D&D 5E Analysis on attack odds and the influence of (some) game elements on said odds

[Aeloric1976]

My preliminary thoughts about your case are the following:

The chance of proning an opponent on an unarmed strike (UnStrAtk) equals the chance to get a hit plus the chance to get a crit (that is, the chance to get a non miss on an unarmed strike) (1-MissC/UnStrAtk) times the chance to get a prone (ProneC) on an unarmed strike. In abbreviated terms:

ProneC/UnStrAtk = (1-MissC/UnStrAtk) x ProneC

Say you score a non miss (hit or crit, it doesn't matter for the proning attempt) 50% of the time and a success on a proning attempt 50% of the time. You'll hit and prone your opponent 25% of the time.

If you have 2 UnStrAtks and a prone attempt on each hit. Only the first successful prone attempt matters. After you successfully prones an opponent you don't have to prone it again. To calculate the chance of getting at least one successful prone in two Unarmed Strikes you have to first calculate the chance of getting no prones in one UnStrAtk, wich is (1 - (ProneC/UnStrAtk)), and then the chance of getting no prones in 2 attack, wich is (1 - (ProneC/UnStrAtk))². Now you determine the chance of getting one prone by subtracting the previous result from 1, in other words, the chance of getting at least one successful prone is 100% minus the chance of getting no prones. In abbreviated terms:

1ProneC/2UnStrAtks = 1 - (1 - (ProneC/UnStrAtk))²

Knowing your overall chance of getting at least one prone, you must determine the chance of getting the prone either on the first attempt or on the second attempt.

On the first is already known: ProneC/UnStrAtk

On the second is what's left from (1ProneC/2UnStrAtks) - (ProneC/UnStrAtk)

Using the values that I used in the first example above (50% for a non miss and 50% success for a proning check), you get that 1ProneC/2UnStrkAtks = 1 - (1-25%)² = 1 - (75%)² = 1 - 56,25% = 43,75%

So, if in 43,75% of the times you get at least one prone, 56,25% you'll get just 2 UnStrAtk and 2 regular attacks (Unarmed or otherwise).

The chance of getting a prone on the first UnStrAtk is 25%

The chance of getting a prone on the second is 43,75% - 25% = 18,75%

My end results:

1) 56,25% you'll get just 2 UnStrAtks with no prones and 2 regular attacks (Unarmed or otherwise) (calculate dpr normally for all attack with no Comb Adv)

2) 25% of the time you'll get a prone on the first UnStrAtk and 3 attacks with Combat Advantage (1 Unarmed and 2 Unarmed or not) (apply Comb Adv on the last 3 attacks)

3) 18,75% of the time you'll get a prone on the second attempt and 2 attacks with Combat Advantage (2 Unarmed or not) (apply Combat Advantage on the last 2 attacks)

I think it is correct, please revise my findings and we can correct any mistakes.

 

[FrogReaver]

My preliminary thoughts about your case are the following:

The chance of proning an opponent on an unarmed strike (UnStrAtk) equals the chance to get a hit plus the chance to get a crit (that is, the chance to get a non miss on an unarmed strike) ((1-MissC/UnStrAtk) times the chance to get a prone (ProneC) on an unarmed strike. In abbreviated terms:

ProneC/UnStrAtk = (1-MissC/UnStrAtk) x ProneC

Say you have a non miss 50% of the time and a success on a prone attempt 50% of the time. You'll hit and prone your opponent 25% of the time.

If you have 2 UnStrAtks and a prone attempt on each hit. Only the first successful prone attempt matters. After you successfully prones an opponent you don't have to prone it again. To calculate the chance of getting at least one successful prone in two Unarmed Strikes you have to first calculate the chance of getting no prones in one UnStrAtk, wich is 1 - (ProneC/UnStrAtk), and then the chance of getting no prones in 2 attack, wich is (1 - (ProneC/UnStrAtk))². Now you determine the chance of getting one prone by subtracting the previous result from 1, in other words, the chance of getting at least one successful prone is 100% minus the chance of getting no prones. In abbreviated terms:

1ProneC/2UnStrAtks = 1 - (1 - (ProneC/UnStrAtk))²

Knowing your overall chance of getting at least one prone, you must determine the chance of getting the prone on the first attempt and on the second.

On the first is already known: ProneC/UnStrAtk

On the second is the rest from (1ProneC/2UnStrAtks) - (ProneC/UnStrAtk)

Using the values that I used in the first example above (50% for a non miss and 50% success for a proning check), you get that 1ProneC/2UnStrkAtks = 1 - (1-25%)² = 1 - (75%)² = 1 - 56,25% = 43,75%

So, if in 43,75% of the times you get at least one prone, 56,25% you'll get just 2 UnStrAtk and 2 regular attacks (Unarmed or otherwise).

The chance of getting a prone on the first UnStrAtk is 25%

The chance of getting a prone on the second is 43,75% - 25% = 18,75%

My end results:

1) 56,25% you'll get just 2 UnStrAtks with no prones and 2 regular attacks (Unarmed or otherwise) (calculate dpr normally for all attack with no Comb Adv)

2) 25% of the time you'll get a prone on the first UnStrAtk and 3 attacks with Combat Advantage (1 Unarmed and 2 Unarmed or not) (apply Comb Adv on the last 3 attacks)

3) 18,75% of the time you'll get a prone on the second attempt and 2 attacks with Combat Advantage (2 Unarmed or not) (apply Combat Advantage on the last 2 attacks)


I think it is correct, please revise my findings and we can correct any mistakes.

Are you sure about the bolded part? It seems strange to me that we can apply the DPR of a non advantage attack to an attack we know must have hit or must have missed.

 

[Aeloric1976]

formula is ideal. If you want to give an example use a level 12 monk with max dex with 65% to hit the enemy and 60% chance for the flurry of blows attack to prone the enemy.

I used different values, but I think it ends up being the same, I mean, the values just help visualizing the thinking behind the results.

 

[FrogReaver]

So my minds cleared up a bit on the problem.

I have 4 attacks. I will calculate the "advantage weighted" DPR of each attack.

The first attacks DPR is normal as there is no chance of getting advantage.
The second attack has a weight dpr of (chance first attack prones) * advD + (chance first attack doesn't prone) * D
The third attack has a weighted DPR of (chance at least 1 of the attacks prone) * AdvD + (Chance no attack prones) * D
The fourth attack is identical to the third.

Using the 50% chance to hit and 50% chance to prone and an average damage of 10 this means:
1. DPR of attack 1 = .5*10 = 5 DPR
2. DPR of attack 2 = (.25) * (.75) * 10 + (.75) * .5 * 10 = 5.625 DPR
3. DPR of attack 3 = (.4375) * .75 * 10 + .5625 * .5 * 10 = 6.09375 DPR
4. DPR of attack 4 = (.4375) * .75 * 10 + .5625 * .5 * 10 = 6.09375 DPR

Total DPR = 22.8125

Do your calculations total up to this amount for the 50% and 50% calculation you did?\
*note I ignored crits for the time being.

 

[Aeloric1976]

Are you sure about the bolded part? It seems strange to me that we can apply the DPR of a non advantage attack to an attack we know must have hit or must have missed.

You're right about "2" and "3". In these you apply full dmg (distributing hits and crits proportionately) either on the first or second attack because you know either the first or the 2nd attack was a hit or crit with a successful prone.

About "1" I still think that you calculate dpr normally because you don't know if it was a non miss without prone ou if it was a miss, so you must calculate dpr normally without applying combat advantage.

So say we use the same figures (50%) and 15 dmg when hitting (45% of the time) and 25 dmg when critting (5% of the time).

1) 56,25% of your turns (Bonus Action Attacks plus Action Attacks) you'll do normal DPR with no Comb Advantage. These are the no proning events (nothing is said about the hitting or critting, just the proning) DPR = ((45%x15)+(5%x25)) x 4 = (6,75 + 1,25) x 4 = 32,00

2) 25% of your turns, you'll do ((45x15)+(5x25))/50 + 3 x ((66,25% x 15) + (9,75% x 25)) = 16 + 3 x (9,9375 + 2,4375) = 16 + 3 x (12,375) = 53,125

3) 18,75% of your turn, you'll do ((45%x15)+(5%x25)) + ((45x15)+(5x25))/50 + 2 x ((66,25% x 15) + (9,75% x 25)) = 8 + 16 + 24,75 = 48,75


Adding all thesse figures you get

(56,25% x 32) + (25% x 53,125) + (18,75% x 48,75) = 18 + 13,28 + 9,14 = 40,42

 

[Aeloric1976]

See next message

 

[Aeloric1976]

So my minds cleared up a bit on the problem.

I have 4 attacks. I will calculate the "advantage weighted" DPR of each attack.

The first attacks DPR is normal as there is no chance of getting advantage.
The second attack has a weight dpr of (chance first attack prones) * advD + (chance first attack doesn't prone) * D
The third attack has a weighted DPR of (chance at least 1 of the attacks prone) * AdvD + (Chance no attack prones) * D
The fourth attack is identical to the third.

Using the 50% chance to hit and 50% chance to prone and an average damage of 10 this means:
1. DPR of attack 1 = .5*10 = 5 DPR
2. DPR of attack 2 = (.25) * (.75) * 10 + (.75) * .5 * 10 = 5.625 DPR
3. DPR of attack 3 = (.4375) * .75 * 10 + .5625 * .5 * 10 = 6.09375 DPR
4. DPR of attack 4 = (.4375) * .75 * 10 + .5625 * .5 * 10 = 6.09375 DPR

Total DPR = 22.8125

Do your calculations total up to this amount for the 50% and 50% calculation you did?\
*note I ignored crits for the time being.

Here you didn't take crits into account, as per your note, but I think your logic is sound about the rest.

I just expressed the same logic in a more analytical fashion.

 

[FrogReaver]

Well I was making that problem much harder than I needed to be...

Thanks for the help!

 

[Aeloric1976]

Well I was making that problem much harder than I needed to be...

Thanks for the help!

We helped each other! Thanks!