# Calculating "Hybrid Probability" When Two Probabilities Affect the Same Event

by Affection
Tags: affect, event, hybrid probability, probabilities
 P: 1 In this "minesweeper grid:" ...I can calculate the probability of the square with the question mark being a mine at 2 in 3 using just the box numbered "3," or 1 in 2 using just the box numbered "4." It seems to me that knowing this, I should be able to calculate a "hybrid" probability that is somewhere between 0.5 and 0.67. Any ideas on how I can do this, or what one would even call such a calculation? Thanks!! --Jon
 Sci Advisor P: 3,085 Abstractly you have 3 events: E = there is a mine in the square with the question mark A = there is a 3 in the square you showed B = there is a 2 in the other square you showed. You have stated the "conditional probabilities" P(E given A) = 2/3 P(E given B) = 1/2 The probability you are asking about is P( E given (A and B) ) As far as terminology goes, I don't know any special name for this probability. (Minesweeper enthusiasts may have a name for it that I don't know about.) For an arbitrary set of events E,A,B, there is no general formula for calculating this probability using only the values of P(E given A) and P(E given B). You can write various expressions to calculate P(E given (A and B)) but they would include terms like P( E and A and B) and P(A and B). In the particular situation you gave, you can calculate P(E given (A and B)) by a combinatorial analysis.
 HW Helper P: 6,180 I believe the way to do it is to enumerate all possible configurations for the bombs. In your case there is a total of 7 possible bomb configurations based on the numbers you have. Of those configurations 6 have a bomb at your question mark (probability 6/7), but only 1 configuration has a bomb left of the question mark (probability 1/7). Your best bet with the information given, is to click the square left of the question mark.
P: 187

## Calculating "Hybrid Probability" When Two Probabilities Affect the Same Event

You are comparing different things. You can't just count the different arrangements because they are not equally likely. There are 4 squares of importance, the question mark, the one to its left, and the two above it if you are basing your decision on the placement of the 3 and the 4 and the number of bombs they are already touching (the 2 above the 3 doesn't matter for the square of interest in this case). There are three possibilities for those four squares: bomb to the left and two bombs above, bomb in the question mark and directly above it, or bomb in the question mark and the second square above it. The problem is that these are not equally likely because one scenario involves 3 bombs while two of them contain only two bombs.

Think of the game setup. Every square is either assigned a bomb with probability x or not with a probability 1-x where x is determined by the game maker but I assume it to be rather small. So my first scenario above occurred in the initial setup with probability xxx(1-x). The other two occurred with probability xx(1-x)(1-x) each. Given some small x, it is much more likely that the bomb is at the question mark than not. If you work out the calculations, the probability that there is a bomb on your question mark is 2(1-x)/(2-x) while the probability that there is not is only x/(2-x). Notice that the sum of these two probabilities is in fact one, meaning there either is or is not a bomb in your square. You should work it out for fun.

I think that the appropriate question to be asking is "given 4 squares and a small probability of any one square containing a bomb, is more likely that those 4 squares contain 2 bombs or 3 bombs".
 P: 4,568 Hey Affection and welcome to the forums. I agree with I Like Serena's comment in that you would need to do some kind of enumeration simply for the fact that you would need to take in all of the relevant neighbourhood information to get the most accurate value. The algorithm would not be straight-forward due to the nature of the system since it has a high amount of variation and because of the nature of how neighbourhood information affects the overall probability calculations.
 P: 187 No, you can't just enumerate the arrangements and assume that they are equally likely because they are, in general, not. The OP asked about using the information given by the existing 3 and 4. In that case, the possible arrangements are not equally likely because they don't contain the same number of bombs. I would agree that adding the existing 2 to the information set would be advisable. In that case there are 7 possible arrangements which all contain three bombs so they are equally likely, but that is a coincidence. It doesn't always happen in this game. As an exercise, consider what happens if the game maker assigned a bomb randomly to a square with probability 0.9 instead of a small probability. In that case it would be more likely that the original four squares of interest contained three bombs rather than two and it would be more likely that your question mark was not a bomb. The answer to the question depends on the parameters of the game.
HW Helper
P: 6,180
 Quote by alan2 No, you can't just enumerate the arrangements and assume that they are equally likely because they are, in general, not
Without information about the total number of bombs, I do believe they are equally likely.

Every configuration that has the (unspecified) total number of bombs would be equally likely.
As it is, we only look at the configurations that satisfy the numbers given.
These configuration will still be equally likely (without information about the total number of bombs).
 P: 187 I'm assuming that everyone knows the rules of the game. The original post referred to the four squares adjoining the 4 and the 3. There are 3 ways to satisfy these conditions. One way contains 3 mines, two ways contain 2 mines. We don't know how many mines there are in those 4 squares but we can find the probability of each configuration and they are not equally likely. Consider a simplified version of the same game. I have 2 cups, cup 1 and cup 2. Into each cup I either place a coin with probability x or I don't place a coin with probability (1-x). Now I ask you "does cup 1 contain a coin"? If you enumerate the possible arrangements and consider them equally likely you would say the arrangements in the two cups could be (00),(01),(10), or (11). Two of four have a coin in cup 1 so it's 50-50. That's not true unless x=1/2. Now I tell you that x=0.1. Then the probability that cup 1 contains a coin is (0.1)(0.9)+(0.1)(0.1)=0.1. The probability that it doesn't is (0.9)(0.9)+(0.1)(0.9)=0.9. If I tell you that x=0.9 then the probabilities are reversed and now the probability that cup 1 contains a coin is 0.9. Now, as in minesweeper, I give you more information. I look and tell you that at least one of the cups contains a coin. I have reduced the possible arrangements to (01), (10), and (11). The probability that cup 1 contains a coin is not 2/3 because the arrangements are not equally likely. In fact, the probability that cup 1 contains a coin is 1/(2-x) and the probability that it does not is (1-x)/(2-x). If x is 0.1 then you only have slightly better than 50-50 chance that cup 1 contains a coin. If x=0.9 then it is over 90% certain that cup 1 contains a coin. This results from the arrangements not being equally likely.
 Sci Advisor P: 3,085 alan2, That's a convincing argument. Doesn't it also invalidate the assertions in the original post about the probabilities 1/2 and 2/3 ?
 P: 187 Yes Stephen. The point is that the probability that one calculates can only depend on the information give. If the OP based the decision only on the existence of the 4, then it's a coin flip. If the OP based the decision only on the existence of the 3 then it's 2/3. But more information is necessarily better and adding an additional piece of information is not necessarily additive, it is frequently more restrictive, so adding the information about the existence of the 3 actually invalidates the assumptions drawn from the existence of the 4 because now there exist arrangements which are not equally likely..
 Sci Advisor P: 3,085 If you had just said "yes", I would have forgotten about this thread. But you say that a decision based only on the existence of the 4 is a coin flip. Is that because of the particular way that mines are randomized in the game Minesweeper? For example, this would say that there aren't any Minesweeper implementations that tend to place the mines in a row farther to the right.
 P: 187 Sorry Stephen. I don't know if you've every played the game. The 4 means that that square must touch exactly 4 mines, either side to side or corner to corner. It is already touching 3 and there are only two remaining squares where the last mine could be. In the absence of any other information, it could be in either of those with equal probability. It's the addition of information, say the 3, that changes it.
P: 3,085
 Quote by alan2 It's the addition of information, say the 3, that changes it.
Being a Bayesian at heart, I'm sympathetic to the idea of informtion changing probability . (Strictly speaking, I should say "information changing the probability space under consideration".) But I'm trying to precisely understand your objection the analysis to the combinatorial analysis which takes into the account the information from the square with 3 and the square with 4. You object to considering combinations with two bombs and combinations with three bombs as being equally likely. As I understand this objection it is based on the fact that bombs are initially placed by a process such that "place a bomb in square (x,y)" is an independent realization of a random variable for each square (x,y) and that the probability of initially placing a bomb in a square is some constant value p for each square. It this correct? Is p a known value in the game of Minesweeper?
 P: 187 You are correct Stephen. The reason that the assumption of equally likely events based solely on the information from the 3 and the 4 is incorrect is that the 3 possible ways to fill the relevant 4 squares with bombs to satisfy the conditions are not equally likely. One of the possible configurations contains 3 bombs while 2 configurations contain 2 bombs. Your assessment of the initial configuration is correct. I would guess from experience (my daughter likes this game) that p is on the order of 0.1-0.2. So any configuration of 4 squares that contains 3 bombs is necessarily less likely than any configuration which contains 2 bombs. Interestingly, if the OP considered the 2 above the 3 in the original analysis, then there would be seven squares of interest. All seven possible configurations of these seven squares which meet the conditions would contain 3 bombs and be equally likely. But we can't just assume things are equally likely. And, of course, as we progress through the game, we add new information with each move but that never assures us of equally likely outcomes for a given choice. Sometimes they are, sometimes they're not.
 Sci Advisor P: 3,085 Taking p as the probability that a bomb is initially placed in a square, lets solve the problem posed in the original post. We use only the information implied in this diagram: $$\begin{bmatrix} * & A & e & e \\ * & B & 3 & e \\ C & Q & m & e \\ m & 4 & m & e \\ e & e & e & * \end{bmatrix}$$ where 'm' indicates a square with a mine, 'e' indicates square known to be empty and A,B,C,Q and the '*' are squares in unknown condition. We consider the possible assignments of mines to squares A,B,C,Q subject to the requirements that there are exactly two mines in the set of squares A,B,Q and there is exactly one mine in the set of squares C,Q. The assignments and their probability (as members of a probability space consisting of all possible assigments of mines to those squares, given no other information). $$\begin{bmatrix} A & B & C & Q & prob\\ m & m & m & e & p^3(1-p)\\ e & m & e & m & (1-p)^2 p^2 \\ m & e & e & m & (1-p)^2 p^2 \end{bmatrix}$$ So restricting consideration to the probability space defined by the above events, we have that the probability that Q contains a mine is: $$\frac{2(1-p)^2p^2}{2(1-p)^2 p^2 + p^3(1-p) }$$ [Edited to fix typo pointed out by alan2 ] but remember, I've never played Minesweeper in my life.
 P: 187 You understand it. You did make one small error in the numerator of your final expression. It should be (1-p)^2, not (1-p^2). Make that correction, and you find that the probability that Q contains a mine is 2(1-p)/(2-p) as I pointed out in my initial post above. You did the same thing that I did but I really like your cool use of LaTex. I have to learn to do that. That was actually kind of fun even though I didn't think that I would create a fuss. I think that it points out a common error in probabilities applied to even simple situations.
 P: 187 P.S. I notice that this was the very first post ever by the original poster. I hope we didn't scare him away.