First to 1000 rolls wins

Question

You and your friend each have a standard $6$-sided die with sides numbered $1, 2, \ldots 6$, every side has an equal probability of arising in a random roll.

• You throw the cube first* the number that land, is the amount of times your friend need to throw his cube.
• The sum of all the numbers that landed from the amount of times your friend threw, is the amount of times you need to throw your dice.
• The sum of all the numbers that landed from the amount of times your friend threw, is the amount of times you need to throw your dice.
• And then it goes and goes.

What is the probability that you will be the first one who need to throw more than $1000$ throws. For example:

1. You threw 3

2. Your friend threw 3 times and got: 2,5,6. The sum is 13.

3. You threw $13$ times and got: $5,2,3,4,6,1,2,5,3,4,1,3,6$. The sum is $45$.

4. Your friend threw $45$ times and got: ..... The sum is $X$

5. You threw $X$ times and got: .... The sum is $Y$ .... What is the probability that you threw before your friend more than $1000$ throws.

*I think it would be interesting the same question but you throw second.

Answer 1

The probability of getting $s$ as the sum of rolling $m$ times one die with $t$-faces (or rolling $m$ of such dice) is given by the number of ways to get that sum, divided by number of different $m$-tuples. That is we consider the "times" (dice) labelled, both for counting the ways to give $s$ and for the total of possible outcomes $t^m$.

Let's call it $P(s,m;t)$. That is the same either for player $A$ and $B$.

Now consider the sequence of the sums $s_{1},s{2},\cdots$.
We must have $$ \left\{ \matrix{ 1 \le s_{\,1} \le t \hfill \cr s_{\,1} \le s_{\,2} \le t\,s_{\,1} \quad \Rightarrow \quad 1 \le s_{\,2} \le t^{\,2} \hfill \cr \quad \vdots \hfill \cr s_{\,n - 1} \le s_{\,n} \le t\,s_{\,n - 1} \quad \Rightarrow \quad 1 \le s_{\,n} \le t^{\,n} \hfill \cr} \right. $$

Calling $q$ the limit you put at $1000$, you are asking which is the probability that $$ s_{\,2n} < q \le s_{\,2n + 1} $$ versus the viceversa.

The game is a Markov chain, with states $1 \cdots q$ , state $q$ an absorbing barrier, and transition probability $P(s_{n+1},\, s_{n};\, t)$.
We shall then compute the probability of reaching the barrier in $2n+1$ vs. $2n$ steps.

Concerning $P(s,\,m;\,t)$, in this other post it is extensively explained that, calling $$ \eqalign{ & N_b (s,r,m) = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ {\rm 0} \le {\rm integer}\;x_{\,j} \le r \hfill \cr x_{\,1} + x_{\,2} + \; \cdots \; + x_{\,m} = s \hfill \cr} \right. = \cr & = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ {\rm 1} \le {\rm integer}\;x_{\,j} \le r + 1 \hfill \cr x_{\,1} + x_{\,2} + \; \cdots \; + x_{\,m} = s + m \hfill \cr} \right. \cr} $$ it is expressible as $$ N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad = \sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s} {r}\, \leqslant \,m} \right)} {\left( { - 1} \right)^k \left( \begin{gathered} m \hfill \\ k \hfill \\ \end{gathered} \right)\left( \begin{gathered} s + m - 1 - k\left( {r + 1} \right) \\ s - k\left( {r + 1} \right) \\ \end{gathered} \right)} $$

Thus $$ \eqalign{ & P(s,\,m;\,t)\quad \left| {\;1 \le t} \right.\quad = {1 \over {t^{\,m} }}N_b (s - m,t - 1,m) = \cr & = {1 \over {t^{\,m} }}\sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{{s - m} \over r}\, \le \,m} \right)} {\left( { - 1} \right)^k \left( \matrix{ m \hfill \cr k \hfill \cr} \right)\left( \matrix{ s - 1 - k\,t \cr s - m - k\,t \cr} \right)} \cr} $$

Given the nature of the process and the formulation for $P$, I do not see that the answer might be formulated analytically, not even in a asymptotic way.

Answer 2

Method:

Let $D(i, j)$ be the probability that a total of $i$ is thrown with $j$ dice. This can be calculated recursively using

$$D(i,j)=\frac{1}{6}\sum^6_{k=1} D(i-k, j-1)$$

Let $P_i(k)$ be the probability that the first player to throw exactly $k$ dice is player $i$.

We know $P_1(1)=1$ and $P_2(1)=0$.

Then if $\{a, b\}=\{1,2\}$, we can calculate $P_1$ and $P_2$ recursively using

$$P_a(i)=\sum_{j<i}P_a(j)6^{-j}(D(i, j)+6^{-2j}(D(i, j)+6^{-2j}(\cdots)))+P_b(j)(D(i, j)+6^{-2j}(D(i, j)+6^{-2j}(\cdots)))$$

$$=\sum_{j<i}(6^{-j}P_a(j)+P_b(j))(D(i, j)+6^{-2j}(D(i, j)+6^{-2j}(\cdots)))$$

$$=\sum_{j<i}(6^{-j}P_a(j)+P_b(j))\left(D(i,j)\sum^\infty_{k=0}6^{-2jk}\right)$$

$$=\sum_{j<i}(6^{-j}P_a(j)+P_b(j))\left(\frac{D(i,j)}{1-6^{-2j}}\right)$$

Note for our problem, if $t$ is the target number of throws (in this case $1000$), we need to exclude all $j$ for which $t\le j$ when we calculate the final probability as the game is over once a score $\ge j$ is thrown.

Also note that the sum is finite and many values of $j$ can be ignored as $D(i, j)=0$ for examples such as $i=7,j=1$.

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Explanation of the recurrence to calculate $P_a(i)$:

Let $E_i(k)$ be the event that player $i$ is the first to throw exactly $k$ dice.

Let $S_i(k)$ be the event that player $i$ scores a sum of $k$ on a particular throw of the dice.

Then we have:

$$P_a(i)=\sum_{j<i} P_a(j)P(\text{after $E_a(j)$, $S_a(j)$ or $S_b(j)$ until $S_b(i)$}\,|\,E_a(j))+P_b(j)P(\text{after $E_b(j)$, $S_a(j)$ or $S_b(j)$ until $S_a(i)$}\,|\,E_b(j))$$

The events whose probabilities are summed over in the above formula are mutually exclusive and you should be able to see that all possible ways of reaching $E_a(i)$ are taken into account.

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For target number of throws $1000$ (or more) the probability is $0.7069979408847353$.

Online JS editor which you can run the programs in: https://js.do

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JS program which uses a recursive method to calculate the answer (note that variables dice[i][j], P1[i], P2[i] correspond to $D(i, j), P_1(i), P_2(i)$ respectively in the above explanation):

<script>

const win = 1000; //target number of throws

var dice = []; //2D array. dice[s][d] gives probability of getting a total of 's' from rolling 'd' dice. (calculated below)

for (i = 1; i <= 6 * (win - 1); i++) {

    dice[i] = [];
}

for (i = 1; i <= 6; i++) {

    dice[i][1] = 1.0 / 6.0;
}

for (d = 2; d < win; d++) {
    for (s = d; s <= 6 * d; s++) {

        dice[s][d] = 0;

        for (i = 1; i <= 6; i++) {

            dice[s][d] += (dice[s - i] || [])[d - 1] || 0;
        }

        dice[s][d] /= 6.0;
    }   
}

//above are the calculations done recursively for the 'dice' array

var P1 = []; //An array. P1[i] gives the probability that the first person to throw exactly 'i' dice is the first player.
var P2 = []; //An array. P2[i] gives the probability that the first person to throw exactly 'i' dice is the second player.

//Note the game ends immediately once a score of 'win' or more is thrown for the following calculations of the above arrays. Therefore P[3000] does not use P[2000] to calculate if 'win' = 1000, for example.

P1[1] = 1;
P2[1] = 0;

for (i = 2; i <= 6 * (win - 1); i++) {

    P1[i] = 0;
    P2[i] = 0;

    for (j = Math.ceil(i / 6.0), k = Math.pow(6, j); j < Math.min(i, win); j++, k *= 6) {

        var k2 = dice[i][j] * (1 + 1.0 / (k * k - 1));

        P1[i] += k2 * (P1[j] / k + P2[j]);
        P2[i] += k2 * (P1[j] + P2[j] / k);
    }
}

var P = 0; //Answer. Calculated by summing over P1[i] for all 'i' >= 'win' as done below.

for (i = win; i <= 6 * (win - 1); i++) {

    P += P1[i];
}

document.write(P);

</script>

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JS program which uses a probabilistic method to approximate the answer:

<script>

const win = 1000; //target number of throws

const trials = 10000000 / win; //number of trials

var P = 0; //answer

var d = 1; //number of dice
var turn = 1; //turn counter

var score = 0; //keeps track of score after each throw

for (t = 0; t <= trials; t++){

    d = 1;
    turn = 1;

    while (d < win) {

        score = 0

        for (i = 1; i <= d; i++) {score += 1 + Math.floor(6 * Math.random());} //simulates rolling 'd' dice

        d = score;

        turn++;
    }

    P += turn % 2; //turn counter odd means it's first player's turn
}

P /= trials;

document.write(P);

</script>