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We have a standard deck of cards. Find the expected number of cards after the first $2$ and before the first ace given that the first $2$ appears before the first ace.

Let $X$ be the number of cards satisfying the conditions, $$ E(X) = \sum_{i=1}^{45} E(X \mid \text{first 2 drawn at } i^{th} \text{ position & before 1st ace}) \cdot P(\text{1st 2 drawn at i & before 1st ace}) $$

$$ = \left(\frac{47}{5} \cdot \frac{4}{52}\right) + \left(\frac{46}{5} \cdot \frac{52-8}{52} \cdot \frac{4}{51}\right) + \ldots + \left(\frac{3}{5} \cdot \frac{52-8}{52} \cdot \ldots \cdot \frac{4}{52-45+1}\right) $$

$$ = \frac{47}{5 \cdot 13} + \sum_{i=2}^{45} \frac{4 \cdot (48 - i)}{5 \cdot (52-i+1)} \cdot \prod_{j=0}^{i-2} \left(\frac{52-j-8}{52-j}\right) $$

$$ = \frac{4081}{1170} + \frac{47}{5 \cdot 13} = \frac{379}{90} \approx 4.21 $$

I was told my final answer is wrong but would like to know what is wrong here. Some precisions about what I did above:

  • $E(X \mid \text{first 2 drawn at } i^{th} \text{ position & before 1st ace})$ = $\frac{48-i}5$ by generalizing this Expected number of cards you should turn before finding an ace to a smaller number of cards.
  • $P(\text{1st 2 drawn at i & before 1st ace})=$ probability of having anything but a $2$ or an ace in the first (i-1$)^{th}$ positions.
  • The sum goes to $45$ which the last possibility for the first 2 to appear before an ace.
Kilkik
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  • The right answer is mine*2 so $\frac{379}{45}$ but I don't see why it should be multiplied by 2 – Kilkik Jul 09 '24 at 13:28
  • YOu are asked for the conditional probability which is the ratio of the intersection divided by the probability of the conditioning event. The denominator is the probability of the conditioning event, which is $\frac 12$. – lulu Jul 09 '24 at 13:45
  • Alternate method: There are $8$ significant cards, and general arguments tell us they should be equally spaced (counting the gaps at the Start and at the End). Thus the gaps, $g$ , are computable as $8+9g=52\implies g=\frac {44}9$. Now just consider the possible locations for the first Ace amongst the significant cards. – lulu Jul 09 '24 at 13:50
  • @Lulu if we denote by $Y_i$ the event that first 2 appears at position $i$ and $Z$ the event that 1st 2 appears before 1st ace then what you suggest in your first comment is $E(X |Z)=\sum E(X| Y_i,Z) P(Y_i| Z ) =\sum E(X| Y_i,Z) P(Y_i \cap Z )/2 $ so its half my original answer so its the right answer/4 – Kilkik Jul 09 '24 at 15:29
  • My bad its P(intersection)/ (1/2) and not P(intersection)/ (2)so you're right – Kilkik Jul 09 '24 at 15:41

1 Answers1

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Here is a way to see the difference in a "direct manner". Assume we do not know that the first $2$ comes first, then the first $A$. Let us count the average number of the following random variable "win":

  • the win is the number of cards between the first $2$, and the first $A$, excluding them when counting,
  • and in case the $A$ comes first we "lose" and add a zero win.

The formula for this is... $$ \left(\frac{48-1}{5} \cdot \frac{4}{52}\right) + \left(\frac{48-2}{5} \cdot \frac{52-8}{52} \cdot \frac{4}{52-1}\right) + \left(\frac{48-3}{5} \cdot \frac{52-8}{52}\cdot \frac{51-8}{51} \cdot \frac{4}{52-2}\right) + \\ + \ldots + \left(\frac{48-45}{5} \cdot \frac{52-8}{52} \cdot \frac{51-8}{51}\cdot \ldots \cdot \frac{52-45-8}{52-45}\cdot \frac{4}{52-44}\right) \\ = \sum_{1\le j\le 45} \frac{48-j}{5}\cdot\left(\prod_{0\le k<j-1}\frac{52-k-8}{52-k}\right)\cdot\frac 4{52-j+1} \ . $$ Here,

  • in the first term, the $4/52$ is the probability that a two comes on the first place, followed by the mean time till the first ace shows up, $(48-1)/5$,
  • in the second term, the $(52-8)/52$ is the probability that the first card is neither $2$, nor $A$, and $4/51$ is the probability that a two comes in the second turn, (before any $A$,) all times the mean time till the first ace shows up, $(48-2)/5$,
  • in the third term, the $(52-8)/52$ is the probability that the first card is neither $2$, nor $A$, the $(51-8)/51$ is the probability that the second card is neither $2$, nor $A$, and $4/50$ is the probability that a two comes in the third turn, thus insuring again $2$ before $A$, all times the mean time till the first ace shows up, $(48-3)/5$, and so on.

So the way the sum is built does not use the information that the $2$ comes first, and implements it tacitly.

Since first $2$ comes first before the first $A$ in half of the times, we have the missing factor $1/2$.

dan_fulea
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