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$\begingroup$ "One guy tosses a coin infinite times." Not in this universe… $\endgroup$– AlexisNov 22, 2017 at 22:15
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2$\begingroup$ This 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 $\endgroup$– Łukasz GradNov 22, 2017 at 22:20
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2$\begingroup$ The techniques in the very closely related thread at stats.stackexchange.com/questions/12174/… will make short work of this question: take a look! $\endgroup$– whuber ♦Nov 23, 2017 at 0:55
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$\begingroup$ Work for some subject? $\endgroup$– Glen_bNov 23, 2017 at 3:08
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$\begingroup$ Closely related: stats.stackexchange.com/questions/12174/… and stats.stackexchange.com/questions/305699/…. $\endgroup$– whuber ♦Jul 6, 2018 at 12:39
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1 Answer
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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, 2018 at 12:39