Why don't hydride shifts occur with hydrohalogenation of 4-methyl-cyclohexene? My book (see bottom) implies that hydrohalogenation of 4-methyl-cyclohexene follows this mechanism:

Why not undergo consecutive hydride shifts to furnish a tertiary carbocation?
I would guess that the second step, nucleophilic attack by bromine, is much faster and prevents this from happening. But when I try to confirm this by looking up some examples I see this one at masterorganicchemistry.com:

This implies the following very similar mechanism:

The carbocation in the example is a smaller ring, meaning it should be less stable and even more reactive (less able to perform hydride shifting.) But then again the charge is closer to the tertiary carbon and easier accessible.
Could someone explain what is happening?

Vollhardt, P.; Schore, N. Molecules: Structure and Function, 6th edition; W. H. Freeman: New York, 2014​, p491 exercise 12-6:

Exercise 12-6
Predict the outcome of the addition of HBr to (a) 1-hexene; (b) trans-2-pentene; (c) 2-methyl-2-butene; (d) 4-methylcyclohexene. How many isomers can be formed in each case?
The reason for not having consecutive hydride shifts is that generally, a new carbocation is formed through shift only when it is more stable than the immediately previous one. 

To form a tertiary carbocation, 3 consecutive hydride shifts would have to occur consecutively (when + is at c-4 from the alkyl group, your first product) in which more than one intermediates have same stability (same number of alpha-H). Hydride shifts will not occur for your second product to form tertiary carbocation as the next intermediate is infact less stable (3 alpha-H as compared to 4 alpha-H previously).



Since the ring already has 6 members (least strain) so ring expansion/contraction will not occur either.