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Is it possible to toss a coin $100$ times and get the same result every time?

I would say it is mathematically possible but in reality it is not because one would need to an infinite amount of time

Let me know what you think :)

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    Why would it need an infinite amount of time? And yes, it is possible. – 2017-01-29
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    I don't think this question is really about mathematics. Also, one does not need a infinite amount of time. – 2017-01-29
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    To specify precisely you need to specify something like a "fair coin" rather than just a coin. Consider a coin with a head on both sides ... – 2017-01-29
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    I'm voting to close this question as off-topic because mathematics is not determined by an opinion poll. – 2017-02-01

3 Answers 3

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It is physically possible in a finite time. The probability that it comes out heads or tails is $\frac{1}{2}$ each time. Note that the first outcome is not defined to be heads or tails. Therefore, the probability of getting 100 identical tosses is: $$\frac{1}{2}\times \left(\frac{1}{2}\right)^{99}+\frac{1}{2}\times \left(\frac{1}{2}\right)^{99}=\left(\frac{1}{2}\right)^{99}$$ Therefore, the probability is approximately: $$\approx 1.5777218\times10^{-30}\approx 1.5777218\times10^{-28}\text{ %}$$ From this, we deduce that the expected number of attempts is: $$\approx 6.338253\times 10^{29} \text{ attempts}$$

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    +1 for both your answer and for having scooped out my error :D – 2017-01-29
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    That is what I meant by mathematically possible but physically impossible. In reality your number of attempts starts at t=0 but what if somebody comes at time t and start watching you tossing , by his calculation you will finish earlier than he expected . The order in the probability theory doesn't apply in reality . – 2017-01-29
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One would need to try, on average, about $10^{30}$ times, but you would most likely succeed within a finite abount of time. That being said, it is mathematically possible for you to never succeed, but that probability is vanishingly small, even compared to the probability of getting it right on the first attempt.

In fact, the probability of never succeeding no matter how long you try is more comparable to the probability of throwing a coin indefinitely and getting alternate heads and tails on every throw forever.

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The probability is $0.5$ each time. What do you get?

$$0.5\times 0.5\times 0.5\times \ldots$$

That is

$$\frac{1}{2^{100}} = \frac{1}{1267650600228229401496703205376}$$

That means that it does happen on average once every $10^{30}$ tosses.

Considering that you want to test it: suppose you can too a coin every $2$ seconds (optimistic hypothesis), you'd need then ish $5\cdot 10^{29}$ seconds, that is $10^{22}$ YEARS.

Not an infinite amount of time, but.. well you know.

Edit

This solution is forced: it shall be $2^{-99}$, as Projectilemotion pointed out. In my version it's forced since the very first toss. For example if I do want $100$ heads, and not just $100$ identical tosses.

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    Are you sure about $\frac{1}{2^{100}}$? The first outcome can be both heads or tails. – 2017-01-29
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    @projectilemotion That is indeed right, the first outcome doesn't matter :D Let's say that in my solution I did force the very first outcome too. For example it's correct if I wanted $100$ heads! Thanks anyway! – 2017-01-29
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    You're welcome. Yes, indeed it would be correct if asked for strictly $100$ heads in a row. – 2017-01-29