In Tableau Two, Part One, we talked about a card, the card game Bridge.
We talked about what constituted a Bridge hand, and
we computed commutatorial probabilities for such a hand.
Let's run through the idea of a Bridge hand again.
So the question at hand is the following.
We are going to deal a bridge hand from a standard pack of 52 cards.
Now in bridge, aces in the pack have a particular significance.
And so the question at hand is, are aces distributed across four bridge players,
in fair and equitable fashion.
Does everybody get an ace?
Now to remind ourselves, we are dealing with a pack of 52 cards.
The cards are distinguished in four major suits, called spades,
hearts, diamonds, and clubs.
In each suit we have 13 cards labeled, ace, two, three, four, up to king.
In spades, in hearts, in diamonds and in clubs.
We play bridge by distributing these cards 13 cards a player to four players.
Traditionally called north, south, east and west.
Now when you distribute a card hand like this.
We can attack this problem purely communitorially by counting.
I'll let you have at it.
Our tact here is going to be a little different.
We're going to be a little moe subtle and
edge up to the event at hand by a conditioning argument.
Now to do this, we first should formalize what is our sample space?
What are our idealized outcomes?
Let's begin with this.
The sample space here.
I'm going to allocate 13 cards to north,
13 cards to east, 13 cards to south, 13 cards to west.
We are going to assume a random shuffle.
Which means that all orderings of the 52 cards are equally likely.
Which means that once you do the shuffle, north gets 13 cards, east 13, south 13,
west 13 and if I can tell you what order these 13, 13, 13 cards are arrived at,
then you've got one bridge hand.
Now, the combinatorial setting here says that all permutations of these cards
are equally likely.
Now there's a mind-boggling number of permutations, 52 factorial permutations.
Every permutation corresponds to an atom in our game, and
the chance of that atom is one in 52 factorial.
Now what is a bridge hand here?
Each permutation gives rise to a particular sequence in
which the cards were administered.
But within one players grasp,
the particular ordering of the cards makes little matter.
It only matters which 13 cards he has obtained.
Since there are 13 factorial permutations of the cards for
each player, there are 13 factorial to the power 4 possible
permutations which result in the same bridge hand.
And therefore, the chance of every distinct bridge hand
is 13 factorial to the power of 4 divided by 52 factorial.
It's still an infinitesimally small number.
There's a very vast number of possible bridge hands.
Now, what are the events at hand?
Well the target event is clearly the event where the distribution
of the cards C1 to three set 13, C14 through C26 and
so on is such that in north there
is an ace somewhere in there, in east there's an ace somewhere in there.
In south there's an ace somewhere in there.
And in west, there is an ace somewhere in there.
The aces of hearts, spades, diamonds and clubs are distributed one per player.
Now let's edge up to this target event one step at a time.
Now manifestly, the ace of spades has to go somewhere.
So let's target a minute, let's call it A of spades.
To mean that the ace of spades goes to some player.
Of course, this is certain.
The ace of spaces is just a synonym for the entire sample space.
Every bridge hand will deal the ace of spades to somebody.
Now that's okay, this is just a starting point, that's right.
Let's define a slightly more subtle event.
Ace A of spades and hearts to mean that
the aces of spades and hearts go to two distinct players.
No one player gets both.
Hm, okay this is clearly a sana quinone if everybody is going to get one ace.
Here's a starting point.
Let's define A of spades, hearts,
and diamonds, to represent the event all permutations.
For which the aces of hearts, spades, and diamonds go to three distinct players.
No player gets two of these aces.
And finally, the target event.
Let A of spades, hearts, diamonds, and clubs represent the event
that the four aces each go to a distinct player, everybody gets an ace.
Now, these are clearly events which are relevant for
the problem at hand, but there are certain inclusions in these sets.
And we can very quickly figure out what these inclusions are.
If everybody gets an ace, then clearly the aces of spades,
hearts and diamonds have to go to three separate players, and therefore the last
event is a subset of, is contained in the event A of spades, hearts and diamonds.
But if three players get spades, hearts and diamonds, one a piece.
Then it must be the case that the aces of spades and
hearts go to two distinct players.
And if the aces of spades and hearts go to two distinct players,
then it must be the case that somebody gets ace of spades.
Which is of course the entire sample space.
Oh good, so now we have an inclusion relationship.
Each of these inclusions, one contained in the other.
Let's see if we can parlay this, this understanding into a simple calculation.
So, you're going to utilize a chain rule, you're going to utilize a understanding
of these set relationships, these inclusion relationships, and
an understanding of how a bridge hand is distributed.
Let's start by writing down the probability of the event at hand.
Now the event that every player gets an ace.
Because of the inclusional relationships,
this is the smallest of the three events about it.
And therefore the intersection of the four events is exactly the smallest of them.
And now you look at me and say, oh this is strange.
You want the probability of an event and
now you've complicated it by introducing other things.
But wait, there's a method in this madness.
The events A of spades, hearts, diamonds, clubs.
A of spades, hearts, diamonds.
A of spades, hearts.
A of spades.
Stands in the row of four distinct events.
There happens to be a particular occlusion relationship, which we'll utilize shortly.
But there are four distinct events.
We could just as well have called them A, B, C and D.
We're looking at the conjunction probability of four events.
That suggests a chaining argument.
We can write this as the first event.
Conditioned upon the occurrence of the last three.
The second event conditioned upon the occurrence of the last two.
The third even conditioned upon the occurrence of the last one, and
then the last event.
And so here's what our chain rule for conditional probabilities tells us.
Oh, okay, it looks like it's getting more complicated.
Are we making any progress?
We'll pause for a moment.
There are certain inclusion relationships.
So the event of A of spades, hearts, and
diamonds is contained in both,
A of spades, hearts, and A of spades.
And therefore, the conditional, the side information of the first conditional
probability collapses to simply the smallest of those three sets A of spades,
hearts and diamonds.
Likewise in the second of the conditional probabilities, the event A of spades and
hearts is manifestly contained in the entire sample space, A of spades.
And therefore the intersection is just the smaller of the two,
A of spades and hearts.
Oh, here is our problem specific understanding which allows us to simplify
this complex expression into something which is much more manageable.
I now have systems of conditional probabilities and that question is,
in this setting, are these conditional probabilities easy to evaluate?
Let's pause and take a look at them one at a time.
Let's start at the right.
What is the probability of A of spades.
Now A of spades means that somebody gets ace of spades.
This is equivalent to the entire sample of spades because every permutation
of your cards will put the ace of spaces somewhere.
Somebody will get an ace of spades.
This is certain.
Another way of thinking about this is,
there are 52 locations the ace of spades can go.
Because of the combinatorial symmetry in our probability measure.
The ace of spades is equally likely to be in any one of those 52 locations.
And therefore,
the probability of the ace of spades is just 52 divided by 52, or one.
Okay.
That looked easy.
Let's take a look at the next one.
The probability of the event A of spades and
hearts, given A of spades.
The probability that two distinct players get the aces of spades and
the aces of hearts, one a piece.
Given that, some players got ace of spades.
Now let's see what is given.
The ace of spades has landed somewhere in the picture.
The moment is landed there.
That position is issued, it's forbidden for where the ace of hearts can land up.
Okay, so, one of the 52 locations is now forbidden for the ace of hearts.
But more, we want the ace of hearts to not just land where the ace of spades is but
it can not land in any of the 13 cards of the player
who has picked the ace of spades.
And therefore,
the ace of hearts has got only 39 remaining locations where it can land.
The symmetry of the problem tells you that all those 39 locations are equally likely,
given that they can not be in these 13 locations.
And therefore the probability of A of spade and
hearts, given A of spades, is exactly 39 over 51.
Remember one possibility is left out, it's not available anymore.
And the remaining 12 cannot be occupied if the ace of hearts is to go somewhere else.
Arguing along these lines the probability of ace A of spades,
hearts, diamonds, given A of spades and hearts.
They are given spades and hearts.
Spades is gone somewhere.
Hearts is gone somewhere else.
Those two locations cannot be where the ace of diamonds lands.
So ace of diamonds has got a total of 50 locations.
From 52 you're taking two away.
But of those 50 locations, 13 plus 13, 26 locations are forbidden.
And therefore there are only 26 locations left for
the ace of diamonds to land safely, and so there are 26 out of 50.
And finally, the remaining ace of clubs is going to leave 13 locations left.
Once you've place the other three aces out of 49 locations.
Notice how elegant and simple these conditional probabilities are.
Once you understand what exactly the chance creation mechanism is, and
putting everything together, we opt in the following.
The first conditional probability.
13 possibilities for the ace of clubs out of 14 are remaining,
having placed the aces of spades, hearts and diamonds.
26 possibilities for the ace of diamonds
given that out of the 50 possibilities, given that you placed the aces of spades
and hearts, 39 possibilities for the ace of hearts.
Given out, out of the 50-odd possibilities,
given that you have placed ace of spades.
And all 52 possibilities for ace of spades.
You evaluate this numerically, and it comes out to about 11%.
In a bridge game, very roughly, 10% of the time.
In 10% of bridge hands, you will expect to find that each player has one ace.
You could try doing this problem directly,
using enumeration direct communitorial methods.
And see if you rediscover the answer.
It is not at all hard.
But the mode of thinking is rather different
when you do an enumeration directly along those lines.
The idea here is that a proper identification of site information
can simplify expressions, can simplify computations.