Suppose you have a bag with five counters in it: three red ones and two blue ones.
You're going to draw out two counters in a row, without replacement.
That means you're going to pull out one counter, and then then pull out a second one without
putting the original one back in first.
Let's draw a tree diagram to model this experiment.
The first counter could be one of three red ones, or one of two blue ones.
Now work across the page from left to right for the second counter.
If the first counter is red, then there are only two red counters left for the second
draw.
And if the first counter is blue, then there are three red counters left but only one blue
counter left for the second draw.
And that's all well and good.
But what if I asked you to extend the experiment to drawing out three counters in a row, without
replacement?
Suddenly the tree becomes a bit large and unwieldy, doesn't it?
You could draw it, but there are so many options to try to fit in, and they're all so similar,
it would be very easy to get confused trying to read anything off the diagram.
So what we do instead is draw a simplified tree diagram.
It works like this.
The first counter could be red or blue.
The chance of getting a red one is three out of five, and the chance of getting a blue
one is the other two out of five.
Now the second counter.
If the first one was red, then the second one could still be red or blue.
But since we're not replacing the first counter, the chance of getting a second red one is
now two out of four: there are only four counters left in the bag, and two of them are red.
The other two are blue, so there's the chance of getting a blue counter on the second draw,
if the first one was red.
Okay, so what about if the first counter was blue?
Then the second could be red or blue, and there are three red counters left out of four,
but only one blue left out of four.
Now, the third draw.
If the first two counters were both red, then there are three counters left in the bag:
one red one and two blue ones.
If the first was red and the second blue, or the first blue and the second red, then
in both of these cases, there are three counters left: two red ones and one blue one.
Now think carefully about the final case.
If both of the first two counters are blue, then there are only three red counters remaining.
You are now certain to get a red counter on the next draw.
It's important here to note that with a simplified tree diagram like this, you can't simply count
the outcomes on the right-hand side of the tree.
Although there are seven possible outcomes here, they are not all equally likely!
To calculate the theoretical probability of any individual outcome, you have to multiply
the probabilities of each partial event that got you to that outcome.
Remember to work across the tree from left to right.
You multiply the probabilities that you wrote on the tree, for each branch from the start
on the left along the path to the outcome you're interested in.
Let me show you another example.
Suppose I roll three dice, in a game where the only thing that matters is whether or
not I roll a six.
What's the probability of getting at least two sixes, exactly one six, or no sixes?
In this case, the only thing I care about after each roll is whether I rolled a six.
So the first die could come up as a six, or as something else.
And there's a one in six chance of getting a six, and five in six chance of not getting
a six.
Then the second die, with the same chance for each outcome.
And the third.
Now let's consider the probabilities.
There are four ways through this tree to get at least two sixes.
But the chance is not four out of eight!
Remember, to calculate the probability of each of these outcomes, I have to multiply
the probabilities of each branch along the way from left to right.
So the first way, three sixes, is one sixth times one sixth times one sixth, which is
one in two hundred and sixteen.
The second, two sixes followed by something else, is one sixth times one sixth times five
sixths, or five in two hundred and sixteen.
The third way, a six, not a six, then a six, is the same probability.
And the fourth way, with the not-a-six roll first, is also five in two hundred and sixteen.
So the theoretical probability of getting any one of these four mutually exclusive options
is simply the addition of these fractions, which comes to sixteen out of two hundred
and sixteen, which simplifies to two in twenty-seven, or about seven point four percent.
There are three ways of getting exactly one six, and they each have the same likelihood,
twenty-five out of two hundred and sixteen.
So the theoretical probability of rolling exactly one six is three times this fraction,
seventy-five out of two hundred and sixteen, or twenty-five out of seventy-two, about thirty-four
point seven percent.
And there's only one way in this tree to get no sixes, but it's a fairly common outcome,
because there's actually many different rolls hiding along this one path.
It's five times five times five, or one hundred and twenty-five out of two hundred and sixteen.
That's around fifty-seven point nine percent.
Now imagine how hard it would be to work these out by drawing a complete tree diagram with
all two hundred and sixteen outcomes at the right-hand side!