Mixing
Odds of winning lottery with multiple unique tickets
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If I buy one lottery ticket with odds of $14M$ to one if I buy another ticket with different numbers does this slash the odds to $7M$ to one? if so it follows that if I double my tickets again to $4$ Tickets it will half again to $3.5M$ to one?
So $8$ tickets will be $1.75M$ to one.
$16$ Tickets $rightarrow.875M$ to one
$32 rightarrow .4475$M to one
$64 rightarrow 218750$ to one
$128 rightarrow 109375$ to one
$256 rightarrow 54587$ to one
Something seems wrong here, i am not looking for a exact answer just ballpark working out.
probability probability-theory
asked May 19 '17 at 11:25
Pad ReganPad Regan
164
$endgroup$
add a comment |
$begingroup$
If I buy one lottery ticket with odds of $14M$ to one if I buy another ticket with different numbers does this slash the odds to $7M$ to one? if so it follows that if I double my tickets again to $4$ Tickets it will half again to $3.5M$ to one?
So $8$ tickets will be $1.75M$ to one.
$16$ Tickets $rightarrow.875M$ to one
$32 rightarrow .4475$M to one
$64 rightarrow 218750$ to one
$128 rightarrow 109375$ to one
$256 rightarrow 54587$ to one
Something seems wrong here, i am not looking for a exact answer just ballpark working out.
probability probability-theory
asked May 19 '17 at 11:25
Pad ReganPad Regan
164
$endgroup$
$begingroup$
Your final division by 2 is about a factor 10 too large.
$endgroup$
– TMM
May 19 '17 at 11:40
1
$begingroup$
There is nothing wrong here at all as long as you ensure that the tickets are different from each other. Your chance of winning goes up linearly with the number of tickets you buy. Your expected loss also goes up linearly with the number of tickets you buy.
$endgroup$
– Ross Millikan
Jul 14 '17 at 21:03
add a comment |
$begingroup$
If I buy one lottery ticket with odds of $14M$ to one if I buy another ticket with different numbers does this slash the odds to $7M$ to one? if so it follows that if I double my tickets again to $4$ Tickets it will half again to $3.5M$ to one?
So $8$ tickets will be $1.75M$ to one.
$16$ Tickets $rightarrow.875M$ to one
$32 rightarrow .4475$M to one
$64 rightarrow 218750$ to one
$128 rightarrow 109375$ to one
$256 rightarrow 54587$ to one
Something seems wrong here, i am not looking for a exact answer just ballpark working out.
probability probability-theory
asked May 19 '17 at 11:25
Pad ReganPad Regan
164
$endgroup$
If I buy one lottery ticket with odds of $14M$ to one if I buy another ticket with different numbers does this slash the odds to $7M$ to one? if so it follows that if I double my tickets again to $4$ Tickets it will half again to $3.5M$ to one?
So $8$ tickets will be $1.75M$ to one.
$16$ Tickets $rightarrow.875M$ to one
$32 rightarrow .4475$M to one
$64 rightarrow 218750$ to one
$128 rightarrow 109375$ to one
$256 rightarrow 54587$ to one
Something seems wrong here, i am not looking for a exact answer just ballpark working out.
probability probability-theory
probability probability-theory
asked May 19 '17 at 11:25
Pad ReganPad Regan
164
asked May 19 '17 at 11:25
Pad ReganPad Regan
164
edited May 19 '17 at 15:56
Pad Regan
asked May 19 '17 at 11:25
Pad ReganPad Regan
164
asked May 19 '17 at 11:25
Pad ReganPad Regan
164
asked May 19 '17 at 11:25
Pad ReganPad Regan
164
164
$begingroup$
Your final division by 2 is about a factor 10 too large.
$endgroup$
– TMM
May 19 '17 at 11:40
1
$begingroup$
There is nothing wrong here at all as long as you ensure that the tickets are different from each other. Your chance of winning goes up linearly with the number of tickets you buy. Your expected loss also goes up linearly with the number of tickets you buy.
$endgroup$
– Ross Millikan
Jul 14 '17 at 21:03
add a comment |
$begingroup$
Your final division by 2 is about a factor 10 too large.
$endgroup$
– TMM
May 19 '17 at 11:40
1
$begingroup$
There is nothing wrong here at all as long as you ensure that the tickets are different from each other. Your chance of winning goes up linearly with the number of tickets you buy. Your expected loss also goes up linearly with the number of tickets you buy.
$endgroup$
– Ross Millikan
Jul 14 '17 at 21:03
$begingroup$
Your final division by 2 is about a factor 10 too large.
$endgroup$
– TMM
May 19 '17 at 11:40
$begingroup$
Your final division by 2 is about a factor 10 too large.
$endgroup$
– TMM
May 19 '17 at 11:40
1
1
$begingroup$
There is nothing wrong here at all as long as you ensure that the tickets are different from each other. Your chance of winning goes up linearly with the number of tickets you buy. Your expected loss also goes up linearly with the number of tickets you buy.
$endgroup$
– Ross Millikan
Jul 14 '17 at 21:03
$begingroup$
There is nothing wrong here at all as long as you ensure that the tickets are different from each other. Your chance of winning goes up linearly with the number of tickets you buy. Your expected loss also goes up linearly with the number of tickets you buy.
$endgroup$
– Ross Millikan
Jul 14 '17 at 21:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Nearly but not quite. Start with a simpler case e.g. two dice and you are trying to get at least one 6. What are the odds? A common naive answer is 2/6. However, this is obviously wrong, consider 3, 4, 5, 6, and 7 dice. With 6 dice, getting at least one 6 is likely but certain. With 7 dice, it is not more than certain.
The correct answer can be seen by listing all 36 ways that two dice could fall. You will see that only 11 out of 36 contain a 6 so the odds are only 11/36. The overcounting is due to one of the 36 cases being double 6.
Back to the lottery, this approach is not so practical but fortunately there is an alternative. It is easier to calculate the chances that you don't win with $n$ tickets.
Don't win with 1 ticket: $frac{13999999}{14000000}$.
Don't win with 2 tickets: $(frac{13999999}{14000000})^2$.
Don't win with 3 tickets: $(frac{13999999}{14000000})^3$.
Don't win with n tickets: $(frac{13999999}{14000000})^n$.
Subtract from 1 to get the odds of winning.
With a very low probability such as $frac{1}{14000000}$, the first few will be very close to the naive answers. As $n$ gets bigger, the difference will become more obvious.
Thanks to comments by Losvos and TMM: note that I have accidentally assumed that each ticket is an independent random choice as would be typical with dice. If you are careful not to repeat your selections then the naive logic will be correct. Buy all possible 14000000 tickets and you are sure to win. However, it is very likely that you will win less than the cost of the tickets. Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot.
answered May 19 '17 at 11:36
badjohnbadjohn
4,2771620
$endgroup$
$begingroup$
This can't be right, what if you buy every ticket?
$endgroup$
– Bobson Dugnutt
May 19 '17 at 11:38
$begingroup$
@Lovsovs A good point. My answer is based on the dice case. I'll amend it.
$endgroup$
– badjohn
May 19 '17 at 11:39
$begingroup$
@TMM, Agreed, I have edited it due to Losvos's similar point.
$endgroup$
– badjohn
May 19 '17 at 11:46
$begingroup$
"Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot." - What do you mean? If I somehow mamage to buy all tickets for the next draw, am I not guaranteed the sole jackpot?
$endgroup$
– TMM
May 19 '17 at 11:51
$begingroup$
@TMM It depends on the rules of the specific lottery. Here in the UK at least, it is possible and common that two people buy a ticket with the same numbers. It is quite common that the jackpot is shared by 2 or more people. Quite early on, over 100 people shared the jackpot. I can imagine that many got very excited that they had won over £1000000. They may have calmed down and realised that it might be only £500000 or £333333 but they probably were not expecting just £80000. Nice but but as life changing as a million. I just hope that not too many resigned their jobs.
$endgroup$
– badjohn
May 19 '17 at 12:07
|
show 10 more comments
$begingroup$
The odds are in this case given by
$$text{number of tickets you have}:{text{number of tickets you don't have}}.$$
The important thing is that the denominator is not $text{total number of tickets}$. So if you have $2^n$ tickets and the total number of tickets is $T,$ the odds of you winning will be
$$2^n:T-2^n iff 1:frac{T}{2^n}-1,$$
which is different to your calculations. Let's take a look:
$n = 1$ gives odds of $ 1:frac{T}{2}-1$
$n =2$ gives odds of $1:frac{T}{4}-1,$
and so on, which grows much slower than your solution.
Edit: Here you can see a graph of your odds of winning as a function of the amount of $n$, the number of times you double your number of tickets. As you can see, it does actually grow exponentially, but it is off to an extremely slow start. At around $n=24$, you have bought all the tickets, so here the odds are $1:0,$ which gives the singularity shown.
So the bottom line is: Don't play the lottery! ;)
answered May 19 '17 at 11:44
Bobson DugnuttBobson Dugnutt
8,53831939
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add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Nearly but not quite. Start with a simpler case e.g. two dice and you are trying to get at least one 6. What are the odds? A common naive answer is 2/6. However, this is obviously wrong, consider 3, 4, 5, 6, and 7 dice. With 6 dice, getting at least one 6 is likely but certain. With 7 dice, it is not more than certain.
The correct answer can be seen by listing all 36 ways that two dice could fall. You will see that only 11 out of 36 contain a 6 so the odds are only 11/36. The overcounting is due to one of the 36 cases being double 6.
Back to the lottery, this approach is not so practical but fortunately there is an alternative. It is easier to calculate the chances that you don't win with $n$ tickets.
Don't win with 1 ticket: $frac{13999999}{14000000}$.
Don't win with 2 tickets: $(frac{13999999}{14000000})^2$.
Don't win with 3 tickets: $(frac{13999999}{14000000})^3$.
Don't win with n tickets: $(frac{13999999}{14000000})^n$.
Subtract from 1 to get the odds of winning.
With a very low probability such as $frac{1}{14000000}$, the first few will be very close to the naive answers. As $n$ gets bigger, the difference will become more obvious.
Thanks to comments by Losvos and TMM: note that I have accidentally assumed that each ticket is an independent random choice as would be typical with dice. If you are careful not to repeat your selections then the naive logic will be correct. Buy all possible 14000000 tickets and you are sure to win. However, it is very likely that you will win less than the cost of the tickets. Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot.
answered May 19 '17 at 11:36
badjohnbadjohn
4,2771620
$endgroup$
$begingroup$
This can't be right, what if you buy every ticket?
$endgroup$
– Bobson Dugnutt
May 19 '17 at 11:38
$begingroup$
@Lovsovs A good point. My answer is based on the dice case. I'll amend it.
$endgroup$
– badjohn
May 19 '17 at 11:39
$begingroup$
@TMM, Agreed, I have edited it due to Losvos's similar point.
$endgroup$
– badjohn
May 19 '17 at 11:46
$begingroup$
"Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot." - What do you mean? If I somehow mamage to buy all tickets for the next draw, am I not guaranteed the sole jackpot?
$endgroup$
– TMM
May 19 '17 at 11:51
$begingroup$
@TMM It depends on the rules of the specific lottery. Here in the UK at least, it is possible and common that two people buy a ticket with the same numbers. It is quite common that the jackpot is shared by 2 or more people. Quite early on, over 100 people shared the jackpot. I can imagine that many got very excited that they had won over £1000000. They may have calmed down and realised that it might be only £500000 or £333333 but they probably were not expecting just £80000. Nice but but as life changing as a million. I just hope that not too many resigned their jobs.
$endgroup$
– badjohn
May 19 '17 at 12:07
|
show 10 more comments
$begingroup$
Nearly but not quite. Start with a simpler case e.g. two dice and you are trying to get at least one 6. What are the odds? A common naive answer is 2/6. However, this is obviously wrong, consider 3, 4, 5, 6, and 7 dice. With 6 dice, getting at least one 6 is likely but certain. With 7 dice, it is not more than certain.
The correct answer can be seen by listing all 36 ways that two dice could fall. You will see that only 11 out of 36 contain a 6 so the odds are only 11/36. The overcounting is due to one of the 36 cases being double 6.
Back to the lottery, this approach is not so practical but fortunately there is an alternative. It is easier to calculate the chances that you don't win with $n$ tickets.
Don't win with 1 ticket: $frac{13999999}{14000000}$.
Don't win with 2 tickets: $(frac{13999999}{14000000})^2$.
Don't win with 3 tickets: $(frac{13999999}{14000000})^3$.
Don't win with n tickets: $(frac{13999999}{14000000})^n$.
Subtract from 1 to get the odds of winning.
With a very low probability such as $frac{1}{14000000}$, the first few will be very close to the naive answers. As $n$ gets bigger, the difference will become more obvious.
Thanks to comments by Losvos and TMM: note that I have accidentally assumed that each ticket is an independent random choice as would be typical with dice. If you are careful not to repeat your selections then the naive logic will be correct. Buy all possible 14000000 tickets and you are sure to win. However, it is very likely that you will win less than the cost of the tickets. Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot.
answered May 19 '17 at 11:36
badjohnbadjohn
4,2771620
$endgroup$
$begingroup$
This can't be right, what if you buy every ticket?
$endgroup$
– Bobson Dugnutt
May 19 '17 at 11:38
$begingroup$
@Lovsovs A good point. My answer is based on the dice case. I'll amend it.
$endgroup$
– badjohn
May 19 '17 at 11:39
$begingroup$
@TMM, Agreed, I have edited it due to Losvos's similar point.
$endgroup$
– badjohn
May 19 '17 at 11:46
$begingroup$
"Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot." - What do you mean? If I somehow mamage to buy all tickets for the next draw, am I not guaranteed the sole jackpot?
$endgroup$
– TMM
May 19 '17 at 11:51
$begingroup$
@TMM It depends on the rules of the specific lottery. Here in the UK at least, it is possible and common that two people buy a ticket with the same numbers. It is quite common that the jackpot is shared by 2 or more people. Quite early on, over 100 people shared the jackpot. I can imagine that many got very excited that they had won over £1000000. They may have calmed down and realised that it might be only £500000 or £333333 but they probably were not expecting just £80000. Nice but but as life changing as a million. I just hope that not too many resigned their jobs.
$endgroup$
– badjohn
May 19 '17 at 12:07
|
show 10 more comments
$begingroup$
Nearly but not quite. Start with a simpler case e.g. two dice and you are trying to get at least one 6. What are the odds? A common naive answer is 2/6. However, this is obviously wrong, consider 3, 4, 5, 6, and 7 dice. With 6 dice, getting at least one 6 is likely but certain. With 7 dice, it is not more than certain.
The correct answer can be seen by listing all 36 ways that two dice could fall. You will see that only 11 out of 36 contain a 6 so the odds are only 11/36. The overcounting is due to one of the 36 cases being double 6.
Back to the lottery, this approach is not so practical but fortunately there is an alternative. It is easier to calculate the chances that you don't win with $n$ tickets.
Don't win with 1 ticket: $frac{13999999}{14000000}$.
Don't win with 2 tickets: $(frac{13999999}{14000000})^2$.
Don't win with 3 tickets: $(frac{13999999}{14000000})^3$.
Don't win with n tickets: $(frac{13999999}{14000000})^n$.
Subtract from 1 to get the odds of winning.
With a very low probability such as $frac{1}{14000000}$, the first few will be very close to the naive answers. As $n$ gets bigger, the difference will become more obvious.
Thanks to comments by Losvos and TMM: note that I have accidentally assumed that each ticket is an independent random choice as would be typical with dice. If you are careful not to repeat your selections then the naive logic will be correct. Buy all possible 14000000 tickets and you are sure to win. However, it is very likely that you will win less than the cost of the tickets. Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot.
answered May 19 '17 at 11:36
badjohnbadjohn
4,2771620
$endgroup$
Nearly but not quite. Start with a simpler case e.g. two dice and you are trying to get at least one 6. What are the odds? A common naive answer is 2/6. However, this is obviously wrong, consider 3, 4, 5, 6, and 7 dice. With 6 dice, getting at least one 6 is likely but certain. With 7 dice, it is not more than certain.
The correct answer can be seen by listing all 36 ways that two dice could fall. You will see that only 11 out of 36 contain a 6 so the odds are only 11/36. The overcounting is due to one of the 36 cases being double 6.
Back to the lottery, this approach is not so practical but fortunately there is an alternative. It is easier to calculate the chances that you don't win with $n$ tickets.
Don't win with 1 ticket: $frac{13999999}{14000000}$.
Don't win with 2 tickets: $(frac{13999999}{14000000})^2$.
Don't win with 3 tickets: $(frac{13999999}{14000000})^3$.
Don't win with n tickets: $(frac{13999999}{14000000})^n$.
Subtract from 1 to get the odds of winning.
With a very low probability such as $frac{1}{14000000}$, the first few will be very close to the naive answers. As $n$ gets bigger, the difference will become more obvious.
Thanks to comments by Losvos and TMM: note that I have accidentally assumed that each ticket is an independent random choice as would be typical with dice. If you are careful not to repeat your selections then the naive logic will be correct. Buy all possible 14000000 tickets and you are sure to win. However, it is very likely that you will win less than the cost of the tickets. Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot.
answered May 19 '17 at 11:36
badjohnbadjohn
4,2771620
edited May 19 '17 at 11:47
answered May 19 '17 at 11:36
badjohnbadjohn
4,2771620
answered May 19 '17 at 11:36
badjohnbadjohn
4,2771620
answered May 19 '17 at 11:36
badjohnbadjohn
4,2771620
4,2771620
$begingroup$
This can't be right, what if you buy every ticket?
$endgroup$
– Bobson Dugnutt
May 19 '17 at 11:38
$begingroup$
@Lovsovs A good point. My answer is based on the dice case. I'll amend it.
$endgroup$
– badjohn
May 19 '17 at 11:39
$begingroup$
@TMM, Agreed, I have edited it due to Losvos's similar point.
$endgroup$
– badjohn
May 19 '17 at 11:46
$begingroup$
"Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot." - What do you mean? If I somehow mamage to buy all tickets for the next draw, am I not guaranteed the sole jackpot?
$endgroup$
– TMM
May 19 '17 at 11:51
$begingroup$
@TMM It depends on the rules of the specific lottery. Here in the UK at least, it is possible and common that two people buy a ticket with the same numbers. It is quite common that the jackpot is shared by 2 or more people. Quite early on, over 100 people shared the jackpot. I can imagine that many got very excited that they had won over £1000000. They may have calmed down and realised that it might be only £500000 or £333333 but they probably were not expecting just £80000. Nice but but as life changing as a million. I just hope that not too many resigned their jobs.
$endgroup$
– badjohn
May 19 '17 at 12:07
|
show 10 more comments
$begingroup$
This can't be right, what if you buy every ticket?
$endgroup$
– Bobson Dugnutt
May 19 '17 at 11:38
$begingroup$
@Lovsovs A good point. My answer is based on the dice case. I'll amend it.
$endgroup$
– badjohn
May 19 '17 at 11:39
$begingroup$
@TMM, Agreed, I have edited it due to Losvos's similar point.
$endgroup$
– badjohn
May 19 '17 at 11:46
$begingroup$
"Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot." - What do you mean? If I somehow mamage to buy all tickets for the next draw, am I not guaranteed the sole jackpot?
$endgroup$
– TMM
May 19 '17 at 11:51
$begingroup$
@TMM It depends on the rules of the specific lottery. Here in the UK at least, it is possible and common that two people buy a ticket with the same numbers. It is quite common that the jackpot is shared by 2 or more people. Quite early on, over 100 people shared the jackpot. I can imagine that many got very excited that they had won over £1000000. They may have calmed down and realised that it might be only £500000 or £333333 but they probably were not expecting just £80000. Nice but but as life changing as a million. I just hope that not too many resigned their jobs.
$endgroup$
– badjohn
May 19 '17 at 12:07
$begingroup$
This can't be right, what if you buy every ticket?
$endgroup$
– Bobson Dugnutt
May 19 '17 at 11:38
$begingroup$
This can't be right, what if you buy every ticket?
$endgroup$
– Bobson Dugnutt
May 19 '17 at 11:38
$begingroup$
@Lovsovs A good point. My answer is based on the dice case. I'll amend it.
$endgroup$
– badjohn
May 19 '17 at 11:39
$begingroup$
@Lovsovs A good point. My answer is based on the dice case. I'll amend it.
$endgroup$
– badjohn
May 19 '17 at 11:39
$begingroup$
@TMM, Agreed, I have edited it due to Losvos's similar point.
$endgroup$
– badjohn
May 19 '17 at 11:46
$begingroup$
@TMM, Agreed, I have edited it due to Losvos's similar point.
$endgroup$
– badjohn
May 19 '17 at 11:46
$begingroup$
"Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot." - What do you mean? If I somehow mamage to buy all tickets for the next draw, am I not guaranteed the sole jackpot?
$endgroup$
– TMM
May 19 '17 at 11:51
$begingroup$
"Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot." - What do you mean? If I somehow mamage to buy all tickets for the next draw, am I not guaranteed the sole jackpot?
$endgroup$
– TMM
May 19 '17 at 11:51
$begingroup$
@TMM It depends on the rules of the specific lottery. Here in the UK at least, it is possible and common that two people buy a ticket with the same numbers. It is quite common that the jackpot is shared by 2 or more people. Quite early on, over 100 people shared the jackpot. I can imagine that many got very excited that they had won over £1000000. They may have calmed down and realised that it might be only £500000 or £333333 but they probably were not expecting just £80000. Nice but but as life changing as a million. I just hope that not too many resigned their jobs.
$endgroup$
– badjohn
May 19 '17 at 12:07
$begingroup$
@TMM It depends on the rules of the specific lottery. Here in the UK at least, it is possible and common that two people buy a ticket with the same numbers. It is quite common that the jackpot is shared by 2 or more people. Quite early on, over 100 people shared the jackpot. I can imagine that many got very excited that they had won over £1000000. They may have calmed down and realised that it might be only £500000 or £333333 but they probably were not expecting just £80000. Nice but but as life changing as a million. I just hope that not too many resigned their jobs.
$endgroup$
– badjohn
May 19 '17 at 12:07
|
show 10 more comments
$begingroup$
The odds are in this case given by
$$text{number of tickets you have}:{text{number of tickets you don't have}}.$$
The important thing is that the denominator is not $text{total number of tickets}$. So if you have $2^n$ tickets and the total number of tickets is $T,$ the odds of you winning will be
$$2^n:T-2^n iff 1:frac{T}{2^n}-1,$$
which is different to your calculations. Let's take a look:
$n = 1$ gives odds of $ 1:frac{T}{2}-1$
$n =2$ gives odds of $1:frac{T}{4}-1,$
and so on, which grows much slower than your solution.
Edit: Here you can see a graph of your odds of winning as a function of the amount of $n$, the number of times you double your number of tickets. As you can see, it does actually grow exponentially, but it is off to an extremely slow start. At around $n=24$, you have bought all the tickets, so here the odds are $1:0,$ which gives the singularity shown.
So the bottom line is: Don't play the lottery! ;)
answered May 19 '17 at 11:44
Bobson DugnuttBobson Dugnutt
8,53831939
$endgroup$
add a comment |
$begingroup$
The odds are in this case given by
$$text{number of tickets you have}:{text{number of tickets you don't have}}.$$
The important thing is that the denominator is not $text{total number of tickets}$. So if you have $2^n$ tickets and the total number of tickets is $T,$ the odds of you winning will be
$$2^n:T-2^n iff 1:frac{T}{2^n}-1,$$
which is different to your calculations. Let's take a look:
$n = 1$ gives odds of $ 1:frac{T}{2}-1$
$n =2$ gives odds of $1:frac{T}{4}-1,$
and so on, which grows much slower than your solution.
Edit: Here you can see a graph of your odds of winning as a function of the amount of $n$, the number of times you double your number of tickets. As you can see, it does actually grow exponentially, but it is off to an extremely slow start. At around $n=24$, you have bought all the tickets, so here the odds are $1:0,$ which gives the singularity shown.
So the bottom line is: Don't play the lottery! ;)
answered May 19 '17 at 11:44
Bobson DugnuttBobson Dugnutt
8,53831939
$endgroup$
add a comment |
$begingroup$
The odds are in this case given by
$$text{number of tickets you have}:{text{number of tickets you don't have}}.$$
The important thing is that the denominator is not $text{total number of tickets}$. So if you have $2^n$ tickets and the total number of tickets is $T,$ the odds of you winning will be
$$2^n:T-2^n iff 1:frac{T}{2^n}-1,$$
which is different to your calculations. Let's take a look:
$n = 1$ gives odds of $ 1:frac{T}{2}-1$
$n =2$ gives odds of $1:frac{T}{4}-1,$
and so on, which grows much slower than your solution.
Edit: Here you can see a graph of your odds of winning as a function of the amount of $n$, the number of times you double your number of tickets. As you can see, it does actually grow exponentially, but it is off to an extremely slow start. At around $n=24$, you have bought all the tickets, so here the odds are $1:0,$ which gives the singularity shown.
So the bottom line is: Don't play the lottery! ;)
answered May 19 '17 at 11:44
Bobson DugnuttBobson Dugnutt
8,53831939
$endgroup$
The odds are in this case given by
$$text{number of tickets you have}:{text{number of tickets you don't have}}.$$
The important thing is that the denominator is not $text{total number of tickets}$. So if you have $2^n$ tickets and the total number of tickets is $T,$ the odds of you winning will be
$$2^n:T-2^n iff 1:frac{T}{2^n}-1,$$
which is different to your calculations. Let's take a look:
$n = 1$ gives odds of $ 1:frac{T}{2}-1$
$n =2$ gives odds of $1:frac{T}{4}-1,$
and so on, which grows much slower than your solution.
Edit: Here you can see a graph of your odds of winning as a function of the amount of $n$, the number of times you double your number of tickets. As you can see, it does actually grow exponentially, but it is off to an extremely slow start. At around $n=24$, you have bought all the tickets, so here the odds are $1:0,$ which gives the singularity shown.
So the bottom line is: Don't play the lottery! ;)
answered May 19 '17 at 11:44
Bobson DugnuttBobson Dugnutt
8,53831939
edited May 19 '17 at 12:00
answered May 19 '17 at 11:44
Bobson DugnuttBobson Dugnutt
8,53831939
answered May 19 '17 at 11:44
Bobson DugnuttBobson Dugnutt
8,53831939
answered May 19 '17 at 11:44
Bobson DugnuttBobson Dugnutt
8,53831939
8,53831939
add a comment |
add a comment |
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$begingroup$
Your final division by 2 is about a factor 10 too large.
$endgroup$
– TMM
May 19 '17 at 11:40
1
$begingroup$
There is nothing wrong here at all as long as you ensure that the tickets are different from each other. Your chance of winning goes up linearly with the number of tickets you buy. Your expected loss also goes up linearly with the number of tickets you buy.
$endgroup$
– Ross Millikan
Jul 14 '17 at 21:03