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Suppose you have two coins: one is a normal coin (with one "heads" side and one "tails" side), and the other is a modified coin with two "heads" sides. The coins are indistinguishable apart from looking at each side.

If you chose one coin at random and, without looking at it, flipped it, there are two "possibilities". If you see a "tails" side, then you know you picked the normal coin. However, if you see a "heads" side, then you have no way of knowing whether you picked the normal coin or the modified coin, and you basically have a 50% chance of correctly guessing your coin. Thus, the certainty of your choice is 100% if you see "tails" and 50% if you see "heads".

However, what if you repeated the process multiple times? You keep your coin and continue to flip it. If you see "tails" even once, then you can stop because you now know which coin you have with 100% certainty. However, if you continue to see "heads" after each flip, you would begin to grow more certain that you have the modified coin. Of course, there is also the chance that you're really unlucky, and you have the normal coin but have just never flipped a "tails". The probability of this occurring would surely decrease after each flip, but this leads to my question.

How do multiple trials affect one's certainty of their choice? For example, you observe "heads" on your first flip. You are 50% certain that you have the modified coin. You flip "heads" again. You are 50 + $x$% certain that you have the modified coin. And so on, and so forth.

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Let $A$ denote the event of choosing the unfair coin and $B$ the event that the first toss is a head, then you have $$\mathbb{P}(A \mid B ) =\frac{\mathbb{P}(B \cap A)}{\mathbb{P}(A)}= \frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$$ and $$\mathbb{P}(A^C \mid B) = \frac{ \mathbb{P}(A^C \cap B)}{\mathbb{P}(A)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$$, provided that you choose each coin with probability $\frac{1}{2}$.
Now with $B_n$ tossing $n$ tails in the first $n$ tosses.

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