What are the odds of drawing the same card $3$ times in a row in a $4$ card deck ( $3$ of the same card and $1$ joker )

arithmeticprobability

I made the question simple but there are $2$ things that i'd like to know:

In a deck of $4$ randomly shuffled cards with $3$ aces and $1$ joker, what are the odds of drawing $3$ aces in a row, and what are the odds of the joker being the last of those cards (in the same conditions, so $4$ cards randomly shuffled, $3$ of which are aces and $1$ is a joker).
And if they are any different, why ( with the steps or method for calculating them would be better ).
It's a dumb question probably, but i feel like i am missing something or doing something wrong
Thanks

EDIT:
So to clear some doubts, what i mean by drawing is taking a card out of the deck and not placing it back, so every time i draw i find myself with $1$ less card in the deck i am drawing from.
Also, for the first part of the question, i want to know the odds of drawing $3$ aces in a row and what it takes to calculate that.
The second part, is referring to how many odds i have of having the joker as the last card in the deck and why is that different from saying drawing $3$ aces in a row (If there is any difference). Am i being clear? Sorry if i am not. I will clarify further if needed.

Best Answer

Drawing(but not replacing) from a shuffled deck of $4$ cards consisting of $3$ aces and a joker, what is the probability that

  • The first three are aces.
  • The last one to be drawn is the joker. Are they different? If yes, why?

\begin{align*} P(\text{the joker is the last card to be drawn})&=P(\text{three aces are drawn one after the other})\\ &=P(\text{first is an ace})\cdot P(\text{second is also an ace})\cdot P(\text{third is also an ace})\\ &=\frac34\cdot \frac 23\cdot \frac 12=\frac14\\\end{align*} They are not different because drawing three cards in a row with a certain probability leaves the joker as the last card definitely.

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