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I've been trying to calculate the odds of winning our national lottery, but there is a rule in the permutations and I want to know if there's a formula that works for it.
In the South African lottery, the ball set consists of numbers 1 through 49.
A single lottery ticket/game has 6 of your chosen numbers.
So how many possible combinations of numbers are there?
The problem is that in your series of 6 numbers, a number cannot be <= less than or equal to the previous number.
In other words the formula isn't 49 ^ 6 49 to the power of 6 nor is it 49 x 48 x 47 x 46 x 45 x 44.
Thanks Rfwoolf ( talk) 11:39, 25 September 2009 (UTC)
The above answers are correct, but I feel that a little explanation might be in order. You have 49 balls and you draw 6 of them. You have 49 to choose from for your first ball, 48 for the second, 47 for the third, 46 for the fourth, 45 for the fifth, and 44 for the sixth. That gives 49 × … × 44 ways. But you're not interested in the order that they came out in. So, how many ways can you re-arrange the order that the six balls came out in? Well you have six balls to choose for the first ball to come out, five for the second, four for the third, three for the fourth, two for the fifth, and one for the sixth. There are 6 × 5 × … × 1 orders in which the six balls could have been drawn. So in total there are
ways of choosing six balls from 49 when the order doesn't matter. ~~ Dr Dec ( Talk) ~~ 22:57, 25 September 2009 (UTC)
with the size of the Lotto jackpots recently who would want to play anyway? even the mathematically stupid don't want to play for "only" R3 million. Of course the above answers demonstrate WHY playing the lotto is a surefire way to lose money, your chances are slim-to-none. Zunaid 22:41, 26 September 2009 (UTC)
If those familiar would comment on this, it would be helpful. Thanx! DRosenbach ( Talk | Contribs) 12:35, 25 September 2009 (UTC)
How would one begin showing whether or not there exists an isomorphism between the space of polynomials: and , the space of infinite sequences in the reals? I can see how the polynomials, which must have a maximum degree for any polynomial, would be isomorphic to the space of sequences with only finitely many nonzero terms, with the obvious bijection between coefficients and terms in the sequence, and this must be contained in , but I have a feeling the latter may have an uncountable basis whilst the former has the obvious countable basis ei=(0,0,...,1,0,0,...) with the 1 in the i-th position - but how would I show such a thing if so?
Otherlobby17 ( talk) 15:51, 25 September 2009 (UTC)
An isomorphism is a structure-preserving bijection. But which structure do you want to preserve? Addition of polynomials? Multiplication by scalars? If just those two, then it's vector-space isomorphism. Then there's multiplication of polynomials by each other, making it an algebra isomorphism. The set of sequences with only finitely many nonzero terms is algebra-isomorphic to the algebra of polynomials in an obvious way (see Cauchy product). But if you want to include sequences with infinitely many nonzero terms, then there's the question of how you could multiply them in a way that corresponds to multiplication of polynomials. I don't have an answer to that one. But now let's look at your mention of dimension. The vector space spanned by the countable basis is the set of all finite linear combinations of the basis vectors. If there's no countable basis for the vector space of all sequences of reals, then there's no vector-space isomorphism. And then a fortiori no algebra isomorphism. Right now I don't have that answer either—it seems as if there should be some obvious way to show there's no countable basis. Maybe later.... Michael Hardy ( talk) 16:33, 25 September 2009 (UTC)
Summary for the OP. Now you have: a combinatorial proof, following the lines of
EmilJ's post. It is the most general, for it also works for ; it exploits the fact that linear indipendence of a special family of sequences may be established just from the combinatorics of the collection of their supports. To exhibit such a family you also have some hints above. You then have a couple of topological proofs, that also come from more general facts: (I) no complete metric topological linear space has countable dimension, and (II) non-separable topological linear spaces have uncountable dimension. A Hilbert space proof is also available, using the isomorphism of with , where the uncountable dimension is more evident. Finally, a simple linearly independent uncountable family is given by exponential sequences; to show the independence you have an asymptotic order argument and a purely algebraic proof. In fact, it is now clear that you can make a proof out of virtually any result vaguely linked with the topic. --
pma (
talk) 12:33, 27 September 2009 (UTC)
Is there a simple example of a sequence of random variables that converge in mean to some random variable or constant, but don't converge in mean square? 128.237.235.59 ( talk) 19:57, 25 September 2009 (UTC)
Mathematics desk | ||
---|---|---|
< September 24 | << Aug | September | Oct >> | September 26 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
I've been trying to calculate the odds of winning our national lottery, but there is a rule in the permutations and I want to know if there's a formula that works for it.
In the South African lottery, the ball set consists of numbers 1 through 49.
A single lottery ticket/game has 6 of your chosen numbers.
So how many possible combinations of numbers are there?
The problem is that in your series of 6 numbers, a number cannot be <= less than or equal to the previous number.
In other words the formula isn't 49 ^ 6 49 to the power of 6 nor is it 49 x 48 x 47 x 46 x 45 x 44.
Thanks Rfwoolf ( talk) 11:39, 25 September 2009 (UTC)
The above answers are correct, but I feel that a little explanation might be in order. You have 49 balls and you draw 6 of them. You have 49 to choose from for your first ball, 48 for the second, 47 for the third, 46 for the fourth, 45 for the fifth, and 44 for the sixth. That gives 49 × … × 44 ways. But you're not interested in the order that they came out in. So, how many ways can you re-arrange the order that the six balls came out in? Well you have six balls to choose for the first ball to come out, five for the second, four for the third, three for the fourth, two for the fifth, and one for the sixth. There are 6 × 5 × … × 1 orders in which the six balls could have been drawn. So in total there are
ways of choosing six balls from 49 when the order doesn't matter. ~~ Dr Dec ( Talk) ~~ 22:57, 25 September 2009 (UTC)
with the size of the Lotto jackpots recently who would want to play anyway? even the mathematically stupid don't want to play for "only" R3 million. Of course the above answers demonstrate WHY playing the lotto is a surefire way to lose money, your chances are slim-to-none. Zunaid 22:41, 26 September 2009 (UTC)
If those familiar would comment on this, it would be helpful. Thanx! DRosenbach ( Talk | Contribs) 12:35, 25 September 2009 (UTC)
How would one begin showing whether or not there exists an isomorphism between the space of polynomials: and , the space of infinite sequences in the reals? I can see how the polynomials, which must have a maximum degree for any polynomial, would be isomorphic to the space of sequences with only finitely many nonzero terms, with the obvious bijection between coefficients and terms in the sequence, and this must be contained in , but I have a feeling the latter may have an uncountable basis whilst the former has the obvious countable basis ei=(0,0,...,1,0,0,...) with the 1 in the i-th position - but how would I show such a thing if so?
Otherlobby17 ( talk) 15:51, 25 September 2009 (UTC)
An isomorphism is a structure-preserving bijection. But which structure do you want to preserve? Addition of polynomials? Multiplication by scalars? If just those two, then it's vector-space isomorphism. Then there's multiplication of polynomials by each other, making it an algebra isomorphism. The set of sequences with only finitely many nonzero terms is algebra-isomorphic to the algebra of polynomials in an obvious way (see Cauchy product). But if you want to include sequences with infinitely many nonzero terms, then there's the question of how you could multiply them in a way that corresponds to multiplication of polynomials. I don't have an answer to that one. But now let's look at your mention of dimension. The vector space spanned by the countable basis is the set of all finite linear combinations of the basis vectors. If there's no countable basis for the vector space of all sequences of reals, then there's no vector-space isomorphism. And then a fortiori no algebra isomorphism. Right now I don't have that answer either—it seems as if there should be some obvious way to show there's no countable basis. Maybe later.... Michael Hardy ( talk) 16:33, 25 September 2009 (UTC)
Summary for the OP. Now you have: a combinatorial proof, following the lines of
EmilJ's post. It is the most general, for it also works for ; it exploits the fact that linear indipendence of a special family of sequences may be established just from the combinatorics of the collection of their supports. To exhibit such a family you also have some hints above. You then have a couple of topological proofs, that also come from more general facts: (I) no complete metric topological linear space has countable dimension, and (II) non-separable topological linear spaces have uncountable dimension. A Hilbert space proof is also available, using the isomorphism of with , where the uncountable dimension is more evident. Finally, a simple linearly independent uncountable family is given by exponential sequences; to show the independence you have an asymptotic order argument and a purely algebraic proof. In fact, it is now clear that you can make a proof out of virtually any result vaguely linked with the topic. --
pma (
talk) 12:33, 27 September 2009 (UTC)
Is there a simple example of a sequence of random variables that converge in mean to some random variable or constant, but don't converge in mean square? 128.237.235.59 ( talk) 19:57, 25 September 2009 (UTC)