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, Leucippus Jun 3 '16 at 2:42

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  • $\begingroup$ 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? $\endgroup$ – joriki Jun 2 '16 at 21:16
  • $\begingroup$ 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 $\endgroup$ – Abstrac Tor Jun 3 '16 at 1:31
  • $\begingroup$ c) I said "without returning them" so that means without replacement $\endgroup$ – Abstrac Tor Jun 3 '16 at 1:32
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    $\begingroup$ @AbstracTor Adjust your post, please. People shouldn't have to look in the comments to find out what you meant. $\endgroup$ – Graham Kemp Jun 3 '16 at 2:47
  • $\begingroup$ Done, also that's how the problem was stated originally in my language and I just translated it in English. $\endgroup$ – 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.)

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

  • $\begingroup$ I merely translated the problem form my mother tongue to English (word to word) but you got it right. $\endgroup$ – Abstrac Tor Jun 3 '16 at 13:04
  • $\begingroup$ 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. $\endgroup$ – Abstrac Tor Jun 3 '16 at 13:40
  • $\begingroup$ And (2,2) can't be 0 $\endgroup$ – Abstrac Tor Jun 3 '16 at 13:42

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