Classic Probability Case...Fair or Unfair

Probability wasn’t even the point, but I don’t seem to be able to pull you through it to what actually was. I have no divine resources available to me that are not also potentially available to anyone else. I’ll try not to bother you any more unless you initiate the exchange.

[quote]squating_bear wrote:

[quote]TigerTime wrote:
You have an 83.3% chance of getting your coke for a nickel and a 16.6% chance of it costing you a dime. [/quote]
Nope.

Sorry - I’m just being an asshole, but there is a chance that it takes 3, 4 or more tries before you get anything. You assumed it could only take two or less. Moral of the story - probabilities are assholes, they are even more annoying than me[/quote]

I’m building off of Pat’s assumption of a fix regiment, which, regardless of how complex the regiment is, it must still be fixed.

Given that the exact regiment has not been established, this is the best guess as it assumes the simplest mechanism for coke distribution.

You are being an ass hole, not because you’re right, but because you’re assuming a greater level of complexity in this coke machine and passing it off as the natural conclusion, which is wrong.

Maybe, in this scenario, the machine is meant to be more complex than, ‘coke, coke, coke, coke, coke, blank, etc.’ but it’s ultimately irrelevant because so long as the machine adheres to the ‘one in six chance’ rule, your odds of getting your coke for a nickel is still 83.3%. Making the machine complicated enough to contain the possibility of multiple failures doesn’t change this on a case-by-case basis and all you’ve succeeded in doing is change the ‘16.6% chance for a dime’ to ‘16.6% chance for more than a nickel’/‘16.6% chance of failure (assuming you don’t plan on multiple attempts)’.