Need very simple reasoning without any calculating

Question

4 cards are drawn from a 32 card deck. """Let $p_1$ the probability that they are all King such that if at least 1 of them is heart.""" edit Let $p_1$ the probability that they are all kings if at least one of them is a heart Let $p_2$ the probability that they are all King

The mysterious question is to prove that $p_1=p_2 \iff$ the draw is with returns

I don't know how to answer the question without doing any calculations, so I think I'm missing a very simple reasoning

My answer Let X be the number of kings and Y the number of hearts. We search $$ p_1=P((X=4)|(Y\ge 1))=\frac{P((X=4)\cap (Y\ge 1))}{P(Y\geq 1)}$$

With no return drawing $(X=4)\subset (Y\ge 1)$, then

$p_1=\frac{P(X=4)}{1-P(Y=0)}$=${\frac 1{\binom{32}{4}}}\over {1-\frac{\binom {24}{4}}{\binom{32}{4}}}$=$\frac 1{ \binom{32}{4}-\binom {24}{4}}$ and clairely $p_2=P(X=4)=\frac 1{\binom{32}{4}}\neq p_1$

With return drawing $p_1=\frac{P((X=4)\cap (Y\ge 1))}{P(Y\geq 1)}=\frac{P(X=4)-(P(X=4)\cap (Y=0 ))}{1-P(Y=0)}$ but $Cardinal (P(X=4)\cap (Y=0 ))=3^4$ then $p_1=\dfrac{4^4-3^4}{32^4-24^4}=\frac 1{8^4}$ and clairely $p_2=\frac 1{8^4}=p_1$

Answer 1

If the draw is without returns, the probability $p(♡|4K)=1$ since you cannot draw four different kings without drawing the king of hearts, and so $$p_1=p(♡,4K)=p(♡|4K)p(4K)=p(4K)=p_2.$$

So the premise of the "mysterious question" that $p_1=p_2$ if and only if the draw is with returns is false.

Your use of conditional probabilities is also wrong. $p(\color{red}{(X=4)}|\color{green}{(Y\ge 1)})$ means "the probability that we have $\color{red}{\text{four kings}}$ given that we have $\color{green}{\text{one or more hearts}}$". What the question asks about is $p(Y\ge1|X=4)$, the probability that given four kings, one of them is a heart, which is $p(♡|4K)$ in my notation, which must be 1 if drawing without returns.

If drawing with returns, $p(Y\ge1|X=4)=1-(3/4)^4={175\over256}\approx0.68$. Here the $(3/4)^4$ is the probability of four draws with no heart. (Since each draw of cards is independent, the chance of drawing one or more hearts, $p(Y\ge1)$, is actually the same as the chance of drawing a king of hearts given that we've drawn four kings, $p(Y\ge1|X=4)$.)

So with returns $$ p_1=p(Y\geq1|X=4)p(X=4)={175\over256}{1\over8^4}\neq p(X=4)=p_2={1\over8^4}. $$

Edit: Apparently "such that if at least one of them is a heart" is supposed to indicate drawing kings when one or more of the cards is a heart. In that case the conclusion is obvious, since the probability of drawing a king is independent of the probability of drawing a heart. $p(K|♡)=p(K)=1/8$ and $p(k,♡)=p(K)p(♡)=1/32$.

Answer 2

I take $p_1$ to be the probability that the drawn cards are all Kings given that at least one of them is a heart. As stated, $p_2$ is the probability that the drawn cards are all Kings.

Then letting $X$ be the number of kings, $Y$ the number of hearts, if the cards are drawn without replacement then $X=4$ if and only if $X=4$ and $Y\geq 1$ (that is, we can only draw four kings if one of them is the king of hearts), so we have \begin{align} p_2 &= P(X=4) \\ &= P(X=4\mid Y\geq1) \\ &= P(X=4\mid Y\geq1) P(Y\geq1) \\ &= p_1 P(Y\geq1). \end{align}

Hence $p_1 = p_2$ only if $P(Y\geq 1) = 1,$ which is false. It is not possible that $p_1 = p_2$ if the cards are drawn without replacement.

Drawing with replacement, on each draw there is a $1/8$ probability of a king and a $1/4$ probability of a heart, each event independent of the other and independent of the occurrence of a king or a heart on any other draw. From this you should be able to form an argument that the events $X=4$ and $Y\geq1$ are independent without doing any explicit calculations.