Chances of winning a raffle The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraProbability of winning a prize in a raffleDoes the order in which prizes are allocated to raffle tickets matter?Chances of winning a raffle when winning tickets are returned to bucket each time.Probability of winning a prize in a raffle (that each person can only win once)Chances of winning a raffle?Probability of winning a prize in a raffle (each person can only win once)percentage of winning this kind of raffleRaffle percentagesPercentage chance of bringing home a raffle prizeFinding odds of a prize draw
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Chances of winning a raffle
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraProbability of winning a prize in a raffleDoes the order in which prizes are allocated to raffle tickets matter?Chances of winning a raffle when winning tickets are returned to bucket each time.Probability of winning a prize in a raffle (that each person can only win once)Chances of winning a raffle?Probability of winning a prize in a raffle (each person can only win once)percentage of winning this kind of raffleRaffle percentagesPercentage chance of bringing home a raffle prizeFinding odds of a prize draw
$begingroup$
If there are 75 tickets and one 1st prize the chance is 1/75 x 100 in percentage terms. If the number of tickets doubles to 150 but there are now two 1st prizes (first 2 tickets drawn) do the odds of winning a 1st prize increase, decrease or stay the same?
I think the maths is 1/150 x 100 = 0.66666 recurring
Plus
1/149 x 100 = 0.67114
= 1.33781
Compared with
1/75 x 100 = 0.33333 recurring.
So yes better odds. Do I have my maths right?
probability percentages
asked Apr 8 at 8:14
Charlie WestropeCharlie Westrope
1
New contributor
Charlie Westrope is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If there are 75 tickets and one 1st prize the chance is 1/75 x 100 in percentage terms. If the number of tickets doubles to 150 but there are now two 1st prizes (first 2 tickets drawn) do the odds of winning a 1st prize increase, decrease or stay the same?
I think the maths is 1/150 x 100 = 0.66666 recurring
Plus
1/149 x 100 = 0.67114
= 1.33781
Compared with
1/75 x 100 = 0.33333 recurring.
So yes better odds. Do I have my maths right?
probability percentages
asked Apr 8 at 8:14
Charlie WestropeCharlie Westrope
1
New contributor
Charlie Westrope is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
What is mean by "first 2 tickets drawn"?
$endgroup$
– drhab
Apr 8 at 8:39
add a comment |
$begingroup$
If there are 75 tickets and one 1st prize the chance is 1/75 x 100 in percentage terms. If the number of tickets doubles to 150 but there are now two 1st prizes (first 2 tickets drawn) do the odds of winning a 1st prize increase, decrease or stay the same?
I think the maths is 1/150 x 100 = 0.66666 recurring
Plus
1/149 x 100 = 0.67114
= 1.33781
Compared with
1/75 x 100 = 0.33333 recurring.
So yes better odds. Do I have my maths right?
probability percentages
asked Apr 8 at 8:14
Charlie WestropeCharlie Westrope
1
New contributor
Charlie Westrope is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If there are 75 tickets and one 1st prize the chance is 1/75 x 100 in percentage terms. If the number of tickets doubles to 150 but there are now two 1st prizes (first 2 tickets drawn) do the odds of winning a 1st prize increase, decrease or stay the same?
I think the maths is 1/150 x 100 = 0.66666 recurring
Plus
1/149 x 100 = 0.67114
= 1.33781
Compared with
1/75 x 100 = 0.33333 recurring.
So yes better odds. Do I have my maths right?
probability percentages
probability percentages
asked Apr 8 at 8:14
Charlie WestropeCharlie Westrope
1
New contributor
Charlie Westrope is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 8:14
Charlie WestropeCharlie Westrope
1
New contributor
Charlie Westrope is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 8:14
Charlie WestropeCharlie Westrope
1
New contributor
Charlie Westrope is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 8:14
Charlie WestropeCharlie Westrope
1
asked Apr 8 at 8:14
Charlie WestropeCharlie Westrope
1
1
New contributor
Charlie Westrope is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Charlie Westrope is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Charlie Westrope is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
What is mean by "first 2 tickets drawn"?
$endgroup$
– drhab
Apr 8 at 8:39
add a comment |
1
$begingroup$
What is mean by "first 2 tickets drawn"?
$endgroup$
– drhab
Apr 8 at 8:39
1
1
$begingroup$
What is mean by "first 2 tickets drawn"?
$endgroup$
– drhab
Apr 8 at 8:39
$begingroup$
What is mean by "first 2 tickets drawn"?
$endgroup$
– drhab
Apr 8 at 8:39
add a comment |
1 Answer
1
active
oldest
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$begingroup$
The probability of winning in the first setting is $frac175$, that is true.
As I understand the game, you cannot win both prices in the second setting. The probability of having the first ticket that is drawn is
$$p(textfirst ticket) = frac1150$$
The probability of being the person with the second ticket drawn is
$$p(textsecond ticket) = frac149150 cdot frac1149 = frac1150$$
because you must be one of the 149 persons not having the first ticket and the one person winning when the second ticket is drawn.
Putting it together you have a probability
$$ frac1150 + frac1150 = frac175$$
so your chances stay the same.
You could also think of it like this: The probability of not winning is
$$p(textneither the first nor the second ticket) = frac149150 cdot frac148149 $$
so the probability of winning becomes
$$ 1 - frac149150 cdot frac148149 = 1 - frac7475 = frac175. $$
answered Apr 8 at 8:34
Fritz HefterFritz Hefter
264
$endgroup$
$begingroup$
Many thanks. The key question is if I have one ticket out of 75 with 1 prize are my chances the same with one ticket out of 150 and two first prize draws. I think I understand the answer to be the chances are identical. Many thanks.
$endgroup$
– Charlie Westrope
Apr 8 at 8:48
$begingroup$
@CharlieWestrope If you buy one ticket from $n$ in total and among the $n$ tickets there are $k$ that are linked with a prize then the probability on winning a prize is $frackn$. Every ticket with a prize has probability $frac1n$ to become the ticket you buy. So the total probability that one of those tickets will become the ticket you buy is $frac1n+frac1n+cdots+frac1n=frackn$.
$endgroup$
– drhab
Apr 8 at 10:14
$begingroup$
@drhab Well... I interpreted it like this: They draw a ticket. The probability that it is yours is $frac1150$. Then they draw another one. This leads to the argument I gave.
$endgroup$
– Fritz Hefter
Apr 8 at 11:40
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
The probability of winning in the first setting is $frac175$, that is true.
As I understand the game, you cannot win both prices in the second setting. The probability of having the first ticket that is drawn is
$$p(textfirst ticket) = frac1150$$
The probability of being the person with the second ticket drawn is
$$p(textsecond ticket) = frac149150 cdot frac1149 = frac1150$$
because you must be one of the 149 persons not having the first ticket and the one person winning when the second ticket is drawn.
Putting it together you have a probability
$$ frac1150 + frac1150 = frac175$$
so your chances stay the same.
You could also think of it like this: The probability of not winning is
$$p(textneither the first nor the second ticket) = frac149150 cdot frac148149 $$
so the probability of winning becomes
$$ 1 - frac149150 cdot frac148149 = 1 - frac7475 = frac175. $$
answered Apr 8 at 8:34
Fritz HefterFritz Hefter
264
$endgroup$
$begingroup$
Many thanks. The key question is if I have one ticket out of 75 with 1 prize are my chances the same with one ticket out of 150 and two first prize draws. I think I understand the answer to be the chances are identical. Many thanks.
$endgroup$
– Charlie Westrope
Apr 8 at 8:48
$begingroup$
@CharlieWestrope If you buy one ticket from $n$ in total and among the $n$ tickets there are $k$ that are linked with a prize then the probability on winning a prize is $frackn$. Every ticket with a prize has probability $frac1n$ to become the ticket you buy. So the total probability that one of those tickets will become the ticket you buy is $frac1n+frac1n+cdots+frac1n=frackn$.
$endgroup$
– drhab
Apr 8 at 10:14
$begingroup$
@drhab Well... I interpreted it like this: They draw a ticket. The probability that it is yours is $frac1150$. Then they draw another one. This leads to the argument I gave.
$endgroup$
– Fritz Hefter
Apr 8 at 11:40
add a comment |
$begingroup$
The probability of winning in the first setting is $frac175$, that is true.
As I understand the game, you cannot win both prices in the second setting. The probability of having the first ticket that is drawn is
$$p(textfirst ticket) = frac1150$$
The probability of being the person with the second ticket drawn is
$$p(textsecond ticket) = frac149150 cdot frac1149 = frac1150$$
because you must be one of the 149 persons not having the first ticket and the one person winning when the second ticket is drawn.
Putting it together you have a probability
$$ frac1150 + frac1150 = frac175$$
so your chances stay the same.
You could also think of it like this: The probability of not winning is
$$p(textneither the first nor the second ticket) = frac149150 cdot frac148149 $$
so the probability of winning becomes
$$ 1 - frac149150 cdot frac148149 = 1 - frac7475 = frac175. $$
answered Apr 8 at 8:34
Fritz HefterFritz Hefter
264
$endgroup$
$begingroup$
Many thanks. The key question is if I have one ticket out of 75 with 1 prize are my chances the same with one ticket out of 150 and two first prize draws. I think I understand the answer to be the chances are identical. Many thanks.
$endgroup$
– Charlie Westrope
Apr 8 at 8:48
$begingroup$
@CharlieWestrope If you buy one ticket from $n$ in total and among the $n$ tickets there are $k$ that are linked with a prize then the probability on winning a prize is $frackn$. Every ticket with a prize has probability $frac1n$ to become the ticket you buy. So the total probability that one of those tickets will become the ticket you buy is $frac1n+frac1n+cdots+frac1n=frackn$.
$endgroup$
– drhab
Apr 8 at 10:14
$begingroup$
@drhab Well... I interpreted it like this: They draw a ticket. The probability that it is yours is $frac1150$. Then they draw another one. This leads to the argument I gave.
$endgroup$
– Fritz Hefter
Apr 8 at 11:40
add a comment |
$begingroup$
The probability of winning in the first setting is $frac175$, that is true.
As I understand the game, you cannot win both prices in the second setting. The probability of having the first ticket that is drawn is
$$p(textfirst ticket) = frac1150$$
The probability of being the person with the second ticket drawn is
$$p(textsecond ticket) = frac149150 cdot frac1149 = frac1150$$
because you must be one of the 149 persons not having the first ticket and the one person winning when the second ticket is drawn.
Putting it together you have a probability
$$ frac1150 + frac1150 = frac175$$
so your chances stay the same.
You could also think of it like this: The probability of not winning is
$$p(textneither the first nor the second ticket) = frac149150 cdot frac148149 $$
so the probability of winning becomes
$$ 1 - frac149150 cdot frac148149 = 1 - frac7475 = frac175. $$
answered Apr 8 at 8:34
Fritz HefterFritz Hefter
264
$endgroup$
The probability of winning in the first setting is $frac175$, that is true.
As I understand the game, you cannot win both prices in the second setting. The probability of having the first ticket that is drawn is
$$p(textfirst ticket) = frac1150$$
The probability of being the person with the second ticket drawn is
$$p(textsecond ticket) = frac149150 cdot frac1149 = frac1150$$
because you must be one of the 149 persons not having the first ticket and the one person winning when the second ticket is drawn.
Putting it together you have a probability
$$ frac1150 + frac1150 = frac175$$
so your chances stay the same.
You could also think of it like this: The probability of not winning is
$$p(textneither the first nor the second ticket) = frac149150 cdot frac148149 $$
so the probability of winning becomes
$$ 1 - frac149150 cdot frac148149 = 1 - frac7475 = frac175. $$
answered Apr 8 at 8:34
Fritz HefterFritz Hefter
264
answered Apr 8 at 8:34
Fritz HefterFritz Hefter
264
answered Apr 8 at 8:34
Fritz HefterFritz Hefter
264
answered Apr 8 at 8:34
Fritz HefterFritz Hefter
264
264
$begingroup$
Many thanks. The key question is if I have one ticket out of 75 with 1 prize are my chances the same with one ticket out of 150 and two first prize draws. I think I understand the answer to be the chances are identical. Many thanks.
$endgroup$
– Charlie Westrope
Apr 8 at 8:48
$begingroup$
@CharlieWestrope If you buy one ticket from $n$ in total and among the $n$ tickets there are $k$ that are linked with a prize then the probability on winning a prize is $frackn$. Every ticket with a prize has probability $frac1n$ to become the ticket you buy. So the total probability that one of those tickets will become the ticket you buy is $frac1n+frac1n+cdots+frac1n=frackn$.
$endgroup$
– drhab
Apr 8 at 10:14
$begingroup$
@drhab Well... I interpreted it like this: They draw a ticket. The probability that it is yours is $frac1150$. Then they draw another one. This leads to the argument I gave.
$endgroup$
– Fritz Hefter
Apr 8 at 11:40
add a comment |
$begingroup$
Many thanks. The key question is if I have one ticket out of 75 with 1 prize are my chances the same with one ticket out of 150 and two first prize draws. I think I understand the answer to be the chances are identical. Many thanks.
$endgroup$
– Charlie Westrope
Apr 8 at 8:48
$begingroup$
@CharlieWestrope If you buy one ticket from $n$ in total and among the $n$ tickets there are $k$ that are linked with a prize then the probability on winning a prize is $frackn$. Every ticket with a prize has probability $frac1n$ to become the ticket you buy. So the total probability that one of those tickets will become the ticket you buy is $frac1n+frac1n+cdots+frac1n=frackn$.
$endgroup$
– drhab
Apr 8 at 10:14
$begingroup$
@drhab Well... I interpreted it like this: They draw a ticket. The probability that it is yours is $frac1150$. Then they draw another one. This leads to the argument I gave.
$endgroup$
– Fritz Hefter
Apr 8 at 11:40
$begingroup$
Many thanks. The key question is if I have one ticket out of 75 with 1 prize are my chances the same with one ticket out of 150 and two first prize draws. I think I understand the answer to be the chances are identical. Many thanks.
$endgroup$
– Charlie Westrope
Apr 8 at 8:48
$begingroup$
Many thanks. The key question is if I have one ticket out of 75 with 1 prize are my chances the same with one ticket out of 150 and two first prize draws. I think I understand the answer to be the chances are identical. Many thanks.
$endgroup$
– Charlie Westrope
Apr 8 at 8:48
$begingroup$
@CharlieWestrope If you buy one ticket from $n$ in total and among the $n$ tickets there are $k$ that are linked with a prize then the probability on winning a prize is $frackn$. Every ticket with a prize has probability $frac1n$ to become the ticket you buy. So the total probability that one of those tickets will become the ticket you buy is $frac1n+frac1n+cdots+frac1n=frackn$.
$endgroup$
– drhab
Apr 8 at 10:14
$begingroup$
@CharlieWestrope If you buy one ticket from $n$ in total and among the $n$ tickets there are $k$ that are linked with a prize then the probability on winning a prize is $frackn$. Every ticket with a prize has probability $frac1n$ to become the ticket you buy. So the total probability that one of those tickets will become the ticket you buy is $frac1n+frac1n+cdots+frac1n=frackn$.
$endgroup$
– drhab
Apr 8 at 10:14
$begingroup$
@drhab Well... I interpreted it like this: They draw a ticket. The probability that it is yours is $frac1150$. Then they draw another one. This leads to the argument I gave.
$endgroup$
– Fritz Hefter
Apr 8 at 11:40
$begingroup$
@drhab Well... I interpreted it like this: They draw a ticket. The probability that it is yours is $frac1150$. Then they draw another one. This leads to the argument I gave.
$endgroup$
– Fritz Hefter
Apr 8 at 11:40
add a comment |
Charlie Westrope is a new contributor. Be nice, and check out our Code of Conduct.
Charlie Westrope is a new contributor. Be nice, and check out our Code of Conduct.
Charlie Westrope is a new contributor. Be nice, and check out our Code of Conduct.
Charlie Westrope is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
What is mean by "first 2 tickets drawn"?
$endgroup$
– drhab
Apr 8 at 8:39