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I have been going through and doing some (non-assessed) homework questions, but am getting really stuck on conditional probability. The following problem is one that I simply cannot get my head around.

Question: Die A has four red and two blue faces, and die B has two red and four blue faces. One of the dice is selected at random for use.

i). What is the probability of red being thrown?

ii). If the first two throws resulted in red, what is the probability of red for the third throw?

I was able to get i), no trouble, it came out as 1/2. The second part has me entirely lost though.

Guest
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  • The probability of ii) is 1/2 because they are independent random variables, i.e. every throw is independent of each other. – Masacroso Aug 11 '16 at 10:29
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    you can use Bayes theorem, you need to find the probability that the red die is in use, so it is P(red die in use given two reds) = P(we select red die and get two reds) / P(2 reds) – Cato Aug 11 '16 at 10:30
  • @ Masacroso the third throw is using the same die as in 1 and 2, and therefore the 2 throws give us information, in an extreme case the die would give one color only, making it a 100% chance the red die is in place, and a 100% chance of red again – Cato Aug 11 '16 at 10:44
  • I think P(red dice in use) = (1/2 x 2/3 x 2/3) / (1/4) = (8 /9), so P(red) = (8/9) x (2/3) + (1/9) x (1/3) = 17 / 27 – Cato Aug 11 '16 at 10:45
  • I understand @AndrewDeighton but the estimation is very weak. In any case we need to apply a hypothesis test to know it significance. – Masacroso Aug 11 '16 at 10:50
  • @Masacroso No need to test anything, there are enough numbers given to calculate the true probability instead of trying to infer anything statistically. – kviiri Aug 11 '16 at 11:12
  • @kviiri I dont think so... if you throw the dice tree times instead of just two the "true" probability changes! xD – Masacroso Aug 11 '16 at 15:04
  • @Masacroso Yes it does, what's your point? The more consecutive reds you throw the more likely you are to be using die A. – kviiri Aug 11 '16 at 17:34
  • @kviiri taken from wikipedia: In the Bayesian view, a probability is assigned to a hypothesis[...] You are doing an estimation. Read my previous comments. – Masacroso Aug 11 '16 at 17:39
  • @Masacroso No, I'm not estimating. We already know the probabilities involved so there's no need to do that. Both dice have 1/2 chance to be picked, A has 2/3 chance per throw to be red and B has 1/3 cahnce to be red. Those are all given in the assignment. – kviiri Aug 11 '16 at 17:44
  • @kviiri any bayesian probability is an estimation. You are not estimating the probability for each dice to be red... you are estimating the probability to pick one dice more than other. – Masacroso Aug 11 '16 at 17:48
  • @Masacroso In real life statistics maybe, but here we know the initial values and can calculate the needed probability with absolute precision. It's no more "estimate" than 1+1=2. Disagree if you will, but my math is sound. – kviiri Aug 11 '16 at 17:52
  • @kviiri no, your math is far to be sound xD Imagine that instead of red-red you take red-blue, then the estimation will be completely different, you see? The probability to take red-blue is not so different, taking in account the estimation when you take red-red. This is the reason that for any kind of estimation we need some criteria (test the hypothesis and percent of confidence) to evaluate the significance of the estimation. – Masacroso Aug 11 '16 at 17:58
  • @Masacroso I don't see how that would invalidate math, and to be honest I am not at any liability to teach you. Do some Monte Carlo tests or something if you don't believe me. Math doesn't need your approval to work. – kviiri Aug 11 '16 at 18:01

3 Answers3

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For part two, start by considering the question: "What is the probability that two throws result in two reds?"

Since Die A has a $2/3$ probability of providing red, and Die B has a $1/3$ probability of providing red. So:

$P(\text{2 reds}\ |\ A) = (2/3) \times (2/3) = 4/9$

$P(\text{2 reds}\ |\ B) = (1/3) \times (1/3) = 1/9$

The probability for receiving two reds in the first place is half the probability of scoring two reds with die A plus half the probability of scoring two reds with die B, since both dice get selected with equal probability:

$P(\text{2 reds}) = (1/2) \times (4/9) + (1/2) \times (1/9) = (4/18) + (1/18) = 5/18$

Now that we know $P(\text{2 reds})$ and $P(\text{2 reds}\ |\ A)$, we can use Bayes' theorem to compute $P(A\ |\ \text{2 reds})$:

$P(A\ |\ \text{2 reds}) = \frac{P(\text{2 reds}\ |\ A) \times P(A)}{P(\text{2 reds})} = \frac{(4/9) \times (1/2)}{5/18} = \frac{4/18}{5/18} = 4/5$

So we're using die A with $4/5$ probability, die B with $1/5$ probability. Finally, the probability for getting the third red can be calculated:

$P(\text{red}\ |\ \text{2 reds}) = P(A\ |\ \text{2 reds}) \times (2/3) + P(B\ |\ \text{2 reds}) \times (1/3)\\ =(4/5) \times (2/3) + (1/5) \times (1/3) = (8/15) + (1/15) = 9/15 = 3/5$

kviiri
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Hints:

Let $R_{i}$ denote the event that throw $i$ results in red. Let $A$ denote the event that $A$ is selected for use. Let $B$ denote the event that $B$ is selected for use. You are looking for $\Pr\left(R_{3}\mid R_{1}\cap R_{2}\right)$

  • $\Pr\left(R_{3}\mid R_{1}\cap R_{2}\right)=\frac{\Pr\left(R_{1}\cap R_{2}\cap R_{3}\right)}{\Pr\left(R_{1}\cap R_{2}\right)}$

  • $\Pr\left(R_{1}\cap R_{2}\right)=\Pr\left(R_{1}\cap R_{2}\mid A\right)\Pr\left(A\right)+\Pr\left(R_{1}\cap R_{2}\mid B\right)\Pr\left(B\right)$

  • $\Pr\left(R_{1}\cap R_{2}\cap R_{3}\right)=\Pr\left(R_{1}\cap R_{2}\cap R_{3}\mid A\right)\Pr\left(A\right)+\Pr\left(R_{1}\cap R_{2}\cap R_{3}\mid B\right)\Pr\left(B\right)$

drhab
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There is a hidden assumption in the statement "one of the dice is selected randomly before use": the same dice is used in all three throws. If the dice is chosen individually before each throw, as some might interpret it, the throws would be independent and the answer to (ii) $\frac12$, as I had it in an earlier version of the answer.

Here is a derivation of the correct answer that is a little more intuitive. Two reds have shown up. The probability that dice A caused it is $\left(\frac23\right)^2=\frac49$, while the probability that dice B caused it is $\left(\frac13\right)^2=\frac19$. In other words, dice A is 4 times more likely to have been selected than dice B.

The probability that red comes up on the third throw is then the sum of $\frac45\times\frac23$ (for dice A) and $\frac15\times\frac13$ (for dice B), which evaluates to $\frac35$.

Parcly Taxel
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  • I disagree, if you had a die with 1 million red faces and 1 blue faces and the other with 1 million blue faces and on red - then if I chose one at random and told you it came up red, it's pretty obvious it's going to be the die with a million red faces, with a massive degree of cetainty, then my next roll is red again – Cato Aug 11 '16 at 10:32
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    @AndrewDeighton I have rewritten my answer again to explain the source of confusion with this question. – Parcly Taxel Aug 11 '16 at 11:50