# Riddle #3

OK, most of these things depend on ambiguity in the question, or else they’d be trivial. I think the following is challenging even when expressed clearly without ambiguity (hopefully)…

An eccentric multi-billionaire with an interest in psychology and probability invites you to participate in a challenge.

He has taken a number of cards, and had a number printed on each of them. On the reverse side of the card, there is another number - twice the value of whatever is on the front.

The cards are put in a bag, and you are invited to draw one. You see that it shows 53. He will pay you that sum (in dollars), or allow you to pay a 10% premium (i.e. \$5.30) to flip the card over and accept whatever sum is on the back, instead.

Do you pay to flip the card?

Further info:

• All rules are explained/guaranteed to you up front - the ‘flip’ offer was not in any way conditional upon whatever card/value you initially pulled, but rather was explained to you before you pulled the card.
• You have no particular idea as to the range of values printed on the cards. The experimenter is a billionaire with a rather cavalier attitude towards money. You will only be allowed to do this experiment once, so any knowledge you gain won’t be applicable to future tries.
• You can’t peek into the bag, feel the type on the back of the card, or any other ‘cheating’ solution like that. Looking at the card in front of you, you have no idea whether it was originally the ‘front’ or the ‘back’.
• The fact that the value you pulled ended in an odd digit is not in any way significant. The billionaire tells you that, subject to the range of the experiment that he had in mind, he had values randomly generated, down to fractions of a penny, then took the original value, and the doubled value, and rounded them to the nearest whole dollar (i.e. the ‘unrounded’ number on the reverse side could be anything from 26.5 to 26.999999, or 106 to 107.999999)
• The billionaire is honest, and, after the experiment, will allow you to fully examine everything (the cards, his number generator and it’s output, etc)

I do flip. But I also note that someone with more risk aversion than me wouldn’t flip.

EDIT: But I am wrong.

Flip.

Um, how is this a riddle? It all depends on whether you want to take the risk. You always come out ahead.

And I posit that the man running this show is not an eccentric multi-billionaire, but rather a deranged millionaire who looks a lot like John Hodgman.

Does your answer change if you are not shown the amount you’ve pulled, but, rather, the offer is made blind?

You’d still have money in your pocket at the end of the experiment, regardless of the outcome - the 10% ‘flip fee’ (always based on what was on the front of the card as you originally pulled it) is well less than what you will get at the end of the experiment regardless of whether your flip had a favorable outcome or not…

Matthew - the riddle/enigma of it is that all you’re doing is getting the opposite side of a blind pull from the bag.

Twisting it a little further, let’s say you pull a card from the bag, blindfolded, and have to decide which way to put it on the table (i.e. so a given side becomes the front). You can freely flip it in your hands as often as you’d like while blindfolded*. You flip it a few times, put it down, and take off your blindfold. Now you still have the 10% flip offer. Is it worth it then?

• Again assuming that there is nothing detectable about the values by touch - i.e. you can’t feel raised print or anything of that nature.

I agree with Matthew, what’s the riddle here? There’s nothing to reason out or figure out. It’s just “do you feel lucky?”.

Flip. It’s free money. For a 10% premium, the worst that could happen is that the value is 50% of the current value which still puts you ahead.

Let’s see… If ‘x’ is the face value, then the expected value of not flipping is simply ‘x’.

The expected value of flipping would be ((1/2)(x*2) + (1/2)(x / 2)) - (x / 10) = 23x/20

So yeah, flip.

Edit: Reversed upon further thought; see below.

The counter argument (not that I’m decided either way), is that it’s illogical to pay a premium to do something you could do freely moments early (i.e. flip the card with your hand while you were reaching into the bag), and for which you have, in the interim, gathered no new information that’s pertinent to the decision.

You always flip because 53 is an odd number.

Ah, but flipping the card while it’s in the bag doesn’t actually have any consequence whatsoever.

Because I trust Phil when he says he’s trying to be unambiguous, you get a decent statistical edge by flipping. If you’re right, you gain 100% of the card’s face value (whatever you saw first), if you’re wrong you only lose 50%.

I see the math solution (15% advantage to flip).

But that’s in conflict with what I see to be the logical solution. Flipping merely inverts an effectively ‘unknown’ state - either of the two unknown states should have equivalent expected values, therefore, paying a premium for one over the other is a poor decision.

So I guess maybe enigma is a better description of the problem than riddle. While I can’t poke a clean hole in the math argument, neither can I in the logical argument, and the two are at odds.

It feels vaguely reminiscent of that physics problem about whether the cat in the box is dead or not, and somehow he is simultaneously both until we know the answer (I’m sure I’m butchering my summary of that one)

BTW, the riddle/enigma is a rephrasing of one I found here http://www.wiskit.com/marilyn/flipping.html, which I tried to set clearer parameters on (and introduced the idea of paying a premium to flip, to make that a non-trivial decision).

Bolded the error. Look at it this way: In your example, if you flip and you’re right, you gain \$47.70 (+\$53.00 -\$5.30) over what you would have had anyway. If you’re wrong, you lose \$26.80 (-\$21.50 -\$5.30). And the general trend is always the same.

The chance of either occurrance is the same, but if you win you’ll win more than you could have lost.

Here’s a similar problem, on the linked page:

Suppose we are presented with the opportunity to open our wallets. Whoever has more money has to give it to the other guy.

A simple analysis suggests that you have a 50/50 chance of winning, and if you do, you’ll gain more money than when you lose. So you should take the bet. But the same analysis suggests that I too should take the bet, and it’s a zero-sum game, so it can’t be advantageous to both of us!

I’m not disputing that, mathematically, it makes perfect sense to flip. But logically, it should make no difference, with the premium making the flip a net negative. If the math is right, where’s the error in the logical argument (or vice versa)?

Hmmm, I think my error is in not accounting for the non-uniform distribution of the numbers, and we don’t know what the distribution is besides that for every ‘x’ there’s a ‘2x’.

Breaking it down by equally-probable cases instead:

1. I pick ‘x’ and stay: I win ‘x’.
2. I pick ‘x’ and switch: I win ‘2x’.
3. I pick ‘2x’ and stay: I win ‘2x’.
4. I pick ‘2x’ and switch: I win ‘x’.

Then it’s clear that there’s no difference between staying or switching. The introduction of the premium penalizes switching though, so then it becomes better to stay.

Yes, there is no advantage in flipping. It has already been explained above why, and you either see it or you don’t.

(For the children, do not turn this into another airplane on a conveyor belt discussion)