Probability that these characters win a game.

Question

I have 11 characters, $[2,3,4,5,6,7,8,9,10,11,12]$, and they all play a game.

Game Description:

All players stand at the start line $n$ spaces away from finish line. Two fair dice are rolled. The two results on each of the dice are summed up, and it gives a result that is equal to the names of one of the characters. The owner of that number moves forward one space. The winner is the character who gets to the finish line first.

For example, on a turn, the dice gives a result of $[3,4]$. Since $3+4=7$, the character 7 moves forward.

If the dice gives $[6,2]$, the character 8moves forward.

What are the chances that each character wins?

Basically I want the chances of winning for all the characters, with respect to other players. By respect to other players, I mean that one can win it quicker than another. And I want the probabilities to be in terms of $n$, for example, $P(x) = \frac{1}{36^{n}}$

Images of a state midgame:

Here, character 6has won the game, and in this case, $n=9$, since that is the number of spaces a character has to move in order to win.

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Sources:

• The game originated from my school
• The image state example was created by me using Microsoft Excel
• The question is a challenge set by myself. I need help basically.

Answer 1

Let $p_k$ be the probability that you roll a $k$. For example, $p_7=6/36$. I assume you're able to calculate these.

We are only interested in a pairwise probability of winning, so let's calculate the probability that $A \in \{2,\dots,12\}$ wins vs $B \in \{2,\dots,12\}$. (For example, $A=7, B=6$.) When analyzing this game, we only care about the rolls that come up $A$ and $B$, all others can be ignored. We'll define the conditional probabilities given $A$ or $B$ was rolled. $$ r_A = \frac{p_A}{p_A+p_B} \qquad r_B = \frac{p_B}{p_A+p_B} $$

Define $f(u, v)$ as the probability that $A$ wins when $A$ is $u$ steps away from the finish, and $B$ is $v$ steps away from the finish. The aim is to calculate $f(n, n)$.

Consider the state $(u,v)$. If we roll an $A$, we go to state $(u-1,v)$. If we roll a $B$, we go to state $(u,v-1)$. Thus, we can write the following relationship for $f$: $$ f(u,v) = \begin{cases} r_Af(u-1,v)+r_Bf(u,v-1) & \text{if } u \geq 1, v \geq 1 \\ 1 & \text{if } u = 0 \\ 0 & \text{if } v = 0 \end{cases} $$ So the goal now is to solve this recurrence relation. I solved it by writing it out for low values of $u$ and $v$. (There is probably a more clever way to do it.) $$ f(u,v) = \begin{cases} r_A^u\left[ 1 + \sum_{i=1}^{v-1}\left[ \left(\prod_{j=0}^{i-1} (u+j)\right)\frac{1}{i!}r_B^i \right] \right] & \text{if } u \geq 0, v \geq 2 \\ r_A^u & \text{if } u \geq 0, v = 1 \\ 0 & \text{if } u \geq 0, v = 0 \\ \end{cases} $$ You can prove this by induction. I will leave it as an exercise for the reader.

Here is the table of probabilities for $A=7, B=6$. $u$ goes across the columns, $v$ goes down the rows. $f(n,n)$ are on the diagonals. For example, $A$ has a 72% chance to win when $n=20$.

Plotting $f(n,n)$ we can see that as expected, the probability that $7$ wins goes to 1.

Answer 2

The chance for each character to move one space towards finish line is different from others.
The chance for 2 to move is $1/36$ , for 3 is $2/36$ , for 4 is $3/36$ , for 5 is $4/36$ , for 6 is $5/36$ , for 7 is $6/36$ , for 8 is $5/36$ , for 9 is $4/36$ , for 10 is $3/36$ , for 11 is $2/36$ and for 12 is $1/36$.
Now ,for simplicity consider that there are only 2, 3, 4 characters.
Consider that we are finding the probability that 2 wins : P(2 wins) = $\sum_{0}^{n}$ { $(1/36)^n$ * $(2/36)^i$ * $(3/36)^j$ }
Where 'i' and 'j' varies from 0 to n as depicted by sigma.
Now, you can extend this as per your need.
This is my first answer. Please help to improve it. Thanks!

Answer 3

Say a horse wins in 9 moves, as in your diagram. Horse 7 moves 6/36 of the time.

If Horse 7 wins at move 40, then it moved in 8 of the first 39 moves; and on move 40. Also, no-one else won by moving on nine of the other 31 moves.

To do 'no-one else won' properly, you need to run a ten-horse race, pretending that Horse 7 was not there.
To get the odds for the ten-horse race, you need to know the odds for all the nine-horse races ... but before that you need the eight-horse races ...

There are 2047 different races to get straight, including 11 one-horse races (easy), 55 two-horse races and one eleven-horse race.

I can't see how to run the algorithm in Excel, but I wrote a program in Matlab. I got these results:

$$\begin{array}{c|llllll} n&H7&H6,H8&H5,H9&H4,H10&H3,H11&H2,H12\\ 1 & 0.167 & 0.139 & 0.111 & 0.0833 & 0.0556 & 0.0278\\ 2 & 0.221 & 0.164 & 0.113 & 0.0691 & 0.0337 & 0.00936\\ 4 & 0.293 & 0.187 & 0.105 & 0.0467 & 0.0136 & 0.0013\\ 8 & 0.382 & 0.201 & 0.083 & 0.0226 & 0.00266 & 3.44e-05\\ 16 & 0.489 & 0.197 & 0.0518 & 0.00612 & 0.000137 & 3.69e-08\\ 32 & 0.613 & 0.171 & 0.0217 & 0.00057 & 5.28e-07 & -\\ 64 & 0.747 & 0.122 & 0.00443 & 6.79e-06 & - & -\\ 128 & 0.874 & 0.0629 & 0.000234 & 1.42e-09 & - & -\\ 256 & 0.964 & 0.0182 & 8.8e-07 & - & - & -\\ 512 & 0.997 & 0.00173 & - & - & - & - \end{array}$$ As usual, $8.8e-07$ means $8.8*10^{-7}=0.00000088$

For example, to win a race to 128, there will be about $6\times128=768$ moves. Of these, Horse 7 would tend to get between 118 and 138 moves, while both Horses 6 and 8 get between 96 and 116 moves. There is still a chance though that 6 or 8 - or one of the others - beats Horse 7.