Sanderson M. Smith

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PROBABILITY PROBLEM (A REAL SITUATION)Last night (May 22, 2002) I witnessed the following situation during a local televised newscast. Fifteen car keys were placed in a box. One of them would start a new car that was parked nearby. By a randomly determined order, 15 contestants were provided the opportunity to pick a key and then attempt to start the car with it. If the key started the car, the contestant won the car.

Keys were not placed back in the box if they failed to start the car.That is, if the first contestant's key failed to start the car, then the second contestant picked from a set of 14 keys, etc.It turned out that the first 14 keys chosen did not start the car, meaning that the last remaining key would (and did) start the car. The 15th contestant who could only choose one key won the car.

My initial thought was that the chance of such a thing happening was quite small. I wondered what picking position would be the most ideal one in terms of likelihood of winning the car. My gut reaction was that I would probably choose to pick 4th or 5th if I had the choice. I began to do a few computations.

The probability that the

firstcontestant would win the car is 1/15, or about 6.67%.For the

secondcontestant to win, the first contestant would have to pick incorrectly, and then the second contestant would have to pick the correct key. The probability that the second contestant would win is (14/15)(1/14), or about 6.67%.For the

thirdcontestant to win, the first and second contestants would need to pick incorrectly, and then the third contestant would have to pick the correct key. The probability that the third contest would win is (14/15)(13/14)(1/13) = 1/15, or about 6.67%.

Yup! You got it! Each contestant has a 6.67% chance of winning the car. The last contestant who simply must test the remaining key if the previous 14 haven't started the car isjust as likely to win as any other contestant.===========================================

Now, what would happen if a key was placed back in the box if it didn't start the car? Is it still true that each contestant would have an equal probability of winning the car?A bit of reflection should have you concluding that the answer is a definite NO.

The probability that the

firstcontestant would win is still 1/15, or 6.67%.The

secondcontestant could win only if the first contestant picked the wrong key and the second picked the correct key. Since the keys are replaced, the probability that the second contestant would win is (14/15)(1/15) = 6.22%.The probability that the

thirdcontestant would win is (14/15)(14/15)(1/15) = 5.81%.The probability that the

Nthcontestant would win is (14/15)^{N-1}(1/15).Here are the probabilities for the 15 contestants:

# 123456789101112131415Prob % 6.67

6.22

5.81

5.42

5.06

4.72

4.41

4.11

3.84

3.58

3.34

3.12

2.91

2.72

2.54

The sum of the probabilities in the table is 64.47%. In other words, with key replacement the probability that the car will be won by one of the 15 contestants is 64.47%. The probability that it will not be won by any of them is 100% - 64.47% = 35.53%.

"Fate laughs at probabilities."

-LYTTON BULWER"A fool must now and then be right, by chance."-WILLIAM COWPER

"Chance favors prepared minds."-LOUIS PASTEUR

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