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You wish to test out two slot machines (Machine 1 and 2). One is "good", i.e. offers better chances of winning, and the other is "bad". You do not know which is which as both are identical, the probability of playing either the "good" or "bad" is 50%.

The probability of winning on the “good” machine is 1/2 and the probability of winning on the “bad” machine is 1/3. What is the probability that Machine 2 is “Good” given you won playing on Machine 1?

ANSWER CHOICES

  • 0.3
  • 0.4
  • 0.5
  • 0.6
  • 0.7

MY SOLUTION SO FAR...

1# Since each machine is equally likely to be the “good” machine we can express this as...

P(M1 is Good)=P(M2 is Bad)=1/2

P(M1 is Good)=P(M2 is Bad)=1/2

P(M1 is Bad)=P(M2 is Good)=1/2

P(M1 is Bad)=P(M2 is Good)=1/2

2# We have also been told the probability of winning for each type of machine

P(Win on M1 | M1 is Good)=1/2

P(Win on M1 | M1 is Bad)=1/3

P(Win on M1 | M1 is Good)=1/2

P(Win on M1 | M1 is Bad)=1/3

3# We can use these probabilities to calculate the probability of losing for each type of machine as well:

P(Loose on M1 | M1 is Good)=1/2

P(Loose on M1 | M1 is Bad)=2/3

P(Loose on M1 | M1 is Good)=1/2

P(Loose on M1 | M1 is Bad)=2/3

4# The probability of M1 being good given that you won on M1 is:

P(M1 is Good | Win on M1)= P(M1 is Good) * [P(Win on M1 | M1 is Good)/P(Win on M1)]= (1/2*1/2) / (1/2*1/2 + 1/3*1/2)=0.6

5# The probability of M1 being bad given that you won on M1 is:

P(M1 is bad | Win on M1)=P(Mi is bad) * [P(Win on M1|M1 is bad)/P(Win on M1)] = (1/2*1/3) / (1/2*1/2 + 1/3*1/2) = 0.4

I don't know how to jump from machine 1's probability to machine 2...

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    $\begingroup$ Note that Machine 2 is good if and only if Machine 1 is bad. Hence any statement about Machine 2 can be translated into a statement about Machine 1. $\endgroup$ – Shagnik Aug 10 '16 at 10:28
  • $\begingroup$ That can't be true - you could pull both Machine 1 and 2 levers simultaneously or one after the other and still win both times... $\endgroup$ – CodeMonkey Aug 10 '16 at 10:32
  • $\begingroup$ @CodeMonkey What you said has nothing to do with what Shagnik said. It is true that machine $1$ is only good if machine $2$ is bad. That doesn't change the fact that you can win on both. $\endgroup$ – 5xum Aug 10 '16 at 10:34
  • $\begingroup$ Maybe my "any" was misleading. Exactly one of the machines is good. Hence Machine 2 being good is equivalent to Machine 1 being bad. So what you need to do is compute the conditional probability of Machine 1 being bad, which is what you have already done. $\endgroup$ – Shagnik Aug 10 '16 at 10:36
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Let's rephrase the question:

What is the probability that machine 2 is good given that you won playing on machine 1?

To the equivalent question:

What is the probability that machine 1 is bad given that you won playing on machine 1?

Let $A$ denote an event of machine 1 being bad.

Let $B$ denote an event of winning on machine 1.

$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac12\times\frac13}{\frac12\times\frac13+\frac12\times\frac12}=\frac25$

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There's a lot of typing going on, so let's just introduce some notation:

  • $M_1$... "Machine $1$ is good".
  • $M_2$... "Machine $2$ is good".
  • $W_1$... "Win on machine $1$".
  • $W_2$... "Win on machine $2$".

Now we can describe everything shortly. For example, losing on machine $2$ is simply $\neg W_2$, since losing is the negation of winning.

Now, you need to calculate the probablity that machine $2$ is good given that you won on machine $1$.

So, you need to calculate

$$P(M_2|W_1)$$

Using Bayes off the bat, you get that $$P(M_2|W_1) = P(W_1|M_2) \cdot \frac{P(M_2)}{P(W_1)}$$

So, what do we have out of the three numbers you need?

Well, $P(W_1|M_2)$ is the probability that you will win on machine $1$ if machine $2$ is good. This is simply the probability of winning on a bad machine, which is $\frac13$.

$P(M_2)$ is even simpler to calculate, it's simply $\frac12$.

The last you need to calculate is $P(W_1)$, i.e. the probability that you will win on machine $1$. To do that, let me first just give you a hint:

Remember the law of total probability.

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  • $\begingroup$ My biggest problem was solving for P(W1|M2)...which may take a while to understand... $\endgroup$ – CodeMonkey Aug 10 '16 at 10:47
  • $\begingroup$ @CodeMonkey $P(W_1|M_2)$ is simple. Just read the expression out loud "The probability of winning on machine $1$ given that I know that machine $2$ is good." Now, if you know that machine $2$ is good, then you also know that machine $1$ is bad. So the probability of winning on it is $\frac13$, because that's the probability of winning on a bad machine. $\endgroup$ – 5xum Aug 10 '16 at 10:53
  • $\begingroup$ @CodeMonkey In other words, the event $M_2$ ("machine $2$ is good") is the same as the event $\neg M_1$ (negation of "machine $1$ is good", i.e. "machine $1$ is bad") $\endgroup$ – 5xum Aug 10 '16 at 10:55

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