I largely do not understand your question. To clear up things  a bit: Gleason's theorem is a theorem about functions on the unit sphere in 3 dimensional space, and only has a connection with quantum mechanics in a somewhat indirect manner. Gleason's theorem uses no QM in the proof.

What Gleason's theorem states goes as follows: let f(e) be a function from the unit sphere to the [0,1] interval, and further, whenever e1, e2 and e3 are a triplet of orthogonal unit vectors on the sphere, let there hold

f(e1) + f(e2) + f(e3) = 1             (1)

then f(e) is continuous. From this follows, as a corollary, that there exists a positive hermitian matrix rho with trace 1 such that

f(e) = (e, rho e)

so that such a function corresponds to a density matrix.

Now, continuity means that the value of f(e) changes little when e changes little. This is clearly not compatible with the assumption that f(e) only takes values 0 and 1. So clearly, a function f(e) defined as above cannot only take values 0 and 1, and must, indeed, be given by a density matrix.

Now what do these things correspond to in QM? The e are one dimensional projectors on the corresponding unit vectors (or more precisely, rays), that is, they are observables. A set of e1, e2 and e3, all of which are orthogonal, corresponds to a complete set of commuting observables, since projectors on orthogonal vectors commute.

A state now should assign to every observable, that is, to every e, a probability of the projector's having the value 1. Since projectors are observables which only permit values of 0 and 1, this is all that is needed.

Assume now a deterministic state. This must tell us unambiguously whether, in that state, e is true or not, that is, whether it will be one or zero. It thus corresponds to a function f(e) which only takes values 1 or 0.

Such a function then cannot satisfy (1), by Gleason's theorem. In fact, that there is no function f(e) having only 0 and 1 values follows without the whole (difficult) proof of Gleason: rather, a simpler argument due to Kochen and Specker is enough.

But why should it satisfy (1)? It follows from QM (and from experiment) that whenever I measure one such triplet simultaneously (which is possible, by commutativity), then one and only one is 1, the others being zero. But from this knowledge alone (1) does not follow: (1) makes the stronger statement that the probability of measuring e1 to be true, is independent of which other commuting observables it is measured in conjunction with.

Thus, according to (1), e1 has the same probability when measured as part of the orthogonal set e1, e2 and e3 as measured as part of e1, e2' and e3'. This cannot be checked, since e2', say, does not commute with e2 and e3. Ad far as I can see, this is the issue of contextuality.

My question pertains to how he derives of these three orthogonal events; if they have  equal mod squared amplitudes that they must have equal probabilities. Is it because the values of e1 e2 and e3 are the same, and thus as there is no diference in their values at all then f(e1), f(e2) and f(e3) are the same. Or that if e1 gives outcome spin up, then e2 and e3 both give spin down, and conversely e2 gives spin down(not spin up) then ~BOTH(e1 and e2 give down) ie equals P(e1 v e2 being spin up) by equal truth values in the triplet state. And iff it a a probability function, P(e1) spin up + P(e2) spin up, thus we get P(spin down) on any particle1> P(spin up) on particle 2, >spin (up) particle 3 and for all permuations.

How does he get to

(A)P(spin down on particle 1) on particle one>p(spin up on particle1 ),

presumably there is contextuality going on that allows one to

(B) say P(spin up on particle 1)= P(spin up on particle 2)

Read my earlier comment carefully. Gleason's theorem is about functions on the unit sphere with values on the [0,1] interval. It does not ``derive'' events e1, e2 and e3. It *assumes* that for all orthogonal triplets e1, e2, e3, the sum

f(e1) + f(e2) + f(e3) = 1

Given this property, and the fact that the values of f are between 0 and 1, he obtains that f is continuous in e, and hence equal to the usual QM form for a density matrix. The latter is non-trivial, but mathematically straightforward. It is the continuity proof that is mathematically tricky.

Perhaps I was looking for the closest example of concrete events where it could be applied to give  a particular case. Would a simple example is given by three entangled particles sigma x particle 1 +sigma y + sigma z. I presume one can characterise this being the quantum state if ally three spin 1 particles with 1/sqrt3 amplitudes for each of three eigenstates entangled so that its always the case that one of the particles gets 1, every time but its always a different particle, so that we have as possible state of three particle entanglements (10-1), (0-1 1),( -1 1 0 )or some such and then one

can identify Pr(1) on the first particle in spin x, as counterfactually having  the same probability or to be identified with Pr(-1) on the first particle on the -x direction or some such by (conditional logic counterfactual truth values) or equal truth values/rotation  on the ciricle(all bases exist),

So Pr(1)x first particle=Pr(1)-x first particle , now define -x as a new context call it 2

by entanglement i in this  new context where the first particle is rotated around to absis  -x call itcontext 2 if the same entanglement correlation is preserved  (if this preserves the same correlation) this must result in necessarily (1 0) on particles 2, 3 , so by necessarily equal truth values (by equal truth values) via entanglement P(1,0) particle 2, particle 3 in context 2=p(-1) on particle 1 in context 2, and by subsitution P(1) x particle 1 context 1= P(1,0) context 2. Then by a non contextuality assumption P(1,0) particle 2,3 context 2=P(1,0) particle, 2,3 context 1=p(1) particle 2 context 1, but (1,0)2,3 only in context 1 occurs iff -1 occurs on particle 1 in context 1 so P(-1) particle1x context1 = (1,0)2,3 context 1=P(1)particle 1x. That gives an equality of two orthogonal equal amplitude events for the first particle, I dont know if this is a concrete special case, that is similar to gleason and i dont know if the qm is right, its not my speciality but I presume that this might be analogous to it, I presume then one can either rotate the other particle or consider another rotation context (i use rotations which makes it similar to the sphere model i gues) where particle 1 must come out zero if it comes out 1 in first context, where the entanglement is preserved, find that entangled set of results for other two particles, use non-contextuality, move it to the first context, and then map that to event on for of particle 1 whose  measurement is truth value equivalent to which would be P(0)  so  particle 1 if it comes out 1 in the first context, and then use that to show thus P(1)particle 1.(which I presume is not explicitly called this in his proof)  is equal P(-1 0 ) on particles, 2, 3 on context 1, necessarily  which by non-contextuality of probability is equal to now P(1)=P(o)=P(-1)=1/3 and then likewise for the other three. I am not sure if this runs particularly if one cannot rotate it to get all three different result eigenvalues.

Maybe this argument does not work in singlet state for some reason with two spin 1/2 particle both measured in the same direction; it would be difficult to give a concrete example without entanglement with a spin 1 particle as i presume one cannot measure in -x and whilst holding fixed that one is still measuring in x on the same particle for the orthogonal direction (unless some of these rotated angles bases commute with each other?), and . So maybe it he didnt explicitely use entanglement but I think the concrete example would amount to something like that