# How can we only use 8192 shots for an experiment with 14 or more qubits?

Let's say you want to do an experiment with 14+ qubits. You apply some arbitrary unitary operator $$U \in (\mathbb{C}^2)^{\otimes n} \times (\mathbb{C}^2)^{\otimes n}$$ to the state $$|\psi\rangle \in (\mathbb{C}^2)^{\otimes n}$$. That is

$$U|\psi \rangle = |\phi \rangle$$

We can take $$|\psi \rangle = |0\rangle^{\otimes n}$$ to fits with current quantum computing setting. Now, if we do this experiment with $$2^{13} = 8192$$ shots, how is this enough to build up the statistical distribution as we have more than $$2^{14}$$ slots to distribute them to. If your output state $$|\phi\rangle$$ is particular eigenstate, says $$|0110\cdots 1 \rangle$$, then this many shots is more than enough. But if $$|\phi\rangle$$ is in a $$2^n$$ superposition state, then how is this enough? We wouldn't have enough experimental data to build up any statistical distribution of the probabilities.

In IBM system, you have the max shots of 8192, but you can do experiment up to 16 qubits device. If I apply Hadamard gate to all the qubit to put them in equal distribution, and making measurement, I wouldn't be able to read out the probabilities as I haven't done enough measurement yet. I mean, I can repeat the experiment with 8192 shots a bunch of times and add up the result to build my statistical distribution. I understand that the 8912 is probably because they don't want people to submit a million shots if it's not needed, like when only an experiment with 2 qubit. For 16 qubits is not a big deal to run the experiment multiple time with 8192 shots and add them up, but what about bigger experiment? I don't have access to their larger system but anyone know if their larger system have more shots allocated? For instance, their 53 qubits system.

Reading out all the probabilities for all the possible output bit strings isn't common in quantum computing. The ideal case is to induce an interference effect that will allow your result to be read out with just one shot. Though that isn't something most algorithms achieve, they nevertheless use only $$O(1)$$ shots, or some other complexity that is far less than $$O(2^n)$$.