# Finding the distribution of a random vector in a conditional probability problem [closed]

Players A and B are playing a game of drawing coins from two boxes without returning/replacing them. Box1 has three coins with values 0, 1 and 2 and Box2 has two coins with values 1 and 2. In the game, each draw Player A chooses one of the two boxes randomly (with 1/2 chance) and Player B draws a coin from that box. After two draws, the drawn coins go to one of the players - if the sum is greater than 2, Player A gets the coins or else Player B gets the coins.

Yi - random variable which signifies the outcome of i-th draw (i = 1,2)

I need to find the distribution of the random vector (Y2,Y1)

I can find out the probabilities of the first draw with the formula for total probability, but I'm having trouble with the second.

Can anyone help me?

## closed as unclear what you're asking by Graham Kemp, choco_addicted, user91500, Shailesh, LeucippusJun 3 '16 at 2:42

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• Three things aren't clear to me: a) You didn't say anything about how Player $A$ chooses a box -- without knowing that we can't know the distribution of the outcomes. b) Does Player $A$ choose one box for both draws, or is it a separate choice for each draw? c) Are the draws with or without replacement? – joriki Jun 2 '16 at 21:16
• a) He chooses it randomly, so there's 1/2 chance he chooses either one b) Again, he chooses either box1 or box1 with 1/2 chance – Abstrac Tor Jun 3 '16 at 1:31
• c) I said "without returning them" so that means without replacement – Abstrac Tor Jun 3 '16 at 1:32
• @AbstracTor Adjust your post, please. People shouldn't have to look in the comments to find out what you meant. – Graham Kemp Jun 3 '16 at 2:47
• Done, also that's how the problem was stated originally in my language and I just translated it in English. – Abstrac Tor Jun 3 '16 at 12:58

You were a bit abstruse, but I think you meant that one of the boxes is chosen randomly each draw and a coin extracted without replacement.

So let $S_1, S_2$ be the events the (initially) two coin box is chosen each draw, and $T_1,T_2$ the complements (the initially three coin box).   Then the Law of Total Probability comes into play:

(Note: Due to symmetry you only have three cases to consider.)

\begin{align} \mathsf P(Y_1{=}x, Y_2{=}y) =& ~ \mathsf P(Y_1{=}x,Y_2{=}y, S_1,S_2)+\mathsf P(Y_1{=}x,Y_2{=}y, S_1,T_2)+\mathsf P(Y_1{=}x,Y_2{=}y, T_1,S_2)+\mathsf P(Y_1{=}x,Y_2{=}y, T_1,T_2) \\[2ex] =&~ \begin{cases}? & :(x,y)\in\{(0,0),(1,1)\}\\ ? & :(x,y)\in\{(0,2),(1,2),(2,0),(2,1)\}\\ ? & : (x,y)\in\{(0,1),(1,0)\} \\ 0 & : (x,y)=(2,2)\textsf{ or otherwise}\end{cases} \end{align}

• I merely translated the problem form my mother tongue to English (word to word) but you got it right. – Abstrac Tor Jun 3 '16 at 13:04
• Also, (0,0) is not symmetric to (1,1). The coins in Box1 are 0,1 and 2. The coins in Box2 are 1 and 2. – Abstrac Tor Jun 3 '16 at 13:40
• And (2,2) can't be 0 – Abstrac Tor Jun 3 '16 at 13:42