Rolling a Fair 6-Sided Die Until Two Consecutive Rolls Have the Same Value

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Sophie Culhane
Sophie Culhane 2020-10-2
评论： Sophie Culhane 2020-10-5
My task is to simulate rolling a fair 6-sided die until two consecutive rolls have the same value and write a function that approximates the expected value EV. I believe I have most of the program figured out until I reach 'same number' in which I get lost. Please help me find out what to put in place of 'same number' to get my program to work. Here is my block of code:
function EV = hw20(NTrials)
%
%
rng('shuffle)
TotalNRolls = 0;
for Trial = 1:NTrials
die = randi(6);
TotalNRolls = TotalNRolls + 1;
while die == 'same number'
TotalNRolls = TotalNRolls + 1;
end
end
EV = TotalNRolls/NTrials;

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回答（2 个）

Ameer Hamza
Ameer Hamza 2020-10-2
编辑：Ameer Hamza 2020-10-2
Try this code
function EV = hw20(NTrials)
%
%
rng('shuffle')
TotalNRolls = 1;
last_die = inf; % any number other then 1 to 6
while true
die = randi(6);
if die == last_die || TotalNRolls==NTrials
break;
else
last_die = die;
end
TotalNRolls = TotalNRolls + 1;
end
EV = TotalNRolls/NTrials;
end
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Ameer Hamza
Ameer Hamza 2020-10-2
Thanks, Steven. Your approach also makes sense for this problem. It is just my guess that NTrials is the maximum number of trials.
Sophie Culhane
Sophie Culhane 2020-10-5
We have not yet learned 'break' therefore I cannot use that in my program. Thank you for the responses so far.

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Alan Stevens
Alan Stevens 2020-10-2
I think the above will produce a probability, rather than an expected value. Try
NTrials = 1000;
N = zeros(NTrials,1);
for Trial = 1:NTrials
TotalNRolls = 1;
die = randi(6);
keepgoing = true;
while keepgoing == true
oldie = die;
die = randi(6);
if die == oldie
keepgoing = false;
end
TotalNRolls = TotalNRolls + 1;
end
N(Trial) = TotalNRolls;
end
EV = sum(N)/NTrials;

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