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One guy tosses a coin infinite times. The coin is fair, so chances of seeing a head (H) or a tail (T) are equal. What is the probability that he observes a TT before a HT?

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closed as off-topic by whuber Jul 6 '18 at 12:45

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Wouldn't the probability of observing TT before HT be 50%?

Since the tosses are iid and the coin is fair, prob(TT) = .5*.5=.25 and prob(HT) = .5*.5=.25

You can think of the infinite sequence of coin tosses as an infinite sequence of pairs of coin tosses, where each of the 4 possible pairs (TT, TH, HT, HH) occurs with equal probability. So then it comes down to what's the probability of observing one of these 4 pairs before another, and since they are equally likely, it should be 50%.

Thoughts?

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  • $\begingroup$ The reasoning is incorrect and so is the answer. Take a look at some sequences. Notice (as pointed out in a comment to the question) that unless the first two tosses are TT, you're inevitably going to get an HT before a TT appears, so the correct answer must be $1/4.$ $\endgroup$ – whuber Jul 6 '18 at 12:39

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