Date: 12/17/97 at 16:53:39 From: Dave Kugelstadt Subject: Penny Toss Dear Dr. Math: My son is in the 7th grade. I try to help him grasp the concepts of advanced math because I believe that a firm understanding of math will do a great deal to advance his quality of life, as it has mine. Usually I am very good but I am not quite sure about this one. His "Problem of the week 6" goes as follows... "Three people each toss a penny at the same time. What is the probability that two people get the same side of the penny and the other person gets the opposite side?" Since there was no specification as to which individual got what, I reasoned that there are only two possible combinations, all of them get the same side or two of them get one side and one gets the other. With this I decided that once I determined the odds of all getting the same side I could subtract that from 100% to get the chances of any two getting the same side. Assuming all else is equal, there is a 1/2 chance that any one person gets any one particular side. Since there are three people I calculated .5 x .5 x .5 = .125 or 12.5% chance that all would get the same side. That leaves 87.5% chance that any two people would get the same side if all three tossed their pennies at the same time (or even not at the same time, I suppose). Before I tell my son this is it I would like to know if my reasoning is sound. Thanks, Dave K.
Date: 12/17/97 at 19:25:35 From: Doctor Tom Subject: Re: Penny Toss Close, but not quite. If there are only 3 pennies, it's easy just to list the possibilities: HHH HHT HTH HTT THH THT TTH TTT where the first column represents the result for the first person, etc. So there are 8 equally likely ways the experiment can come out, 6 of which have two faces the same. Thus the probability is 6/8 = 75%. You were on the right track, but the .5*.5*.5 is the probability that all three throw heads (and also the probability that all three throw tails). So there is a 12.5% chance that all three throw heads, a 12.5% chance that all throw tails, and hence, a 100% - 12.5% - 12.5% = 75% chance that all three flips aren't the same. I don't know if your kid knows about combinations (like "6 choose 2" - the number of ways of picking 2 things from a set of 6), but if so, that's a good way to work it. There are 3 flips, so there are 2^3 = 8 (2 cubed = 8) possible outcomes, and the favorable outcomes are if there is 1 head of the three or 2 heads. (If there are 0 or 3 heads, it's an unfavorable outcome.) The number of favorable outcomes is thus (3 choose 1) + (3 choose 2) = 3 + 3 = 6 favorable out of 8 possible, or a probability of 6/8, or 75%. This last method is far more powerful in terms of answering more complex problems, but you have to know how to count combinations. For example, if the problem were, "7 people flip 1 penny each. What is the probability that there are at least 4 heads tossed?" Well, there are 2^7 = 128 outcomes, and there are: (7 choose 4) + (7 choose 5) + (7 choose 6) (7 choose 7) favorable outcomes. If you work it out, this is 35+21+7+1 = 64 favorable of 128, giving a probability of 64/128, or 1/2. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 12/17/97 at 19:28:16 From: Doctor Anthony Subject: Re: Penny Toss You are partly but not completely right. The easiest way is to consider the probability of NOT getting two of one and one of the other. This would be either HHH or TTT The chance of HHH is (1/2)^3 and of TTT is also (1/2)^3 = 1/8 + 1/8 = 1/4 So the chance of two of one and one of the other is 3/4 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.