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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3"One guy tosses a coin infinite times." Not in this universe… – Alexis Nov 22 '17 at 22:15
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2This usually is solved with Markov Chains but in this case it's easy, note that if you ever roll an H then you will never observe TT before HT – Łukasz Grad Nov 22 '17 at 22:20
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2The techniques in the very closely related thread at https://stats.stackexchange.com/questions/12174/time-taken-to-hit-a-pattern-of-heads-and-tails-in-a-series-of-coin-tosses/12178#12178 will make short work of this question: take a look! – whuber Nov 23 '17 at 00:55
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Work for some subject? – Glen_b Nov 23 '17 at 03:08
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Closely related: https://stats.stackexchange.com/questions/12174/time-taken-to-hit-a-pattern-of-heads-and-tails-in-a-series-of-coin-tosses/12178#12178 and https://stats.stackexchange.com/questions/305699/occurrence-of-at-least-1-ht-and-hh-in-sequences-of-4-coin-flips-not-equally-like. – whuber Jul 06 '18 at 12:39
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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?
Amazonian
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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 $1/4.$ – whuber Jul 06 '18 at 12:39