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I want to generate uniformly distributed random numbers between 1 and 26 with a die: Is this correct:

I have assembled the following algorithm using the Monte Carlo Method:

{1, 2, 3, 4, 5, 6} {7, 8, 9, 10, 11, 12} {13, 14, 15, 16, 17, 18} {19, 20, 21, 22, 23, 24} {25, 26, 0, 0, 0, 0} {0, 0, 0, 0, 0, 0}

First step: Throw the die. If the die is $x$, I choose the $x$-th set.

Second step: Throw the die. If the dice is $y$, I choose the $y$-th element of the chosen set.
If the element is zero -> return to first step and repeat
Else -> return the chosen element as the result.

Since the two steps are independent of each other, each outcome has probability 1/36, and hence, the algorithm has equally distributed output of a number between 1 and 26.

Is this reasoning correct and if not, what should I change? Is there an even easier way?

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  • $\begingroup$ If "zero -> repeat" means "repeat from the first step", then yes, it is uniformly distributed in 1 to 26, since it is equivalent to rolling a "36 sided die" and re-rolling if you get 27 or higher. However, I do not see how you are going to do encryption with this, it seems that you are generating a random sequence of letters instead. $\endgroup$
    – Lucas Prates
    Commented Aug 1, 2021 at 12:05
  • $\begingroup$ How does this encrypt text? It looks like it just generates random numbers. $\endgroup$
    – whuber ♦
    Commented Aug 1, 2021 at 13:48
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    $\begingroup$ hey there, thanks for your comments, I have a given text, say: "hello", I take the h transform it into 8, then I role the dice two times to get an evenly distributed number between 1 and 26, say for example 2 and hence the encrypted letter will be 10 which is j, I will do this with all letters of the text. The result will be a sheet with the encrypted text and a sheet with the generated random numbers. If I want to decrypt I subtract the generated number on the sheet, say here 2 from the encrypted letter, say 10 and I get 10 - 2 = 8 = h $\endgroup$
    – Lukas-Santo Puglisi
    Commented Aug 1, 2021 at 14:28
  • $\begingroup$ but you are right, I will edit my question since it is not clear how I will use the above discribed algorithm to encrypt a text $\endgroup$
    – Lukas-Santo Puglisi
    Commented Aug 1, 2021 at 15:09
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    $\begingroup$ The first duplicate answers your question. The other duplicates are very closely related and include additional methods. $\endgroup$
    – whuber ♦
    Commented Aug 1, 2021 at 15:36

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