Picking the Same BallDate: 07/07/97 at 01:19:58 From: Guy Subject: Probability I have a jar with 377 individually numbered balls in it. One ball is selected at random each day. What are the odds I will randomly select the same ball two days in a row? I know the odds for selecting a given numbered ball on one day are 1 in 377. I'm stumped on how to compute the probability of selecting the same numbered ball on two consecutive days. So far, I have selected one ball per day for 7 days. Wouldn't the praobablity of selecting the same ball on two consecutive days increase if I continued to do this for 30 days? Thank you for your help! Guy Date: 07/09/97 at 14:14:54 From: Doctor Daniel Subject: Re: Probability Hi Guy, I'm not 100 percent sure I understand your problem, so I'll answer a couple of different problems: 1) Suppose I pick one ball today and one ball tomorrow. What's the probability they're the same? Answer: 1/377. This is actually fairly easy, though a little counter-intuitive. Perhaps you could think of it slightly differently; suppose instead of 377 balls, you had a set of dice with 377 faces. Then, really, the question is what's the probability that if I roll 2 of them, they'll both come up on the same face? Well, suppose that the first one comes up 1. Then there's a 1/377 chance that the second one will too. Similarly, if you get a 344 on the first one, there's a 1/377 chance the second one will. So it should be clear that no matter what the first die comes up, the second one has a 1/377 chance of also coming up that number. So it's clear that the chance that the two dice agree is 1/377. This is also the chance that the ball I pick today is the same as the one I pick tomorrow. 2) Suppose I pick one ball today, one ball tomorrow, one ball the next day, and so on, for X days. What's the probability that I have two identical balls come up on consecutive days? This is a good time to introduce a really important rule in probability theory, which is called the law of the complement when people want to be formal. Basically, here it is: The probability that something happens is exactly 1 minus the chance that it doesn't happen. (Yes, I know. That's something obvious, AND you already knew it. Give me a second here!) So I could just as easily ask this question: What's 1 minus the probability that all of the pairs of consecutive balls are different? Now, we know that there are X-1 pairs of consecutive balls (for example, if X = 3, there's the first and second ball, and the second and third ball). We also know that no pair of consecutive balls has any influence on any other pair, when it comes to them being the same. That is, the chance that the second and third balls are the same is 1/377, no matter whether the first and second balls are the same or the 457th and 458th ball are the same. They're _independent_. So the chance that, say, the second and third balls are different AND the fourth and fifth balls are different is 376/377 * 376/377, because the probability that two independent events happen is the product of their probabilities. The probability that the first and second balls are different, the second and third are different, and so on, to the X-1 and Xth balls being different is: 376/377 * 376/377 * ... * 376/377 = (376/377)^(X-1) So the probability that there's a duplicate ball gotten consecutively in the first X days is: 1-(376/377)^(X-1). So, you're right; if you keep doing this for 30 days, the chance that you've gotten the same ball consecutively is: 1-(376/377)^29 = .0741 This is not a particularly good chance; you've only got a 7.4 percent chance of having it happen. It turns out it's not until you've gone for 261 days that it will be 50 percent likely to have happened. I hope this helps answer your question! Good luck. -Doctor Daniel, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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