0:31

So, let's look at the sum of two dice.

Â We now want to roll two fair dice.

Â And I to to sum, and those of you who like gambling, may know the game of creps,

Â and that's where exactly this imported, one roll two dice.

Â So, what are the possible outcomes in this sum of two dice?

Â Yeah, one plus one is the smallest number, that's the two.

Â And it goes all the way to 12, six plus six is the largest sum.

Â So, the basic outcomes in that sum, is two, three, four,

Â five all the way to 12, together, our sample space capital S.

Â And now, we can think of events, what's the probability of an 11 and the 12?

Â Or 12?

Â Or what's the probability of a seven?

Â Let's now move over to a spreadsheet that I've prepared for you, where we can

Â look at the possible combinations and the probability of the sums that are possible.

Â 1:30

So here, we have our spreadsheet where we want to look now at the sum of two dice.

Â So here in the left table, I have in the left column the six possible

Â numbers that the first dice may show, and here in the top row,

Â the six possible numbers, that's a second I may shop.

Â So six times six, notice here we have 36 combinations.

Â And in the table, I wrote for you the sum of the two dice.

Â So here, one plus one is a two.

Â Here, three plus one is a four.

Â Now notice, the different numbers show up different many times.

Â There's only one way to obtain a two, one plus one.

Â But for example, there are three ways to obtain a four,

Â there are also three ways to obtain a ten.

Â And so, this now shows that these probabilities will be different.

Â We have 36 possible combinations.

Â So now here on the right, I looked at the 11 possible sums two, three,

Â four, five, all the way to 12.

Â And in this column called counts, I wrote down how many

Â times does the number on the left, the sum, show up in that table?

Â So, two only shows up once.

Â The four shows up three times, just like the ten.

Â And notice now, the most common number is the seven.

Â Six, there are six combinations to roll a sum of seven.

Â Six plus one, five plus two, blah, blah, blah, until one plus six.

Â Since we have 36 combinations in the next column,

Â I calculated the probability of the seven numbers for you.

Â Six out of 36.

Â One-sixth is the probability that the sum will increase as seven.

Â This is the number that most common, and those of you who like to play craps,

Â you of course know that two and 12 are the least likely numbers.

Â Only one in 36 chance for each of them.

Â That's less than 3%.

Â So, we see here the numbers are different, but by counting the possible combinations,

Â and the total number of combinations we can calculate the probabilities.

Â 4:10

What have you seen?

Â The sample space has 11 elements.

Â And here is now a popular mistake that I see a lot from my students.

Â They think we have 11 possible numbers.

Â So, the probability of all of them is one divided by 11.

Â But we just saw on the spreadsheet that's not the case, why?

Â Because there are several ways to get to seven.

Â X1, five, two.

Â But there's only one way to get a one.

Â There's just a one, one or there's only one way to get a 12, six plus six.

Â So, here's the danger the people sometimes think.

Â My sample space has many elements, probabilities are one divided by them.

Â That's here not correct.

Â The assumption that all probabilities are the same, that's not correct.

Â 5:26

What are now the possible outcomes?

Â One, one.

Â One, two, one, three,four, four.

Â Four, five, four, six, five,one, all the way to six, six.

Â Now we see our sample space is not 11, it has 36 pairs,

Â number on the first die, number on the second die,

Â six times six 36 combinations, and they are now all equally likely.

Â The probability of one, one is the same as the two on the first, and

Â the five on the second.

Â 6:15

Now, there's a little more work to think about, what's a probability of four?

Â Four have three ways as a sum of occurring.

Â First a one, then a three, first a three, then a one, or a 2 2.

Â So, there are now three of these pairs that sum up to four.

Â Three divided by 36.

Â And that is exactly the probability we previously solved in the spreadsheet,

Â when we solved those three combinations, three in 36 of one-twelfth.

Â Now, let's look at different way of calculating this probabilities,

Â 6:56

notice the following when the first dice rolls, and

Â shows up the one, that does not affect the second dice.

Â It's not like the first die calls up the second die and says, dude,

Â I showed the one, you better don't show one, because otherwise,

Â the dumb humans get confused.

Â No, this probability of a one is still the same

Â as it was before I rolled the first die.

Â So, that means the outcome of the first die

Â does not affect the outcome of the second die.

Â This is the concept of independence or statistical independence.

Â More generally, two events are called Independent.

Â If Event A occurring, or

Â not occurring, does not affect the probability of Event B occurring.

Â 7:50

Why do we like this concept?

Â Because now, it's very easy for

Â me to calculate the probability of the intersection of two events.

Â Because now, if A and B are independent,

Â the probability of A and to B, A intersected with B,

Â it's just probability of A times the probability of B.

Â And now, I can use this with my two dice,

Â and ask what's the probability of first a one and then a five?

Â Probability of one, one in six, probability over five, one in six.

Â One-sixth times one-sixth is one-thirty-sixth.

Â And so, now I can finally show you a second way of calculating

Â the probability of the four.

Â Four, again, happens if you see first a one, then a three, first a two,

Â then another two, first a three, then another one.

Â Those three basic outcomes, or

Â if you want to think of them as single outcome events, are disjoint.

Â That means I'm allowed to add the probability of one,

Â three plus a probability of a two, two plus a probability of a three, one.

Â Each of them now I can calculate my

Â multiplication rule one-sixth times one-sixth for the first pair.

Â It's the same for the second pair, it's the same for the third and last pair, so

Â I get one in 36, plus one in 36, plus one divided by 36, wallah, we are back

Â to three divided by 36, one-twelfth the number we saw on the spreadsheet.

Â 9:54

I gave you first look at a concept that is very important in probability,

Â the independence of events, and I showed you a cool consequence of that concept,

Â namely the multiplication for independent events.

Â We will see this in action in future lectures, so

Â I hope you will be back for more fun with intuitive probabilities.

Â Thanks for your attention.

Â