- #1

aperson

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- TL;DR Summary
- I'm trying to write the general formula to obtain the probability of getting a certain number(e.g. all the combinations to obtain 4) or less by throwing two six-faced dices.

In a board game, I need to reach a certain place in a board divided into boxes, I move by throwing 2 six-faced dices.

If my goal is 4 boxes away I need to obtain at least a combination of numbers that sums to 4, but any higher number is also a favorable outcome.

I want to calculate the probability of reaching a precise distance with a single dice throw.

This is my procedure:

two dices have 36 possible combinations, and the likeliness of a numerical outcome Is measured by the number of combinations that sums to it, divided by all possible combinations.

e.g. of all 36 possible combinations, there are 3 that sum to four (1 and 3, 3 and 1, 2 and 2) so the probability of having 4 as an outcome is equal to 3/36.

But to reach a place in the game board that is 4 boxes away I can also roll 5, 6, 7, 8, 9, 10, 11, or 12 so I calculate the number of all possible combinations (36) and subtract the number of combinations(1) that give 2 as an outcome and the number of combinations(2) that give 3 as an outcome:

36 - (1+2) =

36 - 3 = 33

so the possibility of reaching a place 4 boxes apart with a single throw of dice is 33/36.

This is what I have obtained:

2 boxes apart: any combination 36/36

3 boxes apart: any combination but one 35/36

4: 33/36

5: 30/36

6: 26/36

7: 21/36

8: 15/36

9: 10/36

10: 6/36

11: 3/36

12: 1/36

Is this correct?

I don't have much of an education on probability and I would like to write a general formula to embody this procedure but I don't know where to start, could you help me?

If my goal is 4 boxes away I need to obtain at least a combination of numbers that sums to 4, but any higher number is also a favorable outcome.

I want to calculate the probability of reaching a precise distance with a single dice throw.

This is my procedure:

two dices have 36 possible combinations, and the likeliness of a numerical outcome Is measured by the number of combinations that sums to it, divided by all possible combinations.

e.g. of all 36 possible combinations, there are 3 that sum to four (1 and 3, 3 and 1, 2 and 2) so the probability of having 4 as an outcome is equal to 3/36.

But to reach a place in the game board that is 4 boxes away I can also roll 5, 6, 7, 8, 9, 10, 11, or 12 so I calculate the number of all possible combinations (36) and subtract the number of combinations(1) that give 2 as an outcome and the number of combinations(2) that give 3 as an outcome:

36 - (1+2) =

36 - 3 = 33

so the possibility of reaching a place 4 boxes apart with a single throw of dice is 33/36.

This is what I have obtained:

2 boxes apart: any combination 36/36

3 boxes apart: any combination but one 35/36

4: 33/36

5: 30/36

6: 26/36

7: 21/36

8: 15/36

9: 10/36

10: 6/36

11: 3/36

12: 1/36

Is this correct?

I don't have much of an education on probability and I would like to write a general formula to embody this procedure but I don't know where to start, could you help me?