# Joint Probability problem

#### fiffette

##### New Member
Hi again! Have a bit of a problem with this question and would gratefully appreciate some help.
There are 4 cards in a game, numbered 1 to 4. A player randomly draws one card at a time (and without replacement). The game ends when the numbers on the drawn cards sum to 3 and above.
Let X be a random variable - The sum of numbers on the drawn cards.
Let Y be a RV - The number of cards the player had drawn.
Find the joint probability of X & Y.

Well, my struggle is mostly in which method do I calculate the probabilities. I know that X can take the values 3,4,5,6 and Y can take 1,2.
The sample space is 8 I think, because there are 8 possible scenarios of taking cards out until we stop. Am I right to think this way?
However when putting the probabilities in the table, they sum up to 6/8 only. What am I overlooking?

Thanks a bunch!

#### Dason

Hard for us to say when you don't show us what you came up with for the joint probability distribution.

#### BGM

##### TS Contributor
It is a good start to know that the support of the joint distribution is a subset of the product set of individual supports.

So in your case there are at most $$2 \times 4 = 8$$ support points. As what Dason suggested, show us the joint pmf calculated by you on those points (list it or tabulate it) and we can check which part goes wrong.

#### fiffette

##### New Member
My apologies. Here goes-

P(x=3,y=1)= 1/8 (because that would mean choosing the 3 card so there's only one way to do that)
P(x=3,y=2)=2/8 (either taking card '1' out and then card '2' or the other way around)
P(x=4, y=2)=2/8
P(x=5,y=1);P(x=6,y=1)=0
P(x=5,y=2)=2/8
P(x=6,y=2)=1/8

And I just realized I found my mistake. Thank you all!