# Basic question (not mutually exclusive)

1. Jun 18, 2010

### njama

1. The problem statement, all variables and given/known data

I am doing some research about cards, and here it is:

$$P((B_1 \cap B_2) \cup (B_1 \cap B_3) \cup (B_2 \cap B_3) \cup (B_1 \cap B_2 \cap B_3)$$

where $$B_{i}, i=1,..,3$$ is opponent i have AA(aces) or KK(kings).

The expression above means two or more players have AA or KK at same time.

Now let's say there are 3 players.

2. Relevant equations

$$P(A \cup B \cub C) = P(A)+P(B)+P(C)-P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$

3. The attempt at a solution

Now I noticed something:

$$(B_2 \cap B_3) \cup (B_1 \cap B_2 \cap B_3) = (B_2 \cap B_3)$$

is it valid to do this and write the expression above as:

$$P((B_1 \cap B_2) \cup (B_1 \cap B_3) \cup (B_2 \cap B_3))=$$

$$=P(B_1 \cap B_2)+P(B_1 \cap B_3)+P(B_2 \cap B_3)-$$

$$-P((B_1 \cap B_2) \cap (B_1 \cap B_3)) - P((B_1 \cap B_2) \cap (B_2 \cap B_3))-P((B_1 \cap B_3) \cap (B_2 \cap B_3))+$$

$$+P((B_1 \cap B_2) \cap (B_1 \cap B_3) \cap (B_2 \cap B_3))=$$

$$=P(B_1 \cap B_2)+P(B_1 \cap B_3)+P(B_2 \cap B_3)-2*P(B_1 \cap B_2 \cap B_3)$$

Is it valid?

Now because the probability is same whether 1st opponent and second have AA or KK or 1st and 3rd have AA or KK:

$$=3*P(B_1 \cap B_2)-2*P(B_1 \cap B_2 \cap B_3)$$

2. Jun 18, 2010

### Staff: Mentor

The reason for this is that $(B_1 \cap B_2 \cap B_3) = \emptyset$

It's not possible for three players to have 2 aces each or 2 kings each since there are only 4 of each of these cards in a standard deck of 52 cards.
If it produces correct results, it would have to be valid, but you're making things more complicated than they need to be. For example, this probability would be zero.
$$P((B_1 \cap B_2) \cap (B_1 \cap B_3) \cap (B_2 \cap B_3))= 0$$

Again, it's not possible for all three players to have 2 aces each or 2 kings each.

3. Jun 18, 2010

### njama

Ok, you misread my first post:

where $$B_{i}, i=1,..,3$$ is opponent i have AA(aces) or KK(kings).

It says OR. So it is possible that 3 players have 4 aces 2 kings.

Regards.

4. Jun 18, 2010

### Staff: Mentor

It depends on the meaning of "or." In mathematics contexts, "or" also includes "and." If you say that a card hand contains aces or kings, that doesn't preclude a hand that contains two aces and two kings. If you said that card hand contains either aces or kings, then you are using the "exclusive or" sense of the word.

Some of your events, as you are defining them, are mutually exclusive. For example, player 1 has two aces and two kings, player 2 has two aces, and player 3 has one ace and two kings.

With this setup,
$$P(B_1 \cap B_2 \cap B_3) = 0$$

5. Jun 18, 2010

### njama

Wait just a minute...

C_ could be i opponent have aces
D_ could be i opponent have kings

So in this case:

$$P(B_1 \cap B_2 \cap B_3)=P((C_1 \cup D_1) \cap (C_2 \cup D_2) \cap (C_3 \cup D_3))$$

this could be solved afterwards... so it is certainly not true that $$P(B_1 \cap B_2 \cap B_3) = 0$$

Last edited: Jun 18, 2010
6. Jun 18, 2010

### Staff: Mentor

For one of these you probably mean that opponent i has kings.
??? What does that mean?
If B1, B2, and B3 are defined as they were in post 1, I showed you how it could happen that this probability is zero.

Here's my example again.
B1: Player 1 has 2 aces and 2 kings.
B2: Player 2 has 2 aces.
B3: Player 3 has 1 ace and 2 kings.

Each of these events has a nonzero probability, but the intersection of the events is empty, hence the probability of the intersection of these events is zero.

7. Jun 18, 2010

### njama

Mark44 I understand what you say to me.

I meant to write:

C_ i opponent have aces
D_ i opponent have kings

$$P(B_1 \cap B_2 \cap B_3)=P((C_1 \cup D_1) \cap (C_2 \cup D_2) \cap (C_3 \cup D_3))$$

I count the probability afterward.

$$P(B_1 \cap B_2 \cap B_3)=\frac{6*C(4,2)C(2,2)C(4,2)}{C(50,2)C(48,2)C(46,2)}$$

I look for the probability that would gave me exactly three players with more than QQ, that is players with AA or KK.

Yes, your assumption is correct, but that is not what I need.

I am asking whether my calculations are right.

Thank you.

8. Jun 18, 2010

### Staff: Mentor

I think that you are assuming that
$$P(A \cap B)=P(A)*P(B)$$

This is true only if events A and B are independent.

The more general formula is
$$P(A \cap B)=P(A | B) P(B)$$

$$P(B_1 \cap B_2 \cap B_3)=P((C_1 \cup D_1) \cap (C_2 \cup D_2) \cap (C_3 \cup D_3)) = \frac{6*C(4,2)C(2,2)C(4,2)}{C(50,2)C(48,2)C(4 6,2)}$$

is a general formula that does not give the correct results for certain events B1, B2, and B3, which should make you think that something is wrong.

I have provided events that satisfy your conditions, but the probability of the intersection of these events is zero. That value does not match the formula you have.

9. Jun 19, 2010

### njama

Mark44 your events are not possible.

You are playing on single table with 3 opponents.

It is not possible that 1st opponent - 2 aces, 2nd opponent - 2 aces, 3rd opponent - 2 aces because in the suit there are 4 aces and 4 kings.

The game is Texas Hold'em, where are dealed 2 cards pre-flop to all players.

Due to it is not possible that 6 aces would came up.

But still I will prove you:
$$P(B_1 \cap B_2 \cap B_3)=P((C_1 \cup D_1) \cap (C_2 \cup D_2) \cap (C_3 \cup D_3))=$$

after long long calculations:

$$=6*P(C_1 \cap C_2 \cap D_3)-6*P(C_1 \cap C_2 \cap D_2 \cap D_3)=$$
since is not possible that 2nd player have AA and KK at same time:
$$=6*P(C_1 \cap C_2 \cap D_3)$$

Last edited: Jun 19, 2010
10. Jun 19, 2010

### Staff: Mentor

You are missing my point. Taken separately, each event is possible and has a nonzero probability.
B1: Player 1 has 2 aces and 2 kings.
B2: Player 2 has 2 aces.
B3: Player 3 has 1 ace and 2 kings.

But all three events cannot happen in a single hand. That's why, for these events,
$$P(B_1 \cap B_2 \cap B_3) = 0$$

And that's why the formula you have developed is incorrect, for the events I have defined. The formula you have assumes that the events are independent. My example shows a situation in which they are not.

If events A and B are independent,
$$P(A \cap B) = P(A) P(B)$$

Otherwise,
$$P(A \cap B) = P(A | B) P(B)$$

Look up independence in probability theory and conditional probability.

11. Jun 19, 2010

### njama

You're right for the independence. But just look at the other formula with Cs and Ds. Now you can put everything you want in there still will be correct (the last formula which I derived).

Regards.

12. Jun 19, 2010

### Staff: Mentor

I didn't realize that only 2 cards per player were being dealt until you said so in post #9. Let me think about this for a bit.

13. Jun 19, 2010

### Staff: Mentor

OK, now that I understand that each player has only two cards, this formula seems reasonable to me.
$$P(B_1 \cap B_2 \cap B_3)=P((C_1 \cup D_1) \cap (C_2 \cup D_2) \cap (C_3 \cup D_3)) = \frac{6*C(4,2)C(2,2)C(4,2)}{C(50,2)C(48,2)C(4 6,2)}$$

14. Jun 20, 2010

### njama

Thank you and sorry for being not clear enough.