Finding My Car Key on the Second TryDate: 05/29/2000 at 19:43:04 From: Peter J. Stubbs Subject: Probability I'm hoping you can explain this for my daughter, Jessica (age 13). There are seven different car keys in a box, including mine. I'm going to randomly remove one key at a time, and then try to start my car. If the key doesn't start the car, I'll discard the key. We know that the probability is one in seven that I'll retrieve my own car key on the first try. But what is the probability that I'll get my key on the second try? In Ask Marilyn (Parade Magazine, 05/28/00), Marilyn Vos Savant stated that the probability was still one in seven. My daughter insists that it is one in six. If the first key is known to be wrong, and is discarded, then there are six keys left and one favorable outcome, so one in six. Can you explain Marilyn's answer? Before any key is chosen, it seems to me equally likely that my key will be drawn on the first as the second, or even the last try. But once a key is chosen, and known to be wrong, doesn't that change the picture? Is this similar to the Monty Hall problem? Date: 05/30/2000 at 03:19:09 From: Doctor Floor Subject: Re: Probability Hi, Peter, Thanks for writing. This is not similar to the Monty Hall problem. You can read about Monty Hall in the Dr. Math FAQ: http://mathforum.org/dr.math/faq/faq.monty.hall.html The probability that the second key is your key contains two elements: The first key is not your key AND the second key is. Your daughter forgot the first element. The probability that the first key is not your key is 6/7. After that the probability that the second key is your key is 1/6. These two combine to 6/7 * 1/6 = 1/7. Your intuition that before starting it is equally likely that the first, second, third, or last try gives your key is indeed correct. I hope this clears it up. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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