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

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.

  • $\begingroup$ The odds are 1:1 and 1:3, respectively. $\endgroup$ – user Jul 7 '20 at 14:28
  • $\begingroup$ 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? $\endgroup$ – saulspatz Jul 7 '20 at 14:31
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    $\begingroup$ 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? $\endgroup$ – Devansh Kamra Jul 7 '20 at 14:32
  • $\begingroup$ i mean 3 aces and then the joker, so i draw 4 cards in total $\endgroup$ – Alessandro Valentino Jul 7 '20 at 14:32
  • $\begingroup$ How is that different from drawing three Aces? After you draw three Aces, only the Joker is left. $\endgroup$ – 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

  • 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.

  • $\begingroup$ This case is valid when he isn't placing the card he drew back in the deck. When he places the card he draws back in the deck, the probability to draw an ace remains the same everytime. Your answer is correct but the user hasn't stated the question completely. $\endgroup$ – Devansh Kamra Jul 7 '20 at 14:37
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    $\begingroup$ Sure, but I assumed the semantic difference between drawing and replacing. $\endgroup$ – Sameer Baheti Jul 7 '20 at 14:39
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    $\begingroup$ That is valid. But I still feel that user must have stated it in the question. Anyways +1 $\endgroup$ – Devansh Kamra Jul 7 '20 at 14:40
  • $\begingroup$ I edited my question, you can try to edit your answer based on my edits if they can help $\endgroup$ – Alessandro Valentino Jul 7 '20 at 15:01
  • $\begingroup$ @AlessandroValentino Done! $\endgroup$ – Sameer Baheti Jul 7 '20 at 15:16

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