I have a probability of $frac{24}{800000}$ to win with one lottery ticket. What if I buy two tickets?












1












$begingroup$


Assuming there's a lottery with 800,000 tickets and 24 of these tickets contain a win, my chance to win (if I buy only one ticket) is $frac{24}{800,000}$ or $frac{1}{33,333.overline{3}}$, right?



But what are my chances to win if I buy two tickets?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    HInt: compute the probability of NOT winning.
    $endgroup$
    – leonbloy
    Dec 1 '16 at 16:19










  • $begingroup$
    $33000$ is a really poor approximation to $100000/3$.
    $endgroup$
    – TonyK
    Dec 1 '16 at 16:20












  • $begingroup$
    @TonyK: Oh yeah. Obviously it's 33,333.$overline{3}$
    $endgroup$
    – 4242
    Dec 1 '16 at 16:25
















1












$begingroup$


Assuming there's a lottery with 800,000 tickets and 24 of these tickets contain a win, my chance to win (if I buy only one ticket) is $frac{24}{800,000}$ or $frac{1}{33,333.overline{3}}$, right?



But what are my chances to win if I buy two tickets?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    HInt: compute the probability of NOT winning.
    $endgroup$
    – leonbloy
    Dec 1 '16 at 16:19










  • $begingroup$
    $33000$ is a really poor approximation to $100000/3$.
    $endgroup$
    – TonyK
    Dec 1 '16 at 16:20












  • $begingroup$
    @TonyK: Oh yeah. Obviously it's 33,333.$overline{3}$
    $endgroup$
    – 4242
    Dec 1 '16 at 16:25














1












1








1





$begingroup$


Assuming there's a lottery with 800,000 tickets and 24 of these tickets contain a win, my chance to win (if I buy only one ticket) is $frac{24}{800,000}$ or $frac{1}{33,333.overline{3}}$, right?



But what are my chances to win if I buy two tickets?






probability







share|cite|improve this question











$endgroup$




Assuming there's a lottery with 800,000 tickets and 24 of these tickets contain a win, my chance to win (if I buy only one ticket) is $frac{24}{800,000}$ or $frac{1}{33,333.overline{3}}$, right?



But what are my chances to win if I buy two tickets?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '16 at 16:33







4242

















asked Dec 1 '16 at 16:14









42424242

84




84












  • $begingroup$
    HInt: compute the probability of NOT winning.
    $endgroup$
    – leonbloy
    Dec 1 '16 at 16:19










  • $begingroup$
    $33000$ is a really poor approximation to $100000/3$.
    $endgroup$
    – TonyK
    Dec 1 '16 at 16:20












  • $begingroup$
    @TonyK: Oh yeah. Obviously it's 33,333.$overline{3}$
    $endgroup$
    – 4242
    Dec 1 '16 at 16:25


















  • $begingroup$
    HInt: compute the probability of NOT winning.
    $endgroup$
    – leonbloy
    Dec 1 '16 at 16:19










  • $begingroup$
    $33000$ is a really poor approximation to $100000/3$.
    $endgroup$
    – TonyK
    Dec 1 '16 at 16:20












  • $begingroup$
    @TonyK: Oh yeah. Obviously it's 33,333.$overline{3}$
    $endgroup$
    – 4242
    Dec 1 '16 at 16:25
















$begingroup$
HInt: compute the probability of NOT winning.
$endgroup$
– leonbloy
Dec 1 '16 at 16:19




$begingroup$
HInt: compute the probability of NOT winning.
$endgroup$
– leonbloy
Dec 1 '16 at 16:19












$begingroup$
$33000$ is a really poor approximation to $100000/3$.
$endgroup$
– TonyK
Dec 1 '16 at 16:20






$begingroup$
$33000$ is a really poor approximation to $100000/3$.
$endgroup$
– TonyK
Dec 1 '16 at 16:20














$begingroup$
@TonyK: Oh yeah. Obviously it's 33,333.$overline{3}$
$endgroup$
– 4242
Dec 1 '16 at 16:25




$begingroup$
@TonyK: Oh yeah. Obviously it's 33,333.$overline{3}$
$endgroup$
– 4242
Dec 1 '16 at 16:25










1 Answer
1






active

oldest

votes


















1












$begingroup$

Your chances to win at least once are roughly twice as high if you buy two tickets. Not exactly twice, because there is a very small chance both tickets will win, but this is small enough to ignore in an approximation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So that would mean the chance that I will win is now roughly $frac{1}{16,666}$? And if I buy 10 tickets, will it be $frac{1}{3,333}$? That doesn't sound right to me.
    $endgroup$
    – 4242
    Dec 1 '16 at 16:30










  • $begingroup$
    That is, roughly, correct. However, if you buy 10 tickets, the probability of more than one victory from those 10 becomes much larger, so it's actually significantly less than $frac{1}{3,333}$.
    $endgroup$
    – vadim123
    Dec 1 '16 at 16:35










  • $begingroup$
    @4242 the exact calculation can be done via routine counting methods. If we want the probability of not winning after having bought $k$ different lottery tickets it will be $frac{binom{800000-24}{k}}{binom{800000}{k}}$ where $binom{n}{r}$ is the binomial coefficient $frac{n!}{r!(n-r)!}$. The probability of winning at least once will be one minus that amount. For ten tickets that evaluates to approximately $0.0003$, similar to $1/3333$ but not exactly.
    $endgroup$
    – JMoravitz
    Dec 1 '16 at 16:44











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1 Answer
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active

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1












$begingroup$

Your chances to win at least once are roughly twice as high if you buy two tickets. Not exactly twice, because there is a very small chance both tickets will win, but this is small enough to ignore in an approximation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So that would mean the chance that I will win is now roughly $frac{1}{16,666}$? And if I buy 10 tickets, will it be $frac{1}{3,333}$? That doesn't sound right to me.
    $endgroup$
    – 4242
    Dec 1 '16 at 16:30










  • $begingroup$
    That is, roughly, correct. However, if you buy 10 tickets, the probability of more than one victory from those 10 becomes much larger, so it's actually significantly less than $frac{1}{3,333}$.
    $endgroup$
    – vadim123
    Dec 1 '16 at 16:35










  • $begingroup$
    @4242 the exact calculation can be done via routine counting methods. If we want the probability of not winning after having bought $k$ different lottery tickets it will be $frac{binom{800000-24}{k}}{binom{800000}{k}}$ where $binom{n}{r}$ is the binomial coefficient $frac{n!}{r!(n-r)!}$. The probability of winning at least once will be one minus that amount. For ten tickets that evaluates to approximately $0.0003$, similar to $1/3333$ but not exactly.
    $endgroup$
    – JMoravitz
    Dec 1 '16 at 16:44
















1












$begingroup$

Your chances to win at least once are roughly twice as high if you buy two tickets. Not exactly twice, because there is a very small chance both tickets will win, but this is small enough to ignore in an approximation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So that would mean the chance that I will win is now roughly $frac{1}{16,666}$? And if I buy 10 tickets, will it be $frac{1}{3,333}$? That doesn't sound right to me.
    $endgroup$
    – 4242
    Dec 1 '16 at 16:30










  • $begingroup$
    That is, roughly, correct. However, if you buy 10 tickets, the probability of more than one victory from those 10 becomes much larger, so it's actually significantly less than $frac{1}{3,333}$.
    $endgroup$
    – vadim123
    Dec 1 '16 at 16:35










  • $begingroup$
    @4242 the exact calculation can be done via routine counting methods. If we want the probability of not winning after having bought $k$ different lottery tickets it will be $frac{binom{800000-24}{k}}{binom{800000}{k}}$ where $binom{n}{r}$ is the binomial coefficient $frac{n!}{r!(n-r)!}$. The probability of winning at least once will be one minus that amount. For ten tickets that evaluates to approximately $0.0003$, similar to $1/3333$ but not exactly.
    $endgroup$
    – JMoravitz
    Dec 1 '16 at 16:44














1












1








1





$begingroup$

Your chances to win at least once are roughly twice as high if you buy two tickets. Not exactly twice, because there is a very small chance both tickets will win, but this is small enough to ignore in an approximation.






share|cite|improve this answer









$endgroup$



Your chances to win at least once are roughly twice as high if you buy two tickets. Not exactly twice, because there is a very small chance both tickets will win, but this is small enough to ignore in an approximation.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 '16 at 16:26









vadim123vadim123

76.4k897191




76.4k897191












  • $begingroup$
    So that would mean the chance that I will win is now roughly $frac{1}{16,666}$? And if I buy 10 tickets, will it be $frac{1}{3,333}$? That doesn't sound right to me.
    $endgroup$
    – 4242
    Dec 1 '16 at 16:30










  • $begingroup$
    That is, roughly, correct. However, if you buy 10 tickets, the probability of more than one victory from those 10 becomes much larger, so it's actually significantly less than $frac{1}{3,333}$.
    $endgroup$
    – vadim123
    Dec 1 '16 at 16:35










  • $begingroup$
    @4242 the exact calculation can be done via routine counting methods. If we want the probability of not winning after having bought $k$ different lottery tickets it will be $frac{binom{800000-24}{k}}{binom{800000}{k}}$ where $binom{n}{r}$ is the binomial coefficient $frac{n!}{r!(n-r)!}$. The probability of winning at least once will be one minus that amount. For ten tickets that evaluates to approximately $0.0003$, similar to $1/3333$ but not exactly.
    $endgroup$
    – JMoravitz
    Dec 1 '16 at 16:44


















  • $begingroup$
    So that would mean the chance that I will win is now roughly $frac{1}{16,666}$? And if I buy 10 tickets, will it be $frac{1}{3,333}$? That doesn't sound right to me.
    $endgroup$
    – 4242
    Dec 1 '16 at 16:30










  • $begingroup$
    That is, roughly, correct. However, if you buy 10 tickets, the probability of more than one victory from those 10 becomes much larger, so it's actually significantly less than $frac{1}{3,333}$.
    $endgroup$
    – vadim123
    Dec 1 '16 at 16:35










  • $begingroup$
    @4242 the exact calculation can be done via routine counting methods. If we want the probability of not winning after having bought $k$ different lottery tickets it will be $frac{binom{800000-24}{k}}{binom{800000}{k}}$ where $binom{n}{r}$ is the binomial coefficient $frac{n!}{r!(n-r)!}$. The probability of winning at least once will be one minus that amount. For ten tickets that evaluates to approximately $0.0003$, similar to $1/3333$ but not exactly.
    $endgroup$
    – JMoravitz
    Dec 1 '16 at 16:44
















$begingroup$
So that would mean the chance that I will win is now roughly $frac{1}{16,666}$? And if I buy 10 tickets, will it be $frac{1}{3,333}$? That doesn't sound right to me.
$endgroup$
– 4242
Dec 1 '16 at 16:30




$begingroup$
So that would mean the chance that I will win is now roughly $frac{1}{16,666}$? And if I buy 10 tickets, will it be $frac{1}{3,333}$? That doesn't sound right to me.
$endgroup$
– 4242
Dec 1 '16 at 16:30












$begingroup$
That is, roughly, correct. However, if you buy 10 tickets, the probability of more than one victory from those 10 becomes much larger, so it's actually significantly less than $frac{1}{3,333}$.
$endgroup$
– vadim123
Dec 1 '16 at 16:35




$begingroup$
That is, roughly, correct. However, if you buy 10 tickets, the probability of more than one victory from those 10 becomes much larger, so it's actually significantly less than $frac{1}{3,333}$.
$endgroup$
– vadim123
Dec 1 '16 at 16:35












$begingroup$
@4242 the exact calculation can be done via routine counting methods. If we want the probability of not winning after having bought $k$ different lottery tickets it will be $frac{binom{800000-24}{k}}{binom{800000}{k}}$ where $binom{n}{r}$ is the binomial coefficient $frac{n!}{r!(n-r)!}$. The probability of winning at least once will be one minus that amount. For ten tickets that evaluates to approximately $0.0003$, similar to $1/3333$ but not exactly.
$endgroup$
– JMoravitz
Dec 1 '16 at 16:44




$begingroup$
@4242 the exact calculation can be done via routine counting methods. If we want the probability of not winning after having bought $k$ different lottery tickets it will be $frac{binom{800000-24}{k}}{binom{800000}{k}}$ where $binom{n}{r}$ is the binomial coefficient $frac{n!}{r!(n-r)!}$. The probability of winning at least once will be one minus that amount. For ten tickets that evaluates to approximately $0.0003$, similar to $1/3333$ but not exactly.
$endgroup$
– JMoravitz
Dec 1 '16 at 16:44


















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