probability Jim Unregistered Posts: 56 Threads: 19 Joined: Jan 1970 10-05-2004, 11:20 AM Hi, folks. I need some facts checked. Please tell me what you get for probability and odds for picking 10 out of 10 winners on a football ballot if all games are a 50/50 toss-up and what would the probability and odds be if 10 teams were favored at 90% chance of winning and you chose all favorites. If you get the same numbers I did, I will assume that is correct. Verification from two or more of you would be even better. It's always good to know that I have reliable help on the HP Forum Ron Ross Unregistered Posts: 673 Threads: 20 Joined: Oct 2008 10-05-2004, 11:26 AM I get an answer of 0.00098 or 98/100,000 for a 50/50 chance. The odds are much better at 90% favorites and I get: 35% chance of winning. Larry Corrado Unregistered Posts: 99 Threads: 9 Joined: Apr 2013 10-05-2004, 11:56 AM Jim: Since you are requiring 10 independent events in each case, the overall probability is simply the product of the individual probabilities. For the 50% case the overall probability is (0.5)^10 or 0.000977 acording to my HP-11C. For the 90% case, (0.9)^10 = 0.349. This reminds me of the problem our math instructor gave us, about the British avaitors who were flying bombing missions during World War II. Supposedly they had a 20%(!) chance of being shot down on any one mission. The question asked was what is the probability of being shot down after five missions. Some of the students came up with 100% (5 x 20%). Larry Mike (Stgt) Unregistered Posts: 858 Threads: 80 Joined: Feb 2009 10-05-2004, 12:00 PM The chance is always 50% -- it hapens or not Ciao.....Mike Jim Unregistered Posts: 56 Threads: 19 Joined: Jan 1970 10-05-2004, 06:15 PM Thanks, Ron and Larry. Your numbers corroborate mine. Karl Schneider Unregistered Posts: 1,792 Threads: 62 Joined: Jan 2005 10-05-2004, 11:52 PM The answers that others have given are correct; this is a simple problem. A tougher question is, what are the chances of getting, say, 8 out of 10 right? The binomial distribution is as follows: ```P(n,k,p) = C(n,k)* pk * (1-p)(n-k) P is the overall probability n is the number of trials k is the specified number of successes p is the probability of success of each trial C(n,k) is the statistical combination function ``` Now, if there are n successes in n trials (get 'em all correct), this reduces to ```P(n,n,p) = pn ``` If you have a 65% chance of picking each of 10 teams correctly, your chances of getting 8 correct are: ```P(0.65, 10, 8) = C(10,8) * 0.658 * (1-0.65)(10-8) = 0.175652953 ``` -- KS « Next Oldest | Next Newest »

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