Two men consequently flip a «fair» coin (possible outcomes are Head and Tail).
A winner is defined to be a player who produces a Head first.
What is the probability that the first player (he begins a game) will be a winner? Let X be a random variable which shows a number of flips in the series $TTT…H$
which describes one session of this game. What is the expectation E(X)?
my solution for first :
$$p=\frac{1}{2}+\frac{1}{2}(1-p).$$ $p=2/3$.
what about second part ?
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Between this and your prior question it looks like you are just posting your homework for us to do for you. This question is a routine application of the Geometric Distribution – lulu Dec 16 '19 at 19:25
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not homework and improve my skills and try for some problems – Dec 16 '19 at 19:28
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thoese problems in exam from universities lasts years – Dec 16 '19 at 19:30
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For the second part the reasoning is similar to that of the first part.
Let $E$ be the expected number of throws.
$$E=\frac12+\frac12(E+1),$$
since after the first throw either you stop (head, probability $1/2$) or you continue (tail, probability $1/2$) and in this second case your new expected number is $E+1$.
Thus $E=2$.
Alberto Saracco
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