Mixing
UK Lottery Odds Calculation Error
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
I found the probability topic in stats and mechanics to be very interesting, and I attempted to try using it to calculate the odds of winning the UK National Lottery, but failed. My calculation was (1/59*1/58*1/57*1/56*1/55*1/54). The reason was that in the NL there are 6 balls dropped, and the possible outcomes range from 1 to 59. I reasoned that since all 6 need to match, I could assume that each outcome could be treated as an isolated selection, and that my first ball match odds were 1/59 as a result, then if the first ball matches (which it must), the next is 1/58 and so on. The odds I calculated were orders of magnitude less likely than the correct value. What is wrong with my attempt, and why is the correct formula quoted as "59!/(6!*(59-6)!)"?
probability statistics lotteries
asked Jul 25 at 16:56
Droi Rea
286
add a comment |Â
up vote
5
down vote
favorite
I found the probability topic in stats and mechanics to be very interesting, and I attempted to try using it to calculate the odds of winning the UK National Lottery, but failed. My calculation was (1/59*1/58*1/57*1/56*1/55*1/54). The reason was that in the NL there are 6 balls dropped, and the possible outcomes range from 1 to 59. I reasoned that since all 6 need to match, I could assume that each outcome could be treated as an isolated selection, and that my first ball match odds were 1/59 as a result, then if the first ball matches (which it must), the next is 1/58 and so on. The odds I calculated were orders of magnitude less likely than the correct value. What is wrong with my attempt, and why is the correct formula quoted as "59!/(6!*(59-6)!)"?
probability statistics lotteries
asked Jul 25 at 16:56
Droi Rea
286
2
Your numbers can come up in $6!$ different orders. "It could be you!" but probably it won't be.
– Lord Shark the Unknown
Jul 25 at 17:02
For example, if you chose numbers 10, 13, 14, 23, 47, 50, the balls could drop in the sequence 13, 50, 47, 10, 24, 14, or 14, 47, 50, 10, 13, 22, or any of several other sequences, and you would still win. Every single one of those sequences has a certain probability of occurring, and that probability is what your calculation finds: the first drop must be a particular ball (not any of the other 5 you selected), and so forth.
– David K
Jul 25 at 17:22
3
First ball odds are 6/59, not 1/59.
– Kevin
Jul 25 at 20:11
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I found the probability topic in stats and mechanics to be very interesting, and I attempted to try using it to calculate the odds of winning the UK National Lottery, but failed. My calculation was (1/59*1/58*1/57*1/56*1/55*1/54). The reason was that in the NL there are 6 balls dropped, and the possible outcomes range from 1 to 59. I reasoned that since all 6 need to match, I could assume that each outcome could be treated as an isolated selection, and that my first ball match odds were 1/59 as a result, then if the first ball matches (which it must), the next is 1/58 and so on. The odds I calculated were orders of magnitude less likely than the correct value. What is wrong with my attempt, and why is the correct formula quoted as "59!/(6!*(59-6)!)"?
probability statistics lotteries
asked Jul 25 at 16:56
Droi Rea
286
I found the probability topic in stats and mechanics to be very interesting, and I attempted to try using it to calculate the odds of winning the UK National Lottery, but failed. My calculation was (1/59*1/58*1/57*1/56*1/55*1/54). The reason was that in the NL there are 6 balls dropped, and the possible outcomes range from 1 to 59. I reasoned that since all 6 need to match, I could assume that each outcome could be treated as an isolated selection, and that my first ball match odds were 1/59 as a result, then if the first ball matches (which it must), the next is 1/58 and so on. The odds I calculated were orders of magnitude less likely than the correct value. What is wrong with my attempt, and why is the correct formula quoted as "59!/(6!*(59-6)!)"?
probability statistics lotteries
asked Jul 25 at 16:56
Droi Rea
286
edited Jul 25 at 23:04
asked Jul 25 at 16:56
Droi Rea
286
asked Jul 25 at 16:56
Droi Rea
286
asked Jul 25 at 16:56
Droi Rea
286
286
2
Your numbers can come up in $6!$ different orders. "It could be you!" but probably it won't be.
– Lord Shark the Unknown
Jul 25 at 17:02
For example, if you chose numbers 10, 13, 14, 23, 47, 50, the balls could drop in the sequence 13, 50, 47, 10, 24, 14, or 14, 47, 50, 10, 13, 22, or any of several other sequences, and you would still win. Every single one of those sequences has a certain probability of occurring, and that probability is what your calculation finds: the first drop must be a particular ball (not any of the other 5 you selected), and so forth.
– David K
Jul 25 at 17:22
3
First ball odds are 6/59, not 1/59.
– Kevin
Jul 25 at 20:11
add a comment |Â
2
Your numbers can come up in $6!$ different orders. "It could be you!" but probably it won't be.
– Lord Shark the Unknown
Jul 25 at 17:02
For example, if you chose numbers 10, 13, 14, 23, 47, 50, the balls could drop in the sequence 13, 50, 47, 10, 24, 14, or 14, 47, 50, 10, 13, 22, or any of several other sequences, and you would still win. Every single one of those sequences has a certain probability of occurring, and that probability is what your calculation finds: the first drop must be a particular ball (not any of the other 5 you selected), and so forth.
– David K
Jul 25 at 17:22
3
First ball odds are 6/59, not 1/59.
– Kevin
Jul 25 at 20:11
2
2
Your numbers can come up in $6!$ different orders. "It could be you!" but probably it won't be.
– Lord Shark the Unknown
Jul 25 at 17:02
Your numbers can come up in $6!$ different orders. "It could be you!" but probably it won't be.
– Lord Shark the Unknown
Jul 25 at 17:02
For example, if you chose numbers 10, 13, 14, 23, 47, 50, the balls could drop in the sequence 13, 50, 47, 10, 24, 14, or 14, 47, 50, 10, 13, 22, or any of several other sequences, and you would still win. Every single one of those sequences has a certain probability of occurring, and that probability is what your calculation finds: the first drop must be a particular ball (not any of the other 5 you selected), and so forth.
– David K
Jul 25 at 17:22
For example, if you chose numbers 10, 13, 14, 23, 47, 50, the balls could drop in the sequence 13, 50, 47, 10, 24, 14, or 14, 47, 50, 10, 13, 22, or any of several other sequences, and you would still win. Every single one of those sequences has a certain probability of occurring, and that probability is what your calculation finds: the first drop must be a particular ball (not any of the other 5 you selected), and so forth.
– David K
Jul 25 at 17:22
3
3
First ball odds are 6/59, not 1/59.
– Kevin
Jul 25 at 20:11
First ball odds are 6/59, not 1/59.
– Kevin
Jul 25 at 20:11
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
17
down vote
accepted
Firstly, note the answer you obtained is $frac53!59!$, while the probability you should have obtained is $frac53!6!59!$ (you've stated this fraction upside down, where it gives the number of possible outcomes). The reason you're wrong by a factor of $6!$ is because the order in which the numbers fall is irrelevant. This is the combinations-permutations distinction; in technical terms, $frac59!53!6!=^59C_6$ while $frac59!53!=^59P_6$.
answered Jul 25 at 17:02
J.G.
13.1k11424
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
17
down vote
accepted
Firstly, note the answer you obtained is $frac53!59!$, while the probability you should have obtained is $frac53!6!59!$ (you've stated this fraction upside down, where it gives the number of possible outcomes). The reason you're wrong by a factor of $6!$ is because the order in which the numbers fall is irrelevant. This is the combinations-permutations distinction; in technical terms, $frac59!53!6!=^59C_6$ while $frac59!53!=^59P_6$.
answered Jul 25 at 17:02
J.G.
13.1k11424
add a comment |Â
up vote
17
down vote
accepted
Firstly, note the answer you obtained is $frac53!59!$, while the probability you should have obtained is $frac53!6!59!$ (you've stated this fraction upside down, where it gives the number of possible outcomes). The reason you're wrong by a factor of $6!$ is because the order in which the numbers fall is irrelevant. This is the combinations-permutations distinction; in technical terms, $frac59!53!6!=^59C_6$ while $frac59!53!=^59P_6$.
answered Jul 25 at 17:02
J.G.
13.1k11424
add a comment |Â
up vote
17
down vote
accepted
up vote
17
down vote
accepted
Firstly, note the answer you obtained is $frac53!59!$, while the probability you should have obtained is $frac53!6!59!$ (you've stated this fraction upside down, where it gives the number of possible outcomes). The reason you're wrong by a factor of $6!$ is because the order in which the numbers fall is irrelevant. This is the combinations-permutations distinction; in technical terms, $frac59!53!6!=^59C_6$ while $frac59!53!=^59P_6$.
answered Jul 25 at 17:02
J.G.
13.1k11424
Firstly, note the answer you obtained is $frac53!59!$, while the probability you should have obtained is $frac53!6!59!$ (you've stated this fraction upside down, where it gives the number of possible outcomes). The reason you're wrong by a factor of $6!$ is because the order in which the numbers fall is irrelevant. This is the combinations-permutations distinction; in technical terms, $frac59!53!6!=^59C_6$ while $frac59!53!=^59P_6$.
answered Jul 25 at 17:02
J.G.
13.1k11424
answered Jul 25 at 17:02
J.G.
13.1k11424
answered Jul 25 at 17:02
J.G.
13.1k11424
answered Jul 25 at 17:02
J.G.
13.1k11424
13.1k11424
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862620%2fuk-lottery-odds-calculation-error%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Your numbers can come up in $6!$ different orders. "It could be you!" but probably it won't be.
– Lord Shark the Unknown
Jul 25 at 17:02
For example, if you chose numbers 10, 13, 14, 23, 47, 50, the balls could drop in the sequence 13, 50, 47, 10, 24, 14, or 14, 47, 50, 10, 13, 22, or any of several other sequences, and you would still win. Every single one of those sequences has a certain probability of occurring, and that probability is what your calculation finds: the first drop must be a particular ball (not any of the other 5 you selected), and so forth.
– David K
Jul 25 at 17:22
3
First ball odds are 6/59, not 1/59.
– Kevin
Jul 25 at 20:11