PMF of the turn T at which the 4th ace is drawn

Question

Cards are drawn one by one from a shuffled 52-card deck. Find the PMF of the turn T at which the 4th ace is drawn.

Solution given:
The event T=t occurs if no ace is drawn in turns t+1 to 52 and an ace is drawn in turn t. Let $A_i$ be the event that an ace is drawn in turn i. By the multiplication rule, $$\begin{align}P(T=t) &= P(A_t\cap A^c_{t+1}\cap\dots\cap A^c_{52})\\ &= P(A^c_{52})P(A^c_{51}|A^c_{52})\dots P(A^c_{t+1}|A^c_{t+2}\cap \dots\cap A^c_{52})\\ &= \frac{48}{52}\cdot\frac{47}{51}\dots\frac{t-3}{t+1}\cdot\frac{t-4}{t}\end{align}$$, where $t\geq 4$ and $P(T<4)=0$

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Why is $P(A^c_{52})=\frac{48}{52}$, $P(A^c_{51}|A^c_{52})=\frac{47}{51}$, etc.?

Answer 1

In a standard deck of $52$ cards, there are $4$ aces and $52 - 4 = 48$ non-aces.

Assuming each card is equally-likely to be drawn in each turn, the marginal probability of a non-ace card appear in a particular turn is $48/52$

Given that there is a non-ace card appear in another turn, we can exclude that card in the remaining deck of $52 - 1 = 51$ cards, and leaving $48 - 1 = 47$ non-ace cards, therefore the conditional probability is $47/51$. Here we exclude the card as we assume the question that the card drawn are not put back to the deck (draw without replacement), i.e each unique card can appear once only.

Answer 2

Consider the conditional event $A_k^c|A_{k+1}^c\ldots A_{52}^c$, it means

• in the last $52-(k+1)+1=52-k$ draws $52-k$ cards have already been drawn

• so that there are only $k$ cards left to be drawn

• also within these $k$ cards there are still $4$ aces present, since they were not drawn till now and can't be drawn now as well

• leading to $k-4$ cases favorable to the event out of total number of mutually exclusive and equally likely $k$ possible outcomes,

• by classical definition, the probability is $\frac{k-4}{k}$.

• Likewise, other probabilities can also be computed.

Answer 3

I think you can derive the pdf by considering a couple examples and finding the pattern.

The range of $t$ is $4, 5, ..., 52$. For $t=4$,

$P(T=4) = \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49}$

For t = 6, the sixth card must be an ace and the other 3 must be observed anywhere between the first and fifth card. Let "A" be an ace and "N" be a non-ace. So, we might get something like

$\quad \quad \quad ANAANA$

and the probability of this particular outcome would be

$\frac{4}{52} \cdot \frac{48}{51} \cdot \frac{3}{50} \cdot \frac{2}{49} \cdot \frac{47}{48} \cdot \frac{1}{47} = \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49}$

There are $\binom{5}{3}$ ways of finding 3 slots for the aces in the first $5$ cards, so

$P(T=6) = \binom{5}{3} \cdot \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49}$

I think the final pdf would be

$P(T=t) = \frac{\binom{t-1}{3}}{\binom{52}{4}} \quad \text{for} \; t = 4, 5, ..., 52$