Board game probabilities

[Vanessa23]

If you are given two dice and a gameboard how would you figure out the probability of landing on each of the individual squares in one game?

Since there are two dice with 6 faces you can make a few statements without much math:
0 possibility of landing on first square,
(1/6)*(1/6) probability of landing on squares 2 and 3.

However, after that the probability of landing on a square between 4 and 12 is related to both the probability of rolling that sum as well as the probability of landing on a previous square AND rolling the subsequently necessary sum.
For example, to land on square 5 you could roll a (1,4) OR (2,3) OR {(1,1) AND (1,2)} OR {(1,2) AND (1,1)}
In the example I did not include (4,1) or (3,2) because it would have resulted in landing on the same square, yet {(1,1) AND (1,2)} would have been due to landing on square 2 and then square 5, whereas {(1,2) AND (1,1)} would have resulted in landing on square 3 and then square 5.

Which equation can you use to determine the probability of landing on each space of the board? THANK YOU for any thoughts, suggestions or resources.

 

[HallsofIvy]

If you are given two dice and a gameboard how would you figure out the probability of landing on each of the individual squares in one game?

Since there are two dice with 6 faces you can make a few statements without much math:
0 possibility of landing on first square,
(1/6)*(1/6) probability of landing on squares 2 and 3.

I think you had better us a little math! Yes, since the smallesst number you can roll with two die is (1,1)= 2, you cannot land on square 1: probability 0. Yes, the only way you can land on the second square is to roll (1,1)= 2: probability (1/6)(1/6)= 1/36. But you can land on the third square by rolling either (1,2)= 3 or (2,1)= 3. Probability (1/6)(1/6)+ (1/6)(1/6)= 1/18.

However, after that the probability of landing on a square between 4 and 12 is related to both the probability of rolling that sum as well as the probability of landing on a previous square AND rolling the subsequently necessary sum.
For example, to land on square 5 you could roll a (1,4) OR (2,3) OR {(1,1) AND (1,2)} OR {(1,2) AND (1,1)}
In the example I did not include (4,1) or (3,2) because it would have resulted in landing on the same square, yet {(1,1) AND (1,2)} would have been due to landing on square 2 and then square 5, whereas {(1,2) AND (1,1)} would have resulted in landing on square 3 and then square 5.

Which equation can you use to determine the probability of landing on each space of the board? THANK YOU for any thoughts, suggestions or resources.

Are you going to count ways of going completely around the board and then landing on the square? In that case, the probabilities of square 1 and 2 that you calculated earlier are wrong! The first thing you need to do is specify your question more clearly. Asking "what is the probability of landing on a square in one turn" is one thing. Asking "what is the probability of landing on a square in two turns" is another thing. Asking "what is the probability of landing on a square in one or two turns" is yet a third question.

 

[Vanessa23]

I want to know the probability of landing on each individual square if you go around the board once. Therefore, the probability of landing on squares one and two are correct because you would not go back around the board after you pass the last square.
The tricky part is that after square 2 you can land on the rest of the squares after one roll or multiple rolls depending on how many squares you advance per turn.
This is shown if I clarify the example I already used about landing on square 5. I could either roll once and get a (1,4) or (2,3) however, I could also roll the dice on the first turn and get a (1,1) to land on square 2 and then on the second roll get a (1,2) or (2,1) to end up on square 5. Thus, landing on square 5 is based on the probability of rolling a sum of 5 and also on the probability of landing on square 2 and rolling a 3 and the probability of landing on square 3 and rolling a 2. Once you get to higher squares the combinations of probabilities of landing on that square increase rapidly. So is there a set of equations that can be used to simplify this? Is there an easier way of finding out the total probability of landing on square 25 besides adding up all the ways you could possibly land on square 25?

 

[CRGreathouse]

Is there an easier way of finding out the total probability of landing on square 25 besides adding up all the ways you could possibly land on square 25?

I don't actually think it's even as easy as the hard way you mention!

Consider square 4. You can get there with a 4 from square 0, or with a 2 from square 2. But it's not fair to take P(4) + P(2) = 3/36 + 1/36 = 1/9, since it's not certain that you'll ever land on square 2! You need to take the chance of landing on square 2 times the chance of rolling a 2 and add in the chance of rolling a four: P(2)P(2) + P(4) = 1/36 * 1/36 + 3/36 = 109/1296.

I suggest making an array of length N (where N is the number of squares on the board), setting P[0] = 1, and for i from 1 to N-1 setting P = P[i - 12] * 1/36 + P[i - 11] * 2/36 + ... + P[i - 2] * 1/36. Of course for the first dozen you'll need to ignore the entries that go below 0.

 

[CRGreathouse]

I was able to code the above in two lines (82 characters) in Pari:

v=vector(99);v[1]=1.;
for(i=2,#v,v[i]=sum(j=2,min(i-1,12),v[i-j]*min(13-j,j-1)/36))

Obviously the size and runtime depend on the number of squares on the board; I used 99 squares, which took too little time to record (0ms). If I use exact math instead of floating-point, I get amusing fractions. Here's the exact probability of landing on the last (98th) square:

 2592534218952623615698757249804540065914930328447881003644498582188380328513
-----------------------------------------------------------------------------
18147739541668636280463618532168272792698436402026524209529776843597142818816

 

[Vanessa23]

Consider square 4. You can get there with a 4 from square 0, or with a 2 from square 2. But it's not fair to take P(4) + P(2) = 3/36 + 1/36 = 1/9, since it's not certain that you'll ever land on square 2! You need to take the chance of landing on square 2 times the chance of rolling a 2 and add in the chance of rolling a four: P(2)P(2) + P(4) = 1/36 * 1/36 + 3/36 = 109/1296.

I guess I didn't make it clear in my first post, but I understand that is how you would find the probabilities for squares higher than 4.

I suggest making an array of length N (where N is the number of squares on the board), setting P[0] = 1, and for i from 1 to N-1 setting P = P[i - 12] * 1/36 + P[i - 11] * 2/36 + ... + P[i - 2] * 1/36. Of course for the first dozen you'll need to ignore the entries that go below 0.[/QUOTE]

Unfortunately, I had my only statistics class 2 years ago and don't really know what you just suggested. I don't know what Pari is either. I was trying to set this up in excel...? So I am guessing that "i" is the particular square that you want to find the probability for and N is the total number of squares. But which equation would you use in P[i-12] etc?
I thought it might include a binomial distribution, but even that got complicated.

 

[CRGreathouse]

Actually, Excel is a good idea. That wouldn't be hard to set up.

Just do the first 13 by hand, then set up a formula and drag it as far as needed; something like (in A14): =a1/36+a2*2/36+...+a12/36

 

[Pennybags]

Here's an Excel file using the number of spaces (40) on my favorite game board. Critiques are welcome; please let me know if I've made any errors. Contrary to the above post, this is the formula I arrived at for A14:
=A2*(1/36)+A3*(2/36)+A4*(3/36)+A5*(4/36)+A6*(5/36)+A7*(6/36)+A8*(5/36)+A9*(4/36)+A10*(3/36)+A11*(2/36)+A12*(1/36)

One slightly counter-intuitive result: the next-lowest probability spaces following the most-probable seventh, are the 12th, 13th, and 14th. This despite the fact that 14 is easily achieved by rolling two consecutive sevens! (Among several other high-probability combinations).

Also, one expected result: beginning around the 23rd space, the initial probability of landing on any given *distant* space is right around 14.28% (1 in 7). So, on this particular game board, a token more than about halfway around the board cannot be accurately predicted to have a bias toward landing on any particular space during its next transit.

Of course, these results do not take into account many other means by which a token may be moved, which (depending upon the game) may include spaces with special instructions, card draws, and actions of other players.