Finding My Car Key on the Second Try

Date: 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/