ECE302 -- Proj1

Steven Lee & Jonathan Lam; Prof. Keene; 1/29/21

Contents

See accompanying PDF.

http://files.lambdalambda.ninja/reports/20-21_spring/ece302_proj1.lam_lee.pdf

clc; clear;
N = 100000;			% number of trials
D = 3;				% number of dice to take the sum of
T = 3;				% number of trials (in fun method)
S = 6;				% number of skills
F = 6;				% number of faces on the die

1a

q1a_exact = 1 / 6^D;

rolls = roll_dice([N D], F);	% random simulation
q1a_est = sum(sum(rolls, 2) == 18) / N;

print_res('1a', q1a_exact, q1a_est);

1b

q1b_exact = 1 - (1/216)^0*(215/216)^T;
rolls = roll_dice([N*T D], F);
three_roll_sums = reshape(sum(rolls, 2) == 18, N, T);
q1b_est = sum(sum(three_roll_sums, 2) >= 1) / N;

print_res('1b', q1b_exact, q1b_est);

Question 1b: Exact: 0.013825; Estimate: 0.013810; % Err: 0.106243%

1c

For both this question and the 1d: Both the exact (analytical) and simulated values are 0.000000 to 6 digits of precision; this is due to the probability being a small number raised to the sixth power. If we set S to a smaller value (e.g., S=1 or S=2), we see that the exact and simulated values match.

q1c_exact = (q1b_exact)^S;
rolls = roll_dice([N*T*S D], F);
three_roll_sums = reshape(sum(rolls, 2) == 18, [], T);
ability_scores = reshape(sum(three_roll_sums, 2) >= 1, [], S);
q1c_est = sum(sum(ability_scores, 2) == S) / N;

print_res('1c', q1c_exact, q1c_est);

Question 1c: Exact: 0.000000; Estimate: 0.000000; % Err: 100.000000%

1d

P(all <= 9) - P(all <= 9 && none == 9)

q1d_exact = ((81/216)^T - ((81-25)/216)^T)^S;

rolls = roll_dice([N*T*S D], F);
three_roll_sums = reshape(sum(rolls, 2), [], T);
ability_scores = reshape(max(three_roll_sums, [], 2) == 9, [], S);
q1d_est = sum(sum(ability_scores, 2) == S) / N;

print_res('1d', q1d_exact, q1d_est);

Question 1d: Exact: 0.000000; Estimate: 0.000000; % Err: 100.000000%

2a

D_T = 1;			% number of dice trolls use
F_T = 4;			% number of faces on troll die
D_K = 2;			% number of dice Keene uses
F_K = 2;			% number of faces on Keene die

% average hitpoints for trolls
q2a_exact = (1*1 + 2*1 + 3*1 + 4*1) / 4;
rolls = roll_dice([N, D_T], F_T);
q2a_est = mean(sum(rolls, 2));
print_res('2a: avg. troll hp', q2a_exact, q2a_est);

% average damage from fireball
q2a_exact = (2*1 + 3*2 + 4*1) / 4;
rolls = roll_dice([N, D_K], F_K);
q2a_est = mean(sum(rolls, 2));
print_res('2a: avg. fb dmg', q2a_exact, q2a_est);

% probability of dealing more than 3 damage with a fireball
q2a_exact = 1/4;
rolls = roll_dice([N, D_K], F_K);
q2a_est = sum(sum(rolls, 2) > 3) / N;
print_res('2a: P(dmg > 3)', q2a_exact, q2a_est);

Question 2a: avg. troll hp: Exact: 2.500000; Estimate: 2.496060; % Err: 0.157600%
Question 2a: avg. fb dmg: Exact: 3.000000; Estimate: 3.002010; % Err: 0.067000%
Question 2a: P(dmg > 3): Exact: 0.250000; Estimate: 0.251620; % Err: 0.648000%

2b

q2b_exact = ones(1, 4) / 4;
rolls = roll_dice([N D_T], F_T);
q2b_est = histcounts(sum(rolls, 2), 'BinMethod', 'integers') / N;
fprintf('2b: Histogram of Troll HPs (hp={1,2,3,4})\n');
q2b_exact
q2b_est

q2b_exact = [0.25 0.5 0.25];
rolls = roll_dice([N D_K], F_K);
q2b_est = histcounts(sum(rolls, 2), 'BinMethod', 'integers') / N;
fprintf('2b: Histogram of Firebolt Damage (dmg={2,3,4})\n');
q2b_exact
q2b_est

2b: Histogram of Troll HPs (hp={1,2,3,4})

q2b_exact =

    0.2500    0.2500    0.2500    0.2500


q2b_est =

    0.2516    0.2477    0.2506    0.2501

2b: Histogram of Firebolt Damage (dmg={2,3,4})

q2b_exact =

    0.2500    0.5000    0.2500


q2b_est =

    0.2514    0.4993    0.2493

2c

given the trolls, life, we willl evaluate whether or not the fireball can kill the trolls then we do that trial 6 times

%given troll health is 1 or 2, trolls always die, if health is
% 3 then 75% and if health is 4, then 25%

q2c_exact = 0.25*(1/2)^6+0.5*(3/4)^6+0.25;
rolls_troll = reshape(roll_dice([N*6 D_T], F_T), [] ,6);
rolls_fire = sum(roll_dice([N D_K], F_K),2);

diff = rolls_troll - rolls_fire;
maximum = max(diff, [], 2);
q2c_est = 1- sum(maximum > 0)/N;

print_res('2c', q2c_exact, q2c_est);

Question 2c: Exact: 0.342896; Estimate: 0.340680; % Err: 0.646117%

2d

scenarios for 5 dying, one surviving: note that it is only possible that FB=2 or 3 and HP=3 or 4 define the following events: a := HP=4, FB=3, all other HP<=3; P(a) = (1/4)(1/2)(3/4)^5 b := HP=4, FB=2, all other HP<=2: P(b) = (1/4)(1/4)(1/2)^5 c := HP=3, FB=2, all other HP<=2: P(c) = (1/4)(1/4)(1/2)^5 = P(b) P(d) := P(HP=4 ^ 5 dying, one surviving) = P(a) + P(b) P(e) := P(HP=3 ^ 5 dying, one surviving) = P(c) P(f) := P(5 dying, one surviving) = P(a ^ b ^ c) = P(a) + P(b) + P(c) E[HP | 5 dying, 1 surviving] = 4*P(d)/P(f) + 3*P(e)/P(f) (+ 2*0 + 1*0)

Pa = 1/4 * 1/2 * (3/4)^5;
Pb = 1/4 * 1/4 * (1/2)^5;
Pc = Pb;
Pd = Pa + Pb;
Pe = Pc;
Pf = Pa + Pb + Pc;

q2d_exact = 4*Pd/Pf + 3*Pe/Pf;

rolls_T = roll_dice([N*6 D_T], F_T);
rolls_K = roll_dice([N D_K], F_K);

% in simulation, need to select the cases where exactly 5 die
troll_hp = reshape(sum(rolls_T, 2), [], 6);
fire_dmg = sum(rolls_K, 2);
matches = troll_hp - fire_dmg;
matches_where_1_troll_wins = sum(matches > 0, 2) == 1;
matches = matches(matches_where_1_troll_wins, :);
troll_hp = troll_hp(matches_where_1_troll_wins, :);
q2d_est = mean(troll_hp(matches > 0));

print_res('2d', q2d_exact, q2d_est);

Question 2d: Exact: 3.941818; Estimate: 3.940868; % Err: 0.024094%

2e

D_SoT = 2;			% Sword of Tuition
D_HTD = 1;			% Hammer of Tenure Denial
D_hit = 1;			% to hit Keene with sword or hammer
F_SoT = 6;
F_HTD = 4;
F_hit = 20;

q2e_exact = 1/2*(3.5+3.5) + 1/2*1/2*(2.5);
% calculate extra dice rolls here for simplicity; however, due to
% independence it doesn't make affect the result
rolls_hit = reshape(sum(roll_dice([N*2 D_hit], F_hit), 2), [], 2);
rolls_SoT = sum(roll_dice([N D_SoT], F_SoT), 2);
rolls_HTD = sum(roll_dice([N D_HTD], F_HTD), 2);

SoT_damage = rolls_SoT(rolls_hit(:, 1) >= 11);
HTD_damage = rolls_HTD((rolls_hit(:, 1) >= 11) & (rolls_hit(:, 2) >= 11));
q2e_est = (sum(SoT_damage) + sum(HTD_damage)) / N;

print_res('2e', q2e_exact, q2e_est);

2f

when u play dnd with us :)

% helper function
function print_res(question, exact, est)
	pct_err = abs((est - exact) / exact * 100);
	fprintf('Question %s: Exact: %f; Estimate: %f; %% Err: %f%%\n', ...
		question, exact, est, pct_err);
end

% usage: roll_dice([x y])
function res = roll_dice(shape, faces)
	res = randi(faces, shape);
end

Question 1a: Exact: 0.004630; Estimate: 0.004790; % Err: 3.464000%

Published with MATLAB® R2020b