Probability versus intuition problem: coins You are playing with some friends this summer. The game is simple: a fair coin is picked and tossed. Before each toss, you and your friends bet on either heads or tails coming up. Say you begin playing at 8 in the afternoon, and time flies till 12 while playing and chatting.
However something interesting happened in those 4 hours. You tossed the coin around 200 times, and out of those 200 times, heads came up 180 times, surprisingly, but possible. Once again, one friend is an engineer and he certified the coin as fair. In the $201$st round, just before going home, everyone decides to do an all-in for the last toss. You need to decide where to put all your money, or how to win as much as possible. Which side should you trust?

In the analysis I didn't the result did convince me, but in the end the coin tosses should obey a $X \sim Bi(n=201 , p=\frac{1}{2})$, and we have to find $P(X=181 \mid X \ge 180)$ so the probability should be 
$$
P(Y_{201}=heads)=
\frac{P(X=181)}{P(X \ge 180)}=
\frac{\binom{201}{181}\frac{1}{2}^{201}}{1-F(180)}=
.8908
$$
But shouldn't the last toss be independent of the of history of the game? If another friend arrived just for the last toss, he sees the probability as one half. I'm not sure of how to interpret this situation! Maybe I did something wrong with the binomial or missed some concept.
 A: Forgive me if this is off topic, but for a slightly more applied, "engineer-y" perspective on the dynamics of a real-world coin toss and its intuition I might point you in the direction of this paper: Dynamical Bias in the Coin Toss by Diaconis et al.
It offers an interesting perspective, and presents the intriguing result that for "natural flips" the probability of the coin turning up the face it started on is approximately 0.51, that is, generally speaking, you are more likely to get heads if you flip the coin with heads faced up. 
A: It all depends on whether you trust the engineer. If you do, then you should trust that the final flip is 50/50 and of course, the final flip outcome is independent of all the previous flips unless something very strange is going on.
However, if you assume the coin may or may not be fair, and you're not sure which, then you can put what's called a prior distribution on the chances for heads of the coin, and then after observing 200 flips, you can compute the posterior probability of various chances of heads along with confidence intervals on what you think the chances of heads might be, assuming that your prior distribution is fairly accurate. This is Bayesian statistics and it's not too hard to work out in some cases. This would allow you, for example, to make a bet which is not 1:1 whose expected value for a fair coin would be against you, but for a biased coin according to your calculations, would be in your favor.
The simplest choice of prior is a flat prior, where you assume that any choice of chances of getting heads for the coin is equally likely. Then you would estimate that the chances of getting heads is $0.9$, as pointed out by another answer, and with a little more work you could get a confidence interval centered around $0.9$ that has chances of say $95\% that the true chance of heads for the coin is within that confidence interval.
A: You have more information than just $X \geq 180$.  You know that after $200$ tosses, heads came up $180$ times.  What you should compute is $P(X = 181 \mid E)$, where $E$ is the event that after $200$ tosses, heads comes up exactly $180$ times.  This is different than $P(X = 181 \mid X \geq 180)$.
A: Informally, I think the problem is that you are weighing the probability of two very unlikely events.  Sure, you trust your engineering friend...but he could have made a mistake or this could all be part of an elaborate scheme on his part to manipulate betting.  What is the probability of that?  Well, it's something non-zero...  Think of the alternative!  Look at 200 tosses of a fair coin. We want to count the number of Heads...  Approximating the associated distribution with the normal (mean = 100, st. dev. = $\sqrt{50}$) we see that getting 180 H is more than an 11 $\sigma$ event.  I guarantee that the event that your engineering friend made an error is more probable that an 11 $\sigma$ event.  Therefore, personally, I'd reject the statement that this was a fair coin, and I'd bet on H.
