Assumptions
I'm going to be making the following assumptions, based on what you've already provided:
- 3rd level (since you get a crit on a 19 or 20)
- 16 Strength (no ASI to bump up to 18)
- Fighting a CR 3 creature (for base math)
- Average damage is 7.50 (4.50 from the die +3 Str mod)
- Average crit damage is 12.00 (4.50 per die +3 Str mod)
- Attack bonus is +5 (+2 prof, +3 Str mod)
- The enemy has AC 13 (per the DMG guidelines on page 274)
- DPR calculations are:
- Crit damage = Crit % x average crit hit damage
- Normal damage = Hit % x average hit damage
- hit % = 100 – [miss %] – [crit %]
Note that any flat modifier to the damage total won't change with a crit, since you only double the dice rolled, not the modifiers added.
Champion Fighter
Per the DMG page 274, a CR3 creature has an average AC 13, meaning you need to roll an 8 or higher.
With Advantage
AnyDice can tell us our miss chance and our crit chance. From there, we know our hit chance.
- A normal roll of 1d20 will have a 35% miss, 10.00% crit (1.20 DPR), and 55.00% normal hit (4.13 DPR) for a total DPR of 5.33
- A roll with advantage will have a 12.25% miss, 19.00% crit (2.28 DPR), and 68.75% normal hit (5.16 DPR) for a total DPR of 7.44
- A roll with your DMs +2 rule will have a 25.00% miss, 10.00% crit (1.20 DPR), and a 65.00% normal hit (4.88 DPR) for a total DPR of 6.08
With Disadvantage
AnyDice can tell us our miss chance and our crit chance. From there, we know our hit chance.
- A normal roll of 1d20 will have a 35% miss, 10.00% crit (1.20 DPR), and 55.00% normal hit (4.13 DPR) for a total DPR of 5.33 (unchanged)
- A roll with disadvantage will have a 42.25% miss, 1.00% crit (0.12 DPR), and 56.75% normal hit (4.26 DPR) for a total DPR of 4.38
- A roll with your DMs -2 rule will have a 45.00% miss, 10.00% crit (1.20 DPR), and a 45.00% normal hit (3.38 DPR) for a total DPR of 4.58
Rogue
How does this change affect other classes, specifically those who rely on bonus damage dice? I'm using a rogue for this example since sneak attack is easy enough to calculate, but a paladin falls under the same heading with their smite spells and the like.
We use the same percentages and base damage (assume Dex and a rapier) for our fighter, but we add sneak attack damage. That's 2d6 at level 3, so with advantage we add +2d6 (7) on a hit and +4d6 (14) on a crit. Attacks without advantage don't get sneak attack damage added in, so the disadvantage numbers from above carry over (I know you can get sneak attack damage without advantage, but we'll ignore that for simplicity).
With Advantage
- A normal roll of 1d20 will have a 35% miss, 5.00% crit (1.30 DPR), and 60.00% normal hit (8.70 DPR) for a total DPR of 10
- A roll with advantage will have a 12.25% miss, 9.75% crit (2.54 DPR), and 78.00% normal hit (11.31 DPR) for a total DPR of 13.85
- A roll with your DMs +2 rule will have a 25.00% miss, 5.00% crit (1.30 DPR), and a 70.00% normal hit (10.15 DPR) for a total DPR of 11.45
With Disadvantage
- A normal roll of 1d20 will have a 35% miss, 5.00% crit (1.30 DPR), and 60.00% normal hit (8.70 DPR) for a total DPR of 10
- A roll with disadvantage will have a 42.25% miss, 1.00% crit (0.12 DPR), and 56.75% normal hit (4.26 DPR) for a total DPR of 4.38 (identical to the fighter, as no advantage means no sneak attack)
- A roll with your DMs -2 rule will have a 45.00% miss, 5.00% crit (1.20 DPR), and a 50.00% normal hit (2.25 DPR) for a total DPR of 3.45
Conclusion
With your DMs proposed houserule, the expected DPR for any class is going to be decreased because of the fact that you're still only rolling 1 die, so the chance of a critical hit will not change. The biggest, well, advantage of rolling with advantage is it almost doubles your chance of a crit: 9.75% vs. 5.00% for a normal 20 crit and 19.00% vs. 10% for a champion fighter crit.
Indeed, that simple change reduces the overall expected damage output of the entire party, especially those classes that rely on burst damage in the form of more dice. As you gain in levels and get the extra attack feature, magic items/spells that add damage dice, and class features that change the damage dice done, the gap will only increase.
Answers
Below are the expected number of steps/rolls to get to 6 coins:
- Steps To Go From 2 Coins to 6 Coins = 7.52631
- Steps To Go From 3 Coins to 6 Coins = 4.25833
- Steps To Go From 4 Coins to 6 Coins = 2.45833
- Steps To Go From 5 Coins to 6 Coins = 1.125
Solution
The best way to solve this problem in general is not Anydice, but with a tool called Markov chains. But let's start with the basic probabilities.
First, let's define your state. A state is the number of coins you have currently. So if you have 2 coins, you're in state 2, whereas if you have 3 coins, you are in state 3.
The probability of rolling under any number of a d6 is given by the following: \$\frac {x-1}{6}\$; for example, to roll under a 2, you need to roll a 1, which is a \$\frac 1 6\$ chance, or an \$\frac{x-1}{6} = \frac{2-1}{6} = \frac 1 6\$ chance
The reverse is true. To roll at least any number on a d6, it is "1 - Probability(rolling under that number)"; so to roll at least a 2, the probability is equal to not rolling under a 2. The formula is \$1-\frac{x-1}{6} = \frac{7-x}{6}\$
The probability to roll 2d6 and get at least one under a certain number, is equal to the probability of rolling 2d6 and rolling both with at least that number twice. That is, the only way to "fail" your roll is if both d6's are at least that target number. The math for this is "not Pr(roll at least that number) twice", and the formula is "1 - Pr(roll at least that number)2" which is equal to: \$1 - \left(\frac{7-x}{6}\right)^2\$
The probability to move from state x to state x+1 is the same as the probability of rolling 2d6 with at least one under x; put another way, the probability is the same as not rolling a least x twice.
Solution
Hence you have the Markov chain:
Arrows going from one circle to another represent the probability for jumping to the next state. The arrows from a circle going back to itself represent the probabilities for staying in the same state.
And since there are five states (2 through 6, there is no way to fall back to a previous state), this is the full chain:
Now, from this chain, construct what is called a transition matrix which describes all the state change probabilities in mathematical form.
You can read up more about transition matrices online. Roughly, it charts the probability of going from a state on the left to the state on top. So going from state 2 back to state 2 is 0.6944, while the probability of it going to state 4 is 0, since you can only jump to one state at a time.
Here, we do some math to find the expected number of steps from any state to state 6. We do this by solving the equations that arise from the transition matrix. I won't explain them in much detail, I'll just show the final equations.
Let \$h(i,j)\$ be the expected number of steps from state i to state j.
\begin{align}
h(2,6) &= 1 + (0.694) \times h(2,6) + (0.306) \times h(3,6) \\
h(3,6) &= 1 + (0.444) \times h(3,6) + (0.556) \times h(4,6) \\
h(4,6) &= 1 + (0.25) \times h(4,6) + (0.75) \times h(5,6) \\
h(5,6) &= 1 + (0.111) \times h(5,6) + (0.889) \times h(6,6) \\
h(6,6) &= 0
\end{align}
Answer
The problem is now reduced to a system of equations. Solving these, you get:
\begin{align}
h(2,6) &= 7.52631 \\
h(3,6) &= 4.25833 \\
h(4,6) &= 2.45833 \\
h(5,6) &= 1.125 \\
h(6,6) &= 0
\end{align}
Best Answer
Here's a quick AnyDice program to show the effect of applying the D&D 5e (dis)advantage mechanic (i.e. roll twice, pick lowest/highest) to dice of various size. Based on the summary results, here are the average rolls in each case:
From this table we can see that — just considering the average damage per hit, and ignoring the distribution of results — (dis)advantage is equal to a bonus of about ±0.6 on a d4, ±1.0 on a d6, ±1.3 on a d8, ±1.7 on a d10, ±2.0 on a d12 and ±3.3 on a d20.
(This is lower than the often quoted "advantage ≈ +4 or +5" rule of thumb for d20 since it's based on the average roll rather than the chance of meeting a target DC that's typically somewhere near the middle of the 1–20 range. Due to the way the advantage mechanic works, it has the biggest effect compared to a flat bonus for mid-range target numbers.)
Addendum based on comments: For damage rolls with multiple dice, like a 2d6 greatsword, the magnitude of the effect depends on how you choose to generalize (dis)advantage to rolls using multiple dice. Off the top of my head, I can think of at least the following options for rolling NdX with (dis)advantage:
This AnyDice program shows the average result of each of these methods of rolling NdX with advantage. For example, for 2d6 the averages look like this:
For method #4 the average difference between a normal roll and one with advantage is of course the same as for a single dX roll, while for method #1 it's N times that. Methods #2 and #3 tend to be somewhere in between. In fact, for N = 2 method #3 seems to give results exactly halfway between methods #1 and #4, although this does not continue to hold for higher values of N.