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## Precalculus (2018 edition)

### Course: Precalculus (2018 edition) > Unit 6

Lesson 4: Compound probability of independent events using the multiplication rule- Compound probability of independent events
- Probability without equally likely events
- Independent events example: test taking
- Die rolling probability with independent events
- "At least one" probability with coin flipping
- Free-throw probability
- Three-pointer vs free-throw probability
- Independent probability

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# Die rolling probability with independent events

We hope you're not a gambler, but if you had to bet on whether you can roll even numbers three times in a row, you might want to figure this probability first. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Why can't we get the answer by just multiplying 1/2 by 3?(27 votes)
- Firstly, try it. 3 x 1/2 = 3/2 >1.

You can NEVER have a probability >1, so that should be your red flag. You've made a mistake.

Your mistake? You have to multiply 1/2 by 3 times itself. Not 3. because each time its a likelihood is 1/2, and you do it 3 times.(40 votes)

- If three six-sided dies are rolled, once each one, how can I find the probability of getting three even numbers? The number of possible events would be six times six times six? And, if I want to know the probability of any other event, lets say of getting a 6 in any of the three dies, or a addition wich result is a number minor than 14, how can I calculate it?(12 votes)
- Your first question is basically the same as the one described in the video. Rolling three dice one time each is like rolling one die 3 times. And yes, the number of possible events is six times six times six (216) while the number of favourable outcomes is 3 times 3 times 3. Therefore, the probability is still 1/8 after reducing the fraction, as mentioned in the video.

You can calculate the probability of another event just by finding the total number of outcomes, in this case 216, and counting how many results involve getting one or more 6. There are a total of 91 ways to get 6 in at least one of the three dice. Therefore, the probability is 91/216.

The same applies for an addition which results in a number less than 14. Count how many results are favourable (produce a number less than 14) and divide that by the total number of outcomes to get the probability. If my calculations are correct, there are 91 ways to get a sum less than 14. The probability will again be 91/216.

It can be a hassle to list out and count the individual favourable outcomes. It is important to find some sort of pattern for organized counting. For example, for a sum less than 14 it just so happens that the number of favourable outcomes is 1 + 4 + 9 + 16 + 25 + 36 = 91. It is the sum of the first 6 perfect squares. Noticing these patterns can make counting much easier.(8 votes)

- At0:22we are considering a 6-sided dice... but what if we were doing... a 100 sided dice? How would that change the answer?(5 votes)
- On a 6 sided dice there are 3 possible even numbers (2,4,6)

On a 100 sided dice there are 50 possible even numbers (I won't write them all, but I encourage you to if you don't get it)

3/6 and 50/100 are both the same fraction, 1/2! So the probability of rolling an even number on both of those dice is the same! The number of sides would not change the answer.

HOWEVER, if the dice has an uneven number of even and odd sides (like a 7 or 101 sided dice) the probabilities will be DIFFERENT.

With a 7 sided dice there are 3 even numbers (2,4,6), but 4 odd numbers! (1,3,5,7)

and on a 101 sided dice there are 50 even numbers, and 51 odd numbers.

The probability of getting an even number on the first dice is different than on the second dice!

3/7=/=50/101

So as long as the dice has the same number of even and odd sides (if the dice has an even number of sides) it makes no difference how many sides it has. The probability of rolling an even number on a 2 sided dice, on a 10 sided dice, or even on a 1,000,000 sided dice are all the same!

But if the number of odd and even sides are not equal (if the dice has an odd number of sides) then there is not the same probability! The more sides, the closer the fraction will get to 1/2, but it will never be 1/2!

Hope I helped! :)(4 votes)

- I know that from a mathematician's point of view the first roll doesn't affect the second, but from the point of physics, doesn't it matter which side was up, and which side was down? Because when we use the absolutely same force(in intensity & direction) on a dice with side '1' up over and over again, wont we get the same answer every time? Theoretically? (it's still a math question i guess :D)(4 votes)
- Rolling of dice is what is known as a deterministic chaotic system. In other words, it is an outcome that is not random but is affected by so many variables that it appears to be random. Except in extremely advanced mathematical models (see chaos theory) a deterministic chaotic system cannot readily be distinguished from true randomness.

So, no, the outcome of the roll of dice is not random at all, but it is so chaotic, so complicated and nearly impossible to predict, that it might as well be random.(4 votes)

- if the probability of a roll doesn't affect the probabililty of the next roll, then how come it keeps getting smaller and smaller as you continuously roll the die?(3 votes)
- The probability getting smaller is the probability of
*consecutively*rolling an even number,**as defined before you start**. It's like holding a die and saying "What's the probability that I now roll six even numbers in a row?" This probability will get smaller, the more even numbers you assign in a row.

However, if you roll five even numbers in a row the probability**from then on**of getting another even number is just 1/2 (since half are odd and half are even). The die doesn't "remember" what it rolled before.

Hope this helps!(5 votes)

- What is the probability of getting even numbers when 3 dices se thrown together ?

Please help.

Thank you.(5 votes)- All dices have 3 even numbers and 3 odd numbers so for the first dice, the probability is 3/6, same goes for the second and the third dice.

total=9/18 which is half of the time, you will get even numbers.(1 vote)

- I wonder, if we are rolling a die an even number of times, what is the probability of getting even and odd numbers on a die equal times? (For example, we roll 6 times and get 1, 3, 3, 6, 6, 2 - #odd results = # even results)(3 votes)
- Please note that I do not guarantee that my answer is correct, so if anyone could verify it then that would be great. This also means that you should not assume I am correct without first checking if I did not made any mistakes.

OK on with the answer. Since we do not care about the number returned by a roll besides whether it is even or not, we could simplify this a bit by saying every roll has 2 equal likely possible outcomes, that is even or odd.

Let us say that we make n rolls, where n is an even integer.

Then we have 2^n possible sequences of even and odd.

But how many of those sequences have #odd results = # even results?

Well there is a mathematics function that can help us and it is called Combination. I think Sal has some video's about it here:

https://www.khanacademy.org/math/precalculus/prob_comb/combinations

And I think that wikipedia also says something about it here:

https://en.wikipedia.org/wiki/Combination

Since I cannot do the fancy notation I will use C(n,k) instead of the normal notation where n is the top number and k the bottom number in the normal notation.

Anyway in a sequence where #odd results = # even results we know that there must be n/2 even results. And there are C(n , n/2) possible sequences out of the 2^n sequences where this is true.

So the chance that you get as many odd rolls as that you get even rolls for n rolls if n is even is equal to`C(n,n/2) / 2^n`

,(3 votes)

- Sal....

I have a giga dice with 256 faces, with numbers 1 to 256, one number written on each face...

I take 2 such giga dices and roll them. Whats the probability to get the sum of the values on the two dices as 327? I kno it will b some value n/(256^2) lol. so yes it wud b some value n/66564, eek~(2 votes)- Let me give it a try. If you rolled a 70 or lower on the first die, then you couldn't possibly roll high enough on the second die to reach 327. But then for every number between 71 and 256 (inclusive), there is one pair of rolls that equals 327. If you use the notation (firstroll, second roll) for the event, then the possible ways to roll 327 are: (71, 256), (72, 255), (73, 254)...(255, 70), (256, 71). So there are 256-70 ways to roll 327, or 186 ways, which is the top of the fraction. Then you are correct, there are 256^2 possible rolls, so the total probability is 186/66546, or 0.2%.(7 votes)

- Andrew rolls a die. What is the probability he gets a 4 or an even number?(3 votes)
- 1/2. There are three out of six outcomes work. 4 matches both criteria but should only be counted once.(1 vote)

- What would be the possibility of rolling a, let's say six, with a single die in three trials?(4 votes)
- I'm not sure how Sal would do this, but consider the 6x6x6=216 different scenarios in total when we roll the single die in three trials.

When only the first die is 6 there are 1x5x5=25 different combinations, and there are also 25 combinations each for when only the second die is 6 and when only the third die is 6.

There are also situations where only the first two trials are six, which gives 1x1x5=5 combinations, and another 5+5=10 for when only the last two trials are six, and when only the first and last trial are six.

Last but not least, there is one possibility when all trials are 6, which gives 1x1x1=1 combination.

Finally we add all possibilities up, 25+25+25+5+5+5+1=91, therefore the possibility that we roll at least one six in a case where a single die is rolled for three trials is 91/216.

All of this is based on the assumption that the probability of rolling each number with this die for each trial is the same.

Correct me if I'm wrong but I think that should get you the answer.(1 vote)

## Video transcript

Find the probability of rolling
even numbers three times using a six-sided die
numbered from 1 to 6. So let's just figure out the
probability of rolling it each of the times. So the probability of rolling
even numbers. So even roll on six-sided die. So let's think about
that probability. Well, how many total
outcomes are there? How many possible rolls
could we get? Well, you get one, two, three,
four, five, six. And how many of them satisfy
these conditions, that it's an even number? Well, it could be a 2,
it could be a 4, or it could be a 6. So the probability is the events
that match what you need, your condition for right
here, so three of the possible events are an even roll. And it's out of a total of
six possible events. So there is a-- 3 over 6 is
the same thing as 1/2 probability of rolling
even on each roll. Now they're going to
roll-- they want to roll even three times. And these are all going to
be independent events. Every time you roll, it's not
going to affect what happens in the next roll, despite what
some gamblers might think. It has no impact on what happens
on the next roll. So the probability of rolling
even three times is equal to the probability of an even roll
one time, or even roll on six-sided die-- this thing over
here is equal to that thing times that thing again. All right, that's our first
roll-- we copy and we paste it-- times that thing and then
times that thing again. Right? That's our first roll,
which is that. That's our second roll. That's our third roll. They're independent events. So this is going to be equal to
1/2-- that's the same 1/2 right there-- times 1/2
times 1/2, which is equal to 1 over 8. There's a 1 in 8 possibility
that you roll even numbers on all three rolls. On this roll, this roll,
and that roll.