Math 1100, Section 3 |
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Probability Notes |
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Number 1 |
10/9/2002 |
by J. Madden
Abstract: An example showing how to use the sample space model to find probabilities.
The Chevalier de Mere was a 17th-century French gambler. He bet opponents that he could get a 6 in four rolls of a fair die. If he got a 6 in four throws, he won and his opponent would pay him an agreed sum. If he didn’t, he would pay the agreed sum to his opponent. The Chevalier figured his odds of winning to be 1/6 + 1/6 + 1/6 +1/6 = 2/3, since the chance of rolling a 6 on each throw is 1/6.
The Chevalier's reasoning cannot be correct, because it leads to the conclusion that if the die were rolled six times then the probability of getting a 6 would be 1/6 + 1/6 + 1/6 +1/6 + 1/6 +1/6 = 1, a certainty. But it is not certain that in 6 throws one will get a 6. The Chevalier's reasoning would also lead him to conclude that if a coin were flipped twice, one would surely get a head and also surely a tail. But again this is not what we experience.
We solve this problem by considering all of the possible outcomes. The die might come up with a 2 followed by a 5 followed by a 4 and then a 6. (The Chevalier would win.) We could record this outcome as: 2546. Another possible outcome would be 2222, meaning that each time the die was rolled, a 2 came up. (The Chvalier would lose.) All these outcomes are equally likely. It is no more nor less probable that a person with a fair die would roll 2222 than that she would roll 2546 or any other sequence, such as 5334, or 3661 or 1234.
You might think that 3661 would not be a possible outcome, since if 6 came up on the second roll the game would stop. In practice, that might be the way the players behaved. But if we were to view something like 36 as a possible outcome, we would have to deal with the fact that its probability is different from the probability of, say, 3226. This is even clearer if we think about rolling a 6 on the first try. Since the faces are equally likely, the probability of rolling 6 is the same as the probability of rolling any other number on the die, namely 1/6. But the probability of rolling 3226 is much less.
It does no harm to assume that the die is rolled four times, no matter what happens. Perhaps the rolls are recorded by a judge as they occur, and only revealed to the playes after the fourth roll is completed. This does not change he odds of winning the the game. Conceptualizing the problem in this way has the advantage that all the different outcomes are equally likely.
How many outcomes are there? (See them listed.)
Since there are 4 throws and 6 different possibilities for each throw, there are
6*6*6*6 = 1296
different outcomes.
How many of these outcomes are favorable to the Chevalier? (See them here.)
It's actually easier to count the outcomes that are favorable to the opponent. Since a 6 must be avoided, there are only 5 possibilities on each throw if we want to have the Chevalier lose. Thus, there are
5*5*5*5 = 625
ways for the Chevalier to lose.
Conclusion: The probability that the Chevalier loses is
625/1296, or approximately 48.23%;
the probability that he wins is
1-(625/1296) = 671/1296, or approximately 51.77%.