Mixing
Textbook question about a weird lottery
$begingroup$
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 $largefrac{10^{(4)}}{10^4}$. Why is this so?
probability statistics
edited Jan 31 at 11:04
Gaby Alfonso
1,1931418
asked Jan 15 '13 at 21:35
ithisaithisa
1,12221441
$endgroup$
add a comment |
$begingroup$
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 $largefrac{10^{(4)}}{10^4}$. Why is this so?
probability statistics
edited Jan 31 at 11:04
Gaby Alfonso
1,1931418
asked Jan 15 '13 at 21:35
ithisaithisa
1,12221441
$endgroup$
$begingroup$
By $10^{(4)}$, do you mean $10 cdot 9 cdot 8 cdot 7$?
$endgroup$
– Ron Gordon
Jan 15 '13 at 21:38
$begingroup$
From here, can you see if the operator wins or loses on average?
$endgroup$
– Ross Millikan
Jan 15 '13 at 21:45
add a comment |
$begingroup$
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 $largefrac{10^{(4)}}{10^4}$. Why is this so?
probability statistics
edited Jan 31 at 11:04
Gaby Alfonso
1,1931418
asked Jan 15 '13 at 21:35
ithisaithisa
1,12221441
$endgroup$
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 $largefrac{10^{(4)}}{10^4}$. Why is this so?
probability statistics
probability statistics
edited Jan 31 at 11:04
Gaby Alfonso
1,1931418
asked Jan 15 '13 at 21:35
ithisaithisa
1,12221441
edited Jan 31 at 11:04
Gaby Alfonso
1,1931418
asked Jan 15 '13 at 21:35
ithisaithisa
1,12221441
edited Jan 31 at 11:04
Gaby Alfonso
1,1931418
edited Jan 31 at 11:04
Gaby Alfonso
1,1931418
edited Jan 31 at 11:04
Gaby Alfonso
1,1931418
1,1931418
asked Jan 15 '13 at 21:35
ithisaithisa
1,12221441
asked Jan 15 '13 at 21:35
ithisaithisa
1,12221441
asked Jan 15 '13 at 21:35
ithisaithisa
1,12221441
1,12221441
$begingroup$
By $10^{(4)}$, do you mean $10 cdot 9 cdot 8 cdot 7$?
$endgroup$
– Ron Gordon
Jan 15 '13 at 21:38
$begingroup$
From here, can you see if the operator wins or loses on average?
$endgroup$
– Ross Millikan
Jan 15 '13 at 21:45
add a comment |
$begingroup$
By $10^{(4)}$, do you mean $10 cdot 9 cdot 8 cdot 7$?
$endgroup$
– Ron Gordon
Jan 15 '13 at 21:38
$begingroup$
From here, can you see if the operator wins or loses on average?
$endgroup$
– Ross Millikan
Jan 15 '13 at 21:45
$begingroup$
By $10^{(4)}$, do you mean $10 cdot 9 cdot 8 cdot 7$?
$endgroup$
– Ron Gordon
Jan 15 '13 at 21:38
$begingroup$
By $10^{(4)}$, do you mean $10 cdot 9 cdot 8 cdot 7$?
$endgroup$
– Ron Gordon
Jan 15 '13 at 21:38
$begingroup$
From here, can you see if the operator wins or loses on average?
$endgroup$
– Ross Millikan
Jan 15 '13 at 21:45
$begingroup$
From here, can you see if the operator wins or loses on average?
$endgroup$
– Ross Millikan
Jan 15 '13 at 21:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In order to lose money, the operator must pay out more than $$10,000$, so there must be more than $20$ winning tickets. This is the case if and only if all four digits of the winning number are distinct, in which case there are $4!=24$ winning tickets. (If the winning number has three distinct digits, there are only $binom42cdot2=12$ winning tickets.)
There are $10cdot9cdot8cdot7$ ways to choose $4$ distinct digits, so the desired probability is indeed
$$frac{10cdot9cdot8cdot7}{10^4}=frac{10^{underline 4}}{10^4};.$$
(I prefer the notation $n^{underline k}$ to the notation $(n)_k$, which is presumably the Pochhammer symbol that you intended. Your $10^{(4)}$ generally denotes the rising factorial, $10cdot11cdot12cdot13$.)
Added: The probability of taking loss works out to $0.504$, and if he takes a loss, he loses $4cdot500=2000$ dollars. In all other cases there are at most $12$ winning tickets, so he gains at least $10,000-12cdot500=4000$ dollars. Thus, his expected profit is more than
$$0.496cdot4000-0.504cdot2000=976$$
dollars. With a bit more work one can of course calculate it exactly, but at least we know that the operator isn’t a complete idiot, even though he’s more likely to lose money than to make a profit.
answered Jan 15 '13 at 21:40
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In order to lose money, the operator must pay out more than $$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 $binom42cdot2=12$ winning tickets.)
There are $10cdot9cdot8cdot7$ ways to choose $4$ distinct digits, so the desired probability is indeed
$$frac{10cdot9cdot8cdot7}{10^4}=frac{10^{underline 4}}{10^4};.$$
(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, $10cdot11cdot12cdot13$.)
Added: The probability of taking loss works out to $0.504$, and if he takes a loss, he loses $4cdot500=2000$ dollars. In all other cases there are at most $12$ winning tickets, so he gains at least $10,000-12cdot500=4000$ dollars. Thus, his expected profit is more than
$$0.496cdot4000-0.504cdot2000=976$$
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.
answered Jan 15 '13 at 21:40
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
add a comment |
$begingroup$
In order to lose money, the operator must pay out more than $$10,000$, so there must be more than $20$ winning tickets. This is the case if and only if all four digits of the winning number are distinct, in which case there are $4!=24$ winning tickets. (If the winning number has three distinct digits, there are only $binom42cdot2=12$ winning tickets.)
There are $10cdot9cdot8cdot7$ ways to choose $4$ distinct digits, so the desired probability is indeed
$$frac{10cdot9cdot8cdot7}{10^4}=frac{10^{underline 4}}{10^4};.$$
(I prefer the notation $n^{underline k}$ to the notation $(n)_k$, which is presumably the Pochhammer symbol that you intended. Your $10^{(4)}$ generally denotes the rising factorial, $10cdot11cdot12cdot13$.)
Added: The probability of taking loss works out to $0.504$, and if he takes a loss, he loses $4cdot500=2000$ dollars. In all other cases there are at most $12$ winning tickets, so he gains at least $10,000-12cdot500=4000$ dollars. Thus, his expected profit is more than
$$0.496cdot4000-0.504cdot2000=976$$
dollars. With a bit more work one can of course calculate it exactly, but at least we know that the operator isn’t a complete idiot, even though he’s more likely to lose money than to make a profit.
answered Jan 15 '13 at 21:40
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
add a comment |
$begingroup$
In order to lose money, the operator must pay out more than $$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 $binom42cdot2=12$ winning tickets.)
There are $10cdot9cdot8cdot7$ ways to choose $4$ distinct digits, so the desired probability is indeed
$$frac{10cdot9cdot8cdot7}{10^4}=frac{10^{underline 4}}{10^4};.$$
(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, $10cdot11cdot12cdot13$.)
Added: The probability of taking loss works out to $0.504$, and if he takes a loss, he loses $4cdot500=2000$ dollars. In all other cases there are at most $12$ winning tickets, so he gains at least $10,000-12cdot500=4000$ dollars. Thus, his expected profit is more than
$$0.496cdot4000-0.504cdot2000=976$$
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.
answered Jan 15 '13 at 21:40
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
In order to lose money, the operator must pay out more than $$10,000$, so there must be more than $20$ winning tickets. This is the case if and only if all four digits of the winning number are distinct, in which case there are $4!=24$ winning tickets. (If the winning number has three distinct digits, there are only $binom42cdot2=12$ winning tickets.)
There are $10cdot9cdot8cdot7$ ways to choose $4$ distinct digits, so the desired probability is indeed
$$frac{10cdot9cdot8cdot7}{10^4}=frac{10^{underline 4}}{10^4};.$$
(I prefer the notation $n^{underline k}$ to the notation $(n)_k$, which is presumably the Pochhammer symbol that you intended. Your $10^{(4)}$ generally denotes the rising factorial, $10cdot11cdot12cdot13$.)
Added: The probability of taking loss works out to $0.504$, and if he takes a loss, he loses $4cdot500=2000$ dollars. In all other cases there are at most $12$ winning tickets, so he gains at least $10,000-12cdot500=4000$ dollars. Thus, his expected profit is more than
$$0.496cdot4000-0.504cdot2000=976$$
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.
answered Jan 15 '13 at 21:40
Brian M. ScottBrian M. Scott
460k40517918
edited Jan 15 '13 at 21:58
answered Jan 15 '13 at 21:40
Brian M. ScottBrian M. Scott
460k40517918
answered Jan 15 '13 at 21:40
Brian M. ScottBrian M. Scott
460k40517918
answered Jan 15 '13 at 21:40
Brian M. ScottBrian M. Scott
460k40517918
460k40517918
add a comment |
add a comment |
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$begingroup$
By $10^{(4)}$, do you mean $10 cdot 9 cdot 8 cdot 7$?
$endgroup$
– Ron Gordon
Jan 15 '13 at 21:38
$begingroup$
From here, can you see if the operator wins or loses on average?
$endgroup$
– Ross Millikan
Jan 15 '13 at 21:45