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I am trying to learn how to calculate dice probability, but all I can find on the net are the odds of rolling a given number in a single roll.

An example of what I am looking of is rolling a 4+ on a D6 and the odds on rolling a 5+ on a D6 but you get to reroll it once if its not 5+.

By calculating there probability I can see which has a higher chance of happening.

How does the reroll affect the probability?

How would multiple dice afffect the probability, say I was rolling three dice instead of one?

I have lots of other scenerions the involve rerolling so it would be best if you could explain the method rather than just give me a a formula.

A formula answering my question would be great, but it would be better, if I could understand it myself. So I don't have to keep asking similar questions on here.

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3 Answers 3

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The probability that you roll 4+ is the probability that you roll a 4, a 5, or a 6. Each of these events has probability $\frac{1}{6}$; so, the probability of rolling 4+ is $\frac{3}{6}=\frac{1}{2}$.

For your other scenario, note that the event that you get a 5+ when you are allowed on re-roll can be broken up in to two disjoint events: event $A$, in which you roll a 5+ on the first try; and event $B$, in which you roll a number 1-4 on your first attempt, and either a 5 or 6 on the second.

Then $P(A)=\frac{2}{6}=\frac{1}{3}$, since this is the event that you roll either a 5 or a 6. For $B$, we have $$ P(B)=\frac{4}{6}\cdot\frac{2}{6}=\frac{2}{9}, $$ since you must roll a 1, 2, 3, or 4 on the first attempt and either a 5 or 6 on the second. So, overall, the probability of getting 5+ when you allow one re-roll is $$ P(A\text{ or }B)=P(A)+P(B)=\frac{1}{3}+\frac{2}{9}=\frac{5}{9}. $$ (Note that we have used here that $A$ and $B$ are disjoint possibilities.)

So, you are more likely to get a 5+ with a re-roll allowed than to get a 4+ with no re-roll.

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  • $\begingroup$ Does the amount of dice rolled effect the probability for instance if you were rolling two dice, would i just add the probabilities together ? $\endgroup$
    – Skeith
    Commented Aug 28, 2013 at 14:13
  • $\begingroup$ If you are rolling two dice, and they are independent -- that is, the outcome of one roll does not change the probabilities of specific outcomes of the second -- then you multiply the probabilities. $\endgroup$ Commented Aug 28, 2013 at 16:22
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Assuming a fair six sided dice

The probability of rolling a 4 or more (4, 5 or 6) from a single roll is $\frac{3}{6} = \frac{1}{2}$ as there a three winning results and 6 possibilities.

The probability of rolling a 5 or more (5 or 6) from a single roll is $\frac{2}{6} = \frac{1}{3}$ as there a two winning results and 6 possibilities.

The probability of rolling a 5 or more (5 or 6) if two rolls are allowed is $\frac{1}{3} + \frac{2}{3} \cdot \frac{1}{3}= \frac{5}{9}$ This is the probability of getting a 5 or more on the first roll plus the probability that you don't get a 5 or more of the first roll (so get to roll again) times the probability of getting a 5 or more on the second roll.

Note: the probability of getting at least one 5 or 6 would be the same $\frac{5}{9}$ If you rolled both dice at the same time. The probability that the first dice is 5 or more plus the probability that you need to look at the second dice times the probability its 5 or more.

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Ok so in general, say we have 1 roll. Obviously we have a 1 in 6 chance to roll any number. But think of total number of possibilities. There's only 6 possibilities here. Keep that in mind.

Now roll two dice. We have 36 possibilities. 3 dice 216 possibilities...etc. One thing that might help you understand probability with dice is say we roll 6 dice and want to know the probabilty of rolling a six. Many people think...mistakenly...well, if it's a 1 in 6 chance to roll a 6 on each die, then there is a 100% probability of rolling a 6! Well, that's actually wrong. Since there are $46,656$ possible outcomes, and $5^6 = 15,625$ do not have a 6, well then the probability is only $\frac{46,656 - 15,625}{46,656} = 66.5%$. Now I think you may have a better understanding.

For answering your specific question I think Warren Hill does a good job in his answer above.

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  • $\begingroup$ Erm.... not sure this anything to do with the question. $\endgroup$ Commented Aug 28, 2013 at 14:11
  • $\begingroup$ He asked to be helped to understand any scenario. I think this helps understand all possible scenarios....read the last part of his question $\endgroup$ Commented Aug 28, 2013 at 14:13

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