 means a win by team A. A key idea is that A wins the series if A wins three or more of the remaining six games, because then A has won four or more games in all and B has won at most three games in all. The binomial distribution (x!y)%(2^y) gives the probability of obtaining exactly x heads in y tosses of a fair coin. For six tosses the probabilities are shown in the table (] ,: (2^6) %~ 6 !~ ]) i.7 0 1 2 3 4 5 6 0.015625 0.09375 0.234375 0.3125 0.234375 0.09375 0.015625 We find the probability of three or more heads by summing: ([: +/ 3 }. (2^6) %~ 6 !~ ]) i.7 0.65625 That is the theoretical probability that A wins the series, given that A has won the first game. Verb winpct below simulates y repetitions of tossing a fair coin 6 times and returns the percent of those y repetitions which resulted in 3 or more heads. (Heads is 1, tails is 0.) winpct =: 100 * ] %~ [: +/ 3 <: [: +/ [: ? 2 $~ 6 , ] winpct 2000000 65.6252 winpct"0 [ 5#2000000 65.5622 65.6496 65.599 65.6166 65.6298 On 1/31/2012 3:46 AM, Linda Alvord wrote: > Challenge 5 Super Bowl Supposition PLEASE DO NOT RESPOND UNTIL 2/6/2012 12 > am EST > > > > As the Super Bowl approaches, suppose it will be decided like baseball. Four > of seven games determines a winner. Also suppose that the NFL has won the > first game. > > > > Simulate results of 2000000 series and provide the number of times the NFL > wins in 4 5 6 7 games. If the AFL wins this Extended Super Bowl > Contest, the result is an 8 . Create a 2000000 item list of number of > games necessary to determine a winner and provide a frequency distribution. > > > > fd=: [: /:~ ({. , #)/.~ > > fd (expression for 2000000 trials) > > 4 249561 > 5 374865 > 6 373851 > 7 312603 > 8 689120 > > ]games=:fd n,.2000000$6 > > 4 249301 > 5 376266 > 6 375281 > 7 311189 > 8 687963 > > ]prob=:(4+i.5),. (1{"1 games)%2000000 > > 4 0.124651 > 5 0.188133 > 6 0.18764 > 7 0.155595 > 8 0.343982 > > > > ]+/(1{"1 games)%2000000 > > 1 > > > > Now, confirm that your results are reasonable with a theoretical argument. > > > > Also, enjoy the Super Bowl! > > > > Linda > > > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm [BEST LLC] Bayesian Efficient Strategic Trading LLC The information in this communication and any attachment is confidential and intended solely for the attention and use of the named addressee(s). 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