Winning Probabilities in this "Dice Race"

Factual Questions
1

There is a board with five spaces on it in a row–space 1, space 2, space 3, space 4 and space 5.

There are two players, each with a marker on space one.

They play a game as follows:

Each round, a die is rolled. On a 1, 2, 3, or 4, player one moves his marker up one space. On a 5 or 6, player two moves his marker up 2 spaces. Whichever player’s marker reaches space 5 first wins.

I would think that each player has an even chance of winning–each has a probability of one half of winning a game.

But when I try to work it out, I get a result implying that player two has a slight edge–a slightly higher chance of winning. (Something like 131 out of 243.)

Unfortunately, I don’t have the time to tell you how I worked it out at the moment, But I am wondering what the correct answer is. Is 131/243 close to correct? Or is it (as I would have thought) 1/2?

-FrL-

(Basically: The highest number of rounds there can be in a game before it is decided is 5. We can treat the dice as three-sided. So there are a total of 243 (3^5) die roll sequences that need to be counted as equal possibilities. Every one of those sequences in which there are at least two “three” rolls is a win for player two, and every one in which there are one or fewer “three” rolls is a win for player one. So I figure out how many of the sequences have at least two “three” rolls. There are 10 ways for exactly two threes to be distributed amongst 5 rolls, and each of those cases (i.e. in which there are exactly two threes) there are eight ways the other dice might be, so there are 80 sequences with exactly two threes. There are 10 ways for there to be exactly three threes, and in each of those cases there are four ways the other dice might be, so there are 40 sequences with exactly two threes. By similar reasoning, there are 10 ways there can be exactly four threes, and one way there can be exactly five threes, for a total of 131 sequences in which player two wins.

I suspect that I end up counting a few player-two-winning sequences twice somehow in the above procedure, but I’m not sure.)

(I guess I had time after all)

2

Not very insightful, but a quick Haskell code-up confirms that, yes, Player 2 has probability 131/243 of winning.

For those who care, the (trivial) code:



--p1 (a, b) is the probability of Player 1 moving a squares 
-- (weakly) prior to Player 2 moving b squares
p1 :: (Int, Int) -> Rational
p1 (0, b) = 1
p1 (a, 0) = 0
p1 (a, b) = 2/3 * p1(a-1, b) + 1/3 * p1(a, max (b-2) 0)

The answer being obtained by calculating p1 (4, 4), of course.

3

Incidentally, it could’ve been obvious from the start that the two players don’t have equal chances of winning. There is a fixed finite bound on the length of the game; as you noted, 5 rolls suffice. Continuing, there are essentially 3^5 equiprobable roll sequences. This is an odd number; one player or another must win more of them.

4

Yeah, I noticed that while typing up the explanation. I am still suprised though.

-FrL-

5

I’m guessing that your surprise that the game is not fair is because the two players have an equal expected position after N rolls (since each roll advances a player’s position by 2/3 of one space: +1 with probability 2/3 for 1, +2 with probability 1/3 for 2). So after five turns (enough to guarantee that exactly one player has reached space 5 and so won the game), player 1 and player 2 have, of course, the same mean position: 1+5(2/3)=13/3=4.333.

But in order to compute the win probability, you need to know more about the distribution than just the mean values. Intuitively, player 2’s position has larger fluctuations than player 1’s position, because the variance in his number of moves is (obviously) the same as player 1’s while each of his moves is twice as large. So player 2 will be more likely to be far from his mean position (either ahead or behind) after five moves.

If you are used to thinking of Gaussian random variables, then imagine two random variables with the same mean value (4.333) but different standard deviations. Clearly the one with the larger standard deviation, and so wider tails, has more probability density at >5 than the one with the smaller standard deviation; here player 2’s distribution has twice the standard deviation of player 1’s. Since this is a finite game, if you want you can just compute the whole distribution to see the same thing.

6

It might be a little easier to visualize if you simplified it down to a 3 space board (with both starting on space 1). This means there are only going to be two dice rolls maximum. In order for player one succeed in this game, he needs to win both die rolls.

The chances for him to do that is (4/6)*(4/6) = 0.444

Anything else results in a player two victory.

By subtraction, that means player two’s chances are 0.566.

The more spaces the board has, the closer the game gets to 50/50 odds, but it never gets there.

Note that player one has the advantage on game boards of even spacing, while player two has the advantage on boards of odd spacing.

7

This doesn’t really add much, besides a slightly different view, but since I took the time to do the maths:

There are two possible events for each throw, either player one gets to move, or player two gets to move. 2^5 gives us 32 different sequences. (Not that this matters for this calculation.)
Only these 6 are wins for player one.

11111
11112
11121
11211
12111
21111

The first two are actually a win for player one after four throws, with a probability of 16/81 (48/243), the other four each have a probability of 16/243, sum 112/243. That leaves a probability of player two winning of 131/143.

8

Also, a variation of this problem is typically encountered in many videogames.

Often, the player might have a choice between a weapon that does, for example,
50 points of damage every time, and another weapon that randomly does 30 to 70 points of damage.

Even though both weapons average to the same damage (50),
your actual weapon preference will change dramatically depending on whether you are typically dealing with monsters with 49 points of health or 51 points of health.