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I would like to calculate the following conditional probability: I know that the probability of a BLUE ball being drawn is 0.3. I receive a message from A or B who saw the ball that has been drawn. This person tells me that the ball drawn is BLUE. I know the probabilities:

A observing BLUE and sending BLUE- 0.16 B observing RED and sending BLUE - 0.09 A observing BLUE and sending BLUE- 0.05 B observing RED and sending BLUE- 0.17

I also know the probabilities:

A observing BLUE and sending RED - 0.019 B observing RED and sending RED- 0.323 A observing BLUE and sending RED-0.073 B observing RED and sending RED=0.106

I would like to calculate the conditional probability of the ball drawn actually being BLUE given that I received the message BLUE and the information I know about each player's probability of sending a given message.

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  • $\begingroup$ your statement of the problem is vague. Please consider restating it more clearly. $\endgroup$ – David LeBauer Jul 5 '11 at 19:36
  • $\begingroup$ @Daniel Hint: the situation determines four cases altogether: truth+yellow, etc. Can you work out the chances where blue is involved? $\endgroup$ – whuber Jul 5 '11 at 19:44
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    $\begingroup$ @Daniel Please don't destroy your question only to ask a new variant of it. Instead, you are able to edit your original question (this one) to reflect any improvements or changes you would like to make. $\endgroup$ – whuber Jul 7 '11 at 0:52
  • $\begingroup$ @Daniel That's a great improvement: well done. However, it's difficult to decipher the last four lines. What is "PRIOR"? What do "d" and "h" represent? Something seems to be missing in the last line. What does "~0.45" mean? $\endgroup$ – whuber Jul 7 '11 at 13:45
  • $\begingroup$ @Daniel I'm struggling to make sense of this information. To me, the first four assumptions imply the probability of somebody observing blue is 0.16 + 0.05 = 0.21, because (I presume) the two conditions are mutually exclusive (friends are not enemies) and exhaustive (one is either a friend or an enemy). It therefore appears contradictory to assert that this probability equals 0.3. Also missing from your account is any clear and full distinction between "observing" and "sending". What is the error matrix for this channel? $\endgroup$ – whuber Jul 7 '11 at 14:08
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If Truth/Lie and color are independent and everything is random, then you would have an 18% chance of an honest person seeing blue and a 28% chance of a lier seeing yellow (and therefore claiming to see blue), you also have a 12% chance of lier seeing blue and 42% chance of honest seeing yellow. So the probability of an honest person seeing blue given that the person claimed to see blue is $\frac{0.18}{0.18+0.28} = 0.39$.

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    $\begingroup$ Please read the wiki for the homework tag, especially the section on answering homework questions. $\endgroup$ – whuber Jul 5 '11 at 21:13
  • $\begingroup$ @Daniel, It's probably a good idea for you to comment on why you think this answer is right or wrong. Maybe, you could draw a picture? $\endgroup$ – bill_080 Jul 5 '11 at 21:27
  • $\begingroup$ @Daniel, Your last sentence in your original post reads..."I would like to know the probability of an honest person observing the blue colour given that he told me that the colour is blue." Should that say...."I would like to know the probability of an honest person observing the blue colour given that I was told that the colour is blue". In other words, you don't know if the person telling you is honest or not. $\endgroup$ – bill_080 Jul 5 '11 at 21:41
  • $\begingroup$ @Daniel, If someone tells you the colour is blue, how many different ways/combinations can you interpret that information, given your original post? $\endgroup$ – bill_080 Jul 5 '11 at 22:05
  • $\begingroup$ @Daniel let us continue this discussion in chat $\endgroup$ – bill_080 Jul 5 '11 at 22:12
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If the honest person said "blue", and we are not dealing with the possibility of a misperception (honestly getting confused about colors), then there's a 100% chance it's blue.

But I suspect this isn't the question you intended to ask.

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