# Dice Probablility Question

**URL:** <https://boards.straightdope.com/t/dice-probablility-question/143020>\
**Category:** Factual Questions\
**Created:** [December 16, 2002, 4:43am UTC](https://boards.straightdope.com/t/dice-probablility-question/143020 "2002-12-16T04:43:23Z")\
**Posts on this page:** 7\
**Page:** 1

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**Author:** ![Jude](https://avatars.discourse-cdn.com/v4/letter/j/bc8723/32.png) [@Jude](https://boards.straightdope.com/u/Jude)\
**Post date:** [December 16, 2002, 4:43am UTC](https://boards.straightdope.com/t/dice-probablility-question/143020/1 "2002-12-16T04:43:23Z")

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I want to know how to figure out the following probability problem:

If you roll a single die six times, what is the probability of not rolling a one?

Thanks. I’m not necessarily asking for an answer, I just want to know how to do it.

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**Author:** ![andy\_fl](https://avatars.discourse-cdn.com/v4/letter/a/49beb7/32.png) [@andy\_fl](https://boards.straightdope.com/u/andy_fl)\
**Post date:** [December 16, 2002, 4:52am UTC](https://boards.straightdope.com/t/dice-probablility-question/143020/2 "2002-12-16T04:52:41Z")

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each cast is independent of the previous, so :

(5/6) ^6

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**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\
**Post date:** [December 16, 2002, 4:52am UTC](https://boards.straightdope.com/t/dice-probablility-question/143020/3 "2002-12-16T04:52:45Z")

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The outcomes of each of the six rolls do not affect each other, so they are known as _independent events_. To get the total probability of two independent events, you multiply the individual probabilities. For instance, if event A has a 0.6 probability, and event B has a 0.5 probability, the probability for both A and B happening is 0.5 × 0.6 = 0.3. For more than two events, just mulitply all the probabilities.

Now in your case, you have six events. You don’t roll a one on the first throw. You don’t throw a one on the second throw. Etc. Each of them has a 5/6 probability. 5/6 times itself six times is the same as (5/6)[sup]6[/sup], which will give you your answer.

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**Author:** ![SandyHook](https://avatars.discourse-cdn.com/v4/letter/s/dec6dc/32.png) [@SandyHook](https://boards.straightdope.com/u/SandyHook)\
**Post date:** [December 16, 2002, 4:54am UTC](https://boards.straightdope.com/t/dice-probablility-question/143020/4 "2002-12-16T04:54:20Z")

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The odds of not rolling a 1 in 1 roll is 5/6=.8333

.8333 to the 6th power is 0.3348972

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**Author:** ![Ringo](https://avatars.discourse-cdn.com/v4/letter/r/779978/32.png) [@Ringo](https://boards.straightdope.com/u/Ringo)\
**Post date:** [December 16, 2002, 5:00am UTC](https://boards.straightdope.com/t/dice-probablility-question/143020/5 "2002-12-16T05:00:29Z")

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I _think_ my way towards the answer…

Your probability of _not_ rolling the one on one rol is 5/6. For six roles it should be:

[sup]5[/sup]/[sub]6[/sub] \* [sup]5[/sup]/[sub]6[/sub] \* [sup]5[/sup]/[sub]6[/sub] \* [sup]5[/sup]/[sub]6[/sub] \* [sup]5[/sup]/[sub]6[/sub] \* [sup]5[/sup]/[sub]6[/sub] = 0.3348979766804

So, with rounding you have a 1 in 3 chance of not rolling a one on six throws of the die. Calling **ultrafilter** ; clean up on thread 150647.

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**Author:** ![SC\_Wolf](https://avatars.discourse-cdn.com/v4/letter/s/eb9ed0/32.png) [@SC\_Wolf](https://boards.straightdope.com/u/SC_Wolf)\
**Post date:** [December 16, 2002, 5:09am UTC](https://boards.straightdope.com/t/dice-probablility-question/143020/6 "2002-12-16T05:09:34Z")

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Note that this is the theoretical probability. Actual dice can show variations from theoretical probability, due to flaws in their construction.

However, since these variations typically take hundreds to thousands of rolls to be statisically measurable (I know, I won two high school science fairs with projects based on testing this), it usually doesn’t matter.

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**Author:** ![Ringo](https://avatars.discourse-cdn.com/v4/letter/r/779978/32.png) [@Ringo](https://boards.straightdope.com/u/Ringo)\
**Post date:** [December 16, 2002, 5:12am UTC](https://boards.straightdope.com/t/dice-probablility-question/143020/7 "2002-12-16T05:12:21Z")

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roles? _Egad_, self-banishment to MPSIMS for 20 minutes.