Sanderson M. Smith
Home | About Sanderson Smith | Writings and Reflections | Algebra 2 | AP Statistics | Statistics/Finance | Forum
Chevalier de Mere was a mid-seventeenth century high-living nobleman and gambler who attempted to make money gambling with dice. Probability theory had not been developed, but de Mere made money by betting that he could roll at least one 6 on four rolls of one die. Experience led him to believe that he would win more times than he would lose with this bet. Today we know that the probability of winning this bet is 1 - (5/6)4, or 51.8%.
When folks would no longer bet on this game with de Mere, he created a new game. He began to bet he would get a total of 12 (or a double 6) on twenty-four rolls of two dice. This seemed like a good bet, but he began losing money on it. He asked his good friend Blaise Pascal to analyze this game - and probability theory was born. Pascal came to realize that a "backwards" approach might work on this analysis. He figured that the probability of not rolling a total of 12 in twenty-four rolls is (35/36)24, or about 50.9%. Hence, in the long run, this would be a losing game for de Mere. Pascal got interested in analyzing other gambling games, and got Pierre de Fermat to work with him. It can be said that the formal study of probability was launched by two mathematicians and a gambler. Not surprisingly, Pascal's Triangle is a useful tool in probability theory.
Even today it isn't hard to be fooled by a probability that sounds good. For instance, it is easy to show (see diagram below) that in rolling two dice, a total of 7 is more likely than a total of 6 or a total of 8. Here is a de Mere-like challenge.
I'll bet that I can an 8 and a 6 rolling two dice before you can get two 7's
This sounds pretty good! Would you take this bet? An analysis of this game appears below.
Let W be the event "6 or 8" and let w be the event "the other of 6 or 8." Here then are the three ways I can win the bet. (Remember... I'm betting on the 8 and 6 before two 7's)
Now, any total other than 6, 7, or 8 is meaningless here. Hence, only 16 of the 36 cells shown at the right have significance. (Blue numbers).
The probability of (1) is (6/16)(10/16)(5/11) = 10.65%.
The combined probability of (2) and (3) is (10/16)[1 - (6/11)2] = 43.90%.
My probability of winning is 10.65% + 43.90% = 55.55%.
The catch here is that this game does not specify an order for the 8 and the 6. Hence, there are ten possible ways for me to get a favorable first result - and only six for you.
RETURN TO WRITING HOME PAGE