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Given a standard 52-cards deck:
You are extracting cards from the deck without replacement, until you get an "Ace" for the first time. What is the probability that the next card will be "Ace" too?

I've already seen the following Q&A: Probability of drawing an Ace: before and after
According to that thread, the answer should be: $$\frac{4 \cdot 3}{52 \cdot 51} = \frac{1}{221}$$

But the official solution says that the answer is: $$\frac{4}{52}$$ which doesn't make sense IMHO. They solved it only with intuition or "mind trick", as they wrote..

My calculation:

Assuming that the 1st card is Ace, then: $$\frac{4 \cdot 3}{52 \cdot 51} = \frac{1}{221}$$

Assuming that the 2nd card is Ace, then: $$\frac{(52-4) \cdot 4 \cdot 3}{52 \cdot 51 \cdot 50} = \frac{24}{5525}$$

We notice a pattern here.
Having the 1st Ace at the $k$'th draw, then the probability (for a second Ace after that) is: $$ p_1 = \frac{ {_{52-4}P_{k-1}} \cdot 4 \cdot 3 }{ {_{52}P_{k-1}} \cdot {_{52-k}P_{2}} } $$

We need to consider the least-possible scenario - we draw 48 non-Ace cards, then: $$ p_2 = \frac{48! \cdot 4 \cdot 3}{ {_{52}P_{50}} } = \frac{1}{270725} $$

So, the required probability is: $$\begin{align} p &= p_2 + \sum\limits^{48}_{k=1} p_1 \\ &= p_2 + \sum\limits^{48}_{k=1} \frac{ {_{48}P_{k-1}} \cdot 4 \cdot 3 }{ {_{52}P_{k-1}} \cdot {_{52-k}P_{2}} }\\ &= \frac{1}{270725} + \frac{1696}{20825}\\ &= \frac{1297}{15925}\\ &\cong 0.081444 \end{align} $$

But my answer is far from either the official solution and from the answer in that thread.

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  • $\begingroup$ The question you link to is different. It is asking "what's the probability that the second drawn is an Ace?" The answer to that is certainly $\frac 4{52}$ as there is no difference between the second card drawn and the first. Of course, if you specify that the first card drawn was (or was not) itself an Ace then the probability changes. $\endgroup$ – lulu Oct 9 '15 at 14:35
  • $\begingroup$ @lulu I think that I understand your view on the matter. So what did I calculated? $\endgroup$ – Dor Oct 9 '15 at 15:07
  • $\begingroup$ I think your method ought to work. To be precise: I just wrote a code which first computes $p_k$, the probability that the first A is observed on the $k^{th}$ draw, and then calculates the probability that the next draw is also an A. I believe this is what you were doing as well. My code returned $\frac 1{13}$ as the total. Doesn't convey a lot of insight, I understand. $\endgroup$ – lulu Oct 9 '15 at 15:13
  • $\begingroup$ @lulu Indeed you're right! I posted an answer which shows where were my mistakes and include a code that proves it to be equal to $\frac{1}{13}$. $\endgroup$ – Dor Oct 9 '15 at 16:27
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We use a fairly crude counting approach, in order to rely minimally on intuition. There are $\binom{52}{4}$ equally likely ways to choose the positions of the $4$ Aces. We now count the "favourables."

Maybe the first two Aces are in positions 1 and 2. That leaves $\binom{50}{2}$ for the rest.

Maybe they are in positions 2 and 3. That leaves $\binom{49}{2}$ for the rest.

Maybe they are in positions 3 and 4. That leaves $\binom{48}{2}$ for the rest.

And so on, up to positions 49 and 50, leaving $\binom{2}{2}$ for the rest. That gives probability $$\frac{\binom{50}{2}+\binom{49}{2}+\cdots+\binom{2}{2}}{\binom{52}{4}}.$$ The numerator can be simplified in various ways. My favourite is to count the number of ways to choose $3$ numbers from the first $51$. This can be done in $\binom{51}{3}$ ways. But the smallest number can be $1$, leaving $\binom{50}{2}$ ways to choose the other two. Or the smallest can be $2$, and so on. We get the number of favourables obtained above.

Our probability is therefore $\frac{\binom{51}{3}}{\binom{52}{4}}$. This simplifies to $\frac{1}{13}$.

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  • $\begingroup$ Great argument and very well written. +1. $\endgroup$ – Shailesh Oct 9 '15 at 15:04
  • $\begingroup$ I like this approach! +1. Could you please elaborate regarding how you simplified the numerator? $\endgroup$ – Dor Oct 9 '15 at 15:13
  • $\begingroup$ I think that I understand it now. That is a very cool solution..! $\endgroup$ – Dor Oct 9 '15 at 15:33
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Your Error

First of all Andre has posted a very nice answer. Already Upvoted it. I am trying to look at where all you have gone wrong. Too long for an comment, so posting it here.

Let us look at the last case. You have $48!$ ways of the non-ace cards. Now the remaining cards are all aces anyways. So the question of multiplying this by $4 \cdot 3$ does not arise. Since we do not care about the last 2 cards, the denominator will have to be just $50!$. Similarly, the case before that might need to rectified.

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  • $\begingroup$ Indeed you are right about the last term! In the other 48 terms, the mistake was different. I'll add info in my original post. Thanks! +1 $\endgroup$ – Dor Oct 9 '15 at 16:08
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    $\begingroup$ @Dor Yes, right. A good approach would be you post an answer yourself, instead of editing the post. I'll surely upvote it. $\endgroup$ – Shailesh Oct 9 '15 at 16:18
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Here is a bijection between set of all permutation with the first Ace followed by an Ace and set of all permutations with the first card being an Ace

$$B^nAAX <-> AB^nAX$$

where A stands for an Ace, B for not an Ace, $0\leq n \leq 48$ and $X$ contains $2$ Aces and $48-n$ non-Aces. Hence the probability of an Ace following the first Ace is equail to the probability of the first card being Ace which is equal to $$4/52=1/13$$

Probabilistically, conditioned on the location of the second ace, the first ace is uniformly distributed on locations before the second ace - hence probability of it being the first or the last in that sequence are equal.

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  • $\begingroup$ Nice! I was going to go back and try to explain that the fact my fraction simplified to $\frac{4}{52}$ was not an "accident." You have done it, in the strongest (natural bijection) sense. $\endgroup$ – André Nicolas Oct 10 '15 at 1:27
  • $\begingroup$ I started thinking along the lines of bijection after seeing your result ;). $\endgroup$ – A.S. Oct 10 '15 at 1:30
  • $\begingroup$ Yes, it was kind of shouting look at me, look at me. $\endgroup$ – André Nicolas Oct 10 '15 at 1:32
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Fixing my original calculation

Thanks to Shailesh, we now know that the expression $$ p_2 = \frac{48! \cdot 4 \cdot 3}{ {_{52}P_{50}} } $$ Should be replaced with: $$ p_2 = \frac{48!}{ {_{52}P_{48}} } $$ Because that after extracting 48 non-Ace cards, we'll necessarily have the second card as Ace.

Moreover, I had a tiny mistake in the denominator. Instead: $$ {_{52}P_{k-1}} \cdot {_{52-k}P_{2}} $$ It should be: $$ {_{52}P_{k-1}} \cdot {_{52-(k - 1)}P_{2}} $$ (All the $k$ indexes should be shifted by $(-1)$)

I wrote a Python code which shows that this solution works either:

# Python 2.7.6
import math
from fractions import Fraction

f = math.factorial

def nPr(n,r):
    return f(n) / f(n-r)

p = Fraction(0)

for k in range(1, 49):
    p += Fraction(nPr(48, k-1) * 4 * 3,
                nPr(52, k-1) * nPr(52 - (k-1), 2))

p += Fraction(f(48), nPr(52,48))

print p # Outputs 1/13 :)

Thank you all for your effort! :)

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  • $\begingroup$ Well done, Dor, We all learn with each other's help. $\endgroup$ – Shailesh Oct 9 '15 at 16:30

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