How do I calculate probabilities in a 6/49 numbers game?
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A numbers game called the 49s has 49 numbers (1-49). 6 winning numbers are drawn from 49.
Customers can bet on 1,2,3,4 or 5 numbers at a time. For instance if a customer selects number 8, if number 8 is drawn they win. Likewise if the customer selects numbers 8 & 9, if both numbers are drawn they win.
I would like to know how to calculate the probability of correctly selecting 1,2,3 4 or 5 numbers. I understand that there are 13,983,816 different combinations of 6 numbers from 49, but I don't know how to calculate how many of them contain the number 8 for instance, or how many contain both numbers 8 & 9.
Any help would be greatly appreciated.
Paul
probability
asked Aug 29 at 9:52
Paul Garvock
1
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up vote
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down vote
favorite
A numbers game called the 49s has 49 numbers (1-49). 6 winning numbers are drawn from 49.
Customers can bet on 1,2,3,4 or 5 numbers at a time. For instance if a customer selects number 8, if number 8 is drawn they win. Likewise if the customer selects numbers 8 & 9, if both numbers are drawn they win.
I would like to know how to calculate the probability of correctly selecting 1,2,3 4 or 5 numbers. I understand that there are 13,983,816 different combinations of 6 numbers from 49, but I don't know how to calculate how many of them contain the number 8 for instance, or how many contain both numbers 8 & 9.
Any help would be greatly appreciated.
Paul
probability
asked Aug 29 at 9:52
Paul Garvock
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 29 at 9:55
Hello and welcome to Mathematics SE! It might be easier to think of it in probabilities. If we bet on 1 number what is the probability of getting one right? Now what if we bet on 2 numbers? I think focussing on the different combinations is not the way to go.
– Jan
Aug 29 at 9:55
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A numbers game called the 49s has 49 numbers (1-49). 6 winning numbers are drawn from 49.
Customers can bet on 1,2,3,4 or 5 numbers at a time. For instance if a customer selects number 8, if number 8 is drawn they win. Likewise if the customer selects numbers 8 & 9, if both numbers are drawn they win.
I would like to know how to calculate the probability of correctly selecting 1,2,3 4 or 5 numbers. I understand that there are 13,983,816 different combinations of 6 numbers from 49, but I don't know how to calculate how many of them contain the number 8 for instance, or how many contain both numbers 8 & 9.
Any help would be greatly appreciated.
Paul
probability
asked Aug 29 at 9:52
Paul Garvock
1
A numbers game called the 49s has 49 numbers (1-49). 6 winning numbers are drawn from 49.
Customers can bet on 1,2,3,4 or 5 numbers at a time. For instance if a customer selects number 8, if number 8 is drawn they win. Likewise if the customer selects numbers 8 & 9, if both numbers are drawn they win.
I would like to know how to calculate the probability of correctly selecting 1,2,3 4 or 5 numbers. I understand that there are 13,983,816 different combinations of 6 numbers from 49, but I don't know how to calculate how many of them contain the number 8 for instance, or how many contain both numbers 8 & 9.
Any help would be greatly appreciated.
Paul
probability
asked Aug 29 at 9:52
Paul Garvock
1
asked Aug 29 at 9:52
Paul Garvock
1
asked Aug 29 at 9:52
Paul Garvock
1
asked Aug 29 at 9:52
Paul Garvock
1
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 29 at 9:55
Hello and welcome to Mathematics SE! It might be easier to think of it in probabilities. If we bet on 1 number what is the probability of getting one right? Now what if we bet on 2 numbers? I think focussing on the different combinations is not the way to go.
– Jan
Aug 29 at 9:55
add a comment |Â
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 29 at 9:55
Hello and welcome to Mathematics SE! It might be easier to think of it in probabilities. If we bet on 1 number what is the probability of getting one right? Now what if we bet on 2 numbers? I think focussing on the different combinations is not the way to go.
– Jan
Aug 29 at 9:55
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 29 at 9:55
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 29 at 9:55
Hello and welcome to Mathematics SE! It might be easier to think of it in probabilities. If we bet on 1 number what is the probability of getting one right? Now what if we bet on 2 numbers? I think focussing on the different combinations is not the way to go.
– Jan
Aug 29 at 9:55
Hello and welcome to Mathematics SE! It might be easier to think of it in probabilities. If we bet on 1 number what is the probability of getting one right? Now what if we bet on 2 numbers? I think focussing on the different combinations is not the way to go.
– Jan
Aug 29 at 9:55
add a comment |Â
1 Answer
1
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There are $binom49k$ ways to select $k$ numbers from $49$ numbers, where $binom nk$ is a binomial coefficient. There are $binom6k$ ways to select $k$ numbers from $6$ numbers. Thus the probability of selecting $k$ numbers that are among the $6$ numbers drawn is
$$
fracbinom6kbinom49k=frac6!(49-k)!49!(6-k)!;.
$$
answered Aug 29 at 9:59
joriki
167k10180333
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
There are $binom49k$ ways to select $k$ numbers from $49$ numbers, where $binom nk$ is a binomial coefficient. There are $binom6k$ ways to select $k$ numbers from $6$ numbers. Thus the probability of selecting $k$ numbers that are among the $6$ numbers drawn is
$$
fracbinom6kbinom49k=frac6!(49-k)!49!(6-k)!;.
$$
answered Aug 29 at 9:59
joriki
167k10180333
add a comment |Â
up vote
0
down vote
There are $binom49k$ ways to select $k$ numbers from $49$ numbers, where $binom nk$ is a binomial coefficient. There are $binom6k$ ways to select $k$ numbers from $6$ numbers. Thus the probability of selecting $k$ numbers that are among the $6$ numbers drawn is
$$
fracbinom6kbinom49k=frac6!(49-k)!49!(6-k)!;.
$$
answered Aug 29 at 9:59
joriki
167k10180333
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There are $binom49k$ ways to select $k$ numbers from $49$ numbers, where $binom nk$ is a binomial coefficient. There are $binom6k$ ways to select $k$ numbers from $6$ numbers. Thus the probability of selecting $k$ numbers that are among the $6$ numbers drawn is
$$
fracbinom6kbinom49k=frac6!(49-k)!49!(6-k)!;.
$$
answered Aug 29 at 9:59
joriki
167k10180333
There are $binom49k$ ways to select $k$ numbers from $49$ numbers, where $binom nk$ is a binomial coefficient. There are $binom6k$ ways to select $k$ numbers from $6$ numbers. Thus the probability of selecting $k$ numbers that are among the $6$ numbers drawn is
$$
fracbinom6kbinom49k=frac6!(49-k)!49!(6-k)!;.
$$
answered Aug 29 at 9:59
joriki
167k10180333
answered Aug 29 at 9:59
joriki
167k10180333
answered Aug 29 at 9:59
joriki
167k10180333
answered Aug 29 at 9:59
joriki
167k10180333
167k10180333
add a comment |Â
add a comment |Â
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 29 at 9:55
Hello and welcome to Mathematics SE! It might be easier to think of it in probabilities. If we bet on 1 number what is the probability of getting one right? Now what if we bet on 2 numbers? I think focussing on the different combinations is not the way to go.
– Jan
Aug 29 at 9:55