This week the #puzzle is: Let’s Make a Tic-Tac-Deal! #probabilities

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, or if you roll a sum that you have previously rolled, then 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)?

And for extra credit:

In the actual game, you get five rolls instead of three. But as before, rolling a 2, a 12, or 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?

Let’s Make a Tic-Tac-Deal!

Highlight to reveal (possibly incorrect) solution:

Program

I’m in a hurry this week, so I just monto carloed this one.

 3 rolls, accumulated probability = 0.05806
 4 rolls, accumulated probability = 0.18999
 5 rolls, accumulated probability = 0.36138
 6 rolls, accumulated probability = 0.52770
 7 rolls, accumulated probability = 0.66560
 8 rolls, accumulated probability = 0.76980
 9 rolls, accumulated probability = 0.84440
10 rolls, accumulated probability = 0.89594
11 rolls, accumulated probability = 0.93079
12 rolls, accumulated probability = 0.95404
13 rolls, accumulated probability = 0.96941
14 rolls, accumulated probability = 0.97959
15 rolls, accumulated probability = 0.98632

For the case with 3 rolls, I also did the math directly.

3-4-5	2	3	4	24
6-7-8	5	6	5	150
9-10-11	2	3	4	24
3-6-9	2	5	4	40
4-7-10	3	6	3	54
5-8-11	2	5	4	40
3-7-11	2	6	2	24
5-7-9	4	6	4	96
				452

E.g., for the first row, for every possible roll with 2 dice (there are 36), 2 of them have a sum of 3, 3 of them have a sum of 4 and 4 of them have a sum of 5. These 3 numbers multiplied give 24. All the numbers in the right column, multiplied by 3!/36³, are probabilities. The summed probability is therefore 0.0581275720164609, rounded to 0.058128. This checks out.