**If everyone who ate Chinese food today played their lucky numbers in the lottery, what are the chances at least one of them would win?**"

Both fortune cookies and lottery numbers usually show about 5 numbers that can range from roughly 1 to 50. The probability of picking the first number correctly is 5 out of 50. The probability of picking the second number correctly is 4 out of 49. The probability of picking the third number correctly is ... Multiplying these probabilities together, we can find the total probability of finding the right sequence of numbers

^{1},

P = [(5)! · (50-5)!] / 50! = 4.7×10

^{-7}.That's about one in two million. I generally go out for Chinese food about once per month, which seems like a reasonable amount for most people. Taking that as the average and using the fact that there are 3.0×10

^{8}Americans, we can estimate the number of people that went out for Chinese today,

# of people going for Chinese = (prob. of going out for Chinese) · (total # of people)

= (1 day / 30 days) · (3.0×10

^{8 }people)= 1.0×10

^{7 }people.^{10,000,000}

^{}. Likewise, the probability of everyone picking the wrong number is (1-P)

^{10,000,000}

^{}. The probability that at least one person will win is then just

1 - (1-P)

^{10,000,000}= 0.009.

There's about a 1% chance that if everyone played their lucky fortune cookie numbers at leat one would win.

[1] This is the well known binomial distribution.

[2] I'm assuming the fortune cookie's "lucky numbers" are random and uniformly distributed.

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