Comments on: Easiest $100 I’ll Ever Make

By: Chris

Chris — Tue, 04 Aug 2009 05:20:01 +0000

I agree with Alan. The problem isn’t clear. In your computer simulation, you rule out all cases where you don't pick at least one candy, but, given the wording, you could have had:

if (hand[0] != licorice)
            continue;

i.e. rule out all cases where the one that you put on the table isn’t a licorice candy, and that would give you the 1/3 probability.

By: Alan Green

Alan Green — Sun, 14 Jun 2009 07:34:03 +0000

I reject your speculation that I may be wrong.

The protagonist doesn't need to know which candy is licorice in advance, as the question only deals with the subset of possible outcomes where he puts the licorice on the table. In other words, putting the licorice on the table is a given in the same way that drawing at least one licorice is a given.

The puzzle question says:

Without looking in the bag, I draw two pieces of candy from it, and place one of them, which is licorice, on a table.

I don't see how a straight-forward reading of this question would be that candies are drawn randomly (so drawing at least one licorice must be a given for the question to make sense), but the order the candies are placed on the table is is non-random (so doesn't need to be a given).

In hindsight, I see the sneaky phrase "in the bag" to qualify what is not being looked at. The puzzle authors have crafted a tricksy riddle, dependent upon a particular and peculiar parsing. John Howard would be intrigued and pleased that the puzzle's words have a meaning quite different from their face value. Alternatively, you might argue that the reader should know that different kinds of lollies each have a particular feel, but that is not necessarily true in my experience, and would be an unusual - and tricksy - assumption for a math puzzle.

At this point, I'm going to quote xkcd at myself (http://www.xkcd.com/386/) and apologise to you for the John Howard reference. Sorry.

PS: I calculate 77,456 5-sentimo coins.

By: alastair

alastair — Sun, 14 Jun 2009 02:55:36 +0000

Alan, I agree the wording is not as precise as it could be, and I agree with your reword. But given that there is at least one licorice in the hand, how can the protagonist know which one it is, in order to put it on the table, without selecting it?

To put it another way, if I was randomly selecting a candy from my hand and it had a 100% chance of being licorice, the chance that the other candy in my hand is licorice as well would have to be 100% also (since I must have picked both out of the bag to begin with).

(So maybe we’re both wrong

By: Alan Green

Alan Green — Sun, 14 Jun 2009 00:35:03 +0000

Sorry to say this, but I think your interpretation of the puzzle is only something that someone already familiar with the Monty Hall problem could twist out the puzzle’s words.

To make it work the way you coded, the puzzle would need to be interpreted as:

Four different pieces of candy are placed in a bag. One is chocolate, one is caramel, and two are licorice. Without looking in the bag, I draw two pieces of candy from it, then select a drawn piece that is licorice and place it on a table.

However, the puzzle wording you quoted does not imply selection, rather that the protagonist places a randomly selected piece on the table.

I think the puzzle authors were looking for a Monty Hall variation and messed up, resulting in a riddle rather than a puzzle.

By the way, I carefully did all the math, and I agree with JT, the real answer is indeed 1 in 3.

I hope JT pays you in 5 sentimo coins, unwrapped and unbagged, helpfully strewn around around your desk, over a period of a year. (Your agreement said nothing about HOW the money was to be paid, right?