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I have two very biased coins:

  1. The first coin (C1) lands on heads 95% of the time.
  2. The second coin (C2) lands on heads only 1% of the time.

One of the coins has been selected and I want to determine which coin that is. I can toss the coin as many times as possible and record a series of heads and tails results.

Let's say I want to know with 99.99% confidence, and as soon as possible, which coin is being flipped. How can I do so?

I would like to flip the coin, and after each result either continue or make a decision on which coin I have. I don't need to know how many times to flip before I start.

(Sorry if this is a very basic question that has been asked before. I lack the vocab to know how to search for this problem).

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  • $\begingroup$ I don't have time to answer, but this is an awesome question. $\endgroup$
    – Cam.Davidson.Pilon
    Sep 25, 2020 at 1:15
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    $\begingroup$ Sounds like a task for a sequential probability ratio test which is known to have (on average) the smallest number of trials to reach a decision. $\endgroup$
    – Dilip Sarwate
    Sep 25, 2020 at 1:50
  • $\begingroup$ This feels like a homework question. Is this self-study? $\endgroup$
    – EngrStudent
    Sep 25, 2020 at 2:02
  • $\begingroup$ It's not a homework question. It is related to my work but rewritten using coins. $\endgroup$
    – WW.
    Sep 25, 2020 at 2:54
  • $\begingroup$ There are two kinds of valid answers to this question, depending on what you mean by "as soon as possible:" would that mean (1) you wish to determine a minimal but fixed sample size that will produce a correct decision with a chance of 99.99% or (2) you wish to have a sequential decision procedure that will tell you when to stop flipping, as well as tell you which coin it is with 99.99% accuracy, that yields a minimal expected number of flips? The latter value is going to be less than the former, but in (2) you risk having to flip a long time in rare cases. Which is it? $\endgroup$
    – whuber ♦
    Sep 25, 2020 at 14:11

1 Answer 1

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Say the odds of getting $C_1$ over $C_2$ are, in principle, $1:1$. Then, you flip the coin $n$ times and get $x$ heads. If we call the probabilities of heads $p_1=0.95$ and $p_2=0.01$, then the probability that each coin gives $x$ heads is:

$$P(x|C_1)=p_1^x(1-p_1)^{n-x}$$

$$P(x|C_2)=p_2^x(1-p_2)^{n-x}$$

If we use Bayes theorem, we get the odds that the coin is $C_1$ including the information from the coin tosses:

$$\begin{align} \frac{P(C_1|x)}{P(C_2|x)}&=\frac{P(C_1)\cdot p_1^x(1-p_1)^{n-x}}{P(C_2)\cdot p_2^x(1-p_2)^{n-x}}\\ &=\frac{p_1^x(1-p_1)^{n-x}}{p_2^x(1-p_2)^{n-x}}\\ &=\left(\frac{p_1}{p_2}\right)^x\left(\frac{1-p_1}{1-p_2}\right)^{n-x} \end{align}$$

For example, if you tossed the coin 3 times and had 2 heads, you would have:

$$\frac{P(C_1|x)}{P(C_2|x)}=\left(\frac{0.95}{0.01}\right)^{2}\left(\frac{0.05}{0.99}\right)^{1}\approx456$$

That means the odds that the coin is $C_1$ are $456:1$, which is equivalent to a probability of $\frac{456}{456+1}\approx99.8\%$. That looks like a huge value for such a low number of trials, and the reason is that the probabilities $95\%$ and $1\%$ are very different. If the coins had closer probabilities, the odds would not be so dramatic.

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  • $\begingroup$ Thank you for this answer. It has made me realise that a more accurate description of my problem includes that C1 is much more common than C2. So we could say I have a bag of 100 x C1 and 1 x C2 and I want to detect the C2 coin. $\endgroup$
    – WW.
    Sep 25, 2020 at 3:15
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    $\begingroup$ @WW In that case P(C1) and P(C2) change, but the approach remains the same. $\endgroup$
    – Demetri Pananos
    Sep 25, 2020 at 4:51
  • $\begingroup$ How does Bayes' theorem apply here, since there is no prior distribution?? $\endgroup$
    – whuber ♦
    Sep 25, 2020 at 14:08
  • $\begingroup$ @whuber there is a prior distribution, given by $(P(C_1),P(C_2))$. I assumed the prior odds were 1:1, therefore they cancelled when finding the posterior odds. It was a rather arbitrary assumption though, and the OC's comment indicates that it would be better to use 100:1 prior odds $\endgroup$
    – PedroSebe
    Sep 25, 2020 at 17:43
  • $\begingroup$ There is no prior distribution mentioned in the question. Indeed, it's not necessarily the case that there is any relevant distribution. $\endgroup$
    – whuber ♦
    Sep 25, 2020 at 18:33

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