Joint Probability problem

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Muffins

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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!
 
Muffins said:
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.
Click to expand...

It seems that this question is seriously flawed.
First X can have only the values 3 4, 5, or 6, and Y only 1 or 2.

Let me explain if the first card is 3 or 4 the game is over for the sum is 3 or above in which case Y=1.
In this case X=3 or 4

If the first card is either 1 or 2, then no matter what the second card is the sum will be 3 or above.
In this case Y=2 and X=3, 4, 5, or 6.

If were you, I would review the question to see if it is written correctly.
 
Last edited:
Muffins said:
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!
Click to expand...
The question is a bit ambiguous, as whether to include all (X,Y), or just the values at the end of the game. The difference in the table is whether or not to include X=1 or 2 when Y=1. As you have interpreted it, The joint-probability table as has 2 rows times 4 columns. There are 6 non-zero elements in the table, but not all are equally probable. "Without replacement" makes the probability for a particular 2nd card 1/3. For example, I get the probability P(Y=2,X=3) to be the sum of (1+2) and (2+1), or (1/4)(1/3) + (1/4)(1/3) = 1/6. Does that make sense? I do get the sum of probabilities at the end of the game to be 1.
 
Muffins said:
Managed to find the mistake in my calculations.
Click to expand...

Here are my calculations:
\(\displaystyle \begin{array}{*{20}{c}}
{Y\backslash X}&|& 3&4&5&6&{}\\
1&|& {1/4}&{1/4}&0&0&{}\\
2&|& {1/6}&{1/12}&{1/6}&{1/12}&{}
\end{array}\)
 
Oh, that's a little different from mine.
mine turned to be:
x/y12
31/82/8
41/81/8
502/8
601/8

I thought of it as 'there are 8 ways of taking cards out until the game is over with' - aka: (1+2) or (1+3) or (1+4) or (2+1) or (2+3) or (2+4) or (3) or (4)
so for example-
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)

Why would my way be incorrect?
 
Last edited:
Muffins said:
Oh, that's a little different from mine.
mine turned to be:
x/y12
31/82/8
41/81/8
502/8
601/8

I thought of it as 'there are 8 ways of taking cards out until the game is over with'
so for example-
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)

Why would my way be incorrect?
Click to expand...
The eight ways are not all equally probable.
One quarter of the time, the first card drawn will be a 3 and the game is over.
One quarter of the time, the first card drawn will be a 4 and the game is over.
Thus the probabilities in your first column should be 1/4, 1/4, 0, 0

Half the time, there will be two cards drawn: the probability if the first card having a specific value is 1/4, but for the second card there are only 3 to choose from so the probability is 1/3, and the combined probability for two specific cards is (1/4)(1/3) = 1/12. The 2nd column in your table should have denominators of 12 instead of 8. The numerators are correct: for two cases there are 2 permutations, and for the other cases only one possibility.

ok? That should make your table agree with what pka posted.
 
Thank you so very much both. Thank you Phil, again - this is so much clearer now.
 
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