Azzera filtri
Azzera filtri

Select a random number from a set

222 visualizzazioni (ultimi 30 giorni)
Matt Medley
Matt Medley il 8 Nov 2011
Risposto: Souarv De il 8 Giu 2021
I'm simulating a single blackjack hand and am trying to "draw" a card. I can't use the randi function because I don't want all possibilities to have the same probability of being selected. So what I have done is created a row vector , (x), of all the possible card values. Now I would like draw a random number from this selection for my 'draw' function. Thanks for the help!
x = [1,2,3,4,5,6,7,9,10,10,10,10,11] theCard = randi?
  5 Commenti
Walter Roberson
Walter Roberson il 8 Gen 2021
T1 = randperm(9)
T1 = 1×9
9 5 7 2 1 6 8 4 3
reshape(T1, 3, 3)
ans = 3×3
9 2 8 5 1 4 7 6 3
T2 = randperm(9)
T2 = 1×9
1 6 7 3 8 9 2 5 4
reshape(T2, 3, 3)
ans = 3×3
1 3 2 6 8 5 7 9 4
You can see that the randperm(9) call is picking values randomly, and the reshape is making a 3 x 3 out of it.
Or perhaps you mean that you have created an existing 3 x 3 matrix and you want to take a random value out of it?
T3 = [11 12 13
21 22 23
31 32 33]
T3 = 3×3
11 12 13 21 22 23 31 32 33
T3(randi(numel(T3)))
ans = 22
T3(randi(numel(T3)))
ans = 21
Nour Ahmed
Nour Ahmed il 9 Gen 2021
Yesss thankk youuu

Accedi per commentare.

Risposte (2)

Lucas García
Lucas García il 8 Nov 2011
There are a few ways to do it. For example, using randi to select in which position is the card that you will extract.
pos = randi(length(x));
card = x(pos);
  5 Commenti
N/A
N/A il 29 Gen 2019
here is a way to do it :
for i = 1:5
arrayA(i) = randi(length(A));
arrayB(i) = randi(length(B));
end
Vineeth Krishnan
Vineeth Krishnan il 11 Dic 2020
for i=1:5
arrayA(i) = A(randi(length(A));
arrayB(i)= B(randi(length(B));
end

Accedi per commentare.


Souarv De
Souarv De il 8 Giu 2021
rand_Pos = randperm(length(x),1)
card = x(rand_Pos)

Categorie

Scopri di più su Card games in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by