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PPS: As suggested in one of the other answers, let's assume the Bell state $|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B$ is the input entangled state and that A and B then measure with 60 degrees axis difference, rotated in the $z-x$ plane. For spin-$\tfrac12$ states this gives the Schmidt decomposition: $$ |\psi\rangle = \tfrac14\sqrt6 \,|0\rangle_A|0\rangle_B + \tfrac14\sqrt2 \, |0\rangle_A|1\rangle_B + \tfrac14\sqrt2 \, |1\rangle_A|0\rangle_B -\tfrac14\sqrt6 \,|1\rangle_A|1\rangle_B $$ If A and B repeat the proces $N$ times we end up with $$ \begin{align} |\psi\rangle^{\otimes\,N} &= \Big(\tfrac12\sqrt3 \ \frac{|0\rangle_A|0\rangle_B-|1\rangle_A|1\rangle_B}{\sqrt2} +\tfrac12 \,\frac{|0\rangle_A|1\rangle_B+|1\rangle_A|0\rangle_B}{\sqrt2}\Big)^{\otimes\,N} \\ &= \big(\tfrac12\sqrt3 \ |C_+\rangle + \tfrac12 \,|C_-\rangle\big)^{\otimes\,N} \end{align} $$ where the normalized $|C_+\rangle$ and $|C_-\rangle$ denote the combined terms with matching and non-matching readings in one iteration. After $N$ measurements, creating $2^N$ branches in the superposition, we can then evaluate this tensor product and find those superposition terms where exactly $n$ of the $N$ results have matching outcomes (and $N-n$ have opposite outcomes). There are $\tbinom{N}{n}$ of those terms, and each has amplitude $(\tfrac12\sqrt3)^n\ (\tfrac12)^{N-n}$, and they are all orthogonal in the tensor product space. So if we add them up and call the combined sum term $|\Psi_n\rangle$, then it has amplitude: $$ || \Psi_n || = \binom{N}{n}^{1/2} (\tfrac12\sqrt3)^n\ (\tfrac12)^{N-n} \tag{a} $$ The total superposition can then be written as: $$ |\psi_{\,\text{Tot}}\rangle = |\psi\rangle^{\otimes\,N} = \sum_{n=0}^N \ |\Psi_n\rangle, \tag{b} $$ nicely splitting it up into terms where A and B have matching results in exactly $n$ of the $N$ iterations. Now let's observe that the amplitudes $(a)$ are precisely the square roots of the binomial expansion terms in: $$ \big(\tfrac34 + \tfrac14\big)^N = \sum_{n=0}^N \ \binom{N}{n}\ (\tfrac34)^n\ (\tfrac14)^{N-n} \tag{c} $$ and we know that those terms get a very sharp peaking around $n=\tfrac34 N$, so their square roots will also have a peaking around that $n$ value, somewhat less pronounced by the square root, but in the large-$N$ limit still very strong.

$\color{red}{\large \bf PPS:}$ As suggested in one of the other answers, let's assume the Bell state $|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B$ is the input entangled state and that A and B then measure with 60 degrees axis difference, rotated in the $z-x$ plane. For spin-$\tfrac12$ states this gives the Schmidt decomposition: $$ |\psi\rangle = \tfrac14\sqrt6 \,|0\rangle_A|0\rangle_B + \tfrac14\sqrt2 \, |0\rangle_A|1\rangle_B + \tfrac14\sqrt2 \, |1\rangle_A|0\rangle_B -\tfrac14\sqrt6 \,|1\rangle_A|1\rangle_B $$ If A and B repeat the proces $N$ times we end up with $$ \begin{align} |\psi\rangle^{\otimes\,N} &= \Big(\tfrac12\sqrt3 \ \frac{|0\rangle_A|0\rangle_B-|1\rangle_A|1\rangle_B}{\sqrt2} +\tfrac12 \,\frac{|0\rangle_A|1\rangle_B+|1\rangle_A|0\rangle_B}{\sqrt2}\Big)^{\otimes\,N} \\ &= \big(\tfrac12\sqrt3 \ |C_+\rangle + \tfrac12 \,|C_-\rangle\big)^{\otimes\,N} \end{align} $$ where the normalized $|C_+\rangle$ and $|C_-\rangle$ denote the combined terms with matching and non-matching readings in one iteration. After $N$ measurements, creating $2^N$ branches in the superposition, we can then evaluate this tensor product and find those superposition terms where exactly $n$ of the $N$ results have matching outcomes (and $N-n$ have opposite outcomes). There are $\tbinom{N}{n}$ of those terms, and each has amplitude $(\tfrac12\sqrt3)^n\ (\tfrac12)^{N-n}$, and they are all orthogonal in the tensor product space. So if we add them up and call the combined sum term $|\Psi_n\rangle$, then it has amplitude: $$ || \Psi_n || = \binom{N}{n}^{1/2} (\tfrac12\sqrt3)^n\ (\tfrac12)^{N-n} \tag{a} $$ The total superposition can then be written as: $$ |\psi_{\,\text{Tot}}\rangle = |\psi\rangle^{\otimes\,N} = \sum_{n=0}^N \ |\Psi_n\rangle, \tag{b} $$ nicely splitting it up into terms where A and B have matching results in exactly $n$ of the $N$ iterations. Now let's observe that the amplitudes $(a)$ are precisely the square roots of the binomial expansion terms in: $$ \big(\tfrac34 + \tfrac14\big)^N = \sum_{n=0}^N \ \binom{N}{n}\ (\tfrac34)^n\ (\tfrac14)^{N-n} \tag{c} $$ and we know that those terms get a very sharp peaking around $n=\tfrac34 N$, so their square roots will also have a peaking around that $n$ value, somewhat less pronounced by the square root, but in the large-$N$ limit still very strong.

PPS: As suggested in one of the other answers, let's assume the Bell state $|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B$ is the input entangled state and that A and B then measure with 60 degrees axis difference, rotated in the $z-x$ plane. For spin-$\tfrac12$ states this gives the Schmidt decomposition: $$ |\psi\rangle = \tfrac14\sqrt6 \,|0\rangle_A|0\rangle_B + \tfrac14\sqrt2 \, |0\rangle_A|1\rangle_B + \tfrac14\sqrt2 \, |1\rangle_A|0\rangle_B -\tfrac14\sqrt6 \,|1\rangle_A|1\rangle_B $$ If A and B repeat the proces $N$ times we end up with $$ \begin{align} |\psi\rangle^{\otimes\,N} &= \Big(\tfrac12\sqrt3 \ \frac{|0\rangle_A|0\rangle_B-|1\rangle_A|1\rangle_B}{\sqrt2} +\tfrac12 \,\frac{|0\rangle_A|1\rangle_B+|1\rangle_A|0\rangle_B}{\sqrt2}\Big)^{\otimes\,N} \\ &= \big(\tfrac12\sqrt3 \ |C_+\rangle + \tfrac12 \,|C_-\rangle\big)^{\otimes\,N} \end{align} $$ where the normalized $|C_+\rangle$ and $|C_-\rangle$ denote the combined terms with matching and non-matching readings in one iteration. After $N$ measurements, creating $2^N$ branches in the superposition, we can then evaluate this tensor product and find those superposition terms where exactly $n$ of the $N$ results have matching outcomes (and $N-n$ have opposite outcomes). There are ${\small \begin{pmatrix} N \\ n \end{pmatrix}}$ of those terms, and each has amplitude $(\tfrac12\sqrt3)^n\ (\tfrac12)^{N-n}$, and they are all orthogonal in the tensor product space. So if we add them up and call the combined sum term $|\Psi_n\rangle$, then it has amplitude: $$ || \Psi_n || = {\small \begin{pmatrix} N \\ n \end{pmatrix}}^{1/2} (\tfrac12\sqrt3)^n\ (\tfrac12)^{N-n} \tag{a} $$ The total superposition can then be written as: $$ |\psi_{\,\text{Tot}}\rangle = |\psi\rangle^{\otimes\,N} = \sum_{n=0}^N \ |\Psi_n\rangle, \tag{b} $$ nicely splitting it up into terms where A and B have matching results in exactly $n$ of the $N$ iterations. Now let's observe that the amplitudes $(a)$ are precisely the square roots of the binomial expansion terms in: $$ \big(\tfrac34 + \tfrac14\big)^N = \sum_{n=0}^N \ {\small \begin{pmatrix} N \\ n \end{pmatrix}}\ (\tfrac34)^n\ (\tfrac14)^{N-n} \tag{c} $$ and we know that those terms get a very sharp peaking around $n=\tfrac34 N$, so their square roots will also have a peaking around that $n$ value, somewhat less pronounced by the square root, but in the large-$N$ limit still very strong.

PPS: As suggested in one of the other answers, let's assume the Bell state $|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B$ is the input entangled state and that A and B then measure with 60 degrees axis difference, rotated in the $z-x$ plane. For spin-$\tfrac12$ states this gives the Schmidt decomposition: $$ |\psi\rangle = \tfrac14\sqrt6 \,|0\rangle_A|0\rangle_B + \tfrac14\sqrt2 \, |0\rangle_A|1\rangle_B + \tfrac14\sqrt2 \, |1\rangle_A|0\rangle_B -\tfrac14\sqrt6 \,|1\rangle_A|1\rangle_B $$ If A and B repeat the proces $N$ times we end up with $$ \begin{align} |\psi\rangle^{\otimes\,N} &= \Big(\tfrac12\sqrt3 \ \frac{|0\rangle_A|0\rangle_B-|1\rangle_A|1\rangle_B}{\sqrt2} +\tfrac12 \,\frac{|0\rangle_A|1\rangle_B+|1\rangle_A|0\rangle_B}{\sqrt2}\Big)^{\otimes\,N} \\ &= \big(\tfrac12\sqrt3 \ |C_+\rangle + \tfrac12 \,|C_-\rangle\big)^{\otimes\,N} \end{align} $$ where the normalized $|C_+\rangle$ and $|C_-\rangle$ denote the combined terms with matching and non-matching readings in one iteration. After $N$ measurements, creating $2^N$ branches in the superposition, we can then evaluate this tensor product and find those superposition terms where exactly $n$ of the $N$ results have matching outcomes (and $N-n$ have opposite outcomes). There are $\tbinom{N}{n}$ of those terms, and each has amplitude $(\tfrac12\sqrt3)^n\ (\tfrac12)^{N-n}$, and they are all orthogonal in the tensor product space. So if we add them up and call the combined sum term $|\Psi_n\rangle$, then it has amplitude: $$ || \Psi_n || = \binom{N}{n}^{1/2} (\tfrac12\sqrt3)^n\ (\tfrac12)^{N-n} \tag{a} $$ The total superposition can then be written as: $$ |\psi_{\,\text{Tot}}\rangle = |\psi\rangle^{\otimes\,N} = \sum_{n=0}^N \ |\Psi_n\rangle, \tag{b} $$ nicely splitting it up into terms where A and B have matching results in exactly $n$ of the $N$ iterations. Now let's observe that the amplitudes $(a)$ are precisely the square roots of the binomial expansion terms in: $$ \big(\tfrac34 + \tfrac14\big)^N = \sum_{n=0}^N \ \binom{N}{n}\ (\tfrac34)^n\ (\tfrac14)^{N-n} \tag{c} $$ and we know that those terms get a very sharp peaking around $n=\tfrac34 N$, so their square roots will also have a peaking around that $n$ value, somewhat less pronounced by the square root, but in the large-$N$ limit still very strong.

In your analysis you frequently mention "wave function collapse". But there is no reason to assume such a thing exists. Instead we can assume there is only subjective collapse. By insisting that it occurs you are limiting yourself to "objective collapse theories". You are ruling out the many-worlds theory and probably some similar ones that go by other names.

PS: I see that you now edited the question to address this fourth assumption, about the collapse. Let me add that when you write "thus splitting reality into four branches" that may be too suggestive, the situation is a superposition, which is a normal concept in QM, we never call that a "splitting of reality" in other cases. As for your question about circumventing Bell, we now have, for the 4 branches after the measurements:

$\color{red}{\large \text{(For the mathematics & physics: see example at end!)}}$

In your analysis you frequently mention "wave function collapse". But there is no reason to assume such a thing exists. Instead we can assume there is only subjective collapse. By insisting that it occurs you are limiting yourself to "objective collapse theories". You are ruling out the many-worlds theory and probably some similar ones that go by other names.

(EDIT) PS: I see that you now edited the question to address this fourth assumption, about the collapse. When you write "thus splitting reality into four branches", that may be a bit suggestive. The situation is a superposition, which is a normal concept in QM, we never call that a "splitting of reality" in other cases. As for your question about circumventing Bell, we now have, for the 4 branches after the measurements: