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d20 vs 2d20
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<blockquote data-quote="dregntael" data-source="post: 7429652" data-attributes="member: 6790458"><p>I have wondered the same thing before so I did a quick calculation. I calculated the following situations:</p><p></p><ul> <li data-xf-list-type="ul"> Situation 1: attacker rolls with 1d20+A against passive defence of 10+B, attacker wins on a tie </li> <li data-xf-list-type="ul"> Situation 2: attacker rolls with 1d20+A against active defence of 1d20+B, attacker wins on a tie </li> </ul><p>The results actually surprised me: in situation 1, the chance of success is 0.55 + (A-B)/20 (so the attacker has an advantage of 5%), while in situation 2 the chance of success is 0.525 + (A-B)/20 (so a 2.5% advantage for the attacker). So apart from a very small difference in the constant term, the probabilities are exactly the same! So you can probably safely switch between passive and active defence without affecting the balance at all.</p><p></p><p>Here are my calculations for those who want to check them:[sblock]Let N := A-B</p><p></p><p>P(1d20+A >= 10+B)</p><p>= P(1d20-10+N >= 0)</p><p>= 1/20 * SUM_{k=1..20}(P(k-10+N >= 0))</p><p>= (11+N)/20</p><p>= 0.55 + N/20</p><p></p><p>P(1d20+A >= 1d20+B)</p><p>= P(1d20-1d20+N >= 0)</p><p>= 1/400 * SUM_{k=1..20}(SUM_{l=1..20}(P(k-l+N >= 0)))</p><p>= 1/400 * SUM_{k=1..20}(k+N)</p><p>= 1/400 * (SUM_{k=1..20}(k) + 20*N)</p><p>= 1/400 * (210 + 20*N)</p><p>= 0.525 + N/20</p><p>[/sblock]</p><p></p><p>EDIT: I made a mistake in the calculation of situation 2, the right probability is 0.525 + 0.04875 * (A-B) - 0.00125 * (A-B)^2 if A >= B and 0.525 + 0.05125 * (A-B) + 0.00125 * (A-B)^2 if A < B. Here is <a href="https://www.wolframalpha.com/input/?i=plot(piecewise(((0.525+%2B+0.05125+*+N+%2B+0.00125+*+N%5E2,N%3C0),+(0.525+%2B+0.04875+*+N+-+0.00125+*+N%5E2+,+0%3CN)))+,+0.55%2BN%2F20)+for+N+%3D+-10..10" target="_blank">a plot</a> of the difference using wolfram alpha. You can see that the attacker should prefer a passive defence unless their bonus is a lot lower than the defender's.</p><p></p><p>[sblock]</p><p>P(1d20+A >= 1d20+B)</p><p>= P(1d20-1d20+N >= 0)</p><p>= 1/400 * SUM_{k=1..20}(SUM_{l=1..20}(P(k-l+N >= 0)))</p><p>= 1/400 * SUM_{k=1..20}(max(0,min(20,k+N)))</p><p></p><p>if A >= B (N >= 0):</p><p>[...]</p><p>= 1/400 * SUM_{k=1..20}(min(20,k+N))</p><p>= 1/400 * (SUM_{k=1..(20-N)}(min(20,k+N)) + SUM_{k=(21-N)..20}(min(20,k+N)))</p><p>= 1/400 * (SUM_{k=1..(20-N)}(k+N) + SUM_{k=(21-N)..20}(20))</p><p>= 1/400 * ((20-N)(21-N)/2 + (20-N)*N + 20*N)</p><p>= 1/400 * (210 + (39/2)*N + -(N^2)/2)</p><p>= 0.525 + 0.04875 * N - 0.00125 * N^2 </p><p></p><p>if A < B (N < 0):</p><p>[...]</p><p>= 1/400 * SUM_{k=1..20}(max(0,k+N))</p><p>= 1/400 * (SUM_{k=1..(-N)}(max(0,k+N)) + SUM_{k=(-N)..20}(max(0,k+N)))</p><p>= 1/400 * (SUM_{k=1..(-N)}(0) + SUM_{k=(-N)..20}(k+N))</p><p>= 1/400 * SUM_{k'=0..(20+N)}(k')</p><p>= 1/400 * (20+N)(21+N)/2</p><p>= 0.525 + 0.05125 * N + 0.00125 * N^2</p><p>[/sblock]</p></blockquote><p></p>
[QUOTE="dregntael, post: 7429652, member: 6790458"] I have wondered the same thing before so I did a quick calculation. I calculated the following situations: [LIST] [*] Situation 1: attacker rolls with 1d20+A against passive defence of 10+B, attacker wins on a tie [*] Situation 2: attacker rolls with 1d20+A against active defence of 1d20+B, attacker wins on a tie [/LIST] The results actually surprised me: in situation 1, the chance of success is 0.55 + (A-B)/20 (so the attacker has an advantage of 5%), while in situation 2 the chance of success is 0.525 + (A-B)/20 (so a 2.5% advantage for the attacker). So apart from a very small difference in the constant term, the probabilities are exactly the same! So you can probably safely switch between passive and active defence without affecting the balance at all. Here are my calculations for those who want to check them:[sblock]Let N := A-B P(1d20+A >= 10+B) = P(1d20-10+N >= 0) = 1/20 * SUM_{k=1..20}(P(k-10+N >= 0)) = (11+N)/20 = 0.55 + N/20 P(1d20+A >= 1d20+B) = P(1d20-1d20+N >= 0) = 1/400 * SUM_{k=1..20}(SUM_{l=1..20}(P(k-l+N >= 0))) = 1/400 * SUM_{k=1..20}(k+N) = 1/400 * (SUM_{k=1..20}(k) + 20*N) = 1/400 * (210 + 20*N) = 0.525 + N/20 [/sblock] EDIT: I made a mistake in the calculation of situation 2, the right probability is 0.525 + 0.04875 * (A-B) - 0.00125 * (A-B)^2 if A >= B and 0.525 + 0.05125 * (A-B) + 0.00125 * (A-B)^2 if A < B. Here is [URL="https://www.wolframalpha.com/input/?i=plot(piecewise(((0.525+%2B+0.05125+*+N+%2B+0.00125+*+N%5E2,N%3C0),+(0.525+%2B+0.04875+*+N+-+0.00125+*+N%5E2+,+0%3CN)))+,+0.55%2BN%2F20)+for+N+%3D+-10..10"]a plot[/URL] of the difference using wolfram alpha. You can see that the attacker should prefer a passive defence unless their bonus is a lot lower than the defender's. [sblock] P(1d20+A >= 1d20+B) = P(1d20-1d20+N >= 0) = 1/400 * SUM_{k=1..20}(SUM_{l=1..20}(P(k-l+N >= 0))) = 1/400 * SUM_{k=1..20}(max(0,min(20,k+N))) if A >= B (N >= 0): [...] = 1/400 * SUM_{k=1..20}(min(20,k+N)) = 1/400 * (SUM_{k=1..(20-N)}(min(20,k+N)) + SUM_{k=(21-N)..20}(min(20,k+N))) = 1/400 * (SUM_{k=1..(20-N)}(k+N) + SUM_{k=(21-N)..20}(20)) = 1/400 * ((20-N)(21-N)/2 + (20-N)*N + 20*N) = 1/400 * (210 + (39/2)*N + -(N^2)/2) = 0.525 + 0.04875 * N - 0.00125 * N^2 if A < B (N < 0): [...] = 1/400 * SUM_{k=1..20}(max(0,k+N)) = 1/400 * (SUM_{k=1..(-N)}(max(0,k+N)) + SUM_{k=(-N)..20}(max(0,k+N))) = 1/400 * (SUM_{k=1..(-N)}(0) + SUM_{k=(-N)..20}(k+N)) = 1/400 * SUM_{k'=0..(20+N)}(k') = 1/400 * (20+N)(21+N)/2 = 0.525 + 0.05125 * N + 0.00125 * N^2 [/sblock] [/QUOTE]
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