# Probability of Correctly Guessing Coin Toss when Changing the Guess

Generally when tossing a coin $n$ times, the probability of getting heads or tails are both $\frac{1}{2}$ for each toss, so if you were to guess heads for every toss, you should expect to get $\frac{1}{2}$ of the tosses correct.

Now I also understand that for $n$ tosses to consecutively be all heads, the probability would be $(\frac{1}{2})^n$.

With this in mind, if you were trying to maximize the amount of correct guesses for $n$ coin tosses, should you be changing your guess after each successful prediction?

It would seem to me that after you successfully guess the toss, let's say for example your guess was heads, the probability of it being heads again on the next toss is now $(\frac{1}{2})^2$, right? So intuitively it would seem like changing your guess to tails in this case would yield a higher success rate for the next toss and so on.

Does this logic stand? I'm trying to figure out a way to prove mathematically that this would work (or doesn't work) and I'm coming up short. Any help in trying to figure this out would be greatly appreciated.

• To avoid a biased coin you simply throw on another "rng", like a clock or if you are currently breathing in or out. Apart from that, if you have a fair coin you can safely "guess" all heads and have the same EV as if you guess randomly. See en.wikipedia.org/wiki/Gambler%27s_fallacy – SAJW Oct 9 '16 at 6:55
• @saturatedexpo so it's best to treat each toss as independent? – Nick Zuber Oct 9 '16 at 6:57
• yes, in that case. There are cases where the tosses are not independant. Such as bad implementation of the rng. (like building a machine that replicates tosses, which has been done) – SAJW Oct 9 '16 at 6:59

No that is incorrect. The probability of the second toss does not depend on the outcome of the first toss, therefore it is not a conditional probability problem. So whether you were right or wrong on the first toss the probability of the second toss is ($\frac{1}{2}$). If you were to guess the next two tosses then the probability of getting them both right would be $(\frac{1}{2})^2$ no matter what your guess was for each toss, same goes for the nth case.