# Coin tossing problem

My teacher gave us an homework. I solved it, but I don't think I have the right answer.

PROBLEM

We have three coins identical in appearance.

• Coin A falls on tails and heads with equal probability
• Coin B falls twice as much on tails as heads
• Coin C always falls on tails

We choose a coin at random and toss it. It falls on tails.

What is the probability to get tails on the next toss, if we toss the same coin?

MY TRY

Soient les événements

A = "choose coin A" = {t, h}

B = "choose coin B" = {t, t, h}

C = "choose coin C" = {t}

E, get tails at the second throw


We are looking for $P(E)$. Knowing $P(E|A)=1/2$, $P(E|B) = 2/3$ and $P(E|C)=1$.

\begin{split} P(E)& = P(E|A)P(A) + P(E|B)P(B) + P(E|C)P(C)\\ & = \frac{1}{2}\cdot\frac{1}{3} + \frac{2}{3}\cdot\frac{1}{3} + 1 \cdot\frac{1}{3}\\ & = 13/18 = 0,7 \overline{2} \end{split}

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This doesn't make any use of the fact that you got tails on the first throw. You were probably meant to use Bayes' Theorem to compute $P(A\mid T)$, $P(B\mid T)$, and $P(C\mid T)$, where $T$ is the event that you got tails on the first throw. Then you can compute $P(E\mid T)$ using these probabilities. – David Moews Feb 13 '13 at 23:06
$P(A|T) = \frac{P(T|A) P(A)}{P(T)}$, where $P(A) = 1/2$, $P(A)=1/3$ and $P(T) = 2/3$, is that right? – Justin D. Feb 13 '13 at 23:14
Ok, I was wrong for $P(T)$ – Justin D. Feb 13 '13 at 23:17

What you have to do in this case is update your priors according to your observation. Your first prior distribution for selecting a coin was (I assume) $\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$.

Now that you've seen tails, the probabilities for the coin you're holding change; using Bayes' theorem: $$P(A|tails) = P(tails|A)\frac{P(A)}{P(tails)} = \frac{3}{13}$$ where $P(A)=\frac{1}{3}$, $P(tails)=\frac{13}{18}$ as you calculated, and $P(tails|A) = \frac{1}{2}$

Similarly, you can get $$P(B|tails) = P(tails|B)\frac{P(B)}{P(tails)} = \frac{4}{13}$$ $$P(C|tails) = P(tails|C)\frac{P(C)}{P(tails)} = \frac{6}{13}$$

So your new priors are $\left(\frac{3}{13},\frac{4}{13},\frac{6}{13}\right)$, and $$P(tails|new\ priors) = \frac{1}{2}\cdot\frac{3}{13} + \frac{2}{3}\cdot\frac{4}{13} + 1\cdot\frac{6}{13} = \frac{61}{78}$$

(up to possible calculation errors)

And if you see only more $tails$ in the future, you'll have better and better reason to believe you picked coin $C$, and your probability for the next $tails$ will tend to $1$.

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In the last step, arent you multiplying each $P(C|T)$ by the associated $P(T|C)$? Then adding them? What does that yield? – CogitoErgoCogitoSum Feb 13 '13 at 23:39
$P(X|T)$ is the new probability for having selected coin $X$ after having seen $tails$, so this is just the law of total probability, just like OP did to compute $P(T)$. – Alfonso Fernandez Feb 13 '13 at 23:47
I feel as though you are changing the meaning of the variable to justify the operation. I know that $P(X|T)$ is, I dont need that explained. I need you to explain why you multiply $P(X|T)$ by $P(T|X)$. – CogitoErgoCogitoSum Feb 13 '13 at 23:54
$P(T) = \sum P(T|X)P(X)$ but we are given that the flip yielded tails, so we limit ourselves to a world where tails was yielded. With slight abuse of notation this is $P(T|T) = \sum P(T|X,T)P(X|T)$, where $P(T|X,T)$ represents the probability of getting tails given that we're using coin $X$ and that we got tails last time; the probability of coin $X$ yielding tails is independent of previous observations, so $P(T|X,T)=P(T|X)$, and $P(T|T) = \sum P(T|X)P(X|T)$. – Alfonso Fernandez Feb 14 '13 at 2:08

I guess it is 61/78.

From the probabilities you get a total of 13/18 for a tail answer, but the dependent source is

• A: (3/13)*(3/6))
• B: (4/13)*(4/6)
• C: (6/13)*(6/6)

which is (9+16+36)/(13*6) = 61/78 which is about 78%.

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Well, there's your problem. Its in French.

$P(T|A) = 1/2, P(T|B) = 2/3, P(T|C) = 1/1$

$P(A) = P(B) = P(C) = 1/3$

$P(TA) = P(T|A) P(A) = 1/6$

$P(TB) = P(T|B) P(B) = 2/9$

$P(TC) = P(T|C) P(C) = 1/3$

$P(TA \lor TB \lor TC) = P(TA) + P(TB) + P(TC) = 1/6 + 2/9 + 1/3 = 13/18$

$P(T(A \lor B \lor C)) = P(T) = 13/18$

So when you randomly pick the first coin and then flip it, you stand a 13/18 chance of getting Tails.

This does not necessarily mean that if you throw the same coin a second time the probability is the same. The reason is because you are not randomly picking a (new) coin. These probabilities must get factored out of consideration.

We need to be careful because you - or me - stand a very good chance of the Monty Hall fallacy. I'm probably going to be guilty here in a moment even though I recognize it is a risk.

However, getting a Tails on this flip does not provide us with any new information and we are left not knowing what coin we have - this is effectively the same scenario than randomly picking a new coin: Total uncertainty.

If we kept this one coin for the long haul and tallied up results, we can use frequentist reasoning to deduce which coin we have. If we get a Heads, we know its not coin C and this is new information.

So Im inclined to agree with your reasoning. 13/18 is the probability.

Additional: Im trying to understand the reasoning of the other answers. I thin the reasoning is that they are reversing the conditional probabilities using Bayes Theorem.

$P(TA) / P(T) = P(A|T) = (1/6)/(13/18) = 3/13$ $P(TB) / P(T) = P(B|T) = (2/9)/(13/18) = 4/13$ $P(TC) / P(T) = P(C|T) = (1/3)/(13/18) = 6/13$

Im not sure what the next step ought to be then. We now know that the conditional probability of each coin is given that we got a tails. I dont know what to do with this information though. Multiplying each of these by $P(T)$ is redundant information. Someone multiplied each $P(C|T)$ by the associated $P(T|C)$ and Im not sure what that does for us.

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You are indeed committing the fallacy. For example, getting $tails$ makes it much more likely to have selected coin $C$ - you can see this if you observe that the probability of having selected coin $C$ given that you got $heads$ is $0$, and because it is selected with probability $\frac{1}{3}$, all of those cases "go to" $tails$, while only some of the cases having selected another coin "go to" $tails$. – Alfonso Fernandez Feb 13 '13 at 23:25
But the point is, you do not know whether you have coin C or not. Sticking with the same coin... or randomly changing coins... does not provide you with any new information. The fact that you got a Tails - as opposed to a Heads - does not contribute to your knowledge either. You had to get either one or the other since its mutually exclusive... so how does getting either one change the probabilities? I dont see it. – CogitoErgoCogitoSum Feb 13 '13 at 23:33
Let's call the old (fair) priors $v$, the new priors having seen $heads$ $u$, and the new priors after having seen $tails$ $w$. You agree that $v\ne u$, but claim that $v=w$. if $p$ is the probability of seeing $heads$, then $v=pu+(1-p)w$, because in the random case no information is added. but if $v=w$, then $v=pu+(1-p)v$ implies $u=v$. – Alfonso Fernandez Feb 13 '13 at 23:44
If you tossed the coin a hundred times and got a hundred tails, you would probably conclude that you had coin $C$. Then, you got some information from those hundred flips. So, you should also get some information from just one flip, although not as much as from $100$ flips. – David Moews Feb 14 '13 at 1:01
Alternately for the OP, if one can work out chances of getting a TT (61/108) and TH (17/108). Now given that T has happened, these are the only possibilities, and out of 108 tosses, 61 times TT would happen, 17 times TH would happen, and the remaining tosses are ruled out because of the first "T". So the chances of the second tail are 61/(61+17) = 61/78, which is of course Bayes reasoning. – Macavity Feb 14 '13 at 1:51

The key point to consider here is that you now have new evidence that favours a certain hypothesis, and less others. Here is a write up on dealing with exactly such scenarios.

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