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The mathematics of D&D–Damage and HP
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<blockquote data-quote="Asisreo" data-source="post: 8220344" data-attributes="member: 7019027"><p>Damage is cumulative up to the target's HP, but stops being so afterwards. </p><p></p><p>Let's say a monster is a trial. Each trial is independent of each other and what you're basically asking is: how many dice rolls are needed to equal or exceed this monster's HP. Lets say you have a d6 damage and the monster has 6 HP. </p><p></p><p>Well, the RF of the trials never converges. That is to say: it doesn't matter how many times you've killed the monster before, this new monster is completely independent and it could take 6 rounds or it could take 1, the probability is random though 2 is the most likely amount of rounds. </p><p></p><p>Now, if you were to take a large amount of monsters and killed them all, you'd see the amount of rounds it takes to kill these monsters converge to 2. This is the CMF.</p><p></p><p>But you're not looking for the amount of rounds it takes for you to kill a larger amount of monsters, you're looking for the attack option that is likely to kill the one you're currently fighting faster. </p><p></p><p>You're looking for the attack with the highest probability to kill. </p><p></p><p>Its very rare for you to deal damage exactly equal to your enemy's HP. So when you look at your options, you should consider your probability to kill. </p><p></p><p>Let's say you're a wizard with firebolt but you also have a light crossbow with a +1 for dex meaning they have the same average damage (5.5). Lets say the enemy's AC is 0 (guaranteed to hit) and his HP is 7. A bit higher than either attack's expected damage. Which one should you choose or does it matter? </p><p></p><p>The answer is firebolt, which has a 40% chance to kill compared to the light crossbow's 37.5% chance to kill. </p><p></p><p></p><p>Now, here's an extreme example: </p><p></p><p>Let's compare 55(10d10) vs 56(2d6+49). </p><p></p><p>You might want to jump ahead and say that 2d6+49 is better than 10d10 by all accounts, but that's untrue. </p><p></p><p>If, for instance, the enemy has 57 HP, then its actually better to do 10d10 which has a 43.54% chance of killing rather than 2d6+49 which has a 41.67% chance of killing. </p><p></p><p>In this case, <strong>the option that does less expected damage has a higher probability of killing.</strong></p></blockquote><p></p>
[QUOTE="Asisreo, post: 8220344, member: 7019027"] Damage is cumulative up to the target's HP, but stops being so afterwards. Let's say a monster is a trial. Each trial is independent of each other and what you're basically asking is: how many dice rolls are needed to equal or exceed this monster's HP. Lets say you have a d6 damage and the monster has 6 HP. Well, the RF of the trials never converges. That is to say: it doesn't matter how many times you've killed the monster before, this new monster is completely independent and it could take 6 rounds or it could take 1, the probability is random though 2 is the most likely amount of rounds. Now, if you were to take a large amount of monsters and killed them all, you'd see the amount of rounds it takes to kill these monsters converge to 2. This is the CMF. But you're not looking for the amount of rounds it takes for you to kill a larger amount of monsters, you're looking for the attack option that is likely to kill the one you're currently fighting faster. You're looking for the attack with the highest probability to kill. Its very rare for you to deal damage exactly equal to your enemy's HP. So when you look at your options, you should consider your probability to kill. Let's say you're a wizard with firebolt but you also have a light crossbow with a +1 for dex meaning they have the same average damage (5.5). Lets say the enemy's AC is 0 (guaranteed to hit) and his HP is 7. A bit higher than either attack's expected damage. Which one should you choose or does it matter? The answer is firebolt, which has a 40% chance to kill compared to the light crossbow's 37.5% chance to kill. Now, here's an extreme example: Let's compare 55(10d10) vs 56(2d6+49). You might want to jump ahead and say that 2d6+49 is better than 10d10 by all accounts, but that's untrue. If, for instance, the enemy has 57 HP, then its actually better to do 10d10 which has a 43.54% chance of killing rather than 2d6+49 which has a 41.67% chance of killing. In this case, [B]the option that does less expected damage has a higher probability of killing.[/B] [/QUOTE]
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