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We are given a coin of unknown biasing, and allowed to perform experiments with it. Say we have the following series of outcomes of 20 Bernoulli trials, represented by $X_i, \ i \in \mathbb{N}$ : $$ T, T, T, H, H, H, H, H, H, H, T, H, H, T, T, T, H, H, H, T $$ There are $12 \ H \equiv H_{12}$ and $8 \ T \equiv T_8$.

While I don't have the complete derivation, I know that (and have cross-checked my intermediate form), $$ P\left( X_{21} = T \right) = \frac{8 + 1}{12+8+2} = \frac{9}{22} $$ Then what is the probability that $P\left( \left( X_{21} = T \right) \cap \left( X_{22} = T \right) \right) = P(S)$ (suppose) ?

Clearly $$ P(S) = P\left( X_{21} = T \right) \cdot P \left( X_{22} = T \right) = \frac{9}{22} \cdot P \left( X_{22} = T \right) $$ But what is $P \left( X_{22} = T \right)$? Is it the same as $P\left( X_{21} = T \right)$, as $X_{21}$ hasn't actually happened? Or do we re-calibrate given the new outcome of $T$?

So is $$ P(S) = {\left( \frac{9}{22} \right)}^2 = \frac{81}{484} \text{ or } P(S) = \frac{9}{22} \cdot \frac{10}{23}= \frac{90}{506} $$

Varad Mahashabde
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  • Given only this information, shouldn’t $P(T)$ be $\frac{8}{20}$? – Vishu Apr 23 '20 at 19:30
  • @Tavish Because $\frac{8}{20}$ is the most likely probability of $P(T)$, doesn't mean that $P(X_{next} = T)$ is the same. Other probabilities can also give this distribution even if it is the most likely for $\frac{8}{20}$. Accounting for those gives the formula-esque thing shown in the question. 3blue1brown has an (ongoing series)[https://youtu.be/8idr1WZ1A7Q] about deriving on YouTube. (1/2) – Varad Mahashabde Apr 23 '20 at 19:51
  • @Tavish I tried deriving it myself, but got stuck on an integration step. If you want pointers: Representing the a distribution of $m$✔ and $n$❌ as $(m,n)$, $P(T | (m,n)) = \sum_{0 \leq x \leq 1} x \cdot P(P(T) = x | (m,n))$, and then put a Bayes Uno Reverse card on it and move to integration (2/2) – Varad Mahashabde Apr 23 '20 at 20:01

1 Answers1

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$P(X_{21}=T)=\frac9{22}$ results from applying the rule of succession. At How to prove the rule of succession without calculus? I gave a combinatorial proof that doesn’t require integration. We can apply the same idea to derive $P(X_{21}=T\cap X_{22}=T)$.

Assuming a flat prior for the bias of the coin, we draw one number uniformly randomly from $[0,1]$ for the probability $p$ that a toss yields heads and then $22$ further numbers $u_i$ for the $22$ tosses, with toss $i$ showing heads exactly if $u_i\lt p$. The fact that the first $20$ tosses yielded $12$ heads and $8$ tails means that the rank of $p$ among $p$ and the first $20$ $u_i$ is $9$. If we now draw the next number for the $21$st toss, it’s equally likely to be inserted into the order of those $21$ numbers in any of the $22$ possible slots, $9$ of which are above $p$, whence $P(X_{21}=T)=\frac9{22}$. If it’s inserted above $p$ and we insert another one, there are now $23$ slots and $10$ of them are above $p$, so the probability that we get yet another tails is $P(X_{22}=T\mid X_{21}=T)=\frac{10}{23}$, and thus

$$ P(X_{21}=T\cap X_{22}=T)=P(X_{21}=T)P(X_{22}=T\mid X_{21}=T)=\frac9{22}\cdot\frac{10}{23}=\frac{45}{253}\;. $$

The equation that follows “Clearly” in the question is wrong; the events $X_{21}=T$ and $X_{22}=T$ aren’t independent, so you can’t just multiply their probabilities.

joriki
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