Question 197977
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The digits 0 through 9 represent 10 choices, so there are 10 ways to pick the first number, then 9 ways to pick the second number, and 8 ways to pick the third number, hence there are 10 times 9 times 8 or 720 different numbers.


One in 720


The probability that one ticket won't win is *[tex \LARGE \frac{719}{720}].  But once the first ticket doesn't win, there are only 719 possibilities left, only one of which wins, so the probability the second ticket doesn't win is *[tex \LARGE \frac{718}{719}].  Likewise, the probability the third ticket doesn't win is *[tex \LARGE \frac{717}{718}], and the probability of all three events is the product of the three probabilities.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{719}{720}\right)\left(\frac{718}{719}\right)\left(\frac{717}{718}\right)]


But wait!  There's more!  You did win after all!  You get to do your own arithmetic.


<b><i>Super-Double-Plus Extra Credit</i></b>.  Without using pencil, paper, or calculator, how many tickets would you have to buy so that the chance of winning is equal to the chance of losing?


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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