# Probability problem. There are 16 disks in a box.

Question: There are 16 disks in a box. Five of them are painted red, five of them are painted blue, and six are painted red on one side, and blue on the other side. We are given a disk at random, and see that one of its sides is red. Is the other side of this disk more likely to be red or blue?

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To fully specify the problem, you need to specify how we see that one of the sides is red. The answer would be different if a) we randomly choose a side to look at or b) we can somehow tell (e.g. by looking from a distance) whether one of the sides of a disk is red or not or c) the person who drew the disk gives it to us with a red side showing, possibly according to some strategy such as always showing a red side if possible. – joriki Apr 9 '12 at 20:22
@joriki : "We are given a disk at random, and see that one of its sides is red" is enough for me to believe that I am given a random disk and I see a random side of the disk? I don't think we should over exaggerate over this. – Patrick Da Silva Apr 9 '12 at 20:40
@Patrick: I agree. The problem would be if we were told that one of its sides was red -- then we would be in Monty Hall territory. – TonyK Apr 9 '12 at 20:42
@Patrick: I think it likely that yours is the intended interpretation $-$ that we are equally likely to see each of the $32$ sides $-$ but I also think that it’s important for the OP to understand why we need to make some assumption here. – Brian M. Scott Apr 9 '12 at 20:45
"probabilistic method" has a different meaning and I removed it from the title. Also added probability tag. – zyx Apr 9 '12 at 22:47

There are $5 \times 2 + 6 \times 1 = 16$ red sides and of the $16$ red sides, $10$ have red on the other side, so the probability the other side is red is $\frac{10}{16}=\frac{5}{8} \gt \frac{1}{2}$, and so the other side of this disk more likely to be red.

Alternatively, if you are given a disk at random, then it is more likely to be same colour on both sides (and red and blue have equal probabilities) , so if you see a red side then the other side is more likely than not red.

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It may be useful to explain joriki’s comment. The other answers are all based on the assumption that the $16$ disks are equally likely to be chosen, and that once a disk has been chosen, you’re equally likely to see each side. Another way to say this is that you’re equally likely to see any of the $32$ sides. This is almost certainly the intended interpretation, but it’s not the only possible one, and the answer really does depend on the interpretation of We are given a disk at random, and see that one of its sides is red.

Suppose that the person drawing the disk at random has decided ahead of time to show you a red side if the disk has one. Then you will see a red side if and only if one of the red/red or red/blue disks was drawn. Each of these $11$ disks is equally likely to be the one that you’re shown, so on this interpretation the probability that the other side is blue is $\frac6{11}>\frac12$.

At the other extreme, the person drawing the disk at random might decide to show you a red side only if there’s no alternative. In that case the probability that the other side is blue, given that you’re shown a red side, is $0$: you must be looking at one of the red/red disks.

As you can see, the different possible interpretations affect the reduced sample space implied by your seeing a red side and consequently affect the probability.

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Let $A$ be the random variable which can take on three values : red/red, red/blue, blue/blue (it evaluates to the pair of colors at each side of the disk that I am given). Let $B$ be the random variable that can take on two values : red or blue, and $B$ is the color of the face of the random disk that I see. Therefore,

\begin{gather*} P(A = \text{red/red}) = 5/16, \quad P(A = \text{red/blue}) = 6/16, \quad P(A = \text{blue/blue}) = 5/16 \\ P(B = \text{red}) = 16/32 = 1/2, \quad P(B = \text{blue}) = 16/32 = 1/2. \end{gather*}

Now the probability we wish to compute is $P(A = \text{red/red} \, | \, B = \text{red})$, and elementary probability theory shows that this probability is just $$P(A = \text{red/red} \cap B = \text{red}) / P(B = \text{red}),$$ and the probability on the numerator is just $P(A = \text{red/red}) = 5/16$, so that you expect a red/red disk given a red face with probability $10/16 > 1/2$. The red/red disk is therefore more probable than the red/blue disk, given a disk with a red side.

Hope that helps,

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The probability of $A$ given $B$ is:

$$P(A|B) = \frac{P(A \& B)}{P(B)}$$

Here, $A$ is "the other side of the disk is red" and $B$ is "the side that we can see is red".

$P(A\&B)$ = no. of red disks / no. of disks = 5/16
$P(B)$ = no. of red sides / no. of sides = 16/32

So the probability that the other side is red is $\frac{5/16}{16/32} = 5/8$.

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One could use Baye's Theorem here (though, I think the "reduced sample space" approaches of the other answers are slicker).

Let $RB$ be the event that the red/blue card was chosen and let $A$ be the event that the observed side was red.

We will find $P(RB\mid A)$. By Baye's Theorem: \eqalign{ P(RB\mid A) &={P(A\mid RB) P(RB)\over P(A\mid RB) P(RB)+P(A\mid RB^C) P(RB^C) }\cr &={ (1/2)(6/16)\over (1/2)(6/16)+ (1/2)(10/16) }\cr &={6/16\over16/16 }\cr &=3/8. } ($P(A\mid RB^C)=1/2$, since given that we did not pick the red/blue card, we picked one of the remaining 10 cards, five of which are red and the other five blue).

So the probability that we picked the red/blue card given that the observed side was red is $3/8<1/2$. Thus, this is the less likely outcome.

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there are 5 red discs with red on the other side ... and 6 red discs with blue on the other side ... makes sense that you are more likely to find blue on the other side ... unless you lucky at blackjack

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