bayes theorem probability Jar [closed]

I am trying to come up with different theoretical answers below.

I believe the standard one is based on Bayes theorem, but I am struggling to prove it.

A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pick a coin at random, and toss it 10 times. Given that you see 10 heads, what is the probability that the next toss of that coin is also a heads? Prove it.

Does anyone have any thoughts?

closed as off-topic by LocalVolatility, skoestlmeier, Attack68♦, Lliane, olaker♦Mar 27 at 13:38

• This question does not appear to be about quantitative finance within the scope defined in the help center.
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• I'm voting to close this question as off-topic because its not about quant finance. – LocalVolatility Mar 22 at 20:12

Maybe they expected you to reason along these lines:

$$\frac{1}{2^{10}}= \frac{1}{1024} \approx \frac{1}{1000}.$$

So the odds of fair coin vs non fair coin are about even and the probability of heads is approximately

$$0.5 \times 0.5 + 0.5 \times 1 = 0.75.$$

The above argument can be made more precise, let $$u = \frac{999}{1000}\frac{1}{2^{10}}$$ be the probability that an unbiased coin is selected and 10 heads are observed, $$b = \frac{1}{1000}$$ the probability the biased coin is selected and $$p = u + b$$. Either $$u$$ or $$b$$ happened so the probability of heads on the next throw is

$$P(\textrm{Next throw is heads}) = \frac{1}{2} \frac{u}{p} + \frac{b}{p}.$$

You can verify that $$u \approx b$$ and thus the back of the envelop calculation is not far of and that this leads to the same fraction as the other answers.

• naivy, the expected probabiliyy should be 0.5 adjusted for the bayes factor that there is 0.001 chance the coin is double headed. 0.75 probability looks too high – Kiann Mar 22 at 20:23
• You need to condition on the fact that 10 heads in a row have been observed. Either you selected the one biased coin (prob. $\frac{1}{1000}$) or a unbiased coin gives you 10 heads in a row (prob. $\frac{999}{1000}\frac{1}{2^{10}})$. Both are almost equally unlikely. – Bob Jansen Mar 23 at 9:57

Let $$X_i$$ be the result, where $$X_i=1$$ implies heads and $$X_i=0$$ as tails.

Let $$\theta_j\in\{0.5,1\}$$, where $$\theta_j$$ is the bias for heads.

$$\theta_1=.5$$ and $$\theta_2=1$$.

$$\Pr(\theta_1|X_{1\dots{10}}=1)\propto{\left(\frac{1}{2}\right)^{10}}\frac{999}{1000}=\frac{999}{1000\times{1024}}.$$

$$\Pr(\theta_2|X_{1\dots{10}}=1)\propto{1^{10}}\frac{1}{1000}=\frac{1024}{1000\times{1024}}.$$

$$\Pr(X_{1\dots{10}}=1)=\frac{999+1024}{1000\times{1024}}$$

$$\Pr(\theta_1|X_{1\dots{10}}=1)=\frac{999}{999+1024}=\frac{999}{2023}$$

$$\Pr(\theta_2|X_{1\dots{10}}=1)=\frac{1024}{999+1024}=\frac{1024}{2023}$$

$$\Pr(X_{11}=1|X_{1\dots{10}}=1)=\sum_{j=1}^2\left[\theta_j(1-\theta_j)\right]\Pr(\theta_j|X_{1\dots{10}}=1)$$

$$\Pr(X_{11}=1|X_{1\dots{10}}=1)=\frac{1}{2}\frac{999}{2023}+1\frac{1024\times{2}}{2\times{2023}}=\frac{3047}{4046}\approx{.75}$$

I debated answering this question as it could be viewed as more appropriate for Cross Validated or Mathematics, however, I decided to do so for a couple of reasons directly related to QF.

First, quantitative finance is calculated gambling. Bayesian statistics are coherent. Frequentist statistics are incoherent. A statistic is considered coherent if a fair gamble can be created from it. It vastly exceeds the scope of your question, but if you are pricing a loan or an option then it is technically incorrect to use a Frequentist method, at least for a financial intermediary.

The second reason is that this problem is a discrete form of a real finance problem. Given an unknown parameter and a historical record, what is the probability of a future state of the world?

You need to get a very good grasp on the Bayesian prior distribution, the Bayesian posterior distribution and the Bayesian posterior predictive distribution.

Let (X,Y) the condition the coin is (fair, double headed):

$$P(Y | 10H) = P(10H | Y) \frac{P(Y)}{P(10H)}$$

where

$$P(10H|Y) = 1$$
$$P(Y) = \frac{1}{1000}$$
$$P(10H) = P(10H|X)P(X) + P(10H|Y)P(Y) = 0.5^{10}\frac{999}{1000} + \frac{1}{1000} = \frac{999}{1000*1024} + \frac{1}{1000} = \frac{2023}{1000*1024}$$

$$P(Y | 10H) = 1 * \frac{1}{1000} * \frac{1024*1000}{2023} = \frac{1024}{2023}$$
$$P(X | 10H) = \frac{999}{2023}$$

So the probability of the next coin toss being heads is:

$$0.5 * P(X|10H) + 1 * P(Y | 10H) = 0.5*\frac{999}{2023} + 1 * \frac{1024}{2023} = \frac{3047}{4046}$$