If you have the option to roll 2 separate 10-sided dice or a single 10-sided die that you add 2 to the result with a goal of at least one die getting a specific number or higher (say 6 for the purposes of this question) is one of these options inherently better?
It seems that the single die with a modifier is statistically better, but I'm not sure if that accounts for the 2 dice being more consistent.
On
To get $6$ from $d10+2$ you need to roll at least a $4$, which is $7$ of the $10$ faces. Your chances are $70\%$. To fail to get one die at least $6$ with $2d10$, both of them have to be $1$ to $5$, which is a $\frac 12$ chance for each die. The chance they both are $5$ or less is the product of these, or $\frac 14=25\%$, so your chance of success is $75\%$. In this case it is better to roll two dice.
There is no magic formula which says how many points gives you the same chance as how many extra dice. As you roll more dice, the chance you will get at least one $10$ increases, so it becomes almost guaranteed.
On
Suppose in option (a), you have $k$ $n$-sided dice, and in option (b), you have a single $n$-sided die, and you add a modifier $m$. Furthermore, suppose that you are aiming for a target score of at least $t$ (with $1 \leq t \leq n$).
In case (a), you achieve your target unless each of the $k$ dice is below $t$, which happens with probability
$$ P(\text{all $k$ dice are below $t$}) = \left(\frac{t-1}{n}\right)^k $$
so the probability of success here is
$$ P(\text{success using (a)}) = 1-\left(\frac{t-1}{n}\right)^k $$
In case (b), you achieve your target so long as you roll at least $t-m$, which happens with probability
$$ P(\text{success using (b)}) = \min\left\{1, \frac{n-t+m+1}{n}\right\} $$
As Ross Millikan indicates in his answer, there is no simple formula that tells you which is higher, other than comparing the two quantities above. For example, with $k = 2$ dice, each with $n = 10$ sides, versus a single die with a bonus of $m = 2$, we get
$$ 1-\left(\frac{t-1}{10}\right)^2 \qquad \text{vs} \qquad \min\left\{1, \frac{13-t}{10}\right\} $$
and we find that option (a) is better for targets $t = 4, 5, 6, 7, 8$, but option (b) is better for targets $t = 1, 2, 3, 9, 10$.
So, with one die, you can get 6 or greater by rolling a 4, 5, 6, 7, 8, 9, 10. This is a 70% chance. With two dice, you can get a six or greater if the first dice is six or greater (50%) or the first dice is less than 6 (50%) but the second is greater than six (50%). So, the total in this scenario, since the two dice rolls are independent is 50% + 50%*50% = 75%.
Accordingly, it is more likely to get a 6 or greater with two dice.
Previous Post:
It depends on the number you are trying to get. If you're trying to get 10, there is a 1/10 chance of getting it with one die. But of the 36 possibilities with two dice, you can get 3 chances (5+5, 4+6, 6+4) or a 3/36 chance. The one die would increase your odds.
However, if you are trying to get 6 (as you are in this question), you can get it with a 1+5,2+4,3+3,4+2,5+1, so you have a 5/36 chance, which is greater than the 1/10 you would have from the 1 die.
Am I answering the question you're asking?