# How to get an ace of hearts?

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.

• What are the rules of drawing? – callculus Mar 8 at 18:48
• No replacement and as specified according to the value – KombatWombat Mar 8 at 19:07
• "as specified according to the value" I cannot say that I´ve understood this rule. When you stop and why? – callculus Mar 8 at 19:13
• 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 – KombatWombat Mar 8 at 19:16
• 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. – callculus Mar 8 at 19:22

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.

• 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. – Koen Tiels Mar 8 at 21:40
• @KoenTiels Good catch. Thank your very much. I´ve edited the typo. – callculus Mar 8 at 22:03
• Thank you for your help – KombatWombat Mar 9 at 5:09
• @KombatWombat You´re welcome. – callculus Mar 9 at 11:34
• 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? :( – antkam Mar 9 at 17:32