View Full Version : Dice Questions

Thriondel Half-Elven

10-12-2014, 03:09 AM

I'm not one who thinks probability is a fun time. So I had a question.

A game is played with d6s. A 5 or 6 on a roll (on a single die face) is a success. On 1d6 that would be 33.34% chance of success and 66.68% of failure (is that right?)

If a player rolled 2d6 the chance of success and failure would stay the same correct?

Would they have a better chance at succeeding if they rolled 5d6 over 1d6? or would the odds be the same?

On a side note: As a PC goes up in skill they gain "Ranks" in a skill/trait. The Rank is the # of d6s rolled. So a PC with a Rank 1 in Sword would roll 1d6 when attacking with a sword. While a PC with Rank 4 in Sword would roll 4d6 on an attack.

Both need to roll at least one 5 or 6 to hit. Does the PC with a higher rank have a better chance at succeeding more often and thus being more skilled in that skill/trait?

jpatterson

10-12-2014, 03:45 AM

When you say 2d6, do you mean pool (each individual die is its own result) or summed (2+3=5)? I presume you're talking about dice pools since the rest of your question seems to lean that way.

Here is the thing on individual dice rolled in a pool. They each are their own individual objects with that same 16.6667% chance of rolling any one face. The change here is that probability is all about what numbers are rolled or likely to be rolled, given a certain number of "trials" or total rolls (ie rolling a die 20 times and getting fewer 3s and 5s than most anything indicates it is very likely you're going to roll those soon, assuming the die is "fair").

The real torpedo to "each die is its own probability case" in a dice pool is that you are in effect rolling "a number of times", because the real criteria here is the number that is or isn't rolled is both a function of each die, but also the entire set, so if you roll 5d6, the chances of getting any particular face among the group as a whole, are higher than if you rolled a single die, because you have a larger number of objects each with their own variable probabilities, and by chance alone, one is likely to meet your criteria. Still, each die does retain its own same probability, each having that 17% chance to roll any number - but combined, probability for the pool roll, is entirely a different formula.

That is why it IS an advantage in a game system to be able to roll 5d6 instead of 1, because chance alone will flop one or more dice over to what you want, though being chance, it will also flop it over to 1 or 2, but since low numbers stay the same but higher skill is possible, it isn't so bad if your 5d6 rolls low because you still have others that will hit your target. Failure is "less than x" which has a limited range that can happen, say 1-4 if your TN is 5 or 6 on a d6, and becomes smaller as skill goes up, so if you have 5d6 skill, failure is still 1-4 per die, but since you have 5 or 6 dice to get a success, you're very likely to get one, and unless you're counting exact success vs fail like degree of success (3 fails vs 2 successes), it is weighted in favor of success.

fmitchell

10-12-2014, 10:02 AM

Flip the question around: if you roll 1d6, and you fail if it does not show a 5 or a 6, then you have a 2/3 (67%) chance of failing.

If you have 2d6, and you fail only if neither of them show a 5 or 6, then you have a 2/3 * 2/3 = 4/9 (44%) chance of failing. Your chance of failing just went down, or put another way your chance of success just went up.

For any dice pool, if p is the probability of success on each die (as a number from 0 to 1, e.g. 33% is 0.33), then the probability for at least one success on n dice is (1 - (1-p)^n). As n increases, this probability also increases.

Probability math gets a little tricky sometimes, but fortunately there's some useful Web pages for the beginner, like http://www.edcollins.com/backgammon/diceprob.htm. http://anydice.com/ allows you to test out all (or nearly all) combinations.

DMMike

10-12-2014, 10:26 AM

Probability for dummiesŪ:

Write out all your outcomes. Then create a fraction out of them. Your numerator is your desired outcomes, and the denominator is all outcomes. This works great...for up to maybe three dice.

1d6: 1, 2, 3, 4, 5, 6.

Desired outcomes: 2 (5&6).

Total outcomes: 6.

Odds of desired outcome: 2/6, or 1/3.

2d6 outcomes:

1

1

1

1

1

1

2

2

2

2

2

2

3

3

3

3

3

3

4

4

4

4

4

4

5

5

5

5

5

5

6

6

6

6

6

6

1

2

3

4

5

6

1

2

3

4

5

6

1

2

3

4

5

6

1

2

3

4

5

6

1

2

3

4

5

6

1

2

3

4

5

6

Desired outcomes: 20 (all columns with either a 5, 6, or both.)

Total outcomes: 36

Odds of desired outcome: 20/36, or 5/9.

Thriondel Half-Elven

10-13-2014, 09:44 PM

Thanks fellas. I thought I had the right idea. That http://www.edcollins.com/backgammon/diceprob.htm website was very helpful!

Thriondel Half-Elven

10-14-2014, 01:17 AM

I hope I did the math right. So with a 5 OR 6 needed on 3d6 there are 152 possibilities for success. That means a 70% chance of success. Right?

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