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## 7th grade

### Course: 7th grade > Unit 7

Lesson 3: Compound events and sample spaces# Die rolling probability

We're thinking about the probability of rolling doubles on a pair of dice. Let's create a grid of all possible outcomes. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- why isn't the prob of rolling two doubles 1/36? prob of rolling any number on 1 dice is 1/6 shouldn't you multiply the prob of both dice like in the first coin flip video? I understand the explanation given, but I'm trying to figure out why the same coin logic doesn't work.(25 votes)
- You need to consider how many ways you can roll two doubles, you can get 1,1 2,2 3,3 4,4 5,5 and 6,6 These are 6 possibilities out of 36 total outcomes. The probability for rolling one of these, like 6,6 for example is 1/36 but you want to include all ways of rolling doubles.(50 votes)

- If this was in a exam, that way of working it out takes too long so is there any quick ways so you won't waste time?(17 votes)
- well you can think of it like this. It really doesn't matter what you get on the first dice as long as the second dice equals the first. so the probability of the second equaling the first would be 1/6 because there are six combinations and only one of them equals the first.(30 votes)

- At4:14is there a mathematical reason why the favourable outcomes line up on the diagonal?(19 votes)
- That is a result of how he decided to visualize this. Imagine we flip the table around a little and put it into a coordinate system. Along the x-axis you put marks on the numbers 1, 2, 3, 4, 5, 6, and you do the same on the y-axis. We are interested in rolling doubles, i.e. getting the same on both dice. If we let x denote the number of eyes on the first die, and y do the same for the second die, we are interested in the case y = x. But this is the equation of the diagonal line you refer to.(16 votes)

- At 2.30 Sal started filling in the outcomes of both die.

This video wasn't what i was looking for but some of you might be able to help. I had a question:

"Two dice are rolled, copy and complete the table below"......

Then there was a table which looked exactly like the one Sal drew. But it had been filled out differently:

In the first box (Dice 1 and Dice 2) it had been filled in as 2. Sal wrote 1,1. In another box on my sheet (1 across and three down on Sal's diagram) it had been filled out as 1. I am very confused on how they got this answer so if you understand what I'm talking about please answer. Sorry for the bad explaining!(14 votes)- It's because you aren't supposed to add them together.

The reason for it being 1,1 is because it's supposed to symbolize two dice both equaling 1 so it would be 1,1, with each one meaning the outcome of each die.(3 votes)

- Is there a way to find the probability of an outcome without making a chart?(6 votes)
- Probably the easiest way to think about this would be:

P(Rolling a 1 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296

P(Rolling a 2 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296

P(Rolling a 3 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296

P(Rolling a 4 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296

P(Rolling a 5 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296

P(Rolling a 6 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296

Adding these probabilities together, we get: 6/1296 = 1/216.

So the probability you will roll the same number four times in a row with a fair dice is 1/216.

Does this help.(9 votes)

- I was wondering if there is another way of solving the dice-rolling probability and coin flipping problems without constructing a diagram? Is there a way to find the solution algorithmically or algebraically?(3 votes)
- Definitely, and you should eventually get to videos descriving it. But to show you, I will try and descrive how to do it.

So, since there is an equal chance to roll any number on a six sided die, that means the chance of rolling any one number is one out of 6 or 1/6. You can see that with a diagram. Now, rolling two different numbers in a specific order you can tell with a diagram is 1/36. To find this out through math though you multiply probabilities of events happening if you are looking for both of them happening. so you want to roll x first AND y second. so that's 1/6*1/6, which is 1/36. And you can keep goign with this pattern.

The calculation is a bit different if you are looking for one thing to happen OR another. In simple caes it's just adding, like what are the odds of rolling a 1 OR 2 on a dice? you add the two, which you can see on a diagram.

You still have to be careful, like if a problem asks what the odds of rolling a 1 AND a 2 on ONE die is, you can't roll both so the answer is 0. Or rolling a 1 on one die OR rolling a 2 on another. It's still 1/6 since you are rolling them separately and they don't effect each other. It can get a bit confucing, but most of the time you will be using the more simple cases. If ever in doubt use diagrams to see a few cases of what you're doing to get an intuition(8 votes)

- Find the probablility of the occurance of1on a die if it has one more of its faces marked as 1instead of 6(3 votes)
- Here's how you'd do the problem.

We know that two of the sides have*1.*The rest have other numbers.

Since in total, there are 6 sides, get the amount of sides that have 1 divided by 6 (six possible outcomes) and that's your answer.

So in your case, it's 2/6, which is 1/3.(8 votes)

- Can learners open up a black board like Sals some where and work on that instead of the space in between problems?(4 votes)
- If you're working on a Windows pc, you would need either a touchscreen pc, complete with a stylus pen or a drawing tablet. Then you could download for free the Sketchbook Pro software for Windows and invert the colors. Voila, you have a Khan Academy style blackboard.

If you don't have a pc with a touchscreen or drawing tablet, you can use phones with drawing styluses like the Samsung Note phones to do your work on SketchBook mobile.

Or just, you know, do it on plain lined, graphing, or print paper (^ u^

hope this helped.(4 votes)

- Can someone help me

"If you roll two fair six-sided dice, what is the probability that at least one die shows a 3?"(3 votes)- P(at least one 3)=1-P(no 3s)

There are 5 ways to get no 3s on each die, and so there are 25 ways to get no 3s, and so P(no 3s)=25/36.

So the probability you want is

1-(25/36)=11/36.(6 votes)

- From a well shuffled 52 card's and black are removed from cards find the probability of drawing a king or queen or a red card(3 votes)
- If the black cards are all removed, the probability of drawing a red card is 1; there are only red cards left.

Therefore, the probability of (literally anything) or a red card is also 1.(5 votes)

## Video transcript

Find the probability
of rolling doubles on two six-sided dice
numbered from 1 to 6. So when they're talking
about rolling doubles, they're just saying,
if I roll the two dice, I get the same number
on the top of both. So, for example, a 1
and a 1, that's doubles. A 2 and a 2, that is doubles. A 3 and a 3, a 4 and a 4,
a 5 and a 5, a 6 and a 6, all of those are
instances of doubles. So the event in question
is rolling doubles on two six-sided dice
numbered from 1 to 6. So let's think about all
of the possible outcomes. Or another way to
think about it, let's think about the
sample space here. So what can we roll
on the first die. So let me write this
as die number 1. What are the possible rolls? Well, they're
numbered from 1 to 6. It's a six-sided die, so I can
get a 1, a 2, a 3, a 4, a 5, or a 6. Now let's think about the
second die, so die number 2. Well, exact same thing. I could get a 1, a 2,
a 3, a 4, a 5, or a 6. Now, given these possible
outcomes for each of the die, we can now think of the
outcomes for both die. So, for example, in this--
let me draw a grid here just to make it a little bit neater. So let me draw a line there and
then a line right over there. Let me draw actually
several of these, just so that we could really
do this a little bit clearer. So let me draw a full grid. All right. And then let me draw the
vertical lines, only a few more left. There we go. Now, all of this top row,
these are the outcomes where I roll a 1
on the first die. So I roll a 1 on the first die. These are all of those outcomes. And this would be I run
a 1 on the second die, but I'll fill that in later. These are all of the
outcomes where I roll a 2 on the first die. This is where I roll
a 3 on the first die. 4-- I think you get the
idea-- on the first die. And then a 5 on
the first to die. And then finally, this last
row is all the outcomes where I roll a 6
on the first die. Now, we can go
through the columns, and this first column is where
we roll a 1 on the second die. This is where we roll
a 2 on the second die. So let's draw that out, write
it out, and fill in the chart. Here's where we roll
a 3 on the second die. This is a comma that I'm
doing between the two numbers. Here is where we have a 4. And then here is where
we roll a 5 on the second die, just filling this in. This last column is where we
roll a 6 on the second die. Now, every one of these
represents a possible outcome. This outcome is where we roll
a 1 on the first die and a 1 on the second die. This outcome is where we
roll a 3 on the first die, a 2 on the second die. This outcome is where we
roll a 4 on the first die and a 5 on the second die. And you can see here, there are
36 possible outcomes, 6 times 6 possible outcomes. Now, with this out of the way,
how many of these outcomes satisfy our criteria of rolling
doubles on two six-sided dice? How many of these outcomes
are essentially described by our event? Well, we see them right here. Doubles, well, that's rolling
a 1 and 1, that's a 2 and a 2, a 3 and a 3, a 4 and a 4, a
5 and a 5, and a 6 and a 6. So we have 1, 2, 3, 4, 5, 6
events satisfy this event, or are the outcomes that are
consistent with this event. Now given that, let's
answer our question. What is the probability
of rolling doubles on two six-sided die
numbered from 1 to 6? Well, the probability
is going to be equal to the number of outcomes
that satisfy our criteria, or the number of outcomes
for this event, which are 6-- we just figured
that out-- over the total-- I want to do that pink
color-- number of outcomes, over the size of
our sample space. So this right over here,
we have 36 total outcomes. So we have 36 outcomes,
and if you simplify this, 6/36 is the same thing as 1/6. So the probability
of rolling doubles on two six-sided dice
numbered from 1 to 6 is 1/6.