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So I'm having difficulty understanding a certain idea. Say I have a standard $52$ card deck (fairly shuffled, etc.) and I pull $5$ cards from it. We ask, what is the expected value of the number of aces in this hand.

Obviously we could indicator functions, so we can let $I_i$ denote the indicator function for the $i$th card and let $X = I_1 + \cdots + I_5$ denote the value we are trying to find. In essence we're trying to find $E(X)$. (Indicator function basically means $I_i = 1$ if the $i$th card is an ace and $I_i = 0$ if the $ith$ card is not an ace. So $X$ would count the number of aces.)

I know that expected value doesn't care about dependence, so in particular $E(X) = E(I_1) + \cdots E(I_5)$. And since each $I_i$ is an indicator function, letting $p_i$ be the probability that the $i$th card pulled is an ace, we get $E(X) = p_1 + \cdots + p_5$.

At this point, every solution I've seen says, "The chance of any particular card being an ace is $\frac{4}{52}$ and so you get $5 \cdot \frac{4}{52} = \frac{5}{13}$", but this is precisely where I have a problem. I don't understand how the chance of any card being chosen being an ace has probability $\frac{4}{52}$. I totally understand that the 1st card has probably $\frac{4}{52}$ and so $p_1 = \frac{4}{52}$, but after that, the 2nd card no longer has probability $\frac{4}{52}$ since it's dependent on the first card. If the first card is an ace, the chances becomes $\frac{3}{51}$, if the first card isn't an ace, the chances become $\frac{4}{51}$. In either case, we get something different than $\frac{4}{52}$.

The standard definition in this case makes perfect sense: $$E(X) = \sum_{x = 0}^4 x P(X = x) = \sum_{x = 0}^4 x \cdot \frac{\binom{4}{x} \binom{48}{5-x}}{\binom{52}{5}}$$ This is the birthday problem and obviously my brain is trying to work in this way, which is why it's having difficulty seeing the other way.

Can anyone help explain why there magically is no dependency when we look at it from the perspective of an indicator function, i.e. when $E(X) = \sum p_i$?

RobPratt
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jarb_2222
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    "but after that the 2nd card no longer has probability $\frac4{52}$ to be an ace..." Then can you give me any reason why the second card should have less or more chance to be an ace than the first?... It seems that you are not looking at the plain probability that the second card is an ace but at conditional probabilities (which is unnecessary and confusing here). If you insist on that then let $A$ be the event that the first card is an ace and let $B$ be the event that the second card is an ace. Then $P(B)=P(B|A)P(A)+P(B|A^c)P(A^c)=\frac3{51}\frac4{52}+\frac4{51}\frac{48}{52}=\frac4{52}$. – drhab May 16 '21 at 21:28
  • ooo ok. So equation-wise, that makes sense. I guess maybe I just have to sit with it longer to figure it out. It just intuitively feels wrong, but that that could very well be because my brain is looking at conditional probabilities rather than overall probability. Thank you. – jarb_2222 May 16 '21 at 21:34
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    Just wonder: "what is the probability that the second card is an ace?" This without spending any thoughts on the character of the first card (or any other card). Then the answer is clearly $\frac4{52}$. The same is true for the other $4$ cards. – drhab May 16 '21 at 21:34
  • Nobody says $5\cdot \frac{4}{25}=\frac{5}{13}.$ Maybe you meant $52$ rather than $25?$ – Thomas Andrews May 16 '21 at 21:34
  • @ThomasAndrews Yes, I bet that is just a typo. – drhab May 16 '21 at 21:35
  • Defs a type LOL I'll edit it. Sorry about that. – jarb_2222 May 16 '21 at 21:39
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    This is an example of the power of the fact that expectations are linear. The “standard way” as you say, is fairy messy but doable. The linearity makes the computation seem trivial. – Thomas Andrews May 16 '21 at 21:41
  • @drhab But I guess that's my problem when I think "What is the probability that the second card is an ace", my brain automatically thinks, well at this point there are only 51 cards and it's dependent on the first card. Basically it's the word "second" that throws it off for me. – jarb_2222 May 16 '21 at 21:42
  • @ThomasAndrews I think you're missing the point of the question ._. I fully understand that linarity makes the computation trivial. I'm not understanding the intuition behind it. If I don't understand why I can put an equality sign there it doesn't matter if the equation is computationally trivial. drhab is helping me understand the intuition to understand why I'm allowed to put an equal sign there. – jarb_2222 May 16 '21 at 21:44
  • Since you don’t know what the first card is, the second card can be any of $52$ cards, too. The probability that the first card is the ace of spades is them same as the probability the second card is the ace of spades, even if the two events are not independent. – Thomas Andrews May 16 '21 at 21:44
  • @drhab Do you want to post your response as a solution so I can accept it? Thank you for your help. – jarb_2222 May 16 '21 at 21:49
  • Oh, I understand the question, but I’m just recommending that you build your intuition by doing the problems the hard way first. – Thomas Andrews May 16 '21 at 21:55
  • Thank you @ThomasAndrews drhab already was able to clearly identify where my intuition was wrong before you started responding which gave me the avenue to start correcting my intuition. (I was looking at the conditional probability instead of the total probability as drhab mentioned). Doing the problem "the hard way" doesn't always help. I already understood that method. That wasn't the issue. Thank you for trying though. – jarb_2222 May 16 '21 at 22:00
  • On your request I wrote an answer. – drhab May 17 '21 at 06:07
  • Related: https://math.stackexchange.com/questions/1875045/is-this-lot-drawing-fair/1875055#1875055 – Ethan Bolker May 17 '21 at 15:14
  • see this article for a proof of the principle behind linearity of expectation. – user2661923 May 17 '21 at 15:23

2 Answers2

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Answer on request of the OP.

"but after that the 2nd card no longer has probability $\frac4{52}$ to be an ace..."

Then can you give me any reason why the second card should have less or more chance to be an ace than the first?... It seems that you are not looking at the plain probability that the second card is an ace but at conditional probabilities (which is unnecessary and confusing here).

If you still insist on that approach have a look at the following justification.

Let $A$ be the event that the first card is an ace and let $B$ be the event that the second card is an ace. Then: $$P(B)=P(B|A)P(A)+P(B|A^c)P(A^c)=\frac3{51}\frac4{52}+\frac4{51}\frac{48}{52}=\frac4{52}$$ It must be said though that this is a somewhat artificial/unnatural effort to convince you. Much better is it if your intuition on this all can be enlightened.

If someone is asked to take one by one two cards of a deck and questions are asked about the second card then often (and in good caution) his mind wonders back to the first card that was drawn already and to the fact that both cards were drawn in somewhat different circumstances. If however the cards are put on $52$ distinct spots and questions are asked about the card that happens to be put on spot number $2$ (or any other spot) then this temptation is not present.

So in situations as sketched in your question just wonder: "what is the probability that the second card is an ace?" This without spending any thoughts on the character of the first card (or any other card). Then the answer is clearly $\frac4{52}$. And of course same is true for the other $4$ cards.

I sincerely hope that your intuition is enriched by this.

drhab
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The expected number =$\sum\limits_{k=0}^4kP_k$ where $P_k=S_k\frac{Q_kR_k}{D}= \frac{\binom{4}{k}\binom{48}{5-k}}{\binom{52}{5}}$.

Term explanation for Hand that has $k$ aces. $S_k=\binom{5}{k}$ describes permutations of positions of $k$ aces in $5$ card hand. $Q_k=\binom{4}{k}k!$ describes combinations of $k$ out of $4$ aces in hand. $R_k=\binom{48}{5-k}(5-k)!$ describes possible non-aces in hand. $D=\binom{52}{5}5!$ describes cards available before each card is dealt.

herb steinberg
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