How much do you spend on getting attention?
Please try our simple calculator below: you provide estimates based on a typical outreach effort/cycle, and we will show you how much money you are actually spending to get quality attention.
Unit Economics  Per "Approach"
Guessing λ
Now you should put a guess to λ  your agent's likelihood of getting dedicated attention on a typical approach. Usually, λ is not too far off from 1%. When ready, click on "Estimate"Your Effective Cost of Getting Guaranteed Time and Attention
Here is Vintro's estimate on your onaverage effective cost of getting dedicated attention. Remember, it is dedicated attention that really matters.
Businesses seeking resources have a frontline experience about how hard it is. In the conventional world, both physical or digital, "banging" your head in one direction trying to get someone interested can be excruciatingly hard, let alone convincing them to open up their pockets or wield their influence in your favor. Whether you are seeking investors, promoting adoption for your products and services, or trying to find a partnership, we feel your abysmal pains, recurring setbacks, and beleaguering discombobulation. However, there has been little quantitative studies on the costs of finding resources. Some thirdparties provide incidental statistical figures on the cost of "leadgeneration", yet the qualification of a "lead" is often times as bare and as whimsical as an email without name. When seeking resources as a huntergatherer, what you should really care about is dedicated attention and quality interaction. We think that "dedicated attention and quality interaction" is a much more appropriate definition for a "lead", and Vintro is built precisely to deliver these, guaranteed.*
At Vintro, based on concrete and interpretable probability models (rather than fancy blackbox machine learning), we firmly believe that existing methods of outreach are economically inefficient. Don't think so? Try the calculator above to see how much you are currently spending to get a lead of meaningful quality as described above.
 The cost structure of an approach can be quite diverse, but typical elements include: labor cost, travel expenses, demo cost, overhead, and etc. Try putting a number associated with each type cost.
 Next, put an estimate to your likelihood of success in each unit of approach. Usually, this rate can range between 0.1% and 10%, or at most one orderofmagnitude away from 1%.
 Above all, remember that you should fill in numbers corresponding to how much you spend on making one approach to one desired audience member. As such, you can compare the estimated effective cost with the price of pitching to them directly on Vintro.
Here's why  the simple version
For you, the effective cost (EC) for attaining such meaningful leads is the product of your unit approach cost (UAC) and the reciprocal of your success rate per approach (λ). To some people this formula makes sense, to some it might not. We will explain the model and derivation at the back of this article.
\(\text{EC} = \text{UAC} \cdot \dfrac{1}{λ}\)
In reality, averaged across all of your outreach efforts, λ can be painstakingly low. For example, entrepreneurs play the “numbers game” when seeking investment, likely because of the following rule of thumb: for any prominent investor, 1% of startups get into the door, and among those, 1% get the investment. For another example, companies selling enterprisescale solutions with six to seven figure price tags might be spending seven to eight figures on marketing, sales, and all other sorts of overhead yet fail to get the desired exposure to the decision maker all the time.
If your success rate is 10% (although wildly unreasonable), then your effective cost is TEN TIMES your unit approach cost.
If your success rate is 1%, then your EC is ONEHUNDRED TIMES your UAC.
If the odds are even less in your favor, say 0.1%, then your EC is ONETHOUSAND TIMES your UAC.
Simple takeaway: the math really is not in your favor, so that’s why you should go with Vintro.
Here's why  the not so simple version
Assuming that your that you currently maintain an ongoing sales & outreach effort hoping to approach decision makers. An “approach” is a discrete event, yet time is continuous. Let us establish a time metric where each unit of time constitutes the time needed to complete one “approach”. As your sales efforts go on and on, and as time goes by, “success events” in the form of attention might incrementally yet sporadically begin to “arrive”.
In statistical analysis, the arrival process of events, especially rare events, in a continuous metric space is often modeled by a Poisson Arrival Process. The most famous example involves the temporal and spatial modeling of earthquakes. In our analysis of approach success, we are dealing with a Poisson Arrival Process in onedimensional metric space, and our metric is time. The model can be described as follows:

In each infinitesimal time unit (Δt → 0), the probability of an “arrival” is λΔt.

The events of arrivals in nonoverlapping intervals are independent.

The average rate of arrivals per time unit is λ.
Here is a sample picture of what it looks like to observe a Poisson Arrival Process on a timeline:
Now let us examine the applicability of the Poisson model to the case of getting attention:

In an unit of time when an “approach” is done, attentive leads could come in at any time, including those from previous “approaches”.

Until the company is immensely successful by any standard, previous success likely does not bear on current outreach.

Having spent marketing & outreach budget in large figures, either absolutely or relatively, and over an extended period of time, one should be able to reasonably estimate the success rate of “approaches”. That rate easily translates to the time domain given our definition of the time metric.
Given our confidence in the Poisson model's applicability to understanding the success of "approaches", we shall cover some important conclusions from the Poisson Arrival Process as well as their implications in plain words.
First, the distribution of event counts in a given interval of length \(t\) follows the Poisson distribution with rate \(\lambda t\):
$$\mathbb{P}[N(t)=k]=\dfrac{(\lambda t)^k}{k!}e^{\lambda t}$$
The notation \(N(t)\) denotes the number of "arrivals" in the interval \([0,t]\). By the independent condition, also known as the memoryless property, \( N(s+t)N(s)\sim\text{Poisson}(\lambda t)\).
\(T_1,T_2,T_3,\cdots\) denotes the first, second, and third times of arrival. In general, \(T_m\) denotes the time of the \(m^{th}\) arrival.
\(W_1\) is the first waiting time  time between \(0\) and \(T_1\). \(W_2\) is the second waiting time  time between \(T_1\) and \(T_2\), and so on so forth. The waiting time is especially important for our purposes: it directly relates to how much time would need to pass, or how many approach efforts need to be commissioned, till a moment of attention is to finally arrive. To do so, let us begin by calculating the cumulative distribution function (CDF) of \(W_1\), the first waiting time.
$$\begin{split}\mathbb{P}(W_1\leq t)&=1\mathbb{P}(W_1>t)\\&=1\mathbb{P}[N(t)=0]\\&=1e^{\lambda t}\end{split}$$
Taking derivative with respect to \(t\) gives us the probability density function (PDF):
$$\mathbb{P}(W_1=t)=\lambda e^{\lambda t}\Rightarrow W_1\sim\text{Exp}(\lambda)$$
Now let us look at \(W_2\):
$$\begin{split} &\quad\mathbb{P}(W_2\leq tW_1=s)\\&=1\mathbb{P}(W_2>tW_1=s)\\ &=1\mathbb{P}[N(t+s)N(s)=0W_1=s)\\&=1\mathbb{P}[N(t+s)N(s)=0N(s)=1]\\&=11\mathbb{P}[N(t+s)N(s)=0])\\&=1\mathbb{P}[N(t)=0]=1e^{\lambda t}\ (s \text{ is irrelevant})\end{split}$$
\(s\) being irrelevant tells us that \(\mathbb{P}(W_2\leq tW_1=s)=\mathbb{P}(W_2\leq t)\). In other words, \(W_1\) and \(W_2\) are independent. Furthermore, taking the first derivative yields that \(W_2\) follows the exponential distribution with \(\lambda\) as parameter as well.
Using similar reasoning inductively, we claim that:
 \(W_1,W_2,W_3,\cdots\) are independent and identical distributions (i.i.d.)

For \( k = 1,2,3,\cdots\), it holds that \(W_k\sim\text{Exp}(\lambda)\)
This implies that the expected waiting time between "attention" events is:
\(\mathbb{E}W_k = \int_0^\infty x\cdot \lambda e^{\lambda x}=\dfrac{1}{\lambda}\)
In other words, even if an "attention" event can arrive any moment, you should expect on average to spend on \(1/\lambda\) outreach cycles before finally getting an attentive lead.
\(\lambda\)  Approaches Needed 
10%  10 
5%  20 
1%  100 
0.5%  200 
0.1%  1000 
Yet if you go with Vintro, as long as the person is available in our system, the number is ONE. If not, you get your money back.
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