If we have a biased coin with the probability of tails equals $\frac{7}{10}$, is $\frac{5}{10}$ a better prediction than $\frac{9}{10}$? If we have a biased coin with the probability of getting tails equals $\frac{7}{10}$, is $\frac{5}{10}$ a better prediction than $\frac{9}{10}$? Intuitively it seems to me that $\frac{5}{10}$ should be better than $\frac{9}{10}$, but I am not sure why. So, I would like to understand answer for any legit probabilities instead of $\frac{7}{10}$, $\frac{5}{10}$ and $\frac{9}{10}$.
I am asking this because I've read this blog post, and now I want to make a simple website where 2 persons can input their predictions of some event as probabilities, and it would give them the number how much everyone should bet so that it's fair, but I am not sure I like the author's definition of "fair".
 A: Let's define precisely what one might mean by "better prediction."  Suppose we have a reference coin, whose probability of landing tails is exactly $p = 0.7$.  Further, imagine we employ a betting strategy which comprises the independent toss of a second "strategy" coin, whose probability of landing tails is $\theta$, for some $\theta \in [0,1]$ that we are free to choose, and that we "win" when the outcome of the tosses of both coins agree; i.e., both coins land heads, or both coins land tails.  We lose if the coins disagree.
The question that we are faced with, then, is "which of the two strategies, $\theta = 0.5$, or $\theta = 0.9$, produces a higher expected frequency of wins?"  This is an easy thing to answer:  If $X_1$ is the outcome of the reference coin and $X_2$ the outcome of the "strategy" coin, then clearly $X_1, X_2$ are independent Bernoulli random variables with $$X_1 \sim \operatorname{Bernoulli}(p = 0.7), \\ X_2 \sim \operatorname{Bernoulli}(p = \theta).$$  The desired expectation is equal to $$\begin{align*} \Pr[X_1 = X_2] &= \Pr[X_1 = 0 \mid X_2 = 0]\Pr[X_2 = 0] + \Pr[X_1 = 1 \mid X_2 = 1]\Pr[X_2 = 1] \\ &= \Pr[X_1 = 0]\Pr[X_2 = 0] + \Pr[X_1 = 1]\Pr[X_2 = 1] \\ &= (1-p)(1-\theta) + p\theta \\ &= 0.3(1-\theta) + 0.7 \theta \\ &= 0.3 + 0.4\theta. \end{align*}$$  It immediately follows that this probability is greater when $\theta = 0.9$ than when $\theta = 0.5$; therefore, the second strategy--equivalent choosing to bet on "tails" 9 times out of 10--yields a higher expected frequency of agreement with the reference coin than betting equally on heads versus tails.  In fact, we see that always selecting tails is the best strategy of all, with an expected proportion of agreement being $p = 0.7$.  And it may seem strange at first, but this even outperforms what we might initially think of as the proper strategy of selecting $\theta = 0.7$, which actually does worse than any $\theta > 0.7$.
A little more thought should reveal why this is intuitively true.
