What is the expected number of turns required to get the required roll for Complete Destruction, given no evolutions or other cards?
1 Answer
The probability of getting a Complete Destruction in 1 turn is roughly 20%. So, the expected number of turns is 1/0.2 or 5 turns on average.
The 20% number comes from two sources. First, I found a lengthy discussion at board game geek on this topic, which gives the final result of approximately 20%. http://boardgamegeek.com/thread/1155539/king-tokyo-odds/page/2
Secondly, I wrote a quick Matlab simulation to compute this. The optimal strategy is pretty clearly to take any die that's a duplicate of another die, and reroll it. Using that fact, I get the following table, which performed 100,000 trials:
% of the time, you get 1-6 different numbers
on 6 dice after 3 rolls
1 2 3 4 5 6
Roll 1 : 0.0 2.0 23.2 50.1 23.1 1.6
Roll 2 : 0.0 0.0 3.3 35.2 52.7 8.8
Roll 3 : 0.0 0.0 0.4 17.4 62.3 19.8
Here is the matlab script I used, in case you're interested:
% King of Tokyo, Total Destruction
% Object of total destruction card, is to get 1-6 on 6 different dice
% in three rolls.
NUMTRIALS = 100000;
results.numUniqueRoll1 = zeros(1,6);
results.numUniqueRoll2 = zeros(1,6);
results.numUniqueRoll3 = zeros(1,6);
for ix = 1 : NUMTRIALS
n1 = numel(unique(randi(6, 1, 6)));
results.numUniqueRoll1(n1) = 1 + results.numUniqueRoll1(n1);
n2 = n1 + sum(unique(randi(6,1,6-n1)) > n1);
results.numUniqueRoll2(n2) = 1 + results.numUniqueRoll2(n2);
n3 = n2 + sum(unique(randi(6,1,6-n2)) > n2);
results.numUniqueRoll3(n3) = 1 + results.numUniqueRoll3(n3);
end
fprintf(' %% of the time, you get 1-6 different numbers\n');
fprintf(' on 6 dice after 3 rolls\n');
fprintf(' 1 2 3 4 5 6\n');
fprintf('Roll 1 :');
fprintf('%6.1f', 100*results.numUniqueRoll1/NUMTRIALS);
fprintf('\nRoll 2 :');
fprintf('%6.1f', 100*results.numUniqueRoll2/NUMTRIALS);
fprintf('\nRoll 3 :');
fprintf('%6.1f', 100*results.numUniqueRoll3/NUMTRIALS);
fprintf('\n');