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Hi,

You might also find it useful to look at this page:
  http://fsl.fmrib.ox.ac.uk/fsl/fslwiki/Randomise/Theory#Monte_Carlo_Permutation_Tests
which gives a table that relates the number of permutations to the confidence interval on the p-values, which is one of the most useful ways of determining how many permutations are needed.

All the best,
Mark


On 29 Jun 2014, at 13:59, Anderson M. Winkler <[log in to unmask]> wrote:

Hi,

You can choose the number you'd like, not just 500, 5000 or 10000. If you use -n 8423, it will run 8423 permutations. Unless there are too few subjects or too many ties in the design so that the total number of possible unique shufflings is smaller than what's requested. In that case, randomise will run the max possible number of permutations, even if less than requested.

You don't have to compute the number of permutations by hand, although we provided the formulas and described all this here. More is always better, and in general it's a good idea to use a few thousands.

All the best,

Anderson



On 29 June 2014 13:19, Xinfa Shi <[log in to unmask]> wrote:
Hi,

Is there any other number of permutations except 500 and 5000 to select  ? Could I understand it this way:first compute the number of permutations ,and then if n<500,selecting 500; n>500,selecting 5000 or 10000?

Regards .

Xin-Fa Shi .




2014-06-09 14:23 GMT+08:00 Anderson M. Winkler <[log in to unmask]>:

Hi,
I'm not checking the final value, but the formula is right. It's not necessary to run them all. Use instead 5000 or 10000 permutations (option -n).
Anderson



On 9 June 2014 04:18, SUBSCRIBE FSL Xinfa_Shi <[log in to unmask]> wrote:
Hi,

Now I have 15 subjects in control group and 18 subjects in patient group .If I make a two-sample unpair t-test in randomise ,the number of  permutations is (15+18)! /(15! x 18!) =1037158366 .My god ,that’s too large . Is it right ?

Best regards.

Xinfa Shi .