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Sorry for my lack of knowledge in math, I just thought this was the best place to ask despite my very little knowledge since there is people here that know a lot. So from my understanding, if a coin was flipped in a completely random manner, there is $50\%$ chances of getting heads, or tails. So if a coin was flipped $2$ times, in $3$ out of $4$ scenarios, the result would not be $2$ times heads. Therefore the chances of something happening $n$ times in a row are $\frac{1}{2^n}$? So from that logic, if someone where to bet 10 dollars on tails every time the coin flipped heads n times in a row, would that person win the bet $(100 - (\frac{1}{2^n})\%$ of the times? Or at least win more times than lose?

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The title does not correspond to the text of the post. –  Did Dec 1 '12 at 13:23
For some basic information about writing math at this site see e.g. here, here, here and here. –  Julian Kuelshammer Dec 1 '12 at 13:52
Yes the chance of coin landing heads $n$ times in a row is $1/2^n$. But once it has already happened, it says nothing about the next toss, it's likely to be heads or tails with equal probability 50%. This has been answered on this site before; coins have no memory. (Also, probability of "n+1 heads" is same as "n heads followed by a tail", both of them are $1/2^{n+1} = (1/2^n)(1/2)$.) –  ShreevatsaR Dec 1 '12 at 14:13
Ok I understand. But: If a coin was flipped $\infty$ times, then that means $\frac {1}{2} \infty$ of the times it'd land on heads, and $\frac {1}{2} \infty$ it'd land on tails, but lets say you flip a coin 1 time and get heads, then flip it $\infty$ times, that'd mean its more likely to get heads than tails cause $\frac{1}{2} \infty + 1 \gt \frac{1}{2} \infty$? –  Mr D Dec 1 '12 at 14:46
@André Nicolas, ok I see the difference, thanks! If you care to write an answer I'll choose it as the correct one :) In case you want the points I mean. –  Mr D Dec 1 '12 at 19:01
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