Comments on: A Riddle

By: PhilSteinmeyer.com » Blog Archive » Riddle Answers

PhilSteinmeyer.com » Blog Archive » Riddle Answers — Fri, 14 Apr 2006 22:12:30 +0000

[...] Thanks to all who commented on the riddles yesterday. The first riddle had the highest number of responses to any post I’ve made to date. Here’s my proposed answer to the riddles (paradoxes?) posted yesterday, here and here. [...]

By: Dominik

Dominik — Fri, 14 Apr 2006 13:56:07 +0000

But think of this,as there are three and not two poosible outcomes: You have a 100 % chance of walking away with 53 $. Or, you have a 50 % chance of walking away with 26 $ and a 50 % chance of walking away with 106$. Those are your options.
The question is, are you a gambler or not?

By: PJayTycy

PJayTycy — Fri, 14 Apr 2006 11:22:00 +0000

Damn, after reading the previous responses, I’m not even certain about the best thing to do in the “large number” assumption.

Especially this is shouting at me : “you can flip the card for free as much as you want before putting it on table, then why pay to flip it once more ?”

By: PJayTycy

PJayTycy — Fri, 14 Apr 2006 11:05:27 +0000

(I’ve not read any of the responses)

If you’re looking at the doubled side, flipping the card will get you 40%
If you’re looking at the not-doubled side, flipping will get you 190%

There’s 50% chance for each of them, so if you flip the card, you’ll get 115% return on average. If you can do this experiment a number of times, it’s clearly best to always flip the card.

With only one shot, I’m not sure what to do. I think I would flip the card, based on the above thinking in “large numbers”, although I know it’s not valid in this “one shot” case.

By: Esteban

Esteban — Fri, 14 Apr 2006 03:26:33 +0000

I think the observation about an upper bound is sort of sidestepping the problem. Having to guess at the upper bound “solves” the problem by adding an extra gratuitous constraint. I find it a rather disappointing solution.

The paradox can’t be fully attributed to infinite ranges. If you have an upper bound _and it is known to the player_, you just make all values higher than upper_bound/2 trivial no-flip decisions. At upper_bound/2 and below, the paradox persists, and now infinity is not to blame.

Another way to escape the problem would be noting that the subjective value of money is often not linear with the objective amount. If the player is a broke teenager who wishes nothing more than 49$ to buy the just-released Halo 3, then 53$ means *a lot* more to him than 26.5$, and 106$ wouldn’t make his life that much better.

But I assumed we would abstract these considerations, assuming that the value is linear and that the upper bound doesn’t matter.

Personally, from a rational standpoint I’d get the 53$ and be done with it. Both the flipping-is-better-expected-value and the flipping-is-paying-for-nothing are correct and sound arguments, but I believe the latter is more lucid and more specific to this problem, so I would give it precedence. Very much like when there are both streetlights and a traffic guard directing the traffic, I’ll do what the guard says, even if I have no reason to believe the streetlights are broken.

This case makes me look at the strategy of maximizing expected value as an heuristic rather than an absolute measure of rationality. If I have a consistent model to reason about the problem that is more specific to it, I will use that instead.

By: PhilSteinmeyer.com » Blog Archive » Riddle #2

PhilSteinmeyer.com » Blog Archive » Riddle #2 — Fri, 14 Apr 2006 02:19:07 +0000

[...] Phil Steinmeyer’s rumblings on the game biz, programming, and life « A Riddle [...]

By: Tuck

Tuck — Fri, 14 Apr 2006 01:34:07 +0000

if the values on the card were x and 1000x, the choice would be obvious, who wouldn’t risk 53/2 dollars for the chance at winning 53000. The only reason this problems seems less obvious is because the values are x and 2x, but 2x is still a bigger gain that x/2 is a loss. So flip as long as the multiplier is greater than 1.5

By: Thomas Warfield

Thomas Warfield — Thu, 13 Apr 2006 23:09:10 +0000

As people have stated, mathematically you flip because the expected value is higher.

But ignoring that, psychologically you flip. Think about it… some guy is going to give you money no matter how it works out. It’s just a question of how much. You can’t lose, either way you make money. You might as well go for it.

By: Tom Cain

Tom Cain — Thu, 13 Apr 2006 22:17:42 +0000

I wouldn’t flip. My reason is more psychology than math. Let me change the drawn number to 100 to make it easier. Once I’ve drawn I have $100 of this guy’s money in my pocket. It’s mine. Now the question for me becomes, will I risk $60 of my money to gain $100 on a coin toss? I wouldn’t if he had never given me the initial $100, so I wouldn’t here either. I’d probably go buy a few new games instead.

By: David Dunham

David Dunham — Thu, 13 Apr 2006 22:06:36 +0000

I don’t think I was entirely clear: the problem is not given a random number, you can pay 10% to win double or half. You’ve already picked the random number.