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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. •   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.
• 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? •  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.
• At is there a mathematical reason why the favourable outcomes line up on the diagonal? • 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.
• 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! • Is there a way to find the probability of an outcome without making a chart? • 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.
• can anyone tell me what 1x1 is? • That's weird 🤔. The probability of getting any side on a fair die is 1/6. How come the probability of rolling 2 different dice is the same? During the video my intuition was telling me that the answer should be 1/6 * 1/6. What am I missing here? • The wording of the third sentence of your question is problematic. I will assume you are asking about the probability of rolling doubles on two different dice.

Yes, the probability of rolling any specific sequence of two numbers is 1/6 * 1/6 = 1/36, but there are 6 possible sequences that give doubles: 1,1; 2,2; 3,3; 4,4; 5,5; and 6,6. So the probability of rolling doubles is 6 * 1/36 = 1/6.

A better intuition goes like this: no matter what is rolled on the first die, the probability that the second die matches it is 1/6. Therefore, the overall probability that they match (i.e. rolling doubles) is 1/6.

Have a blessed, wonderful day!
• 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? • 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  