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I have a question about understanding the math to a previous question I had posed (answered successfully). Link at the bottom.

The question is how does it make sense that rolling a 2D6 Die consisting of 111 22 3 for sides has the math work out where the probability to roll TWO 1s is 25% chance and rolling a 1 and a 2 is 33% chance? what is the math here for a laman?

It is also interesting to me that only when rolling 2 die is there a 50% chance of getting 2 circles.

Here is the link (Using AnyDice.com to calculate Non traditional dice combinations)

ABOVE is the original question. BELOW is more math about these dice.

Here is prob for different combinations.  1= circle 2 =triangle and 3- diamond on the die

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  • \$\begingroup\$ Might I recommend math.stackexchange.com? \$\endgroup\$
    – From
    Aug 25, 2023 at 20:17
  • \$\begingroup\$ ill post there as well. This was a question for dice probability and where i originally found such questions proposed. Thanks for the advice! \$\endgroup\$ Aug 25, 2023 at 20:43
  • \$\begingroup\$ are ones circles, are twos triangles, and are threes diamonds? \$\endgroup\$
    – Ruse
    Aug 25, 2023 at 20:51
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    \$\begingroup\$ What do the 1 2 and 3 values of the dice have to do with the Circles, Triangles and Diamonds on the chart? The shapes come out of nowhere. \$\endgroup\$
    – Blckknght
    Aug 25, 2023 at 21:01
  • \$\begingroup\$ What is a "laman" in this context? Google is unhelpful for this term. \$\endgroup\$
    – Brian
    Aug 26, 2023 at 23:47

3 Answers 3

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If you are rolling two six sided dice, there are 36 different combinations of faces that can come up. If several of the faces have the same symbols, many of those combinations will have the same combinations of values, but we can still consider all 36, and count the number of combinations that get us to the results we want.

2d6 1 1 1 2 2 3
1 1, 1 1, 1 1, 1 1, 2 1, 2 1, 3
1 1, 1 1, 1 1, 1 1, 2 1, 2 1, 3
1 1, 1 1, 1 1, 1 1, 2 1, 2 1, 3
2 2, 1 2, 1 2, 1 2, 2 2, 2 2, 3
2 2, 1 2, 1 2, 1 2, 2 2, 2 2, 3
3 3, 1 3, 1 3, 1 3, 2 3, 2 3, 3

The number of ways you can roll two 1s on two of your dice is 9. You have three choices of 1 faces on the first die, and three choices of 1 faces on the second die. 9/36 is 25%.

Counting the number of ways to roll a 1 and a 2 is a little trickier. There are three ways you can roll a 1 on the first die, and two ways to roll a 2 on the second die, which combine to make 6 ways to roll 1,2. But we also need to include the odds of rolling the results in the other order, 2,1, which is another 6 results. So there are 12 ways total. 12/36 is 33%.

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The really Easy Way To Understand This Is With A Table

For small numbers of dice, it's often possible to just exhaustively list out all the possibilities, and organize them into a table. I've done that for you below.

Imagine you have the dice as you've specified, each with 3 1's, 2 2's and 1 3. Imagine one is red and one is green. The numbers running across the top are the possible values of your red die, the numbers running along the left are the green die.

The rest of the numbers are the sums of those numbers-- the possible results of throwing your two dice and adding the red and green values together.

Notice the table has 36 entries, because there are six sides on the red die, and six sides on the green die; six times six is 36.

Now, everywhere inside that table that you see a 2, it is the result of both the red and green die both coming up as 1. If you count them, you will see there are 9 of them; and 9 is one quarter of 36, or 25% of 36.

Similarly, everywhere inside that table that you see a 3, it is the result of both the one of the dice coming up as 1, and the other die coming up as 2. If you count them, you will see there are 12 of them; and 12 is one third of 36, or 33.3% of 36.

$$ \begin{array}{c|lcr} n & \text{1} & \text{1} & \text{1} & \text{2}& \text{2}& \text{3} \\ \hline 1 & 2 & 2 & 2 & 3 & 3 & 4 \\ 1 & 2 & 2 & 2 & 3 & 3 & 4 \\ 1 & 2 & 2 & 2 & 3 & 3 & 4 \\ 2 & 3 & 3 & 3 & 4 & 4 & 5 \\ 2 & 3 & 3 & 3 & 4 & 4 & 5 \\ 3 & 4 & 4 & 4 & 5 & 5 & 6 \end{array} $$

(Purely as an aside, this is a neat kind of way to come up with a die roll that usually comes out low, but very occasionally comes out high. I might use this to model something like the damage of a dart from a game of darts-- usually it's an annoyance, but once in a while it hits something vital and does real damage.)

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First, "percent" means "per one hundred", right? So we're dealing with hundreds. 50% is just another way of writing 0.5, or 1/2.

Half of the outcomes (1/2, 0.5 or 50%) on one roll of your custom d6 are 1, one third (1/3, ~0.33 or ~33%) are 2 and one sixth (1/6, ~0.16 or ~16%) are 3.

  1. The odds of rolling two 1s, for a total score of two (1 + 1 = 2), on two of those custom dice is equal to the odds of rolling 1 on both of them, multiplied - so (1/2) * (1/2), or 0.5 * 0.5, which is 1/4, 0.25 or 25%.

  2. The odds of NOT rolling any 1 is similarly the product of the odds for each die not to show a 1. On a regular die, this would be (5/6) * (5/6), since those would have five sides that are not 1. On your die, however, the odds are (1/2) * (1/2) = 1/4, 0.25 or 25%.

  3. The odds of rolling AT LEAST one 1, so either one or both dice show a 1, is 1.0 (100%, all outcomes) MINUS the odds of not rolling any 1, or 1 - ((1/2) * (1/2)), or 1 - 0.25, that is, 0.75 or 75%.

  4. The odds of rolling EXACTLY ONE 1, so one of the dice is a 1 but the other is something else, is the odds of at least one 1 minus the odds of two 1s, so 0.75 - 0.25 = 0.5, or 50%.

  5. If one die is a 1 and the other is something else, then the other die must be either a 2 or a 3. There are two sides with a 2 and only one side with a 3, so three possible outcomes. Let's simply divide the 50% (0.5, or 1/2) from paragraph 4 by 3: (1/2)/3 = 1/6. This means the odds for a 1 and any one of the three non-1 sides is 1/6, or about 16%.

  6. Since two of the sides are 2s, that means the odds for a 1 and a 2 is 2 * (1/6) = 1/3, or about 33%.

I have no idea what circles, triangles and diamonds are.

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  • \$\begingroup\$ thanks for the response! it was very informative! Slightly above my head but It looks like the math is correct. If i read your answer right... \$\endgroup\$ Aug 25, 2023 at 20:56

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