I have a question about probability. Imagine we have a deck of 40 cards with 4 different suits of ten cards. Define two events:

$A \equiv $ We take a card and it's an ace $\Rightarrow P(A)=\frac{1}{10}$

$B \equiv $ We take a card and it's a spade $\Rightarrow P(B)=\frac{1}{4}$

$P(A\cap B) = \frac{1}{40}$

We can see that, since we return the card each time, these events are independent:

$P(A\cap B) = P(A) \cdot P(B)$

If we now add a joker to the deck, which can take any card value, then:

$P(A) = \frac{5}{41}$

$P(B) = \frac{11}{41}$

$P(A\cap B) = \frac{2}{41}$

And we can see that now the events aren't independent!

$P(A\cap B) \ne P(A)\cdot P(B)$

My question is, why does adding the joker make the events independent? How is it different from just having one more ace of spades? I mean, before adding the joker we already had a card that belonged to both events (the ace of spades)...


  • $\begingroup$ "since we return the card each time"? Just to be clear, do you draw just once from the 40-card deck, or do you draw once, put the card back, and then draw again? $\endgroup$ – David K Feb 15 at 22:29

Note: the rules aren't entirely clear. Is the Joker always interpreted as $A\spadesuit$? If not, what rules determine how it is interpreted?

To be clear: If you added a second $A\spadesuit$ then the events aren't independent either. You'd get $$P(A)=\frac 5{41}\quad P(B)=\frac {11}{41}\quad P(A\cap B)=\frac 2{41}$$ just as before.

Indeed, having doubled the $A\spadesuit$, either directly or by the Joker, you make it so that drawing an Ace is evidence that the card is a spade, and drawing a spade is evidence that it is an Ace.

Specifically, before you draw anything, the probability that a random draw will be a spade is $\frac {11}{41}=0.268$. If I tell you that you have drawn an ace, however, the probability that it is a spade is now $\frac 25=.4$ so drawing an ace is strong evidence that you have drawn a spade.

  • $\begingroup$ Thanks for your reply. The thing I don't understand is why having two A♠ makes the events dependent but having one doesn't. Could you expand a bit on that? $\endgroup$ – Sigma Feb 15 at 13:07
  • $\begingroup$ I don't understand. Having two $A\spadesuit$ makes the events dependent, not independent. $\endgroup$ – lulu Feb 15 at 13:09
  • $\begingroup$ Oops, yeah I meant dependent. $\endgroup$ – Sigma Feb 15 at 13:10
  • $\begingroup$ Just having one A♠ leaves the events independent but adding another one makes them dependent. $\endgroup$ – Sigma Feb 15 at 13:10
  • $\begingroup$ To see, intuitively, that it makes the events dependent note that, before you draw anything, the probability that a random draw will be a spade is $\frac {11}{41}=0.268$. If I tell you that you have drawn an ace, however, the probability that it is a spade is now $\frac 25=.4$. Thus drawing an ace is strong evidence that you have drawn a spade. $\endgroup$ – lulu Feb 15 at 13:11

That the original deck composition creates events for "draw an Ace" and "draw a Spade" that are independent is a very rare and special property of that deck. It's due to the "highy symmetry" of the deck contents: From each of 4 suits and 10 values, take exactly one of each (suit, value)-pair.

If you add another card to it, be it a joker or another Ace of Spades as in lulu's answer, you destroy that symmetry and now get the absolutely common case of dependent events for "draw an Ace" and "draw a Spade". You would also get these events as dependent if you added 3 cards, or removed 4. Only if you added another 40 cards with the same composition as the original deck would you again get the case of independent events.

In other words, in the vast sea of configurations of cards, those 2 events are almost always depedent. You just started in one of the few configurations where it was actually independent. If you add a card, it is very natural that you go to a configuration that corresponds to the 'normal' state: dependent events.


It can be helpful to look at extreme cases. Consider what happens when you add $60$ jokers to the deck: $$P(A)=\frac{64}{100}\\ P(B)=\frac{70}{100}\\ P(A\cap B)=\frac{61}{100}$$ Clearly $P(A\cap B)$ is greater than $P(A)\cdot P(B)$; this is because any given card is more likely than not a joker, so if it's an ace, it's probably also a spade.

Your situation is the same, just with fewer jokers.


We take a card and its an ace

This is not a description of an event to which you can assign probability. What is the probability I will draw a card today and it will be an ace? Maybe I will not draw any cards today. How do you compute the probability of that?

Now introduce a second description, "We take a card and its a spades," and the situation is even more murky. Did we put the first card back and then draw another? Or are the two sentences talking about things that would occur simultaneously, that is we draw just one card from the deck one time, and look at that single card, which might be an ace and might be a spade.

A better way to describe probabilities is to set up exactly the experiment/trial (or experiments/trials) that we will perform, and then specify the results as events. So either there is a single trial ("shuffle these $40$ cards, draw one, and look at it") or two trials ("shuffle these $40$ cards, draw one, look at it, then shuffle the cards again, draw a card, and look at it").

For a double trial, with one joker in the deck, there are not just $41$ possible equally-likely, distinguishable outcomes; there are $41\times41$ of those outcomes. For each card we could draw the first time there are $41$ possible cards we could draw the second time. The number of ways to draw an ace the first time and a spade the second time is $5\times 11$ ways. So if we actually make two draws then the events are independent.


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