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Where is the bomb: How to estimate the probability, given row and column totals?
An inconsistency between the concept of “subindependence” and the chi-square test for independence?How to find probability with multiple pieces of evidence?Probability of independent events given the historyProbability that two random letters from a language will be the same?LLR with Positive and Negative Values vs. Dunning method with Entropy-based CalculationWhat is the probability that a sequence of events completes within a given time interval?Given hit probability and number of projectiles, randomly determine number of hitsProbability of getting the exact same letters in Scrabble 2 turns in a row?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
This question is inspired by a mini-game from Pokemon Soulsilver:
Imagine there are 15 bombs hidden on this 5x6 area:

Now, how would you estimate the probability to find a bomb on a specific field, given the row/column totals?
If you look at column 5 (total bombs = 5), then you might think: Within this column the chance to find a bomb in row 2 is double the chance to find one in row 1.
This (wrong) assumption of direct proportionality, which basically can be described as drawing standard independence-test operations (like in Chi-Square) into the wrong context, would lead to the following estimations:

As you can see, direct proportionality leads to probability estimates over 100%, and even before that, would be wrong.
So I performed a computational simulation of all possible permutations which led to 276 unique possibilities of placing 15 bombs. (given row and column totals)
Here is the average over the 276 solutions:
This is the correct solution, but due to the exponential computational work, I would like to find an estimation method.
My question is now: Is there an established statistical method to estimating this?
I was wondering if this was a known problem, how it is called and if there are papers/websites you could recommend!
probability estimation chi-squared independence games
New contributor
KaPy3141 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 2 more comments
$begingroup$
This question is inspired by a mini-game from Pokemon Soulsilver:
Imagine there are 15 bombs hidden on this 5x6 area:

Now, how would you estimate the probability to find a bomb on a specific field, given the row/column totals?
If you look at column 5 (total bombs = 5), then you might think: Within this column the chance to find a bomb in row 2 is double the chance to find one in row 1.
This (wrong) assumption of direct proportionality, which basically can be described as drawing standard independence-test operations (like in Chi-Square) into the wrong context, would lead to the following estimations:

As you can see, direct proportionality leads to probability estimates over 100%, and even before that, would be wrong.
So I performed a computational simulation of all possible permutations which led to 276 unique possibilities of placing 15 bombs. (given row and column totals)
Here is the average over the 276 solutions:
This is the correct solution, but due to the exponential computational work, I would like to find an estimation method.
My question is now: Is there an established statistical method to estimating this?
I was wondering if this was a known problem, how it is called and if there are papers/websites you could recommend!
probability estimation chi-squared independence games
New contributor
KaPy3141 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Fast and easy approach: For higher number of rows & columns, you could conduct a Monte Carlo simulation, where you would check the random subsample of the possible configurations that is lower then then total number of possibilities. It'd give you an approximate solution.
$endgroup$
– Tim♦
8 hours ago
$begingroup$
Thanks for your comment! Unfortunately I see a dilemma in this approach: If I just randomly place 15 bombs, I need 268 million tries to get just 70 (non-unique!) solutions. -> It is super unlikely that random placement leads to a valid solution. (=correct totals)
$endgroup$
– KaPy3141
7 hours ago
$begingroup$
The dilemma is that, if I now help the simulation, by using rules how to place the subsequent bombs after the first one. Then I would distort the result! (Every field I start produces a solution -> 50:50 bias)
$endgroup$
– KaPy3141
7 hours ago
1
$begingroup$
I dont understand your computational solution. What are the numbers in cells? They certainly don't add up to 100%, it's not PMF. They also dont look like CDF, the right/bottom cell is not 100%
$endgroup$
– Aksakal
4 hours ago
1
$begingroup$
@Aksakal These are the marginal probabilities that any given cell contains a bomb. The numbers add to 15, the number of total bombs on the board.
$endgroup$
– Dougal
3 hours ago
|
show 2 more comments
$begingroup$
This question is inspired by a mini-game from Pokemon Soulsilver:
Imagine there are 15 bombs hidden on this 5x6 area:

Now, how would you estimate the probability to find a bomb on a specific field, given the row/column totals?
If you look at column 5 (total bombs = 5), then you might think: Within this column the chance to find a bomb in row 2 is double the chance to find one in row 1.
This (wrong) assumption of direct proportionality, which basically can be described as drawing standard independence-test operations (like in Chi-Square) into the wrong context, would lead to the following estimations:

As you can see, direct proportionality leads to probability estimates over 100%, and even before that, would be wrong.
So I performed a computational simulation of all possible permutations which led to 276 unique possibilities of placing 15 bombs. (given row and column totals)
Here is the average over the 276 solutions:
This is the correct solution, but due to the exponential computational work, I would like to find an estimation method.
My question is now: Is there an established statistical method to estimating this?
I was wondering if this was a known problem, how it is called and if there are papers/websites you could recommend!
probability estimation chi-squared independence games
New contributor
KaPy3141 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This question is inspired by a mini-game from Pokemon Soulsilver:
Imagine there are 15 bombs hidden on this 5x6 area:

Now, how would you estimate the probability to find a bomb on a specific field, given the row/column totals?
If you look at column 5 (total bombs = 5), then you might think: Within this column the chance to find a bomb in row 2 is double the chance to find one in row 1.
This (wrong) assumption of direct proportionality, which basically can be described as drawing standard independence-test operations (like in Chi-Square) into the wrong context, would lead to the following estimations:

As you can see, direct proportionality leads to probability estimates over 100%, and even before that, would be wrong.
So I performed a computational simulation of all possible permutations which led to 276 unique possibilities of placing 15 bombs. (given row and column totals)
Here is the average over the 276 solutions:
This is the correct solution, but due to the exponential computational work, I would like to find an estimation method.
My question is now: Is there an established statistical method to estimating this?
I was wondering if this was a known problem, how it is called and if there are papers/websites you could recommend!
probability estimation chi-squared independence games
probability estimation chi-squared independence games
New contributor
KaPy3141 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
KaPy3141 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
KaPy3141 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
KaPy3141KaPy3141
211 bronze badge
211 bronze badge
New contributor
KaPy3141 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
KaPy3141 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Fast and easy approach: For higher number of rows & columns, you could conduct a Monte Carlo simulation, where you would check the random subsample of the possible configurations that is lower then then total number of possibilities. It'd give you an approximate solution.
$endgroup$
– Tim♦
8 hours ago
$begingroup$
Thanks for your comment! Unfortunately I see a dilemma in this approach: If I just randomly place 15 bombs, I need 268 million tries to get just 70 (non-unique!) solutions. -> It is super unlikely that random placement leads to a valid solution. (=correct totals)
$endgroup$
– KaPy3141
7 hours ago
$begingroup$
The dilemma is that, if I now help the simulation, by using rules how to place the subsequent bombs after the first one. Then I would distort the result! (Every field I start produces a solution -> 50:50 bias)
$endgroup$
– KaPy3141
7 hours ago
1
$begingroup$
I dont understand your computational solution. What are the numbers in cells? They certainly don't add up to 100%, it's not PMF. They also dont look like CDF, the right/bottom cell is not 100%
$endgroup$
– Aksakal
4 hours ago
1
$begingroup$
@Aksakal These are the marginal probabilities that any given cell contains a bomb. The numbers add to 15, the number of total bombs on the board.
$endgroup$
– Dougal
3 hours ago
|
show 2 more comments
$begingroup$
Fast and easy approach: For higher number of rows & columns, you could conduct a Monte Carlo simulation, where you would check the random subsample of the possible configurations that is lower then then total number of possibilities. It'd give you an approximate solution.
$endgroup$
– Tim♦
8 hours ago
$begingroup$
Thanks for your comment! Unfortunately I see a dilemma in this approach: If I just randomly place 15 bombs, I need 268 million tries to get just 70 (non-unique!) solutions. -> It is super unlikely that random placement leads to a valid solution. (=correct totals)
$endgroup$
– KaPy3141
7 hours ago
$begingroup$
The dilemma is that, if I now help the simulation, by using rules how to place the subsequent bombs after the first one. Then I would distort the result! (Every field I start produces a solution -> 50:50 bias)
$endgroup$
– KaPy3141
7 hours ago
1
$begingroup$
I dont understand your computational solution. What are the numbers in cells? They certainly don't add up to 100%, it's not PMF. They also dont look like CDF, the right/bottom cell is not 100%
$endgroup$
– Aksakal
4 hours ago
1
$begingroup$
@Aksakal These are the marginal probabilities that any given cell contains a bomb. The numbers add to 15, the number of total bombs on the board.
$endgroup$
– Dougal
3 hours ago
$begingroup$
Fast and easy approach: For higher number of rows & columns, you could conduct a Monte Carlo simulation, where you would check the random subsample of the possible configurations that is lower then then total number of possibilities. It'd give you an approximate solution.
$endgroup$
– Tim♦
8 hours ago
$begingroup$
Fast and easy approach: For higher number of rows & columns, you could conduct a Monte Carlo simulation, where you would check the random subsample of the possible configurations that is lower then then total number of possibilities. It'd give you an approximate solution.
$endgroup$
– Tim♦
8 hours ago
$begingroup$
Thanks for your comment! Unfortunately I see a dilemma in this approach: If I just randomly place 15 bombs, I need 268 million tries to get just 70 (non-unique!) solutions. -> It is super unlikely that random placement leads to a valid solution. (=correct totals)
$endgroup$
– KaPy3141
7 hours ago
$begingroup$
Thanks for your comment! Unfortunately I see a dilemma in this approach: If I just randomly place 15 bombs, I need 268 million tries to get just 70 (non-unique!) solutions. -> It is super unlikely that random placement leads to a valid solution. (=correct totals)
$endgroup$
– KaPy3141
7 hours ago
$begingroup$
The dilemma is that, if I now help the simulation, by using rules how to place the subsequent bombs after the first one. Then I would distort the result! (Every field I start produces a solution -> 50:50 bias)
$endgroup$
– KaPy3141
7 hours ago
$begingroup$
The dilemma is that, if I now help the simulation, by using rules how to place the subsequent bombs after the first one. Then I would distort the result! (Every field I start produces a solution -> 50:50 bias)
$endgroup$
– KaPy3141
7 hours ago
1
1
$begingroup$
I dont understand your computational solution. What are the numbers in cells? They certainly don't add up to 100%, it's not PMF. They also dont look like CDF, the right/bottom cell is not 100%
$endgroup$
– Aksakal
4 hours ago
$begingroup$
I dont understand your computational solution. What are the numbers in cells? They certainly don't add up to 100%, it's not PMF. They also dont look like CDF, the right/bottom cell is not 100%
$endgroup$
– Aksakal
4 hours ago
1
1
$begingroup$
@Aksakal These are the marginal probabilities that any given cell contains a bomb. The numbers add to 15, the number of total bombs on the board.
$endgroup$
– Dougal
3 hours ago
$begingroup$
@Aksakal These are the marginal probabilities that any given cell contains a bomb. The numbers add to 15, the number of total bombs on the board.
$endgroup$
– Dougal
3 hours ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
There's no unique solution
I don't think that true discrete probability distribution can be recovered, unless you make some additional assumptions. Your situation is basically a problem of recovering the joint distribution from marginals. It is sometimes solved by using copulas in the industry, for example financial risk management, but usually for continuous distributions.
The special case of independent rows and columns is easy: $P_i^j=P_itimes P^j$
where $P_i$ and $P^j$ are marginals of rows and columns. For instance, row $P_6=3/15=0.2$ and column $P^3=3/15=0.2$, hence the probability that a bomb is in row 6 and column 3 is $P_6^3=0.04$. This is assuming that more than one bomb can be placed in a cell, of course. You actually produced this distribution in your first table.
Discrete Copulas
There are discrete copulas, but they have issues: they're not unique. It doesn't make them useless. So, I'd try applying discrete copulas. You can find a good overview of them in Genest, C. and J. Nešlehová (2007). A primer on copulas for count data. Astin Bull. 37(2), 475–515.
Copulas can be especially useful, as they usually allow to explicitly induce dependence, or to estimate it from data when the data is available. I mean the dependence of row and columns when placing bombs. For instance, it could be the case when if the bomb is one the first row, then it is more likely that it will be one the first column too.
Example
Let's apply Kimeldorf and Sampson copula to your data, assuming again that more than one bomb can be placed in a cell. The copula for a dependency parameter $theta$ is defined as:
$$C(u,v)=(u^-theta+u^-theta-1)^-1/theta$$
You can think of $theta$ as an analog of the correlation coefficient.
Independent
Let's start with the case of weak dependence, $theta=0.000001$, where we have the following probabilities (PMF) and marginal PDFs are shown too on the panels on the right and at the bottom:

You can see how in the column 5 the second row probability does have twice higher probability than the first row. This is not wrong contrary to what you seemed to imply in your question. All probability do add up to 100%, of course, as do the marginals on the panels match the frequencies. For instance, the column 5 in the lower panel shows 1/3 which corresponds to stated 5 bombs out of total 15 as expected.
Positive Correlation
For stronger dependency (positive correlation) with $theta=10$ we have the following:

Negative Correlation
The same for stronger but negative correlation (dependency) $theta=-0.2$:

You can see that all probabilities add up to 100%, of course. Also, you can see how dependency impacts the PMF's shape. For positive dependency (correlation) you get the highest PMF concentrated on the diagonal, while for the negative dependency it is off-diagonal
No more than one bomb in a field
My copula solution will not work here without modification. We need to use copulas that hold the constrain that $P_i^jle 1$
$endgroup$
add a comment
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$begingroup$
There's no unique solution
I don't think that true discrete probability distribution can be recovered, unless you make some additional assumptions. Your situation is basically a problem of recovering the joint distribution from marginals. It is sometimes solved by using copulas in the industry, for example financial risk management, but usually for continuous distributions.
The special case of independent rows and columns is easy: $P_i^j=P_itimes P^j$
where $P_i$ and $P^j$ are marginals of rows and columns. For instance, row $P_6=3/15=0.2$ and column $P^3=3/15=0.2$, hence the probability that a bomb is in row 6 and column 3 is $P_6^3=0.04$. This is assuming that more than one bomb can be placed in a cell, of course. You actually produced this distribution in your first table.
Discrete Copulas
There are discrete copulas, but they have issues: they're not unique. It doesn't make them useless. So, I'd try applying discrete copulas. You can find a good overview of them in Genest, C. and J. Nešlehová (2007). A primer on copulas for count data. Astin Bull. 37(2), 475–515.
Copulas can be especially useful, as they usually allow to explicitly induce dependence, or to estimate it from data when the data is available. I mean the dependence of row and columns when placing bombs. For instance, it could be the case when if the bomb is one the first row, then it is more likely that it will be one the first column too.
Example
Let's apply Kimeldorf and Sampson copula to your data, assuming again that more than one bomb can be placed in a cell. The copula for a dependency parameter $theta$ is defined as:
$$C(u,v)=(u^-theta+u^-theta-1)^-1/theta$$
You can think of $theta$ as an analog of the correlation coefficient.
Independent
Let's start with the case of weak dependence, $theta=0.000001$, where we have the following probabilities (PMF) and marginal PDFs are shown too on the panels on the right and at the bottom:

You can see how in the column 5 the second row probability does have twice higher probability than the first row. This is not wrong contrary to what you seemed to imply in your question. All probability do add up to 100%, of course, as do the marginals on the panels match the frequencies. For instance, the column 5 in the lower panel shows 1/3 which corresponds to stated 5 bombs out of total 15 as expected.
Positive Correlation
For stronger dependency (positive correlation) with $theta=10$ we have the following:

Negative Correlation
The same for stronger but negative correlation (dependency) $theta=-0.2$:

You can see that all probabilities add up to 100%, of course. Also, you can see how dependency impacts the PMF's shape. For positive dependency (correlation) you get the highest PMF concentrated on the diagonal, while for the negative dependency it is off-diagonal
No more than one bomb in a field
My copula solution will not work here without modification. We need to use copulas that hold the constrain that $P_i^jle 1$
$endgroup$
add a comment
|
$begingroup$
There's no unique solution
I don't think that true discrete probability distribution can be recovered, unless you make some additional assumptions. Your situation is basically a problem of recovering the joint distribution from marginals. It is sometimes solved by using copulas in the industry, for example financial risk management, but usually for continuous distributions.
The special case of independent rows and columns is easy: $P_i^j=P_itimes P^j$
where $P_i$ and $P^j$ are marginals of rows and columns. For instance, row $P_6=3/15=0.2$ and column $P^3=3/15=0.2$, hence the probability that a bomb is in row 6 and column 3 is $P_6^3=0.04$. This is assuming that more than one bomb can be placed in a cell, of course. You actually produced this distribution in your first table.
Discrete Copulas
There are discrete copulas, but they have issues: they're not unique. It doesn't make them useless. So, I'd try applying discrete copulas. You can find a good overview of them in Genest, C. and J. Nešlehová (2007). A primer on copulas for count data. Astin Bull. 37(2), 475–515.
Copulas can be especially useful, as they usually allow to explicitly induce dependence, or to estimate it from data when the data is available. I mean the dependence of row and columns when placing bombs. For instance, it could be the case when if the bomb is one the first row, then it is more likely that it will be one the first column too.
Example
Let's apply Kimeldorf and Sampson copula to your data, assuming again that more than one bomb can be placed in a cell. The copula for a dependency parameter $theta$ is defined as:
$$C(u,v)=(u^-theta+u^-theta-1)^-1/theta$$
You can think of $theta$ as an analog of the correlation coefficient.
Independent
Let's start with the case of weak dependence, $theta=0.000001$, where we have the following probabilities (PMF) and marginal PDFs are shown too on the panels on the right and at the bottom:

You can see how in the column 5 the second row probability does have twice higher probability than the first row. This is not wrong contrary to what you seemed to imply in your question. All probability do add up to 100%, of course, as do the marginals on the panels match the frequencies. For instance, the column 5 in the lower panel shows 1/3 which corresponds to stated 5 bombs out of total 15 as expected.
Positive Correlation
For stronger dependency (positive correlation) with $theta=10$ we have the following:

Negative Correlation
The same for stronger but negative correlation (dependency) $theta=-0.2$:

You can see that all probabilities add up to 100%, of course. Also, you can see how dependency impacts the PMF's shape. For positive dependency (correlation) you get the highest PMF concentrated on the diagonal, while for the negative dependency it is off-diagonal
No more than one bomb in a field
My copula solution will not work here without modification. We need to use copulas that hold the constrain that $P_i^jle 1$
$endgroup$
add a comment
|
$begingroup$
There's no unique solution
I don't think that true discrete probability distribution can be recovered, unless you make some additional assumptions. Your situation is basically a problem of recovering the joint distribution from marginals. It is sometimes solved by using copulas in the industry, for example financial risk management, but usually for continuous distributions.
The special case of independent rows and columns is easy: $P_i^j=P_itimes P^j$
where $P_i$ and $P^j$ are marginals of rows and columns. For instance, row $P_6=3/15=0.2$ and column $P^3=3/15=0.2$, hence the probability that a bomb is in row 6 and column 3 is $P_6^3=0.04$. This is assuming that more than one bomb can be placed in a cell, of course. You actually produced this distribution in your first table.
Discrete Copulas
There are discrete copulas, but they have issues: they're not unique. It doesn't make them useless. So, I'd try applying discrete copulas. You can find a good overview of them in Genest, C. and J. Nešlehová (2007). A primer on copulas for count data. Astin Bull. 37(2), 475–515.
Copulas can be especially useful, as they usually allow to explicitly induce dependence, or to estimate it from data when the data is available. I mean the dependence of row and columns when placing bombs. For instance, it could be the case when if the bomb is one the first row, then it is more likely that it will be one the first column too.
Example
Let's apply Kimeldorf and Sampson copula to your data, assuming again that more than one bomb can be placed in a cell. The copula for a dependency parameter $theta$ is defined as:
$$C(u,v)=(u^-theta+u^-theta-1)^-1/theta$$
You can think of $theta$ as an analog of the correlation coefficient.
Independent
Let's start with the case of weak dependence, $theta=0.000001$, where we have the following probabilities (PMF) and marginal PDFs are shown too on the panels on the right and at the bottom:

You can see how in the column 5 the second row probability does have twice higher probability than the first row. This is not wrong contrary to what you seemed to imply in your question. All probability do add up to 100%, of course, as do the marginals on the panels match the frequencies. For instance, the column 5 in the lower panel shows 1/3 which corresponds to stated 5 bombs out of total 15 as expected.
Positive Correlation
For stronger dependency (positive correlation) with $theta=10$ we have the following:

Negative Correlation
The same for stronger but negative correlation (dependency) $theta=-0.2$:

You can see that all probabilities add up to 100%, of course. Also, you can see how dependency impacts the PMF's shape. For positive dependency (correlation) you get the highest PMF concentrated on the diagonal, while for the negative dependency it is off-diagonal
No more than one bomb in a field
My copula solution will not work here without modification. We need to use copulas that hold the constrain that $P_i^jle 1$
$endgroup$
There's no unique solution
I don't think that true discrete probability distribution can be recovered, unless you make some additional assumptions. Your situation is basically a problem of recovering the joint distribution from marginals. It is sometimes solved by using copulas in the industry, for example financial risk management, but usually for continuous distributions.
The special case of independent rows and columns is easy: $P_i^j=P_itimes P^j$
where $P_i$ and $P^j$ are marginals of rows and columns. For instance, row $P_6=3/15=0.2$ and column $P^3=3/15=0.2$, hence the probability that a bomb is in row 6 and column 3 is $P_6^3=0.04$. This is assuming that more than one bomb can be placed in a cell, of course. You actually produced this distribution in your first table.
Discrete Copulas
There are discrete copulas, but they have issues: they're not unique. It doesn't make them useless. So, I'd try applying discrete copulas. You can find a good overview of them in Genest, C. and J. Nešlehová (2007). A primer on copulas for count data. Astin Bull. 37(2), 475–515.
Copulas can be especially useful, as they usually allow to explicitly induce dependence, or to estimate it from data when the data is available. I mean the dependence of row and columns when placing bombs. For instance, it could be the case when if the bomb is one the first row, then it is more likely that it will be one the first column too.
Example
Let's apply Kimeldorf and Sampson copula to your data, assuming again that more than one bomb can be placed in a cell. The copula for a dependency parameter $theta$ is defined as:
$$C(u,v)=(u^-theta+u^-theta-1)^-1/theta$$
You can think of $theta$ as an analog of the correlation coefficient.
Independent
Let's start with the case of weak dependence, $theta=0.000001$, where we have the following probabilities (PMF) and marginal PDFs are shown too on the panels on the right and at the bottom:

You can see how in the column 5 the second row probability does have twice higher probability than the first row. This is not wrong contrary to what you seemed to imply in your question. All probability do add up to 100%, of course, as do the marginals on the panels match the frequencies. For instance, the column 5 in the lower panel shows 1/3 which corresponds to stated 5 bombs out of total 15 as expected.
Positive Correlation
For stronger dependency (positive correlation) with $theta=10$ we have the following:

Negative Correlation
The same for stronger but negative correlation (dependency) $theta=-0.2$:

You can see that all probabilities add up to 100%, of course. Also, you can see how dependency impacts the PMF's shape. For positive dependency (correlation) you get the highest PMF concentrated on the diagonal, while for the negative dependency it is off-diagonal
No more than one bomb in a field
My copula solution will not work here without modification. We need to use copulas that hold the constrain that $P_i^jle 1$
edited 2 hours ago
answered 7 hours ago
AksakalAksakal
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$begingroup$
Fast and easy approach: For higher number of rows & columns, you could conduct a Monte Carlo simulation, where you would check the random subsample of the possible configurations that is lower then then total number of possibilities. It'd give you an approximate solution.
$endgroup$
– Tim♦
8 hours ago
$begingroup$
Thanks for your comment! Unfortunately I see a dilemma in this approach: If I just randomly place 15 bombs, I need 268 million tries to get just 70 (non-unique!) solutions. -> It is super unlikely that random placement leads to a valid solution. (=correct totals)
$endgroup$
– KaPy3141
7 hours ago
$begingroup$
The dilemma is that, if I now help the simulation, by using rules how to place the subsequent bombs after the first one. Then I would distort the result! (Every field I start produces a solution -> 50:50 bias)
$endgroup$
– KaPy3141
7 hours ago
1
$begingroup$
I dont understand your computational solution. What are the numbers in cells? They certainly don't add up to 100%, it's not PMF. They also dont look like CDF, the right/bottom cell is not 100%
$endgroup$
– Aksakal
4 hours ago
1
$begingroup$
@Aksakal These are the marginal probabilities that any given cell contains a bomb. The numbers add to 15, the number of total bombs on the board.
$endgroup$
– Dougal
3 hours ago