WEBVTT
00:00:01.240 --> 00:00:04.270
We will now apply our
multinomial formula for
00:00:04.270 --> 00:00:07.270
counting the number of
partitions to solve the
00:00:07.270 --> 00:00:09.510
following probability problem.
00:00:09.510 --> 00:00:12.690
We have a standard 52-card
deck, which we
00:00:12.690 --> 00:00:14.190
deal to four persons.
00:00:14.190 --> 00:00:18.890
Each person gets 13 cards as,
for example, in bridge.
00:00:18.890 --> 00:00:21.360
What is the probability
that each person
00:00:21.360 --> 00:00:24.710
gets exactly one ace?
00:00:24.710 --> 00:00:27.500
Well, before we start, as
always we will need a
00:00:27.500 --> 00:00:29.170
probability model.
00:00:29.170 --> 00:00:34.700
We deal the cards fairly, and
this is going to be our model.
00:00:34.700 --> 00:00:37.760
But we still need to interpret
our statement.
00:00:37.760 --> 00:00:41.200
To give this interpretation,
let us first think of the
00:00:41.200 --> 00:00:42.640
outcomes of the experiment.
00:00:42.640 --> 00:00:44.250
What are the possible
outcomes?
00:00:44.250 --> 00:00:49.440
An outcome of this experiment is
a partition of the 52 cards
00:00:49.440 --> 00:00:52.720
into the four persons
so that each person
00:00:52.720 --> 00:00:55.360
gets exactly 13 cards.
00:00:55.360 --> 00:01:00.050
Our statement about dealing the
cards fairly will be an
00:01:00.050 --> 00:01:03.680
assumption that all partitions
are equally likely.
00:01:12.180 --> 00:01:15.280
So since all partitions, all
outcomes of the experiment,
00:01:15.280 --> 00:01:18.030
are equally likely, this means
that we can solve a
00:01:18.030 --> 00:01:20.750
probability question
by counting.
00:01:20.750 --> 00:01:24.070
We need to count the number of
elements of our sample space,
00:01:24.070 --> 00:01:27.200
the number of possible outcomes,
and then count the
00:01:27.200 --> 00:01:29.820
number of outcomes that
make the event
00:01:29.820 --> 00:01:32.370
of interest to occur.
00:01:32.370 --> 00:01:34.360
Let us start with the
number of elements
00:01:34.360 --> 00:01:36.680
of the sample space.
00:01:36.680 --> 00:01:39.810
This is the problem that
we just dealt with a
00:01:39.810 --> 00:01:41.020
little while ago--
00:01:41.020 --> 00:01:45.250
the number of outcomes, the
number of partitions of 52
00:01:45.250 --> 00:01:52.015
items into four persons, where
we give 13 cards to person
00:01:52.015 --> 00:01:58.150
one, 13 cards to person two, 13
cards to person three, and
00:01:58.150 --> 00:02:00.700
13 cards to person four.
00:02:00.700 --> 00:02:04.700
The number of possible ways of
doing this is equal to this
00:02:04.700 --> 00:02:07.160
multinomial coefficient.
00:02:07.160 --> 00:02:10.360
So now let us count the number
of outcomes that belong to the
00:02:10.360 --> 00:02:13.980
event of interest, namely the
outcomes where each person
00:02:13.980 --> 00:02:16.250
gets an ace.
00:02:16.250 --> 00:02:19.400
We think of the process of
constructing such an outcome
00:02:19.400 --> 00:02:20.870
as a multi-stage process.
00:02:20.870 --> 00:02:22.740
And we count the number
of choices that we
00:02:22.740 --> 00:02:24.579
have at each stage.
00:02:24.579 --> 00:02:26.890
The process is as follows.
00:02:26.890 --> 00:02:30.390
We first distribute
the four aces.
00:02:30.390 --> 00:02:33.820
We take the ace of spades and
give it to one person.
00:02:33.820 --> 00:02:35.620
In how many ways can we do it?
00:02:35.620 --> 00:02:38.690
We can do it in four ways.
00:02:38.690 --> 00:02:40.660
Then we take the next ace.
00:02:40.660 --> 00:02:43.170
The next ace must be given
to a different person.
00:02:43.170 --> 00:02:46.530
And so at that stage, we have
three different choices about
00:02:46.530 --> 00:02:48.750
who to give that ace to.
00:02:48.750 --> 00:02:51.200
Then we consider the next ace.
00:02:51.200 --> 00:02:54.220
At this point, two persons
already have aces.
00:02:54.220 --> 00:02:57.610
So we have two available
choices for who can
00:02:57.610 --> 00:02:59.760
get the next ace.
00:02:59.760 --> 00:03:02.660
And finally for the last ace,
we do not have any choice.
00:03:02.660 --> 00:03:05.370
We give it to the only remaining
person who doesn't
00:03:05.370 --> 00:03:08.510
yet have an ace.
00:03:08.510 --> 00:03:13.990
Having distributed the four
aces, then we need to somehow
00:03:13.990 --> 00:03:19.590
distribute the remaining 48
cards to the four people.
00:03:19.590 --> 00:03:22.050
But we can do that in
any way we want.
00:03:22.050 --> 00:03:25.950
So all we need to do is to just
partition the 48 cards
00:03:25.950 --> 00:03:28.950
into four subsets of given
cardinalities.
00:03:28.950 --> 00:03:34.680
And this can be done by a number
of ways, which is the
00:03:34.680 --> 00:03:36.190
number of such partitions.
00:03:36.190 --> 00:03:39.350
We have already found
what that number is.
00:03:39.350 --> 00:03:43.135
And it is this particular
multinomial coefficient.
00:03:46.880 --> 00:03:50.440
So the number of ways that we
can distribute the cards so
00:03:50.440 --> 00:03:53.940
that each person gets an ace,
according to the counting
00:03:53.940 --> 00:03:56.870
principle, is going to be the
number of ways that we can
00:03:56.870 --> 00:04:00.150
distribute the aces times the
number of ways that we can
00:04:00.150 --> 00:04:02.060
distribute the remaining
cards.
00:04:02.060 --> 00:04:05.860
The product of this number gives
us the count, gives us
00:04:05.860 --> 00:04:09.510
the cardinality, of the
event of interest.
00:04:09.510 --> 00:04:12.860
We also have the cardinality
of the sample space.
00:04:12.860 --> 00:04:15.680
So the desired probability
can be found by
00:04:15.680 --> 00:04:18.000
dividing these two numbers.
00:04:18.000 --> 00:04:21.230
And the final answer
takes this form.
00:04:24.810 --> 00:04:27.300
Let us now look at the
same problem but
00:04:27.300 --> 00:04:28.910
in a different way.
00:04:28.910 --> 00:04:32.560
Probability problems can often
be solved in multiple ways.
00:04:32.560 --> 00:04:35.120
And some can be faster
than others.
00:04:35.120 --> 00:04:38.690
So we want to look for a smarter
solution that perhaps
00:04:38.690 --> 00:04:42.970
will get us in a faster way
to the desired answer.
00:04:42.970 --> 00:04:45.750
We will use the following
trick.
00:04:45.750 --> 00:04:48.990
We will think about a very
specific way of dealing the
00:04:48.990 --> 00:04:51.230
cards which is the following.
00:04:51.230 --> 00:04:56.210
We take the 52 cards, the card
deck, and stack it so that the
00:04:56.210 --> 00:04:59.180
four aces are at the top.
00:04:59.180 --> 00:05:01.200
So they are first.
00:05:01.200 --> 00:05:05.970
And then we deal those cards
to the players as follows.
00:05:05.970 --> 00:05:12.480
We think of each player having
13 slots of his own.
00:05:12.480 --> 00:05:16.630
And the cards will be
placed randomly into
00:05:16.630 --> 00:05:19.290
the different slots.
00:05:19.290 --> 00:05:24.060
So we can do this one card at a
time, starting from the top.
00:05:24.060 --> 00:05:31.380
We take the first ace and send
it to a random location.
00:05:31.380 --> 00:05:34.050
Then we will take the second
ace, send it to a random
00:05:34.050 --> 00:05:35.930
location, and so on.
00:05:35.930 --> 00:05:39.780
What we want to calculate is the
probability that the four
00:05:39.780 --> 00:05:45.150
aces will end up in locations
or in slots that are
00:05:45.150 --> 00:05:48.320
associated with different
persons.
00:05:48.320 --> 00:05:50.610
So let us calculate
this probability.
00:05:50.610 --> 00:05:52.290
The first ace can go anywhere.
00:05:52.290 --> 00:05:53.710
It doesn't matter.
00:05:53.710 --> 00:05:59.720
For the second ace, it has
51 slots to choose from.
00:05:59.720 --> 00:06:03.120
It's 51 because we started
with 52, but one slot has
00:06:03.120 --> 00:06:06.130
already been taken by
that particular ace.
00:06:06.130 --> 00:06:10.430
So for the ace of hearts,
we have 51 slots
00:06:10.430 --> 00:06:12.910
that it can go to.
00:06:12.910 --> 00:06:18.500
And out of those 51, we have
39 of them that belong to
00:06:18.500 --> 00:06:20.970
people who do not
yet have an ace.
00:06:20.970 --> 00:06:24.820
So this is the probability that
the ace of hearts gets
00:06:24.820 --> 00:06:29.110
placed into a slot that belongs
to a person who is
00:06:29.110 --> 00:06:34.180
different than the person
who got the first ace.
00:06:34.180 --> 00:06:36.570
Now let us consider this ace.
00:06:36.570 --> 00:06:40.890
What is the probability that
this ace will get into a slot
00:06:40.890 --> 00:06:44.620
which belongs to either this
person or that person?
00:06:44.620 --> 00:06:50.330
It has 26 slots in which
this desired
00:06:50.330 --> 00:06:52.220
event is going to happen.
00:06:52.220 --> 00:06:57.280
And it's 26 out of the
50 available slots.
00:06:57.280 --> 00:07:00.300
Finally, let us consider
this ace.
00:07:00.300 --> 00:07:03.410
So having placed that ace and
assuming that it got to a
00:07:03.410 --> 00:07:08.060
different person, what is the
probability now that this ace
00:07:08.060 --> 00:07:12.760
is going to go to this person
who doesn't yet have one?
00:07:12.760 --> 00:07:15.050
The probability of this
happening is the number of
00:07:15.050 --> 00:07:20.060
slots associated with that
person, which is equal to 13
00:07:20.060 --> 00:07:25.580
divided by the number
of slots that this
00:07:25.580 --> 00:07:27.320
card can choose from.
00:07:27.320 --> 00:07:31.710
And the number of slots is 52
minus the 3 slots that have
00:07:31.710 --> 00:07:36.310
already been taken,
so it's 49.
00:07:36.310 --> 00:07:40.280
And so this is the answer
to our problem.
00:07:40.280 --> 00:07:42.909
This expression looks very
different from the expression
00:07:42.909 --> 00:07:45.520
that we derived a
little earlier.
00:07:45.520 --> 00:07:49.210
But you can do the algebra, the
arithmetic, simplify the
00:07:49.210 --> 00:07:52.560
answer, and you will verify that
indeed it's exactly the
00:07:52.560 --> 00:07:53.860
same answer.
00:07:53.860 --> 00:07:58.520
And in case you're curious, the
numerical value turns out
00:07:58.520 --> 00:08:01.130
to be 0.105.
00:08:01.130 --> 00:08:02.880
So there's about 10% [chance]
00:08:02.880 --> 00:08:05.820
that when you deal the cards
in bridge, each one of the
00:08:05.820 --> 00:08:12.100
players is going to end up
having exactly one ace.
00:08:12.100 --> 00:08:16.230
So this was a faster way of
getting to the answer to our
00:08:16.230 --> 00:08:18.860
problem, compared to
the previous one.
00:08:18.860 --> 00:08:21.750
But it raises a legitimate
question.
00:08:21.750 --> 00:08:25.560
Is the way that we dealt the
cards by putting the aces on
00:08:25.560 --> 00:08:29.530
top and then dealing them,
is that way a fair way
00:08:29.530 --> 00:08:31.340
of dealing the cards?
00:08:31.340 --> 00:08:34.990
Is it true that with this way
of dealing the cards all
00:08:34.990 --> 00:08:38.090
partitions are equally likely?
00:08:38.090 --> 00:08:41.120
It turns out that this
is indeed the case.
00:08:41.120 --> 00:08:45.210
But it does require
a bit of thinking.
00:08:45.210 --> 00:08:47.530
Maybe you can see
it intuitively
00:08:47.530 --> 00:08:49.020
that this is the case.
00:08:49.020 --> 00:08:53.290
But if not, then it is something
that one can prove.
00:08:53.290 --> 00:08:55.850
It can be proved formally
as follows.
00:08:55.850 --> 00:09:01.360
One first needs to check that
all permutations, that is all
00:09:01.360 --> 00:09:03.940
possible allocations
of cards into
00:09:03.940 --> 00:09:07.270
slots, are equally likely.
00:09:07.270 --> 00:09:12.420
And because of this, one can
then argue that any possible
00:09:12.420 --> 00:09:16.250
partition into subsets of [13]
00:09:16.250 --> 00:09:18.280
is also equally likely.
00:09:18.280 --> 00:09:22.160
So this is an equivalent way
of dealing the cards to the
00:09:22.160 --> 00:09:25.330
one that we considered earlier,
which was that every
00:09:25.330 --> 00:09:27.710
partition is equally likely.
00:09:27.710 --> 00:09:31.350
Therefore, we did indeed solve
the same problem, and so this
00:09:31.350 --> 00:09:35.650
is a legitimate alternative way
of getting to the answer.
00:09:35.650 --> 00:09:39.140
And of course, it's reassuring
to check that this numerical
00:09:39.140 --> 00:09:41.940
expression agrees with the
numerical expression we had
00:09:41.940 --> 00:09:43.190
derived earlier.