I’m glad this person won’t be endangering innocent drivers

My dad is a retired driving instructor and has come across people similar to this. Undiagnosed learning difficulties are probably part of it, but in many cases the people that struggled also had English being not a native language. Maybe it is the combination of the two in these extreme cases

There are countries where you would need to undergo a psych evaluation if you fail more than x times

With questions along the lines of: “You approach a zebra crossing with 2 pedestrians on it, do you:”

A) Honk your horn

B) Wait for them to cross

C) Run them over

D) Get out and rugby tackle them to the ground

the passing grade is 43/50,

They might just be randomly guessing and hoping that they’ll eventually get it, and thinking that their chances are better than they are.

I think that that’d be…let’s see. Say there are four possible answers for each question. So he’s got a 75% chance of failing any individual question.

$ maxima -q
(%i1) load("distrib")$
(%i2) cdf_binomial(7, 50, .75);
(%o2)                       1.8188415357867314E-19

That should give the probability of failing at most 7 answers out of 50 if there’s a 75% chance of failing any one.

So he’s got something like a 0.000000000000000018188% chance of passing the test by randomly guessing.

His chance of failing a single instance of that test:

(%i3) 1-cdf_binomial(7, 50, .75);

(%o3)                                 1.0

Ah. He has such a ludicrously small chance of passing that Maxima can’t represent it with the current floating point precision.

kagis a bit to figure out how to do this

Okay, apparently Maxima has bigfloats, but they default to only 16 digits of precision; not enough for this. This should give us 200 digits of floating point precision with bigfloats.

(%i4) fpprec:200;

(%o4)                                 200
(%i5) 1-bfloat(cdf_binomial(7, 50, .75));

(%o5) 9.999999999999999998181158464213268688894671703526026636336767389444876177016785501194817697978578507900238037109375b-1

Okay, so now chance of failing 128 tests in a row by randomly guessing:

(%i6) (1-bfloat(cdf_binomial(7, 50, .75)))^128;

(%o6) 9.9999999999999997671882834192983948674103659072385593201782461674117226992783470289641501110105148249790638571335177402867593272110042747272666144926576839664587182158166580514670324207313719393913737b-1

So then his chance of managing to get at least one success out of 128 tests in a row by randomly guessing:

(%i7) 1-(1-bfloat(cdf_binomial(7, 50, .75)))^128;

(%o7) 2.328117165807016051325896340927614406798217538325882773007216529710358498889894851750209361428664822597132406727889957252727333855073423160335412817841833419485329675792686280606086263120236298536839b-17

So he’s got about a 0.0000000000000023% chance of passing at least once in a 128 random-guess-based series of test attempts (assuming, again, that each question has four multiple choice answers). That is, he could keep doing this for the rest of his life and he’s virtually certain not to pass.