Post History
#11: Post edited
by
Michael Hardy
·
2024-07-18T09:17:23Z (5 months ago)
some MathJax usage improvements
Without trial and error, how can I effortlessly deduce all $n, k_i ∈ ℕ ∋ \binom n {k_1, k_2, ..., k_n} =$ given c?
#### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k_i ∈ ℕ ∋ \dbinom n { k_1, k_2, ..., k_i} = c$ ? For example below, $i = 1, \color{limegreen}{c = 4,072,530}$. Rule out trial and error!- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as I prefer to pick unpopular ["numbers to reduce the number of ways you split the prize"](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't let players pick this 2nd 6-tuple. OLG's phone operator replied
>I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k ≤ 20, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires k ≤ 10, because research proves that players dislike picking over 10 integers, which they find inconvenient. Remember, many players buy physical paper tickets.
- #### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k_i ∈ ℕ[]() ∋ \dbinom n { k_1, k_2, \ldots , k_i} = c$ ? For example below, $i = 1, \color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as I prefer to pick unpopular ["numbers to reduce the number of ways you split the prize"](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't let players pick this 2nd 6-tuple. OLG's phone operator replied
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers $k ≤ 20,$ $n$ to satisfy $\dbinom n k =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires $k ≤ 10,$ because research proves that players dislike picking over $10$ integers, which they find inconvenient. Remember, many players buy physical paper tickets.
#10: Post edited
by
Chgg Clou
·
2023-08-26T04:37:14Z (over 1 year ago)
- #### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k_i ∈ ℕ ∋ \dbinom n { k_1, k_2, ..., k_i} = c$ ? For example below, $i = 1, \color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't let players pick this 2nd 6-tuple. OLG's phone operator replied
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k ≤ 20, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires k ≤ 10, because research proves that players dislike picking over 10 integers, which they find inconvenient. Remember, many players buy physical paper tickets.
- #### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k_i ∈ ℕ ∋ \dbinom n { k_1, k_2, ..., k_i} = c$ ? For example below, $i = 1, \color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as I prefer to pick unpopular ["numbers to reduce the number of ways you split the prize"](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't let players pick this 2nd 6-tuple. OLG's phone operator replied
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k ≤ 20, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires k ≤ 10, because research proves that players dislike picking over 10 integers, which they find inconvenient. Remember, many players buy physical paper tickets.
#9: Post edited
by
Chgg Clou
·
2023-08-25T23:13:52Z (over 1 year ago)
Without trial and error, how can I effortlessly deduce all $n, k ∈ ℕ ∋ \binom n {k_1, k_2, ..., k_n} =$ given c?
- Without trial and error, how can I effortlessly deduce all $n, k_i ∈ ℕ ∋ \binom n {k_1, k_2, ..., k_n} =$ given c?
#8: Post edited
by
Chgg Clou
·
2023-08-25T23:13:40Z (over 1 year ago)
Without trial and error, how can I effortlessly deduce all $n, k ∈ ℕ ∋ \binom n {k} =$ given c?
- Without trial and error, how can I effortlessly deduce all $n, k ∈ ℕ ∋ \binom n {k_1, k_2, ..., k_n} =$ given c?
#### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. Rule out trial and error!- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't let players pick this 2nd 6-tuple. OLG's phone operator replied
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k ≤ 20, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires k ≤ 10, because research proves that players dislike picking over 10 integers, which they find inconvenient. Remember, many players buy physical paper tickets.
- #### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k_i ∈ ℕ ∋ \dbinom n { k_1, k_2, ..., k_i} = c$ ? For example below, $i = 1, \color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't let players pick this 2nd 6-tuple. OLG's phone operator replied
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k ≤ 20, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires k ≤ 10, because research proves that players dislike picking over 10 integers, which they find inconvenient. Remember, many players buy physical paper tickets.
#7: Post edited
by
Chgg Clou
·
2023-08-25T10:50:12Z (over 1 year ago)
- #### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
I asked why OLG don't simply let players pick this 2nd 6-tuple. OLG's phone operator replied that>I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k ≤ 20, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires k ≤ 10, because research proves that players dislike picking over 10 integers, which they find inconvenient. Remember, many players still buy physical paper tickets.
- #### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't let players pick this 2nd 6-tuple. OLG's phone operator replied
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k ≤ 20, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires k ≤ 10, because research proves that players dislike picking over 10 integers, which they find inconvenient. Remember, many players buy physical paper tickets.
#6: Post edited
by
Chgg Clou
·
2023-08-25T10:49:18Z (over 1 year ago)
#### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. Rule out trial and error!- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k ≤ 20, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires k ≤ 10, because research proves that players dislike picking over 10 integers, which they find inconvenient. Remember, many players still buy physical paper tickets.
- #### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k ≤ 20, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires k ≤ 10, because research proves that players dislike picking over 10 integers, which they find inconvenient. Remember, many players still buy physical paper tickets.
#5: Post edited
by
Chgg Clou
·
2023-08-25T10:48:59Z (over 1 year ago)
- #### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
>I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already!
- #### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k ≤ 20, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires k ≤ 10, because research proves that players dislike picking over 10 integers, which they find inconvenient. Remember, many players still buy physical paper tickets.
#4: Post edited
by
Chgg Clou
·
2023-08-25T10:43:21Z (over 1 year ago)
- #### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
>I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k, n to satisfy $\dbinom n {k} = 4,072,530$. If this was possible, our actuaries would have actioned this already!
- #### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already!
#3: Post edited
by
Chgg Clou
·
2023-08-25T10:42:07Z (over 1 year ago)
With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ϶ \dbinom n {k} = c$ ? For the example below, $p = 4, 072, 530$. Rule out trial and error.#### Context- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}\color{red}{{/2}} = 4,072,530$.- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
I asked OLG why OLG don't simply let players pick this 2nd 6-tuple. The telephone operator replied that- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k, n to satisfy $\dbinom n {k} = 4,072,530$. If this was possible, our actuaries would have actioned this already!
- #### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. Rule out trial and error!
- ## Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked why OLG don't simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k, n to satisfy $\dbinom n {k} = 4,072,530$. If this was possible, our actuaries would have actioned this already!
#2: Post edited
by
Chgg Clou
·
2023-08-25T10:39:45Z (over 1 year ago)
Without trial and error, how can I efficaciously deduce all $n, k ∈ ℕ ϶ \binom n {k} =$ given c?
- Without trial and error, how can I effortlessly deduce all $n, k ∈ ℕ ∋ \binom n {k} =$ given c?
With an online or computer software, for a given $c ∈ ℕ $, how can I swiftly deduce all natural numbers that $n, k ∈ ℕ ϶ \dbinom n {k} = c$ ? For the example below, $p = 4, 072, 530$.- #### Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}\color{red}{{/2}} = 4,072,530$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked OLG why OLG don't simply let players pick this 2nd 6-tuple. The telephone operator replied that
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k, n to satisfy $\dbinom n {k} = 4,072,530$. If this was possible, our actuaries would have actioned this already!
- With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ϶ \dbinom n {k} = c$ ? For the example below, $p = 4, 072, 530$. Rule out trial and error.
- #### Context
- >[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
- Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}\color{red}{{/2}} = 4,072,530$.
- But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
- >[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
- numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
- I asked OLG why OLG don't simply let players pick this 2nd 6-tuple. The telephone operator replied that
- >I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k, n to satisfy $\dbinom n {k} = 4,072,530$. If this was possible, our actuaries would have actioned this already!
#1: Initial revision
by
Chgg Clou
·
2023-08-25T10:37:06Z (over 1 year ago)
Without trial and error, how can I efficaciously deduce all $n, k ∈ ℕ ϶ \binom n {k} =$ given c?
With an online or computer software, for a given $c ∈ ℕ $, how can I swiftly deduce all natural numbers that $n, k ∈ ℕ ϶ \dbinom n {k} = c$ ? For the example below, $p = 4, 072, 530$.
#### Context
>[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}\color{red}{{/2}} = 4,072,530$.
But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](https://math.codidact.com/posts/289079), I loathe that
>[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
I asked OLG why OLG don't simply let players pick this 2nd 6-tuple. The telephone operator replied that
>I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k, n to satisfy $\dbinom n {k} = 4,072,530$. If this was possible, our actuaries would have actioned this already!