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A card is drawn at random from a pack of $52$. According to the value of this card ($A =1, J=Q=K=10$) as many further cards are drawn. What is the probability that the ace of hearts is among those drawn(including the first card)? [Answer = $5/34$]

I tried splitting it up into 2 parts: Part 1 for the ace: It can either be the ace of hearts at first which is $(1/4)(51/52)$ or ace of hearts second ${^4}C_3\cdot 1/51$. But I can't seem to get the answer please help.

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  • $\begingroup$ What are the rules of drawing? $\endgroup$ – callculus Mar 8 at 18:48
  • $\begingroup$ No replacement and as specified according to the value $\endgroup$ – KombatWombat Mar 8 at 19:07
  • $\begingroup$ "as specified according to the value" I cannot say that I´ve understood this rule. When you stop and why? $\endgroup$ – callculus Mar 8 at 19:13
  • $\begingroup$ So you draw one card if it's an ace you draw one more card and stop if it's a j q or k you draw 10 more and stop $\endgroup$ – KombatWombat Mar 8 at 19:16
  • $\begingroup$ This comment you should write into the question. You´ve omitted this crucial information. You shouldn´t be surprised that nobody can answer your question without these information. $\endgroup$ – callculus Mar 8 at 19:22
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Splitting the problem is a good idea.

1) A heart is drawn first. There a two cases

a) Ace of hearts: $\frac1{52} \quad (\color{blue}{I})$

We are done.

b) An ace of not hearts and then the ace of hearts: $\frac3{52}\cdot \frac1{51} \quad (\color{blue}{II})$

2)

We can draw the number $2$ ($k=2$) and then draw two cards where one of them is the ace of hearts. The probability is

$$\frac4{52}\cdot \frac{\binom{1}{1}\cdot \binom{50}{1}}{\binom{51}{2}}$$

We can sum up the probabilities from $k=2$ to $k=9$

$$\frac4{52}\cdot \left( \sum_{k=2}^9\frac{\binom{1}{1}\cdot \binom{50}{k-1}}{\binom{51}{k}}\right) \quad (\color{blue}{III})$$

3) What is left where we have to draw 10 after the first draw. This is the case when we have drawn a $10$, Jack, Queen or King.

We have $k=10$ and regard the $4$ cases by multiplying it by $4$

$$4\cdot \frac4{52}\cdot \frac{\binom{1}{1}\cdot \binom{50}{9}}{\binom{51}{10}} \quad (\color{blue}{IV})$$

What is left is to sum up the intermediate results: $ (\color{blue}{I})+ (\color{blue}{II})+(\color{blue}{III})+(\color{blue}{IV})$

Especially for the third term a calculator is needed. Here is what I got as the final result. It is the proposed solution.

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    $\begingroup$ There are only three aces of not hearts. The probability in 1b should be $\frac{3}{52} \cdot \frac{1}{51}$. The final result is correct. $\endgroup$ – Koen Tiels Mar 8 at 21:40
  • $\begingroup$ @KoenTiels Good catch. Thank your very much. I´ve edited the typo. $\endgroup$ – callculus Mar 8 at 22:03
  • $\begingroup$ Thank you for your help $\endgroup$ – KombatWombat Mar 9 at 5:09
  • $\begingroup$ @KombatWombat You´re welcome. $\endgroup$ – callculus Mar 9 at 11:34
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    $\begingroup$ Observation: ${50 \choose k-1} / {51 \choose k} = k / 51$ which also has the easy explanation as the ace of hearts being in the next $k$ cards out of $51$. Question: obviously the answer reduces to $5/34$ but why? This is such a simple fraction that I keep thinking there might be an alternate proof... or is it just a coincidence? :( $\endgroup$ – antkam Mar 9 at 17:32

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