主题：【文摘】一道有趣的概率题 -- 天下第一银杏树

N = 1e3; % total # of runs

% behind door # 1 is the car. This configuration is fixed, while your draws are random.

Upick = ceil(3*rand(N,1)); % the door you picked in each of N runs.

P_no_swtch = length(find(Upick==1))/N; % probability of getting the car if you don't switch

SWTCH = [0, 3, 2;

3, 0, 1;

2, 1, 0]; % SWTCH(a,b) denotes the target door as you switch, when you have

% picked door #a, and the host has showed door #b.

P_swtch = 0; % Initialize: probability of getting the car if you do switch

for k = 1:N

host = ceil(3*rand(1)); % the host randomly picks a door, #1->#3

while host == Upick(k) || host ==1 % this door has to be different

host = ceil(3*rand(1)); % from your door, and it has to reveal a goat.

end

NewChoice= SWTCH(Upick(k), host);

if NewChoice == 1 % is a car

P_swtch = P_swtch + 1; % accumulate the hits

end

end

P_swtch = P_swtch/N;

%% Conclusion:

[P_no_swtch, P_swtch]

% == [0.3320, 0.6680], i.e., [1/3, 2/3]. Therefore switching is preferable.

• 再谈第二题