I wanna take a moment to talk about a terrible board game.
(Image links to the English Wikipedia article about the game)
It's one of these mostly-luck-based board games people keep in their shelves because you shouldn't throw plastic pieces into a fireplace, or the "classic" kind of board game, as this comic would put it. For those familiar with the game "Sorry!" or any of the trillion variations world wide, yeah, that one. It's mostly luck based, but there are still some decisions you can make during normal play. So I ended up asking myself the question: Given optimal play, assuming red goes first and the game proceeds clockwise, what is the probability of red, black, yellow and green winning?
First of all, I don't know the answer. It would take a lot of work to figure that out, but I wanted to figure out just how difficult it is. So I wanted to start with simplifying the game. Skipping a lot of details, each player begins with 4 pegs of their color in the "B" fields, rolling a 6 sided die at the beginning of their turn. If it's a 6, they can move one peg to the "A" field and roll again. The goal is to make it all the way around, as the white circle just before the "A" field leads to the "home row" (fields a-d). The crux is you must roll exactly the right number to move into it, so if a peg comes to a halt in front of the home row, you need to roll exactly a 4 to get to d, etc. Of course, the combinatorics of this is utter hell. So let's simplify. The sheer amount of time to even get anywhere close to a win condition makes the game obnoxious to analyze. So, what if we skip the entire race (and sending other player's pieces back to B) mechanics and put A in front of the home row? That also removes what little strategic dimension the game had, so now there is no more decision making. The optimal play is the only play!
This is at least... in principle analyzable. However, even though the game is now even more brain dead to play than the original, I would like to take a moment to show you people the issue with even this "stupid" a game: how many "game states" are there? That is, how many ways are there for pieces to be arranged on the board at a given player's turn? I'm gonna go over the rules this dumbed down version would have in a bit (well not quite, I'll make it even simpler), but I can assure you, basically at any given turn the pegs of each color can be in any configuration. We can ignore the B fields, since they are for all practical purposes just storage spaces for pegs that haven't "entered" the game properly. Black pegs in this simple version can only be on black fields. All black pegs are identical, so the "state" of black's part of the board is a series of 5 yes or no questions: "Is a peg on field A?", "Is a peg on field a?", "...b?", "..c?" and "..d?". Now..
Each player has up to 4 pieces in the game at any moment, meaning at most 4 of these questions are yes.
if a-d are all occupied, then A is empty and that player has won, meaning the game is over. That means that only one color can have the home row full.
Any other configuration can be realized, no matter whose turn it is. Just.. take me word for it, for now.
That means there are 5 configurations with 4 black pegs (either A is empty = win, or one of a-d is empty), 10 configurations with 3 pegs, 10 configurations with 2 pegs, 5 configurations with 1 peg and 1 configuration with 0 pegs, 31 configurations in total (per player) , of which 1 is a "win configuration". (I won't bore you with how I came up with those numbers, you could either write them all out on paper and count, use binomial coefficient magic you may recall from high school, whatever.) If each player can be in any of these configurations but only one player can win that means there are 30×30×30×30=30⁴= 810000 configurations of the board where no player has won. And since it could be the turn of any player for any of those, there are 4 times as many game states, 3240000. You see why I didn't even bother writing down the rules yet? I really don't wanna try look at probabilities for a game with that many possible ways the pieces can be arranged. Screw this, let's make it a two player game.
Now we're down to 1800 non-win game states. This is something I could easily program in Python or something and afterwards very carefully verify by hand. However, I still cannot be bothered to. So, why is it still so many combinations? Because every field in the home row almost doubles the number of possible arrangements a single color's pegs can have. So.. let's make it dumber still. Let's make it minimal. What is the smallest possible game? Well.. let's knock it down a few pegs (geddit? And you thought the math part was the suffering here).
BEHOLD.
With one piece each, there are only 3 configurations per player (peg in B, in A, or in a = win), meaning there are 4 configurations of the board where no player has won, thus 8 non-win game states (since it's either the turn of red or black). Since we don't actually care about the configuration of pieces of the board once a player has won we can add to the above two additional states "red wins" and "black wins", giving us a total of 10. We don't have to worry about whose turn it is because the rules will be such that you can only win during your own turn. All that we need now is names for the 10 different states and the rules of the game. Then, at least, we can determine the odds for red and black for the world's stupidest two player board game.™
Let's denote the board like this: (<position of red's piece>,<position of black's piece>;<who's rolling the die next>). To make it hopefully a bit more readable I will call "0" the potion where the peg is not in the game, and "1" the one where it is on A. Then the game has the following 8 non-win game states plus the 2 win states which I will just name after the players.
(0,0;R) = no peg on the board, red's turn
(0,0;B) = no peg, black's turn
(1,0;R) = red's peg on A, no black peg, red's turn
(1,0;B) = same as above but black's turn
(0,1;R) = black's peg on A, red's turn
(0,1;B) = as above but black's turn
(1,1;R) = red's peg on A, black's peg on A, red's turn
(1,1;B) = as above but black's turn
R = red won
B = black won
As for the rules...
Red begins, the players take turns to roll a 6 sided die. The pegs begin on their respective "out" fields B.
If the player's peg is on B: On a 6, the player is allowed to move their peg from B to A and roll again (see rule 3). Otherwise, it's the next player's turn.
If the player's peg is on A: On a 1, the player can move the piece 1 field (to a), winning the game. On any other number, their turn ends.
That's the entire game! And it is only slightly worse than the original, amazing. Each turn can either increase the left or right number from 0 to 1, make the player who's turn it is win, or change who's turn it is without affecting the board. The likelihood of the game "progressing" is always 1/6 (either roll a 6 or a 1, depending on the context) while the likelihood of the game "stalling" is 5/6. So every possible game can be summarized as a graph of the 10 different game states, with arrows showing which states can lead to which and with what probability. You can tell when in the process I stopped giving a crap about aesthetic.
We can now ask, "what is the probability of red winning?" and get a definite answer with some math. But.. fucking hell it's midnight already? Okay, that has to be enough 'tism for one post. Look forward to a followup (or maybe I'll just edit this post). Stay tuned! FUCK IT WE BALL, I FINISHED THIS SHIT AT 2AM.
Alright so what's gonna be annoying are all those pesky cycles that could mean the game could go on hypothetically forever (just like the real one!), but we can deal with those by starting at the "bottom" of the graph and working our way up. It's pretty clear that the probability of a player winning should only depend on the game's state, so whether it is turn 5 or 105 the probability for red to win when the game is in state (1,1;R) should be the same. The probability of that is some number. We could simply simulate an arbitrary number of games in that state, and intuitively we would expect some fixed percentage of red wins (which we called R) to pop out of that simulation. I won't do much formal mathematics here. There is a 1/6 chance of red winning immediately, and a 5/6 chance of the game changing states. So in almost plain English we know:
[probability of R given (1,1;R)] = 1/6 + 5/6×[probability of R given (1,1;B)] .
In state (1,1;B), there is no chance for R to win in the turn itself, but a 5/6 chance of the state changing back!
[probability of R given (1,1;B)] = 5/6×[probability of R given (1,1;R)].
Putting them together and using that (5/6)² = 25/36 we get
[probability of R given (1,1;R)] = 1/6 + 25/36×[probability of R given (1,1;R)].
Now the same probability appears on both sides! We can simplify and find I'm getting tired of this verbosity, let's write the conventional way mathematicians do for this stuff. They don't write [probability of R given (1,1;R)], they write P(R|(1,1;R)).
11/36 × P(R|(1,1;R)) = 1/6
P(R|(1,1;R)) = 6/11 ≈ 54%.
With the above we can figure out all probabilities for the two game states, and they add up to 1 since a game taking forever is infinitely unlikely (don't worry about it, but it is a fun rabbit hole)
P(R|(1,1;R)) = 6/11 = P(B|(1,1;B)) and P(B|(1,1;R)) = 5/11 = P(R|(1,1;B)).
Moving up in the graph we can now replace the two "solved" game states with their win probabilities, taking into account that reaching these states has a probability of 1/6 itself).
Let's focus on the left half. R appears in both states there, but B only once. We already know that the chances of R and B will add up to one, so let's choose the path of least resistance and try P(B|(1,0;R)). Then we can use the same trick as above!
P(B|(1,0;R)) = 5/6 × P(B|(1,0;B)) = 5/6×( 1/11 + 5/6×P(B|(1,0;R)) )
P(B|(1,0;R)) = 30/121 ≈ 25%
P(R|(1,0;B)) = 91/121. The right side of the graph is much of the same. Multiplying these by 1/6 again means we can delete another entire row from the graph!
Almost done! Now all that is left is computing P(R|(0,0;R)), because this is the state the game actually starts at. As we have seen for P(R|(1,1;R)), if the board is in a state where both players are equally close to winning, the one whose turn it is is (intuitively) at a slight advantage. We expect the same to be true now... let's suffer through this once more.
P(R|(0,0;R)) = 91/726 + 5/6×P(R|(0,0;B)). Once more
P(R|(0,0;R)) = 91/726 + 5/6×( 5/121 + 5/6×P(R|(0,0;R)) )
11/36 P(R|(0,0;R)) = 91/726 + 25/726 = 58/363
P(R|(0,0;R)) = 696/1331 ≈ 52.29% and thus
P(B|(0,0;R)) = 635/1331 ≈ 47.71%.
And there you have it: The probability of red winning in this simplified version of the game is 696/1331 or about 52%. It would be cool to see how less dumbed down versions of the game compare to that, though this "0IQ version" of the game is actually contained in the real deal! I had real instances of the full game play out to the point where both players were just sitting there, waiting for the chance to roll a godforsaken 1 to end the game. And now you know: if you wanna flip a coin to decide who won instead of prolonging your suffering should you ever reach that point in the game, you are only shuffling around a strategic edge of like 2%.
You're welcome.
