What does this mean in my code?

This should be almost at the very beginning in every Python course.

Maybe it will help you look at "Structograms" / "Nassi-Shneiderman diagrams" .

Break your code back into different levels – starting with

 if antwort == "links": elif antwort == "rechts": elif antwort == "vorne": elif antwort == "hinten": else:

Note that "if", the "elif"s and "else" are always exactly the same far indented

(I also put it in again because it looks like this should be in a loop – otherwise "Write again!" would make no sense.)

Then separately write the block to be executed at "left":

 while antwort not in Ja_Nein: # Stilfehler: # zwar ist in diesem konkreten Fall klar, # dass antwort beim ersten Mal niemals in Ja_Nein liegt, # aber dieser Codeblock ist nicht ohne Weiteres wiederverwendbar antwort = input() if antwort == "Ja": elif antwort == "Nein":

In order to be able to use this block within [if Response == "left":], you enter it further than [if Response == "left":]:

 while antwort not in Ja_Nein: # Stilfehler: # zwar ist in diesem konkreten Fall klar, # dass antwort beim ersten Mal niemals in Ja_Nein liegt, # aber dieser Codeblock ist nicht ohne Weiteres wiederverwendbar antwort = input() if antwort == "Ja": elif antwort == "Nein":

I would then proceed from the order in such a way that I then establish the framework for the cases "right", "front" and "backward".

For this block, next follows the block for "Yes" within the block for "left":

 chance = random.uniform(0, 1) if chance <= 0.6: else:

And again the indentations:

 chance = random.uniform(0, 1) if chance <= 0.6: else:

etc.

All that is below one line and is moved further is dependent on this line, up to the first following line, which is only as far as or less far engaged.

You will check in line 25 whether the answer == "left". Even if this is not the case, line 28 is executed – check whether answer == "Yes". Regardless of the answer to this, next check is whether chance <= 0.6.

The if-block of [chance <= 0.6] in line 30 is continued at [else] in line 34 – this is the first line after line 30, which is also far indented – and is formally continued in line 37 with the [elif] – that is also far indented.

But

 if else elif

is inadmissible, of course there is an error message. How should the interpreter/compiler know that this elif belongs to the if in line 28? Finally, it has been taught that indentations and ONLY give indentations, which belonged together and what is subordinate.