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The 10,000 tickets for a lottery are numbered 0000 to 9999. A four-digit winning number is drawn and a prize is paid on each ticket whose four-digit number is any arrangement of the number drawn. For instance, if winning number 0011 is drawn, prizes are paid on tickets numbered 0011, 0101, 0110, 1001, 1010, and 1100. A ticket costs \$1 and each prize is $500.

(b) Assuming that all tickets are sold, what is the probability that the operator will lose money on the lottery?

I'm having trouble with this problem. Clearly different tickets have differing probabilities of winning (that's what (a) was about) but it seems to be very messy to calculate all the different probabilities of winning in different ways.

The answer given by the textbook is $\frac{10^{(4)}}{10^4}$. Why is this so?

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By $10^{(4)}$, do you mean $10 \cdot 9 \cdot 8 \cdot 7$? – Ron Gordon Jan 15 '13 at 21:38
From here, can you see if the operator wins or loses on average? – Ross Millikan Jan 15 '13 at 21:45
up vote 3 down vote accepted

In order to lose money, the operator must pay out more than $\$10,000$, so there must be more than $20$ winning tickets. This is the case if and only if all four digits of the winning number are distinct, in which case there are $4!=24$ winning tickets. (If the winning number has three distinct digits, there are only $\binom42\cdot2=12$ winning tickets.)

There are $10\cdot9\cdot8\cdot7$ ways to choose $4$ distinct digits, so the desired probability is indeed

$$\frac{10\cdot9\cdot8\cdot7}{10^4}=\frac{10^{\underline 4}}{10^4}\;.$$

(I prefer the notation $n^{\underline k}$ to the notation $(n)_k$, which is presumably the Pochhammer symbol that you intended. Your $10^{(4)}$ generally denotes the rising factorial, $10\cdot11\cdot12\cdot13$.)

Added: The probability of taking loss works out to $0.504$, and if he takes a loss, he loses $4\cdot500=2000$ dollars. In all other cases there are at most $12$ winning tickets, so he gains at least $10,000-12\cdot500=4000$ dollars. Thus, his expected profit is more than


dollars. With a bit more work one can of course calculate it exactly, but at least we know that the operator isn’t a complete idiot, even though he’s more likely to lose money than to make a profit.

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