If I have an event where the outcomes aren't uniformly distributed, how would I make the “fairest” bingo...
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If I have an event where the outcomes aren't uniformly distributed, how would I make the “fairest” bingo card out of the events?
Do the 12 lines of a bingo card have equal chance of winning?Limit Theorem applicabilityWhen is the expected number of events per period the inverse of the expected duration between events?Weighted Probability ProblemGambler's Ruin Problem$K$ events that are $(K-1)$-wise Independent but not Mutually/Fully IndependentDetermining the number of values required to find the probabilities of a set of valuesSolving the probability of independent evnts without the complementHow to interpret probability of a nonrepeating event?Listing all possible trebles from multiple sets where you can only select one element of each setthree independent exponential random variables, find P(X<Y<Z), intuition
$begingroup$
I think I should start by clarifying what I mean by "fairest". I mean that the probability of winning with any one line is as close as possible to winning with any of the others. Formally I want to know this:
Lets say you you have 25 independent outcomes to an event: $e_n$ with respective probabilities $p_n$, and $p_1 leqslant p_2 leqslant ... leqslant p_{25}$.
Take a 5x5 grid containing the elements $e_n$, label the straight lines of 5 grid points (horizontal, diagonal, and vertical) by $l_k$ for $1 leqslant k leqslant 12$.
Perform the event, cross out the gridpoint of $e_n$ if outcome $e_n$ occurs, and repeat the event until a full line in the 5x5 grid has been crossed out. if every grid point of $l_k$ has been crossed out, you have "won" with line $l_k$.
For a given 5x5 grid, let $Pl_k$ be the probability of winning with line $l_k$.
Is there a way to find the arrangement of outcomes $e_n$ into a 5x5 grid to minimise: $$sup_{1 leqslant i,j leqslant 12}{|Pl_i - Pl_j|}$$
How would I do this if the events aren't indepenent?
My intuition says that the best arrangement would be one of the 5x5 magic squares, replacing $n$ with $e_n$, but I can't justify that, and I doubt that would work if the events weren't independent.
combinatorics probability-theory
asked 2 days ago
BanbadleBanbadle
15010
$endgroup$
|
show 4 more comments
$begingroup$
I think I should start by clarifying what I mean by "fairest". I mean that the probability of winning with any one line is as close as possible to winning with any of the others. Formally I want to know this:
Lets say you you have 25 independent outcomes to an event: $e_n$ with respective probabilities $p_n$, and $p_1 leqslant p_2 leqslant ... leqslant p_{25}$.
Take a 5x5 grid containing the elements $e_n$, label the straight lines of 5 grid points (horizontal, diagonal, and vertical) by $l_k$ for $1 leqslant k leqslant 12$.
Perform the event, cross out the gridpoint of $e_n$ if outcome $e_n$ occurs, and repeat the event until a full line in the 5x5 grid has been crossed out. if every grid point of $l_k$ has been crossed out, you have "won" with line $l_k$.
For a given 5x5 grid, let $Pl_k$ be the probability of winning with line $l_k$.
Is there a way to find the arrangement of outcomes $e_n$ into a 5x5 grid to minimise: $$sup_{1 leqslant i,j leqslant 12}{|Pl_i - Pl_j|}$$
How would I do this if the events aren't indepenent?
My intuition says that the best arrangement would be one of the 5x5 magic squares, replacing $n$ with $e_n$, but I can't justify that, and I doubt that would work if the events weren't independent.
combinatorics probability-theory
asked 2 days ago
BanbadleBanbadle
15010
$endgroup$
$begingroup$
There are 12 straight lines, not 10.
$endgroup$
– Travis
2 days ago
$begingroup$
Wow that was dumb, was thinking of a 4x4 for some reason. Fixed now, thanks.
$endgroup$
– Banbadle
2 days ago
$begingroup$
On a more basic level, how would you even calculate $Pl_k$? You may be able to calculate the probability of line $l_k$ being filled by $le T$ moves, and from that the prob of the line being filled at exactly the $T$th move, but for the line to actually *win* this would also require the other $11$ lines being not yet filled - and that seems like a really difficult problem.
$endgroup$
– antkam
2 days ago
$begingroup$
If you let $Pl_{i,t}$ be the probability of it being solved after t events, would trying to maximise the limit of $sup({ {Pl_{i,t} / Pl_{j,t}} | Pl_{i,t} leqslant Pl_{j,t}})$ as t tends to infinity be the same as what I've said above? I was hoping there would be some kind of algorithm I was too dumb to find to help with this. I get calculating the probabilities separately would be too hard, but I couldn't see a better way to formulate it.
$endgroup$
– Banbadle
2 days ago
$begingroup$
Actually, reading my last comment over, if the formulation is the same then wouldn't maximising that for any t>4 lead to the solution? Meaning that I'm just minimising the difference between the chance of getting one line after 5 tries and another line after 5 tries. They're then very easy numbers to work with
$endgroup$
– Banbadle
2 days ago
|
show 4 more comments
$begingroup$
I think I should start by clarifying what I mean by "fairest". I mean that the probability of winning with any one line is as close as possible to winning with any of the others. Formally I want to know this:
Lets say you you have 25 independent outcomes to an event: $e_n$ with respective probabilities $p_n$, and $p_1 leqslant p_2 leqslant ... leqslant p_{25}$.
Take a 5x5 grid containing the elements $e_n$, label the straight lines of 5 grid points (horizontal, diagonal, and vertical) by $l_k$ for $1 leqslant k leqslant 12$.
Perform the event, cross out the gridpoint of $e_n$ if outcome $e_n$ occurs, and repeat the event until a full line in the 5x5 grid has been crossed out. if every grid point of $l_k$ has been crossed out, you have "won" with line $l_k$.
For a given 5x5 grid, let $Pl_k$ be the probability of winning with line $l_k$.
Is there a way to find the arrangement of outcomes $e_n$ into a 5x5 grid to minimise: $$sup_{1 leqslant i,j leqslant 12}{|Pl_i - Pl_j|}$$
How would I do this if the events aren't indepenent?
My intuition says that the best arrangement would be one of the 5x5 magic squares, replacing $n$ with $e_n$, but I can't justify that, and I doubt that would work if the events weren't independent.
combinatorics probability-theory
asked 2 days ago
BanbadleBanbadle
15010
$endgroup$
I think I should start by clarifying what I mean by "fairest". I mean that the probability of winning with any one line is as close as possible to winning with any of the others. Formally I want to know this:
Lets say you you have 25 independent outcomes to an event: $e_n$ with respective probabilities $p_n$, and $p_1 leqslant p_2 leqslant ... leqslant p_{25}$.
Take a 5x5 grid containing the elements $e_n$, label the straight lines of 5 grid points (horizontal, diagonal, and vertical) by $l_k$ for $1 leqslant k leqslant 12$.
Perform the event, cross out the gridpoint of $e_n$ if outcome $e_n$ occurs, and repeat the event until a full line in the 5x5 grid has been crossed out. if every grid point of $l_k$ has been crossed out, you have "won" with line $l_k$.
For a given 5x5 grid, let $Pl_k$ be the probability of winning with line $l_k$.
Is there a way to find the arrangement of outcomes $e_n$ into a 5x5 grid to minimise: $$sup_{1 leqslant i,j leqslant 12}{|Pl_i - Pl_j|}$$
How would I do this if the events aren't indepenent?
My intuition says that the best arrangement would be one of the 5x5 magic squares, replacing $n$ with $e_n$, but I can't justify that, and I doubt that would work if the events weren't independent.
combinatorics probability-theory
combinatorics probability-theory
asked 2 days ago
BanbadleBanbadle
15010
asked 2 days ago
BanbadleBanbadle
15010
edited 2 days ago
Banbadle
asked 2 days ago
BanbadleBanbadle
15010
asked 2 days ago
BanbadleBanbadle
15010
asked 2 days ago
BanbadleBanbadle
15010
15010
$begingroup$
There are 12 straight lines, not 10.
$endgroup$
– Travis
2 days ago
$begingroup$
Wow that was dumb, was thinking of a 4x4 for some reason. Fixed now, thanks.
$endgroup$
– Banbadle
2 days ago
$begingroup$
On a more basic level, how would you even calculate $Pl_k$? You may be able to calculate the probability of line $l_k$ being filled by $le T$ moves, and from that the prob of the line being filled at exactly the $T$th move, but for the line to actually *win* this would also require the other $11$ lines being not yet filled - and that seems like a really difficult problem.
$endgroup$
– antkam
2 days ago
$begingroup$
If you let $Pl_{i,t}$ be the probability of it being solved after t events, would trying to maximise the limit of $sup({ {Pl_{i,t} / Pl_{j,t}} | Pl_{i,t} leqslant Pl_{j,t}})$ as t tends to infinity be the same as what I've said above? I was hoping there would be some kind of algorithm I was too dumb to find to help with this. I get calculating the probabilities separately would be too hard, but I couldn't see a better way to formulate it.
$endgroup$
– Banbadle
2 days ago
$begingroup$
Actually, reading my last comment over, if the formulation is the same then wouldn't maximising that for any t>4 lead to the solution? Meaning that I'm just minimising the difference between the chance of getting one line after 5 tries and another line after 5 tries. They're then very easy numbers to work with
$endgroup$
– Banbadle
2 days ago
|
show 4 more comments
$begingroup$
There are 12 straight lines, not 10.
$endgroup$
– Travis
2 days ago
$begingroup$
Wow that was dumb, was thinking of a 4x4 for some reason. Fixed now, thanks.
$endgroup$
– Banbadle
2 days ago
$begingroup$
On a more basic level, how would you even calculate $Pl_k$? You may be able to calculate the probability of line $l_k$ being filled by $le T$ moves, and from that the prob of the line being filled at exactly the $T$th move, but for the line to actually *win* this would also require the other $11$ lines being not yet filled - and that seems like a really difficult problem.
$endgroup$
– antkam
2 days ago
$begingroup$
If you let $Pl_{i,t}$ be the probability of it being solved after t events, would trying to maximise the limit of $sup({ {Pl_{i,t} / Pl_{j,t}} | Pl_{i,t} leqslant Pl_{j,t}})$ as t tends to infinity be the same as what I've said above? I was hoping there would be some kind of algorithm I was too dumb to find to help with this. I get calculating the probabilities separately would be too hard, but I couldn't see a better way to formulate it.
$endgroup$
– Banbadle
2 days ago
$begingroup$
Actually, reading my last comment over, if the formulation is the same then wouldn't maximising that for any t>4 lead to the solution? Meaning that I'm just minimising the difference between the chance of getting one line after 5 tries and another line after 5 tries. They're then very easy numbers to work with
$endgroup$
– Banbadle
2 days ago
$begingroup$
There are 12 straight lines, not 10.
$endgroup$
– Travis
2 days ago
$begingroup$
There are 12 straight lines, not 10.
$endgroup$
– Travis
2 days ago
$begingroup$
Wow that was dumb, was thinking of a 4x4 for some reason. Fixed now, thanks.
$endgroup$
– Banbadle
2 days ago
$begingroup$
Wow that was dumb, was thinking of a 4x4 for some reason. Fixed now, thanks.
$endgroup$
– Banbadle
2 days ago
$begingroup$
On a more basic level, how would you even calculate $Pl_k$? You may be able to calculate the probability of line $l_k$ being filled by $le T$ moves, and from that the prob of the line being filled at exactly the $T$th move, but for the line to actually *win* this would also require the other $11$ lines being not yet filled - and that seems like a really difficult problem.
$endgroup$
– antkam
2 days ago
$begingroup$
On a more basic level, how would you even calculate $Pl_k$? You may be able to calculate the probability of line $l_k$ being filled by $le T$ moves, and from that the prob of the line being filled at exactly the $T$th move, but for the line to actually *win* this would also require the other $11$ lines being not yet filled - and that seems like a really difficult problem.
$endgroup$
– antkam
2 days ago
$begingroup$
If you let $Pl_{i,t}$ be the probability of it being solved after t events, would trying to maximise the limit of $sup({ {Pl_{i,t} / Pl_{j,t}} | Pl_{i,t} leqslant Pl_{j,t}})$ as t tends to infinity be the same as what I've said above? I was hoping there would be some kind of algorithm I was too dumb to find to help with this. I get calculating the probabilities separately would be too hard, but I couldn't see a better way to formulate it.
$endgroup$
– Banbadle
2 days ago
$begingroup$
If you let $Pl_{i,t}$ be the probability of it being solved after t events, would trying to maximise the limit of $sup({ {Pl_{i,t} / Pl_{j,t}} | Pl_{i,t} leqslant Pl_{j,t}})$ as t tends to infinity be the same as what I've said above? I was hoping there would be some kind of algorithm I was too dumb to find to help with this. I get calculating the probabilities separately would be too hard, but I couldn't see a better way to formulate it.
$endgroup$
– Banbadle
2 days ago
$begingroup$
Actually, reading my last comment over, if the formulation is the same then wouldn't maximising that for any t>4 lead to the solution? Meaning that I'm just minimising the difference between the chance of getting one line after 5 tries and another line after 5 tries. They're then very easy numbers to work with
$endgroup$
– Banbadle
2 days ago
$begingroup$
Actually, reading my last comment over, if the formulation is the same then wouldn't maximising that for any t>4 lead to the solution? Meaning that I'm just minimising the difference between the chance of getting one line after 5 tries and another line after 5 tries. They're then very easy numbers to work with
$endgroup$
– Banbadle
2 days ago
|
show 4 more comments
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$begingroup$
There are 12 straight lines, not 10.
$endgroup$
– Travis
2 days ago
$begingroup$
Wow that was dumb, was thinking of a 4x4 for some reason. Fixed now, thanks.
$endgroup$
– Banbadle
2 days ago
$begingroup$
On a more basic level, how would you even calculate $Pl_k$? You may be able to calculate the probability of line $l_k$ being filled by $le T$ moves, and from that the prob of the line being filled at exactly the $T$th move, but for the line to actually *win* this would also require the other $11$ lines being not yet filled - and that seems like a really difficult problem.
$endgroup$
– antkam
2 days ago
$begingroup$
If you let $Pl_{i,t}$ be the probability of it being solved after t events, would trying to maximise the limit of $sup({ {Pl_{i,t} / Pl_{j,t}} | Pl_{i,t} leqslant Pl_{j,t}})$ as t tends to infinity be the same as what I've said above? I was hoping there would be some kind of algorithm I was too dumb to find to help with this. I get calculating the probabilities separately would be too hard, but I couldn't see a better way to formulate it.
$endgroup$
– Banbadle
2 days ago
$begingroup$
Actually, reading my last comment over, if the formulation is the same then wouldn't maximising that for any t>4 lead to the solution? Meaning that I'm just minimising the difference between the chance of getting one line after 5 tries and another line after 5 tries. They're then very easy numbers to work with
$endgroup$
– Banbadle
2 days ago