Deck probability for five cards with different face values What is the probability of a five card hand where all values are different (where aces are 1s) and the highest value is 8?
This should include 32 cards total, with there being 8 different face values. But I'm not sure where to go from here
 A: The first card must be 8, if I understood correctly, so we must choose $1$ from $4$, i.e., there are 4 different choices. For the second card we have $32-4$ different outcomes, for the third we have $32-4\cdot 2$ valid cards, for the four $32-4\cdot 3$ and for the fifth $32- 4\cdot 4$.  And the total kind of hands are $32\cdot 31\cdot 30 \cdot29\cdot 28$. So
$$P=\frac{4}{(32)_5}\prod_{k=1}^{4} (32- 4k)$$
A: To avoid a misinterpretation of the question assume that there are $m\ge 5$ different face values (Aces, 2's, ...) with total $4m$ cards in the deck (32, 52, or whatever). Then


*

*The probability of a five card hand where all values are different is


$$\frac{\binom{m}{5}4^5}{\binom{4m}{5}}$$


*The probability of a five card hand where all values are different and these values are at most $5\le k\le m$ is


$$\frac{\binom{k}{5}4^5}{\binom{4m}{5}}$$


*The probability of a five card hand where all values are different and the highest value is exactly $5\le k\le m$ 


$$\frac{\binom{k-1}{4}4^5}{\binom{4m}{5}}$$
A: *

*Choose 5 distinct ranks, and a suit for each

*Divide by total ways for choosing any 5 cards
Pr = $\dfrac{\binom{8}{5}4^5}{\binom{32}{5}}=\dfrac{256}{899}$
