Correlation between probability when order matters and not for different sample spaces I will ask by example
Let's say we have some set of numbers from $1$ to $9$ with duplications of length $10$, e.g. $S=\{2,1,1,2,4,3,4,8,1,2\}$.
Now we draw out in uniform random way $4$ numbers with replacement from the set.
We want to find $2$ probabilities:
a) The probability that the drawn-out numbers look like $1,2,\_, \_$ where '_' is any number, i.e. the probability that first extracted number from the set is $1$ and second is $2$
b) The probability that at least one of $4$ numbers is $1$ and one is $2$
The question is if it's true that for every sets $\mathbf{S_1,S_2}$. $$Pr(a|S_1)>Pr(a|S_2) \iff Pr(b|S_1)>Pr(b|S_2)$$ And if yes how to prove it
P.s.
I'm getting two formulas for $Pr(a)$ and $Pr(b)$. Let's call $$x=\{\text{getting} \;1\; \text{by random drawn out from the set}\}$$ $$y=\{\text{getting} \;2\; \text{by random drawn out from the set}\}$$ Then:
$Pr(a)=Pr(x)⋅Pr(y)$
$Pr(b)=1-Pr^4(\text{not }x)-Pr^4(\text{not }y)+Pr^4(\text{not }x \text{ and not }y)$
Can I find formula such that $Pr(b)=c⋅Pr(a)$ ?
 A: I'm assuming (from the formulas given), that the drawing is with replacement (I assume that the phrase "with repetitions" is used to indicate that).
The formulas given at the end are correct, and considering that 
$$Pr(\text{not }x)=1-Pr(x),$$
$$Pr(\text{not }y)=1-Pr(y), \text{ and}$$
$$Pr(\text{not }x\text{ and not }y)=1-Pr(x)-Pr(y)$$
(the latter because events $x$ and $y$ are disjoint), we get
$$Pr(b)=1-(1-Pr(x))^4-(1-Pr(y))^4+(1-Pr(x)-Pr(y))^4$$
Wolfram Alpha evaluates the latter to
$$Pr(b)=2Pr(x)Pr(y)\times\left[2(Pr(x)^2+Pr(y)^2) + 3Pr(x)Pr(y)-6(Pr(x)+Pr(y))+6\right]$$
As can be seen, this is not just a function of $Pr(a)=Pr(x)Pr(y)$, the individual $Pr(x), Pr(y)$ values play a role as well.
Finding a counterexample for the question requires playing a bit with the numbers, and one gets:
If in set $S_1$ we have $Pr(x)=0.25, Pr(y)=0.5$, we get
$$Pr(a)=0.5\times0.25=0.125, $$
$$Pr(b)=1-(1-0.25)^4-(1-0.5)^4+(1-0.25-0.5)^4=1-0.75^4-0.5^4+0.25^4 = 0.625$$
However, for a set $S_2$ with $Pr(x)=0.3, Pr(y)=0.4$, we get
$$Pr(a)=0.3\times0.4=0.12, $$
$$Pr(b)=1-(1-0.3)^4-(1-0.4)^4+(1-0.3-0.4)^4=1-0.7^4-0.6^4+0.3^4 = 0.6384$$
So we have a counterexample for the question. It may not be a 100% nice example as one cannot get $Pr(x)=0.25$ for a 10-element set $S_1$, but it shows that the desired conclusion cannot be drawn generally. 
A numerical calculation for the cases $Pr(x),Pr(y)=0.1, 0.2,\ldots,0.9$ (with $Pr(x)+Pr(y) \le 1$ as constraint) should be managable, to test if the assumed equivalence holds for 10-element sets. But as I understood, the real question is more general, so the result for that is negative: The equivalence does not hold.
