# Independence of two decks of cards, each with a different numbers of cards

I am trying to get an intuitive feel for what independence 'is.' My initial guess was that variables are independent if they can't casually affect each other.

I was discussing this on a hacker news thread, and someone gave this counter-example:

Independence is a more strict condition. The probabilities must follow the rule P(A and B) = P(A) * P(B) that is the rule that is used in the article.

Let's pick an easier example. You buy two equal decks of cards while traveling abroad. When you return home, you realize that it may be a deck with 48 cards (123456789JQK)or a deck with only 40 cards (1234567JQK) without the "8" and "9".

To make it simple, suppose that you magically know that you have a 50% chance of having bought a 48 cards deck and a 50% chance of having bought a 40 cards deck.

One of your friend come and start to shuffle both decks, one in each hand, without mixing them.

If your friend picks a card from the right deck, what is the probability that it's a "8"? (spoiler alert: There is a 50% that the deck has only 40 cards (without the "8" and "9") and you can't pick the "8", and there is a 50% that the deck has 48 cards and there is a 1/12 to pick the "8", so the probability is: 1/2 * 0 + 1/2 * 1/12 = 1/24)

If your friend picks a card from the left deck, what is the probability that it's a "8"? (again 1/24)

Are they independent? Whatever he does with the left decks doesn't affect the right deck. To get less relation, you can use two friends, and put them in a different room, and make them pick the card exactly simultaneously. If the events were independent, the probability that both have picked an "8" would be 1/24 * 1/24 = 1/576

But the correct answer is more complicated. There is a 50% that the deck has only 40 cards (without the "8" and "9") and you can't pick the "8", and there is a 50% that the deck has 48 cards and there is a 1/12 to pick the "8" in each one, so the probability is: 1/2 * 0 + 1/2 * 1/12 * 1/12 = 1/288. That is 2x bigger than if they were independent.

I don't have a good sense of why the example's 'correct' answer is correct, or why the two decks can't be treated independently.

• I realized that 1/12 has to be incorporated twice because the example shifts to considering the probability that two people draw the same card, whereas it initially asked what the probability of drawing a certain card would be. Apr 24, 2016 at 16:08
• One thing to keep in mind "causation" generally has a strong time dimensions - the cause must come before the effect. Probability does not have this restriction - so independence need not match with causation. You can go the other way as well. Take two ind rvs $A,X$ which both take values $\{-1,1\}$ with equal probability, set $Y=AX$ so that $X$ "causes" $Y$. But we have $Pr(Y=1)=\frac {Pr (X=1)}{2}+\frac {Pr (X=-1)}{2}=\frac {1}{2}$ and $Pr (Y=1|X=1)=\frac {1}{2} =Pr (Y=1)$. Similarly, $Pr (Y=-1)=Pr (Y=-1|X=-1)=\frac {1}{2}$. This means that $X$ and $Y$ are independent! Apr 24, 2016 at 16:51

You need to calculate $P($Friend I and Friend II both picked 8$)$. Even though the two friends are in different rooms, the events are not independent, because both the decks have the same properties. Thus, if deck 1 has 40 cards, then so does deck 2. Now consider,

$P($Friend I and Friend II both picked 8 $|$ deck has 40 cards)

Now the events $A =$ "Friend I picked 8 given deck has 40 cards" and $B =$ "Friend II picked 8 given deck has 40 cards" are independent because once you know how many cards there are in the deck, the two decks are no longer "related" in any way. Thus,

$P($Friend I and Friend II both picked 8 $|$ deck has 40 cards) = P(Friend I picked 8 $|$ deck has 40 cards) P(Friend II picked 8 $|$ deck has 40 cards) = $(0)(0)$ = 0.

Similarly,

$P($Friend I and Friend II both picked 8 $|$ deck has 48 cards) = P(Friend I picked 8 $|$ deck has 48 cards) P(Friend II picked 8 $|$ deck has 48 cards) = $\dfrac{1}{12}\dfrac{1}{12}$

Putting this together, by the total law of probability

$P($Friend I and Friend II both picked 8) = $P($Deck has 40 cards)$P($Friend I and Friend II both picked 8 $|$ deck has 40 cards) + $P($Deck has 48 cards)$P($Friend I and Friend II both picked 8 $|$ deck has 48 cards) = $\dfrac{1}{2}.0 + \dfrac{1}{2} \dfrac{1}{12} \dfrac{1}{12} = \dfrac{1}{288}.$

• I would clarify/rephrase your initial statement that the events are not independent because the two decks are the same - they have the same properties. Not just that there is $40$ cards in both. So information about one deck is relevant to the other deck. Apr 24, 2016 at 16:19