Genetic linkage greater than 50 centimorgans
Understanding the statistics we use when talking about recombination rate is an important question that is unfortunately too often dismissed in an intro course to evolutionary biology or population genetics and that is misunderstood by many. I have a partial understanding of it but hopefully, I'll manage to write a decent answer.
#Short answer
A recombination rate and a genetic distance (in centiMorgan) are two different things. While the recombination rate is bounded between 0 and 0.5, the genetic distance is bounded between 0 and infinity. There is a one-to-one function from the genetic distance to the recombination rate. For small recombination rate, the recombination rate and the genetic distance take very similar values but for greater values, the recombination rate is much lower than the genetic distance.
#Long answer
There is a tiny bit of math below. These equations are mainly for curiosity as one can understand the answer without understanding the maths behind it.
Definitions of $r$ and $M$
You are getting confused between two different statistics

The rate of recombination $r$ between two loci
$r$ is the probability for two loci to remain in the same gamete after recombination occurred. This probability cannot be greater than 0.5 ($0 \le r \le 0.5$).
The distance in Morgans $M$ (or more commonly in centiMorgans) between two loci. $M$ is the expected number of cross-over that occurs between the two loci.

Morgans and centiMorgans
You will note that I talk in Morgans rather than centiMorgans, which is not typical in the literature, but it helps at conveying the intuition of what it means. If $150$ centiMorgans $=$ $1.5$ Morgans. For two loci distant by 1.5 Morgans, the expected number of cross-over between the two loci is 1.5 (by definition). Below are some more explanations on these two definitions with some drawing :)
Case study with loci A and B
While $M$ and $r$ are closely related, they are not exactly the same thing. Consider the following sequence with the loci A and B
---[A]------------------[B]---

Let's assumed the two loci are very far apart and $M=2$. The probability of having exactly $k$ crossovers is therefore given by a Poisson distribution with rate $M=2$
$$P(k) = \frac{e^{-M}M^k}{k!}$$
Let's say for a given case that $k=1$ (a single cross-over occurred). This cross over is represented by a &quot;/&quot; below
---[A]-------------/----[B]---

Here clearly the two sequences at loci A and B will be separated. Let's say now that $k=2$ (two cross-over occurred).
---[A]---/-----/--------[B]---

Here, even if crossovers have occurred, the two sequences at loci A and B will remain together. Only the sequence in between the two cross-overs will come from the homologous chromosome.
Relationship between $r$ and $M$ - in words
You might see it coming from the previous section. The probability $r$ of these two loci A and B to be separated via recombination is the probability of an odd number of recombination events to occur between them (knowing that $M$ is the expected number of cross-overs).
Relationship between $r$ and $M$ - in equation
Let's calculate first the probability $p_{even}$ that an even number of cross-overs occur. This probability is just
$$p_{even} = \sum_{k = 0}^{\infty}{e^{-M}M^{2k} \over (2k)!}$$
, where I just added the constant $2$ before $k$ at both the numerator and denominator. With some algebra and trig, one can show that
$$p_{even} = \sum_{k = 0}^{\infty}{e^{-M}M^{2k} \over (2k)!} =
e^{-M}\sum_{k = 0}^{\infty}{M^{2k} \over (2k)!} = $$
$$e^{-M} \: cosh(M) = e^{-M}\left ( \frac{e^{M} + e^{-M}}{2}\right ) = 
{1 + e^{-2M}\over 2} $$

As, $r = p_{odd} = 1 - p_{even}$,
$$r = 1 - {1 + e^{-2M}\over 2} = {1 - e^{-2M}\over 2}$$
Here we go! We have our relationship between $r$ and $M$! Let's graph it
#Relationship between $r$ and $M$ - on a graph
I just graphed the above equation in Mathematica (Plot[y = (1 - Exp[-2 M ])/2, {M, 0, 5}])

Here is the same graph but zoomed on lower values of $r$ and $M$ (Plot[y = (1 - Exp[-2 M ])/2, {M, 0, 0.1}])

We clearly see from the graph that for low values of $M$, as $M$ increases $r$ increase quasi linearly ($M≈r$). For greater values of $M$, $r$ still increases but slower and slower until reaching an asymptote / plateau at $\frac{1}{2}$. $r$ is indeed bounded between 0 (when $M=0$) and $\frac{1}{2}$ (when $M=\infty$).
Note, the fact that the sum of probabilities of every even $k$ in a Poisson distribution is always lower or equal to 0.5 is an interesting math fact in itself!