Let's start estimating the probability of a single user receiving repeated faulted machines; we need to define some parameters and assumptions:

### Definitions:

**Probability of a Faulted Machine (Pf):** The chance that any given machine is faulty.
**Number of Replacements (n):** How many times you've received a replacement, including the original purchase. In this case, it's 4 (original + 3 replacements).

Using the binomial probability formula, the probability of receiving *k* faulted machines out of *n* total machines is:

*P(X=k) = (n choose k) × Pf*^{k} × (1-Pf)^{n-k}

However, in my case, I've received faulty machines every single time, so *k = n*. Thus, the formula becomes:

*P(X=n) = Pf*^{n}

If we knew the actual probability *Pf* (i.e., the fraction of machines that are faulty), we could calculate this directly. Lets assume a high number, say 10% of all machines are faulty (*Pf=0.10*), and I've received 4 defective machines in a row:

*P(X=4) = 0.10*^{4} = 0.0001 = 0.01%

This means there's a 0.01% chance of receiving 4 defective machines in a row if 10% of all machines are faulty.

However, there's a nuance: The chance of several users experiencing the same repeated issues suggests that the probability isn't just random bad luck. If multiple users are consistently receiving faulty units, *Pf* might be higher than estimated, there could be systemic issues in certain batches, or the way the refurbished units are handled might be problematic.

To thoroughly assess the situation, we can estimate now the case for multiple users

## New definition:

**j**: Number of users experiencing the repeated fault.

## Scenario:

Four users (**j=4**) each receiving four faulty machines in a row (**n=4**).

### Assumption:

All users get machines from the same pool of refurbished machines, and the probability of getting a faulty machine remains constant across all picks.

## Calculations:

The probability that a single user gets four faulty machines in a row is:

*P*_{1} = Pf^{4}

The probability that four users each get four faulty machines in a row is the fourth power of *P*_{1}:

*P*_{k} = (Pf^{4})^{4} = Pf^{16}

Now, let's solve for *Pf* in the equation *P*_{k} = Pf^{16} given a specific *P*_{k}.

For simplicity, let's use the probability we got for one single user in the initial calculations above, *P*_{k} = 0.0001 (0.01% chance that all four users would get four faulty machines in a row). Then,

*0.0001 = Pf*^{16}

To solve for *Pf*, take the 16th root of both sides:

*Pf = 0.0001*^{1/16}

Calculating this value, *Pf ≈ 0.38* or 38%.

This means that for there to be a 0.01% chance of four users each receiving four faulty machines consecutively from the pool, approximately 38% of the refurbished machines in that pool would need to be faulty, that this a lot!

Of course, we need to keep in mind that this is a simplified model and the actual probability can be influenced by various factors. However, this gives a basic understanding of **how high the faulty rate needs to be**, in order to observe such a number of repeated faulty machines by several users under the defined assumptions.