In a hand of 13 cards, what is the probability that all cards have different values I'm going to go through my thoughts, tell me if I'm right.
There are $52\choose 13$ different hands with 13 cards. Each value has to appear once with 4 suits for each so there are $ 4^{13}$ different hands with each number once.
So i conclude that the probability is $\frac{4^{13}}{52\choose13} \approx 1.1\times 10^{-4}$. I am really bad a probability so any help and/or tips on approaching problems at first year university level with would be appreciated (I can only find very basic or high level probability stuff with no intermediate level)
 A: You appear to have have a correct method to compute the probability (everything was fine except that you wrote $54$ when you presumably meant to write $52$).
I will show another way to derive the same result.
Suppose you fill the hand one card at a time. When you pick the first card from the deck, there are $52$ cards in the deck and any of them can be drawn. That is, you have a $\frac{52}{52}$ chance that the first card you draw will be "correct."
But for the next card, there are only $48$ remaining cards that you can draw without duplicating the rank of he card already in your hand,
and there are $51$ cards to draw from. So you have a $\frac{48}{51}$ chance
that the second card is also "correct," given that the previous card was.
The chance is only $\frac{44}{50}$ that the third card will be "correct,"
given that the previous cards were.
Each time you draw a card, the number of "correct" possibilities for the next draw goes down by $4,$ because you removed one card and because the other three cards of the rank you just drew also become "incorrect" for all subsequent draws.
Finally you have $40$ cards in the deck just before drawing the thirteenth card, but only $4$ of them are "correct," giving you a $\frac{4}{40}$ probability.
So the probability to draw thirteen cards in a row without getting any two of the same rank is
\begin{align}
\frac{52}{52}\cdot&\frac{48}{51}\cdot\frac{44}{50}\cdot\frac{40}{49}\cdot\frac{36}{48}\cdot\frac{32}{47}\cdot\frac{28}{46}\cdot\frac{24}{45}\cdot\frac{20}{44}\cdot\frac{16}{43}\cdot\frac{12}{42}\cdot\frac{8}{41}\cdot\frac{4}{40} \\[0.7ex]
&= \frac{(4\cdot13)(4\cdot12)(4\cdot11)(4\cdot10)(4\cdot9)(4\cdot8)(4\cdot7)(4\cdot6)(4\cdot5)(4\cdot4)(4\cdot3)(4\cdot2)(4\cdot1)}{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43\cdot42\cdot41\cdot40} \\
&= \frac{4^{13} \cdot 13!}{52! / 39!} \\
&= \frac{4^{13}}{\left(\frac{52!}{13!39!}\right)} \\
&= \frac{4^{13}}{\binom{52}{13}}.
\end{align}
I think your way is neater, don't you agree?
