Poll of the Day > If you rolled a 6-sided dice, 6 times, what is the % chance of getting a 6?

(+) Yeah! (+)

I'm trying to understand probability, or at least what math formula I'd use to determine this.

I know the chance of rolling a 6, on 1 roll is about 16%. But how does it scale when rolling it over and over. You clearly don't multiply it by the number of rolls, because something like this can't ever become 100%. But...I don't know what the formula is.

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TomNook posted...

I'm trying to understand probability, or at least what math formula I'd use to determine this.

I know the chance of rolling a 6, on 1 roll is about 16%. But how does it scale when rolling it over and over. You clearly don't multiply it by the number of rolls, because something like this can't ever become 100%. But...I don't know what the formula is.

Look up a dice rolling app on a website. Most have the formula on the page.

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Since the previous rolls do not affect the outcome of the subsequent rolls, it's always a 1in6 chance, no matter the number of rolls.

Just like every flip of a coin is always 50/50.

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Random chance because of random chance.

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Its a 50/50 chance, either you get a 6 or you don't

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SunWuKung420 posted...

Since the previous rolls do not affect the outcome of the subsequent rolls, it's always a 1in6 chance, no matter the number of rolls.

Just like every flip of a coin is always 50/50.

Hmmm, I don't think I explained what I mean well enough and I can't find this on the websites I checked, because the websites only talk about getting the same number of every dice rolled at once.

I know that every single roll will always be 1 in 6. But if you take a 6-sided dice and decide you are going to roll it 100 times. It can't be just a 1 in 6 chance that you will hit a 6 during those 100 rolls. I don't mean 1 in 6 for each roll. I mean the chance of hitting a 6 period, after all is said and done. Not the amount of 6's, or independently factoring each roll. Just getting a 6 at all.

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roughly 1/3 for rolling zero 6s, 40% chance for rolling one 6, 20% chance for rolling 2 6s, 5% chance for three 6s, less than 1% chance for rolling 4 or more sixes.

anydice.com is very useful for figuring die probability out.

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TomNook posted...

Hmmm, I don't think I explained what I mean well enough and I can't find this on the websites I checked, because the websites only talk about getting the same number of every dice rolled at once.

I know that every single roll will always be 1 in 6. But if you take a 6-sided dice and decide you are going to roll it 100 times. It can't be just a 1 in 6 chance that you will hit a 6 during those 100 rolls. I don't mean 1 in 6 for each roll. I mean the chance of hitting a 6 period, after all is said and done. Not the amount of 6's, or independently factoring each roll. Just getting a 6 at all.

It's still 1 in 6. Because

SunWuKung420 posted...

the previous rolls do not affect the outcome of the subsequent rolls

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PMarth2002 posted...

roughly 1/3 for rolling zero 6s, 40% chance for rolling one 6, 20% chance for rolling 2 6s, 5% chance for three 6s, less than 1% chance for rolling 4 or more sixes. So about 2/3 for rolling at least one six over the course of six rolls.

anydice.com is very useful for figuring die probability out.

Thank you. That sounds about right. What is the specific formula for figuring this out if you know, because I need to apply it to other numbers for myself? I just don't even know the terms for looking it up, because there are such a ridiculous amount of probability terminology.

SunWuKung420 posted...

It's still 1 in 6. Because

So flipping a million coins in a row, there is only a 50% chance of getting heads at least once?

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Thank you. That sounds about right. What is the specific formula for figuring this out if you know, because I need to apply it to other numbers for myself? I just don't even know the terms for looking it up, because there are such a ridiculous amount of probability terminology.

The odds of not getting a six is (5/6)^n, n being the number of dice rolls, so the odds of getting at least one six is (1/6)^n.

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PMarth2002 posted...

The odds of not getting a six is (5/6)^n, n being the number of dice rolls, so the odds of getting at least one six is (1/6)^n.

Awesome! Thanks!

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TomNook posted...

Awesome! Thanks!

No problem.

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PMarth2002 posted...

The odds of not getting a six is (5/6)^n, n being the number of dice rolls, so the odds of getting at least one six is (1/6)^n.

The odds of getting at least one 6 would be 1 - (5/6)^n
(and of course, multiple by 100 at the end to turn the decimal into a percentage)

So for six rolls, that would be 66.5%

PMarth2002 posted...

The odds of not getting a six is (5/6)^n

This is true

PMarth2002 posted...

so the odds of getting at least one six is (1/6)^n

This is false

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It's been awhile but I believe your probability is the total number of desired outcomes divided by the total number or possible outcomes for any specific roll.

So in case of rolling 6 dice, your chance of getting no 6's is 5/6 and you roll that 6 times (so (5/6)^6 then you subtract that from the total (6/6 or 1).

So to get the actual answer you solve this equation 1-(5/6)^6 which I'm not going to do.

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Blighboy posted...

This is true

This is false

Yeah I just realized i fucked it up. 1 - (5/6)^n is correct. Sorry tomnook. I was working backwords from 5/6 being the odds of not getting a six, but I had a brain fart on writing the actual formula.

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wolfy42 posted...

It's been awhile but I believe your probability is the total number of desired outcomes divided by the total number or possible outcomes for any specific roll.

So in case of rolling 6 dice, your chance of getting no 6's is 5/6 and you roll that 6 times (so (5/6)^6 then you subtract that from the total (6/6 or 1).

So to get the actual answer you solve this equation 1-(5/6)^6 which I'm not going to do.

Your explanation seems poorly worded, but you have the correct equation in the end.
And I already calculated it anyway :p

As a general rule in probability: It's usually easier to find the probability of something by finding the probability of not getting that result and then subtracting that number from 1.

Yeah there was like no reponses (or no correct ones) when I opened the topic, then I stopped in the middle of posting (may be why it seemed disjointed) to make some tea), still think I posted within like 5 minutes but a buncha people already answered by then.

Oh well, it was just off the top of my head anyway, other people explained it better.

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PMarth2002 posted...

Yeah I just realized i fucked it up. 1 - (5/6)^n is correct. Sorry tomnook. I was working backwords from 5/6 being the odds of not getting a six, but I had a brain fart on writing the actual formula.

Haha, it's all good. I kind of realized the other part of that equation didn't quite work when I started plugging it in.

Fascinating stuff though. Math is always way more fun when you can find real world applications. I made this topic because I was curious about some RPG drop rates and the likelihood of how many runs it would take to get certain things.

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streamofthesky posted...

The odds of getting at least one 6 would be 1 - (5/6)^n
(and of course, multiple by 100 at the end to turn the decimal into a percentage)

So for six rolls, that would be 66.5%

This. (1/6)^n would give you the chance of rolling N sixes in N rolls.

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SunWuKung420 posted...

It's still 1 in 6. Because

Serious question: Do you enjoy being a dumbass?

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https://anydice.com/program/1c3e

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SunWuKung420 posted...

Since the previous rolls do not affect the outcome of the subsequent rolls, it's always a 1in6 chance, no matter the number of rolls.

This (aka 16,7%) is the only correct answer.

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Roughly 66%. I only know this because binomial distributions with the format 1/x chance with x tries equal to that value*.
*Technically they approach 0.63, but I'm bad at math and only know this from a trivial standpoint. Calculus sucks.

You can plug the values into a calculator like this:
https://stattrek.com/online-calculator/binomial.aspx
With the values (.16, 6, 1) you get the follwoing:
P(X>=x) = 0.64.

I am not surprised that PotD doesn't grasp the concept of cumulative distributions.
But yes Marth and streamofthesky came to the right answer. At least someone was paying attention in stats to understand what a binomial distribution was.

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Tha_Man_DS posted...

This (aka 16,7%) is the only correct answer.

That is the chance of rolling a 6 on the Nth roll. That's not what he's asking. He's asking the chance of rolling one 6 in N rolls, which very much is not 1/6 unless N=1. The Gambler's Fallacy is a matter of thinking that the next roll is more likely to be 6 because the last five rolls haven't been, not a matter of thinking that you're more likely to see at least one 6 in 100 rolls than in 1 roll.

Judgmenl posted...
At least someone was paying attention in stats to understand what a binomial distribution was.

Most of my practice with this didn't come from stats so much as from video games. I often calculated these probabilities to figure out how unlucky I'd been to not have seen a certain item drop yet in WoW, as well as using the corresponding logarithmic equation to figure out roughly how many runs a given item could be expected to take. WoW was a long time ago (I quit in July 2009), but these principles still come up often enough in other games that I'll break out the calculator now and then.

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Yea I used to calculate the likelihood to get a Drop in Runescape based on the number of kills I was getting e.g. 1/512 chance to get an Abyssal Whip in a Slayer Task to kill 200 Abyssal Demons.

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6%

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1/6
+(1/6)*(1/6)
+(1/6)*(1/6)*(1/6)
+(1/6)*(1/6)*(1/6)*(1/6)
+(1/6)*(1/6)*(1/6)*(1/6)*(1/6)
+(1/6)*(1/6)*(1/6)*(1/6)*(1/6)*(1/6)

Edit NVM wrong, I'll give this a shot when I'm off work

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I can see why Sunny had to get out of the life insurance game.

Tha_Man_DS posted...

This (aka 16,7%) is the only correct answer.

I hope you two were deliberately taking the piss on this one.

Anyways, TC, an easy shorthand I use for this sort of question is to ask if it's easier to calculate the opposite of what you're trying. In this example, if you roll six dice, what's the % chance you don't get any sixes?

This, conceptually, is much easier to calculate - it's just 5/6 (the odds of not getting a six on a single roll) to the power of the number of rolls (since all of them have to show a not-six result for the condition to be met) - in other words (5/6)^6.

Now that you know the odds of not satisfying your condition (i.e. getting all non-six results), 100% minus that value is the answer you want.

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Yeah 1 - (5/6)^6

I thought everyone was wrong and I was going to be clever, but no, that's it.

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adjl posted...
That is the chance of rolling a 6 on the Nth roll. That's not what he's asking. He's asking the chance of rolling one 6 in N rolls, which very much is not 1/6 unless N=1. The Gambler's Fallacy is a matter of thinking that the next roll is more likely to be 6 because the last five rolls haven't been, not a matter of thinking that you're more likely to see at least one 6 in 100 rolls than in 1 roll.

Sorry to be pedantic, but he's asking for the probability of at least one 6 in N rolls. Figuring out the probability of exactly one 6 (no more, no less) is a different question w/ a different answer, IIRC involving a Combination function/equation.
The 1 - (5/6)^n equation includes the probability of every result that has one or more 6's in the resulting rolls (one 6, two 6's, three 6's, etc...)

tldr: He's asking P(X>=1).

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