If letters 1 and 2 must be equal with p=0.5, and 1 and 3 must be equal with
p=0.5, then letter 1 must be the same as either 2 or 3. Therefore:
Choose a letter.
Make a pair of (letter, (not letter)).
Reverse the pair with p = 0.5
Concatenate your letter and the pair.
Is that what you need?
B.
On Jun 2, 2015, at 8:26 AM, Benjamin Ward (ENV) <b.w...@uea.ac.uk> wrote:
> Dear R-List,
>
> I have a set of possibilities I want to sample from:
>
> bases <- list(c('A', 'C'), c('A', 'G'), c('C', 'T'))
> possibilities <- as.matrix(expand.grid(bases))
>
>> possibilities
> Var1 Var2 Var3
> [1,] "A" "A" "C"
> [2,] "C" "A" "C"
> [3,] "A" "G" "C"
> [4,] "C" "G" "C"
> [5,] "A" "A" "T"
> [6,] "C" "A" "T"
> [7,] "A" "G" "T"
> [8,] "C" "G" "T"
>
> If I want to randomly sample one of these rows. If I do this, I find that it
> is 25% likely that my choice will have an identical first and last letter
> (e.g. [1,] "A" "A" "C"). It is also 25% likely that my choice will have an
> identical first and third letter (e.g. [4,] "C" "G" "C"). It is not likely
> at all that the second and third letter of my choice could be identical.
>
> What I would like to do, is sample one of the rows, but given the constraint
> that the probability of drawing identical letters 1 and 2 should be 50% or
> 0.5, and at the same time the probability of drawing identical letters 1 and
> 3 should be 50%. I am unsure on how to do this, but I know it involves coming
> up with a modified set of weights for the sample() function. My progress is
> below, any advice is much appreciated.
>
> Best Wishes,
>
> Ben Ward, UEA.
>
>
> So I have used the following code to come up with a matrix, which contains
> weighting according to each criteria:
>
> possibilities <- as.matrix(expand.grid(bases))
> identities <- apply(possibilities, 1, function(x) c(x[1] == x[2], x[1] ==
> x[3], x[2] == x[3]))
> prob <- matrix(rep(0, length(identities)), ncol = ncol(identities))
> consProb <- apply(identities, 1, function(x){0.5 / length(which(x))})
> polProb <- apply(identities, 1, function(x){0.5 / length(which(!x))})
> for(i in 1:nrow(identities)){
> prob[i, which(identities[i,])] <- consProb[i]
> prob[i, which(!identities[i,])] <- polProb[i]
> }
> rownames(prob) <- c("1==2", "1==3", "2==3")
> colnames(prob) <- apply(possibilities, 1, function(x)paste(x, collapse = ",
> "))
>
> This code gives the following matrix:
>
> A, A, C C, A, C A, G, C C, G, C A, A,
> T C, A, T A, G, T C, G, T
> 1==2 0.25000000 0.08333333 0.08333333 0.08333333 0.25000000 0.08333333
> 0.08333333 0.08333333
> 1==3 0.08333333 0.25000000 0.08333333 0.25000000 0.08333333 0.08333333
> 0.08333333 0.08333333
> 2==3 0.06250000 0.06250000 0.06250000 0.06250000 0.06250000 0.06250000
> 0.06250000 0.06250000
>
> Each column is one of the choices from 'possibilities', and each row gives a
> series of weights based on three different criteria:
>
> Row 1, that if it possible from the choices for letter 1 == letter 2, that
> combined chance be 50%.
> Row 2, that if it possible from the choices for letter 1 == letter 3, that
> combined chance be 50%.
> Row 3, that if it possible from the choices for letter 2 == letter 3, that
> combined chance be 50%.
>
> So:
>
> If I used sample(x = 1:now(possibilities), size = 1, prob = prob[1,])
> repeatedly, I expect about half the choices to contain identical letters 1
> and 2.
>
> If I used sample(x = 1:now(possibilities), size = 1, prob = prob[2,])
> repeatedly, I expect about half the choices to contain identical letters 1
> and 3.
>
> If I used sample(x = 1:now(possibilities), size = 1, prob = prob[3,])
> repeatedly, I expect about half the choices to contain identical letters 2
> and 3. Except that in this case, since it is not possible.
>
> Note each row sums to 1.
>
> What I would like to do - if it is possible - is combine these three sets of
> weights into one set, that when used with
> sample(x = 1:nrow(possibilities, size = 1, prob = MAGICPROB) will give me a
> list of choices, where ~50% of them contain identical letters 1 and 2, AND
> ~50% of them contain identical letters 1 and 3, AND ~50% again contain
> identical letters 2 and 3 (except in this example as it is not possible from
> the choices).
>
> Can multiple probability weightings be combined in such a manner?
>
>
>
>
> [[alternative HTML version deleted]]
>
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