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Calculate lottery's second prize using combination - lottery probability question.
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up vote
2
down vote
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Assume a lottery game of the following rules:
Picking your numbers:
- Pick a total of 6 different numbers from the lot containing 42 numbers (1 to 42).
Draw:
- Draw 7 balls, with no repetition(i.e: ball drawn is not put back in the lot) from the lot labeled from 1 to 42.
Results:
- If the first 6 balls drawn matches your own 6 numbers (order doesn't matter): Jackpot.
- If 5 of the first 6 balls drawn matches 5 of your numbers (order doesn't matter) and the 7th drawn ball matches your 6th number: second prize.
- If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize.
I'll end it here for not having many other prizes.
If I want to check my chance of winning the jackpot, it's pretty straightforward and looks like a combination $C(42,6)$, so it should be:
$$
frac42cdot41cdot40cdot39cdot38cdot376! = 5,245,786.
$$
So my chance of getting the jackpot is $(frac15,245,786)$
For the third prize it's also a straightforward combination $C(42,5)$, it's equal to:
$$
frac42cdot41cdot40cdot39cdot385! = 850,668.
$$
So third prize probability is equal to $left(frac1850,668right)$
Now I am being stumbled on how to calculate the 2nd prize probability. My memories from school are not helping me enough to get my answer. I know that it should be between the two numbers I got there, however any calculations I am making end ups with a probability much higher than the first prize's.
Could you please verify that my 1st and 3rd prize probabilities are well calculated and help me calculate the 2nd prize probability?
probability combinations lotteries
edited 2 hours ago
amWhy
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asked 3 hours ago
Paul Karam
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up vote
2
down vote
favorite
Assume a lottery game of the following rules:
Picking your numbers:
- Pick a total of 6 different numbers from the lot containing 42 numbers (1 to 42).
Draw:
- Draw 7 balls, with no repetition(i.e: ball drawn is not put back in the lot) from the lot labeled from 1 to 42.
Results:
- If the first 6 balls drawn matches your own 6 numbers (order doesn't matter): Jackpot.
- If 5 of the first 6 balls drawn matches 5 of your numbers (order doesn't matter) and the 7th drawn ball matches your 6th number: second prize.
- If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize.
I'll end it here for not having many other prizes.
If I want to check my chance of winning the jackpot, it's pretty straightforward and looks like a combination $C(42,6)$, so it should be:
$$
frac42cdot41cdot40cdot39cdot38cdot376! = 5,245,786.
$$
So my chance of getting the jackpot is $(frac15,245,786)$
For the third prize it's also a straightforward combination $C(42,5)$, it's equal to:
$$
frac42cdot41cdot40cdot39cdot385! = 850,668.
$$
So third prize probability is equal to $left(frac1850,668right)$
Now I am being stumbled on how to calculate the 2nd prize probability. My memories from school are not helping me enough to get my answer. I know that it should be between the two numbers I got there, however any calculations I am making end ups with a probability much higher than the first prize's.
Could you please verify that my 1st and 3rd prize probabilities are well calculated and help me calculate the 2nd prize probability?
probability combinations lotteries
edited 2 hours ago
amWhy
191k27223436
asked 3 hours ago
Paul Karam
1135
New contributor
Paul Karam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Assume a lottery game of the following rules:
Picking your numbers:
- Pick a total of 6 different numbers from the lot containing 42 numbers (1 to 42).
Draw:
- Draw 7 balls, with no repetition(i.e: ball drawn is not put back in the lot) from the lot labeled from 1 to 42.
Results:
- If the first 6 balls drawn matches your own 6 numbers (order doesn't matter): Jackpot.
- If 5 of the first 6 balls drawn matches 5 of your numbers (order doesn't matter) and the 7th drawn ball matches your 6th number: second prize.
- If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize.
I'll end it here for not having many other prizes.
If I want to check my chance of winning the jackpot, it's pretty straightforward and looks like a combination $C(42,6)$, so it should be:
$$
frac42cdot41cdot40cdot39cdot38cdot376! = 5,245,786.
$$
So my chance of getting the jackpot is $(frac15,245,786)$
For the third prize it's also a straightforward combination $C(42,5)$, it's equal to:
$$
frac42cdot41cdot40cdot39cdot385! = 850,668.
$$
So third prize probability is equal to $left(frac1850,668right)$
Now I am being stumbled on how to calculate the 2nd prize probability. My memories from school are not helping me enough to get my answer. I know that it should be between the two numbers I got there, however any calculations I am making end ups with a probability much higher than the first prize's.
Could you please verify that my 1st and 3rd prize probabilities are well calculated and help me calculate the 2nd prize probability?
probability combinations lotteries
edited 2 hours ago
amWhy
191k27223436
asked 3 hours ago
Paul Karam
1135
New contributor
Paul Karam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Assume a lottery game of the following rules:
Picking your numbers:
- Pick a total of 6 different numbers from the lot containing 42 numbers (1 to 42).
Draw:
- Draw 7 balls, with no repetition(i.e: ball drawn is not put back in the lot) from the lot labeled from 1 to 42.
Results:
- If the first 6 balls drawn matches your own 6 numbers (order doesn't matter): Jackpot.
- If 5 of the first 6 balls drawn matches 5 of your numbers (order doesn't matter) and the 7th drawn ball matches your 6th number: second prize.
- If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize.
I'll end it here for not having many other prizes.
If I want to check my chance of winning the jackpot, it's pretty straightforward and looks like a combination $C(42,6)$, so it should be:
$$
frac42cdot41cdot40cdot39cdot38cdot376! = 5,245,786.
$$
So my chance of getting the jackpot is $(frac15,245,786)$
For the third prize it's also a straightforward combination $C(42,5)$, it's equal to:
$$
frac42cdot41cdot40cdot39cdot385! = 850,668.
$$
So third prize probability is equal to $left(frac1850,668right)$
Now I am being stumbled on how to calculate the 2nd prize probability. My memories from school are not helping me enough to get my answer. I know that it should be between the two numbers I got there, however any calculations I am making end ups with a probability much higher than the first prize's.
Could you please verify that my 1st and 3rd prize probabilities are well calculated and help me calculate the 2nd prize probability?
probability combinations lotteries
probability combinations lotteries
edited 2 hours ago
amWhy
191k27223436
asked 3 hours ago
Paul Karam
1135
New contributor
Paul Karam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
amWhy
191k27223436
asked 3 hours ago
Paul Karam
1135
New contributor
Paul Karam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
amWhy
191k27223436
edited 2 hours ago
amWhy
191k27223436
edited 2 hours ago
amWhy
191k27223436
191k27223436
asked 3 hours ago
Paul Karam
1135
New contributor
Paul Karam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Paul Karam
1135
asked 3 hours ago
Paul Karam
1135
1135
New contributor
Paul Karam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Paul Karam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Paul Karam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
Your logic for case number (3) does not seem correct.
First, you have to draw 6 balls with exactly 5 balls matching 5 numbers from your set of 6 picked numbers. In total There are C(42,6) ways to draw six balls. How many winning combinations do we have? From your set of 6 balls you can pick five matching balls in C(6,5) ways. The sixth number can be any number from the remaining 42-5=37, minus one (your sixth pick). So the total number of winning combinations is $C(6,5)times36$. So the probability is:
$$p_1=frac36times6choose 542choose 6$$
But you have to draw the seventh number as well from the remaining 42-6=36 numbers without hitting your last (sixth) number. That probability is:
$$p_2=frac3536$$
The total proability is:
$$p=p_1times p_2=frac36times6choose 542choose 6timesfrac3536=frac35times6choose 542choose 6=frac15374699$$
You can use a similar logic for case (2).
The probability $p_1$ is the same. For the seventh ball we have 36 choices and there is 1 winning ball between them. So the probability $p_2$ is:
$$p_2=frac136$$
...and the final probability for the second prize is:
$$p=p_1times p_2=frac36times6choose 542choose 6timesfrac136=frac6choose 542choose 6=frac32622893$$
(35 times smaller than the probability for the third prize)
answered 2 hours ago
Oldboy
4,0281325
Great! This explains where most of my mistakes where. Thank you for clarifying.
– Paul Karam
1 hour ago
add a comment |Â
up vote
3
down vote
For $3^rd$ prize, you need to select which 5 numbered balls you will be getting, out of 6 in $^6C_5$ ways (get balls matching to those 5 numbers from 42); and then, select the $6^th$ ad $7^th$ ball in $^36C_2$ ways
(42-6=36, balls not matching any of the 6 numbers)
So, probability of winning $3^rd$ prize is $$frac^6C_5cdot^36C_2^42C_7$$
For $2^nd$ prize,
Among first 6, only 5 match, selected their corresponding numbers in $^6C_5$ ways.
First 5 balls drawn match these 5 numbers.
Now, $6^th$ ball is selected from 36 (42-6=36, it does not match any of 6 numbers); and number corresponding to last ball to be drawn is already known(one left from 6)
So, probability of winning $2^nd$ prize is: $$frac^6C_5cdot^36C_1cdot^1C_1^42C_7$$
Similarly, For $1^st$ prize, $$frac^6C_6cdot^36C_1^42C_7$$
answered 1 hour ago
omega
9032818
2
I think that your calculation for the third prize is also not correct. You are ignoring the fact that you have to draw the seventh ball as well and that ball must not hit your last pick. Just read the condition carefully: If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize. "Nothing else" must include the seventh ball as well.
– Oldboy
1 hour ago
yup, now i guess its correct.
– omega
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Your logic for case number (3) does not seem correct.
First, you have to draw 6 balls with exactly 5 balls matching 5 numbers from your set of 6 picked numbers. In total There are C(42,6) ways to draw six balls. How many winning combinations do we have? From your set of 6 balls you can pick five matching balls in C(6,5) ways. The sixth number can be any number from the remaining 42-5=37, minus one (your sixth pick). So the total number of winning combinations is $C(6,5)times36$. So the probability is:
$$p_1=frac36times6choose 542choose 6$$
But you have to draw the seventh number as well from the remaining 42-6=36 numbers without hitting your last (sixth) number. That probability is:
$$p_2=frac3536$$
The total proability is:
$$p=p_1times p_2=frac36times6choose 542choose 6timesfrac3536=frac35times6choose 542choose 6=frac15374699$$
You can use a similar logic for case (2).
The probability $p_1$ is the same. For the seventh ball we have 36 choices and there is 1 winning ball between them. So the probability $p_2$ is:
$$p_2=frac136$$
...and the final probability for the second prize is:
$$p=p_1times p_2=frac36times6choose 542choose 6timesfrac136=frac6choose 542choose 6=frac32622893$$
(35 times smaller than the probability for the third prize)
answered 2 hours ago
Oldboy
4,0281325
Great! This explains where most of my mistakes where. Thank you for clarifying.
– Paul Karam
1 hour ago
add a comment |Â
up vote
5
down vote
accepted
Your logic for case number (3) does not seem correct.
First, you have to draw 6 balls with exactly 5 balls matching 5 numbers from your set of 6 picked numbers. In total There are C(42,6) ways to draw six balls. How many winning combinations do we have? From your set of 6 balls you can pick five matching balls in C(6,5) ways. The sixth number can be any number from the remaining 42-5=37, minus one (your sixth pick). So the total number of winning combinations is $C(6,5)times36$. So the probability is:
$$p_1=frac36times6choose 542choose 6$$
But you have to draw the seventh number as well from the remaining 42-6=36 numbers without hitting your last (sixth) number. That probability is:
$$p_2=frac3536$$
The total proability is:
$$p=p_1times p_2=frac36times6choose 542choose 6timesfrac3536=frac35times6choose 542choose 6=frac15374699$$
You can use a similar logic for case (2).
The probability $p_1$ is the same. For the seventh ball we have 36 choices and there is 1 winning ball between them. So the probability $p_2$ is:
$$p_2=frac136$$
...and the final probability for the second prize is:
$$p=p_1times p_2=frac36times6choose 542choose 6timesfrac136=frac6choose 542choose 6=frac32622893$$
(35 times smaller than the probability for the third prize)
answered 2 hours ago
Oldboy
4,0281325
Great! This explains where most of my mistakes where. Thank you for clarifying.
– Paul Karam
1 hour ago
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Your logic for case number (3) does not seem correct.
First, you have to draw 6 balls with exactly 5 balls matching 5 numbers from your set of 6 picked numbers. In total There are C(42,6) ways to draw six balls. How many winning combinations do we have? From your set of 6 balls you can pick five matching balls in C(6,5) ways. The sixth number can be any number from the remaining 42-5=37, minus one (your sixth pick). So the total number of winning combinations is $C(6,5)times36$. So the probability is:
$$p_1=frac36times6choose 542choose 6$$
But you have to draw the seventh number as well from the remaining 42-6=36 numbers without hitting your last (sixth) number. That probability is:
$$p_2=frac3536$$
The total proability is:
$$p=p_1times p_2=frac36times6choose 542choose 6timesfrac3536=frac35times6choose 542choose 6=frac15374699$$
You can use a similar logic for case (2).
The probability $p_1$ is the same. For the seventh ball we have 36 choices and there is 1 winning ball between them. So the probability $p_2$ is:
$$p_2=frac136$$
...and the final probability for the second prize is:
$$p=p_1times p_2=frac36times6choose 542choose 6timesfrac136=frac6choose 542choose 6=frac32622893$$
(35 times smaller than the probability for the third prize)
answered 2 hours ago
Oldboy
4,0281325
Your logic for case number (3) does not seem correct.
First, you have to draw 6 balls with exactly 5 balls matching 5 numbers from your set of 6 picked numbers. In total There are C(42,6) ways to draw six balls. How many winning combinations do we have? From your set of 6 balls you can pick five matching balls in C(6,5) ways. The sixth number can be any number from the remaining 42-5=37, minus one (your sixth pick). So the total number of winning combinations is $C(6,5)times36$. So the probability is:
$$p_1=frac36times6choose 542choose 6$$
But you have to draw the seventh number as well from the remaining 42-6=36 numbers without hitting your last (sixth) number. That probability is:
$$p_2=frac3536$$
The total proability is:
$$p=p_1times p_2=frac36times6choose 542choose 6timesfrac3536=frac35times6choose 542choose 6=frac15374699$$
You can use a similar logic for case (2).
The probability $p_1$ is the same. For the seventh ball we have 36 choices and there is 1 winning ball between them. So the probability $p_2$ is:
$$p_2=frac136$$
...and the final probability for the second prize is:
$$p=p_1times p_2=frac36times6choose 542choose 6timesfrac136=frac6choose 542choose 6=frac32622893$$
(35 times smaller than the probability for the third prize)
answered 2 hours ago
Oldboy
4,0281325
edited 1 hour ago
answered 2 hours ago
Oldboy
4,0281325
answered 2 hours ago
Oldboy
4,0281325
answered 2 hours ago
Oldboy
4,0281325
4,0281325
Great! This explains where most of my mistakes where. Thank you for clarifying.
– Paul Karam
1 hour ago
add a comment |Â
Great! This explains where most of my mistakes where. Thank you for clarifying.
– Paul Karam
1 hour ago
Great! This explains where most of my mistakes where. Thank you for clarifying.
– Paul Karam
1 hour ago
Great! This explains where most of my mistakes where. Thank you for clarifying.
– Paul Karam
1 hour ago
add a comment |Â
up vote
3
down vote
For $3^rd$ prize, you need to select which 5 numbered balls you will be getting, out of 6 in $^6C_5$ ways (get balls matching to those 5 numbers from 42); and then, select the $6^th$ ad $7^th$ ball in $^36C_2$ ways
(42-6=36, balls not matching any of the 6 numbers)
So, probability of winning $3^rd$ prize is $$frac^6C_5cdot^36C_2^42C_7$$
For $2^nd$ prize,
Among first 6, only 5 match, selected their corresponding numbers in $^6C_5$ ways.
First 5 balls drawn match these 5 numbers.
Now, $6^th$ ball is selected from 36 (42-6=36, it does not match any of 6 numbers); and number corresponding to last ball to be drawn is already known(one left from 6)
So, probability of winning $2^nd$ prize is: $$frac^6C_5cdot^36C_1cdot^1C_1^42C_7$$
Similarly, For $1^st$ prize, $$frac^6C_6cdot^36C_1^42C_7$$
answered 1 hour ago
omega
9032818
2
I think that your calculation for the third prize is also not correct. You are ignoring the fact that you have to draw the seventh ball as well and that ball must not hit your last pick. Just read the condition carefully: If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize. "Nothing else" must include the seventh ball as well.
– Oldboy
1 hour ago
yup, now i guess its correct.
– omega
1 hour ago
add a comment |Â
up vote
3
down vote
For $3^rd$ prize, you need to select which 5 numbered balls you will be getting, out of 6 in $^6C_5$ ways (get balls matching to those 5 numbers from 42); and then, select the $6^th$ ad $7^th$ ball in $^36C_2$ ways
(42-6=36, balls not matching any of the 6 numbers)
So, probability of winning $3^rd$ prize is $$frac^6C_5cdot^36C_2^42C_7$$
For $2^nd$ prize,
Among first 6, only 5 match, selected their corresponding numbers in $^6C_5$ ways.
First 5 balls drawn match these 5 numbers.
Now, $6^th$ ball is selected from 36 (42-6=36, it does not match any of 6 numbers); and number corresponding to last ball to be drawn is already known(one left from 6)
So, probability of winning $2^nd$ prize is: $$frac^6C_5cdot^36C_1cdot^1C_1^42C_7$$
Similarly, For $1^st$ prize, $$frac^6C_6cdot^36C_1^42C_7$$
answered 1 hour ago
omega
9032818
2
I think that your calculation for the third prize is also not correct. You are ignoring the fact that you have to draw the seventh ball as well and that ball must not hit your last pick. Just read the condition carefully: If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize. "Nothing else" must include the seventh ball as well.
– Oldboy
1 hour ago
yup, now i guess its correct.
– omega
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For $3^rd$ prize, you need to select which 5 numbered balls you will be getting, out of 6 in $^6C_5$ ways (get balls matching to those 5 numbers from 42); and then, select the $6^th$ ad $7^th$ ball in $^36C_2$ ways
(42-6=36, balls not matching any of the 6 numbers)
So, probability of winning $3^rd$ prize is $$frac^6C_5cdot^36C_2^42C_7$$
For $2^nd$ prize,
Among first 6, only 5 match, selected their corresponding numbers in $^6C_5$ ways.
First 5 balls drawn match these 5 numbers.
Now, $6^th$ ball is selected from 36 (42-6=36, it does not match any of 6 numbers); and number corresponding to last ball to be drawn is already known(one left from 6)
So, probability of winning $2^nd$ prize is: $$frac^6C_5cdot^36C_1cdot^1C_1^42C_7$$
Similarly, For $1^st$ prize, $$frac^6C_6cdot^36C_1^42C_7$$
answered 1 hour ago
omega
9032818
For $3^rd$ prize, you need to select which 5 numbered balls you will be getting, out of 6 in $^6C_5$ ways (get balls matching to those 5 numbers from 42); and then, select the $6^th$ ad $7^th$ ball in $^36C_2$ ways
(42-6=36, balls not matching any of the 6 numbers)
So, probability of winning $3^rd$ prize is $$frac^6C_5cdot^36C_2^42C_7$$
For $2^nd$ prize,
Among first 6, only 5 match, selected their corresponding numbers in $^6C_5$ ways.
First 5 balls drawn match these 5 numbers.
Now, $6^th$ ball is selected from 36 (42-6=36, it does not match any of 6 numbers); and number corresponding to last ball to be drawn is already known(one left from 6)
So, probability of winning $2^nd$ prize is: $$frac^6C_5cdot^36C_1cdot^1C_1^42C_7$$
Similarly, For $1^st$ prize, $$frac^6C_6cdot^36C_1^42C_7$$
answered 1 hour ago
omega
9032818
edited 1 hour ago
answered 1 hour ago
omega
9032818
answered 1 hour ago
omega
9032818
answered 1 hour ago
omega
9032818
9032818
2
I think that your calculation for the third prize is also not correct. You are ignoring the fact that you have to draw the seventh ball as well and that ball must not hit your last pick. Just read the condition carefully: If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize. "Nothing else" must include the seventh ball as well.
– Oldboy
1 hour ago
yup, now i guess its correct.
– omega
1 hour ago
add a comment |Â
2
I think that your calculation for the third prize is also not correct. You are ignoring the fact that you have to draw the seventh ball as well and that ball must not hit your last pick. Just read the condition carefully: If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize. "Nothing else" must include the seventh ball as well.
– Oldboy
1 hour ago
yup, now i guess its correct.
– omega
1 hour ago
2
2
I think that your calculation for the third prize is also not correct. You are ignoring the fact that you have to draw the seventh ball as well and that ball must not hit your last pick. Just read the condition carefully: If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize. "Nothing else" must include the seventh ball as well.
– Oldboy
1 hour ago
I think that your calculation for the third prize is also not correct. You are ignoring the fact that you have to draw the seventh ball as well and that ball must not hit your last pick. Just read the condition carefully: If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize. "Nothing else" must include the seventh ball as well.
– Oldboy
1 hour ago
yup, now i guess its correct.
– omega
1 hour ago
yup, now i guess its correct.
– omega
1 hour ago
add a comment |Â
Paul Karam is a new contributor. Be nice, and check out our Code of Conduct.
Paul Karam is a new contributor. Be nice, and check out our Code of Conduct.
Paul Karam is a new contributor. Be nice, and check out our Code of Conduct.
Paul Karam is a new contributor. Be nice, and check out our Code of Conduct.
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