Odds vs. ProbabilityDate: 12/06/96 at 12:06:47 From: Jessica Peace Subject: Calculating odds I'm stuck...and don't know where to begin. Please help. Here is the question: Assume four individuals are each drawing two numbers from a box that contains eight numbers. The numbers in the box are assigned in numerical order, 1-8. Also assume that the four individuals are identified in alphabetical order: A, B, C, and D. Individual 'A' wants to draw numbers 2 & 7; individuals B, C, and D have no preference as to which two numbers they pull from the box. What are the odds, or how do I write the formula to determine the odds, that individual 'A' will draw the two numbers that he wants? Also, what will be the odds if the numbers in the box are grouped by lots of two; i.e. 2 and 7, 1 and 8, 3 and 6, and 4 and 5? Would this mean the odds would be 4:1 that Individual 'A' would draw the lot of 2 and 7? How would the order of drawing from the box affect the odds in both scenarios? Thank you. Jessica Date: 12/06/96 at 15:46:41 From: Doctor Wilkinson Subject: Re: calculating odds First of all, it's important to know how the term "odds" is used, as opposed to "probability". If I toss a coin, it may come down heads or tails. The two possibilities are equally likely (assuming the coin is fair, etc). The probability of it coming down heads is 1/2, because that is one of the two equally likely possibilities. The term "odds" is used usually with "against" and is used to compare the unfavorable with the favorable possibilities (instead of the favorable possibilities with all the possibilities), so the odds against the coin's coming down heads are 1:1. Similarly, if you roll a die, the probability of a 3 is 1/6, since there are six possibilities, one of which is a 3. The odds against a 3 are 5:1, since there are 5 unfavorable possibilities and one favorable. The probability of throwing either a 2 or a 3 is 2/6 or 1/3, and the odds against are 2:1. It is easy to translate from probability to odds, but they are not the same thing. Now that I've confused you thoroughly, let's look at your problem. The other players (B, C, and D) really have nothing to do with it. A picks two numbers out of eight, and there is only one outcome that he wants. So the key to the problem is to figure out how many possible pairs of numbers there are. Then the probability of getting a 2 and a 7 will just be 1 divided by the total number of possibilities, and the odds against will be the ratio of the number of unfavorable possiblities to 1. So how many possibilities are there altogether? If you don't know how to count them, let me know. For the second part of the problem, you have only 4 possibilities, one favorable and 3 unfavorable, so the probability of winning is 1/4, and the odds are 3:1 against. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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