# 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 )

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.

• The odds are 1:1 and 1:3, respectively. – user Jul 7 '20 at 14:28
• What does it mean to say you draw three Aces in a row, the last of which is a Joker? Do you mean two Aces followed by a Joker? – saulspatz Jul 7 '20 at 14:31
• It actually depends whether you are placing the card back in the deck after drawing it or not. Which one are you intending to do? – Devansh Kamra Jul 7 '20 at 14:32
• i mean 3 aces and then the joker, so i draw 4 cards in total – Alessandro Valentino Jul 7 '20 at 14:32
• How is that different from drawing three Aces? After you draw three Aces, only the Joker is left. – saulspatz Jul 7 '20 at 14:34

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