1
$\begingroup$

Sorry if this is a n00b question, I'm just trying to wrap my head around the problem. I am trying to refute conventional wisdom here, so any help is greatly appreciated.

And now for the question:

For example, I flip a coin 100 times and I end up with 50 consecutive heads followed by 50 consecutive tails. How would I calculate the probability or likelihood that the next time I replay the scenario that I will end up with the exact same result? Would the equation be different if we were using a ternary or quaternary event instead of a binary event?

Conventional, non-statistical, wisdom would assume that coming to the same result on consecutive attempts would be incredibly remote, but that the odds of eventually getting an identical result would increase as more attempts are made, i.e. 50/50, 50/50 would be implausible, but 50/50, 49/51, 51/49 ... 50/50 would be plausible.

$\endgroup$
2
$\begingroup$

You have some event with certain probability of success. So the probability that the next time this event happens is the success probability. In your case you have flipped coin 100 times with 50 consecutive heads and 50 consecutive tails. The probability of this event happening is $1/2^{100}$. If you repeat this experiment independent of the first one, the probability that it will happen remains the same. Although you generated 100 random Bernoulli variables with probability $p=1/2$, your random variable of interest is a Bernoulli with probability $p=1/2^{100}$.

The probability that that eventually you get the same result increases with number of trials. This number of trials needed for success is random variable with geometric distribution.

If you use ternary or quartenary event the same holds, as long as your a looking for specific event to happen. Then calculate probability of this event and again you have a Bernoulli random variable with this probability.

$\endgroup$
3
$\begingroup$

The probability of 50 tails followed by 50 heads, assuming balanced iid bernoulli coins, is simply $\prod_{i=1}^{100} P(Event_i=Predicted_i)=\frac 1 2 ^{50} \times \frac 1 2 ^{50} = 2^{-100}$.

50/50 is just as unlikely as 49/51, if order must be preserved.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.