How to combine the probability of liking something with the one of it being liked? How to combine the probability of liking something with the one of it being liked?
I'd like to estimate the probability of a person liking a dish by only having the following two bits of information given:


*

*The person likes 70% (like_rate = 0.7) of all dishes you offer him.

*The dish is liked by 80% (liked_rate = 0.8) of all people that try it.


So I'm looking for a function f(like_rate, liked_rate). I think it has to fulfill the following properties:


*

*f(1.0, ?) ≈ 1.0 (If the person likes everything, the dish does not matter much.)

*f(0.0, ?) ≈ 0.0 (If the person hates everything, the dish does not matter much.)

*f(?, 1.0) ≈ 1.0 (If everybody likes the dish, the person does not matter much.)

*f(?, 0.0) ≈ 0.0 (If nobody likes the dish, the person does not matter much.)


Of course we then run into a problem for the following two cases:


*f(1.0, 0.0)

*f(0.0, 1.0)
For practical consideration these are very unlikely to happen, so maybe their results could just be 0.5.
Another property that should be fulfilled in any case, at least from how I understand it, is:


*f(0.5, 0.5) = 0.5
But how to calculate these?


*f(0.5, 0.0) = ?

*f(0.5, 1.0) = ?

*f(0.0, 0.5) = ?

*f(1.0, 0.5) = ?
There are many possible functions in 3D space that fulfill the given conditions. But is one of them the correct one? If so, which one is it?
 A: At the moment the question is a bit under defined as there is a lot of subjectivity in how one is to interpret it. For instance 


*

*Is the person who likes 70% of dishes from the same population as the people who have already tried it? I.e. is it the case that in general any member of the population likes 70% of the dishes, and in this case 80% who tried it like it?

*If the person is representative of the population, then it seems like a reasonable assumption to say that he has an 80% change of liking it. This however is contrary to the scenarios which you give where they might have a 0% chance of liking it. So that makes me think you do not see them as coming fro the same population.

*If we suppose they are not from the same population, then we need a way to relate their tastes to the populations. If they are completely independent, then knowing that 80% of the population like the dish will not impact this person't likelihood of enjoying it.
So, with all those caveats, I outline below one interpretation of the question.

Reformulating the problem
In my interpretation, I will assume that the person in question comes from the same population as those who have already tried the dish. To make it clear what I mean by this, I will rephrase the problem in terms of tossing a biased coin.
Suppose you have a collection of identical coins, and that each of them is biased in a different way. Since the coins appear identical, if you pick one out at random you do not know the overall probability that it will fall heads. You do, however know from past experience that in general the average coin will land heads 70% of the time.
[Pause: To clarify the analogy: here each coin represents one of your dishes, and landing heads is analogous to the dish being liked. What we are saying is that from past experience a random person from the population will like 70% of your dishes] 
Now you start tossing the coin and after a number of throws observe that in total it has landed heads 80% of the time.
[Pause: i.e. a number of people from your population eat the dish and 80% like it]
You would now like to know: given the prior knowledge about the collection of coins, and the new information about how many of these coin tosses have been heads, what is the probability that the next coin I throw will be a head?

A Bayesian Solution
This problem formulation in terms of combining prior knowledge with new observations is often tackled by turning to Bayesian probability.
I do not have the space, nor authority, to give a complete introduction to Bayesian theory but there are plenty of places for you to read up on it. Instead I give a fairly brief summary to get to the end conclusion.
The Prior
The first thing we need to decide is how strongly we believe that the bias of the coin is exactly 70%. Suppose there is just a single coin, and when we obtained it we'd been told it would land heads up 70% of the time: then our belief would be very strong that this could would land heads up 70% of the time. And just because we'd observed a run of 80% heads, we would still believe that the next toss would have a 70% change of being heads: because of how strong our believe was.
Suppose instead that we'd obtained 5 coins, and been told that they respectively land heads up 50/60/70/80/90% of the time: then if you picked a coin on average, before throwing it once you'd expect it would land heads with 70% chance. And after a few throws you would be far more willing to move away from the 70% assumption.
This demonstrates how subjective the question is, and how dependent it is on your prior belief.
The prior distribution is a probability distribution on the set $[0,1]$ of possible biases. For instance, in the first example above (where we are adamant that the bias is 70%) this prior is
$$p(\theta) = \begin{cases}
1 & \text{if $\theta = 0.7$,} \\
0 & \text{else.}
\end{cases}
$$
In the second example, the prior would be
$$p(\theta) = \begin{cases}
\frac15 & \text{if $\theta \in\{0.5,0.6,0.7,0.8,0.9\}$,} \\
0 & \text{else.}
\end{cases}
$$
Note that both priors have the property that the expected value of the distribution is $0.7$.
To make the problem computationally tractable we will make a very specific assumption about the form of the prior, and suppose that it is Beta distributed with parameters $\alpha, \beta > 0$
$$
p(\theta) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \theta^{\alpha-1}(1-\theta)^{\beta-1}
$$
There is insufficient space to fully justify this choice, but there are a few important points.


*

*The mean is $\alpha / (\alpha + \beta)$; so we will want to choose parameters $\alpha,\beta$ such that this is equal to $0.7$.

*These have a natural interpretation as $\alpha$ being the number of heads that have occurred when throwing coins from this set on previous occassions (i.e before the current experiment), and $\beta$ the number of tails (so that $\alpha + \beta$ is the total number of tosses). In terms of dishes, this would be equivalent to saying of all other dishes you've served, not including the one currently in question there were a total of $\alpha$ that were liked, and $\beta$ that weren't.

*The larger $\alpha, \beta$ are the more certainty (less variance) you have in your prior assumption.
You will have to choose what parameters $\alpha,\,\beta$ make sense in your context. For instance (in the land of dishes, not coins): if you have only ever served 10 dishes, and 7 were liked then you'd use $\alpha = 7,\,\beta = 3$ which would give a moderate prior belief for the next dish. If you were basing it on restaurant review where there were 700 positive review, and 300 negative ($\alpha = 700, \,\beta =300$) that would be a much more confident prior belief.
The Posterior
Now that we have arrived at a prior distribution, we can combine this with the newly observed data to derive the posterior. This is done through Bayes rule by the formula:
$$p(\theta | x) =  \frac{p(x|\theta) p(\theta)}{p(x)}.$$
Here $p(x|\theta)$ is the probability of observing 80% heads, if you were to assume that the coin had a bias of $\theta$. Again, space is insufficient to give full detail, but there are again some important points:


*

*The more data you have (i.e. the more coin tosses that made up the 80% figure) the better your posterior knowledge will be. We will suppose that there were a total of $n$ tosses, $x_1,\ldots, x_n$ and that $x_i = 1$ denotes that coin $i$ was heads. We have from the problem set up that $n^{-1}\sum_i x_i = 0.8$.

*The magic of choosing the prior to be Beta distributed is that the posterior $p(\theta |x)$ is also Beta distributed. In particular, if the prior parameters were $\alpha, \beta$ then the posterior parameters are
$$ \alpha' = \alpha + \sum_i x_i, \qquad \beta' = \beta + n - \sum_i x_i.$$


*This fits with the interpretation we gave before: when we had the prior information a total of $\alpha + \beta$ dishes had been tasted, of which $\alpha$ were liked. Now a further $n$ have been tasted, of which $\sum_i x_i$ are liked. So in total $\alpha + \beta +n = \alpha' + \beta'$ have been tasted, of which $\alpha + \sum_i x_i$ have been liked.


So, in all: combining our prior belief, and our observations the posterior mean for $\theta$, which is the expected probability that the next coin will be a head is
$$\frac{ \alpha + \sum_{i} x_i }{\alpha + \beta + n}$$

A worked example
To make all of this a bit more concrete, and back in the context of your original question, we consider a specific example.
Suppose that you'd come to the conclusion that 70% of your dishes were like after 10 people had tried them and 7 of them had liked their dishes. Then you would choose your prior parameters $\alpha = 7, \, \beta = 3$.
Now suppose that a further 5 people came to eat your new dish, of which 4 enjoyed it (i.e. 80%). Then your posterior prediction for the popularity of the new dish would have parameters $\alpha' = 7 + 4$ and $\beta' = 3 + 1$, so that the posterior probability of liking the dish would be $11/15 \sim 73.3$%.
Suppose instead that it was based on restaurant reviews, and previously you'd had 700 people like the dishes, and 300 not ($\alpha = 700$, $\beta = 300$). Now you check your latest reviews and see that you have a further 100 reviews, of which $80$ were positive. Then the posterior probability would be $780/1100 \sim 71.0$%
So we see that in the second example when there is significantly more evidence to support our initial belief, we are more reluctant to move away from it.

Final Thoughts


*

*Hopefully it is clear from this that the percentages of 70% and 80% on their own are not strong enough to let us update our belief that someone will like the dish. We need not just the percentage, but the weight of evidence behind them.

*Although we've invoked a lot of Bayesian machinery to do this, the final answer does not actually use any of the Bayesian theory itself, and is of itself hopefully quite interpretable.


Finally, we link this back to the specific examples you gave. Hopefully the above motivates you to think of the function $f$ not in terms of $f(\text{likes}, \, \text{liked})$, but rather as $f( (\alpha,\,\beta) \colon \, (n,x) )$, and from the above this would have the formula
$$ f(  (\alpha,\,\beta) \colon \, (n,x) ) = \frac{\alpha + x}{\alpha + \beta + n -x}.$$
Your statements numbered 1-4 still hold, where the extent to which $\approx$ is accurate depends on the level of belief in the priors, and the amount of new evidence.
Statements 5-6 do now have an interpretation which makes perfect sense.
Statement 7 holds, and in fact more generally we have that if $\alpha/(\alpha + \beta) = x/n$ then
$$ f(  (\alpha,\,\beta) \colon \, (n,x) ) = \frac{\alpha}{\alpha + \beta},$$
i.e. if the prior mean and the mean of the evidence are equal, then the posterior mean is also equal.
All the remaining statements are just special cases of the above formula. 
