Let’s Make a Tic-Tac-Deal!¶

Fiddler¶

The game of Tic-Tac-Deal 2.0 has a 3-by-3 square grid with the numbers 3 through 11, arranged as follows:

3 4 5

6 7 8

9 10 11

You start by rolling a standard pair of six-sided dice and add the two numbers rolled. You place an X on the board on the square that contains the sum. If the sum is a 2 or 12, your roll is wasted.

If you have exactly three rolls of the dice, what are your chances of getting three Xs in a row (either horizontally, vertically, or diagonally)?

Solution¶

There are 8 wins :

  1. 3 horizontal.
  2. 3 vertical.
  3. 2 diagonal.

To achieve their respective dice rolls, 3 & 11 have 2 ways, 4 & 10 have 3 ways, 5 & 9 have 4 ways, and 6 & 8 have 5 ways.

There are 6 ways to order 3 different sums.

Here are the probabilities to achieve wins :

$3-4-5$ & $11-10-9$¶

$\frac{2}{36} \times \frac{3}{36} \times \frac{4}{36} \times 6 \times 2 = \boxed{\dfrac{288}{36^3}}$

$6-7-8$¶

$\frac{5}{36} \times \frac{6}{36} \times \frac{5}{36} \times 6 = \boxed{\dfrac{900}{36^3}}$

$3-6-9$ & $11-8-5$¶

$\frac{2}{36} \times \frac{5}{36} \times \frac{4}{36} \times 6 \times 2 = \boxed{\dfrac{480}{36^3}}$

$4-7-10$¶

$\frac{3}{36} \times \frac{6}{36} \times \frac{3}{36} \times 6 = \boxed{\dfrac{324}{36^3}}$

$3-7-11$¶

$\frac{2}{36} \times \frac{6}{36} \times \frac{2}{36} \times 6 = \boxed{\dfrac{144}{36^3}}$

$5-7-9$¶

$\frac{4}{36} \times \frac{6}{36} \times \frac{4}{36} \times 6 = \boxed{\dfrac{576}{36^3}}$

Answer¶

$$\frac{288+900+480+324+144+576}{36^3}=\boxed{\frac{2,712}{36^3} \approx 0.0581}$$

Extra Credit¶

In the actual game, you get five rolls instead of three. But as with rolling a 2 or 12, rolling a number that you have already rolled is a wasted turn.

With five rolls of the dice, what are your chances of getting three Xs in a row, either horizontally, vertically, or diagonally?

Solution¶

I wrote code that did the following :

  1. Create the sample space of two dice rolls.
  2. Set $W = 0$
  3. For each possiblility of $n$ two dice rolls :
    1. Convert to $n$ sums of two dice.
    2. For each of the eight wins :
      • If the win triple is a subset of the set of dice roll sums:
        1. Add a win to $W$.
        2. Continue with the next dice rolls sum possiblity.
  4. Return the value of $W$.

Answer¶

$$\boxed{\frac{21,849,720}{36^5} \approx 0.3613}$$

Rohan Lewis¶

2025.10.13¶

Code can be found here.