A Riddle

I posted this on a forum I frequent, but so far haven’t gotten any answers I’m satisfied with.

—–

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 reach into the bag, feel around a bit, and pull out a card. 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? Why/Why not?

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)

In answering why you did or didn’t pay, don’t rely on something simplistic like “I’m a gambler”, or “I’m risk averse”. Try to figure out mathematically/logically why one or the other choice yields the best expectation.

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32 Responses to “A Riddle”

1. Dan Says:

I’d flip.

By flipping, I’d either get \$100.70 or \$21.20. I have a 50% chance of either, so the average value (if I got to do this multiple times) is \$60.95 — better than \$53.

I think another problem sort of along the same lines is better. You’re on a game show, and are presented with 3 doors. There is a prize behind one. The host asks you to select a door, and you do. However, before opening your door, he opens another door that does not have the prize behind it (and the prize does not move). He then asks you if you want to keep your original choice, or switch to the other door (the one you did not pick and that has not been opened). Justify your choice in terms of probability, it is not a trick question.

2. Mark Says:

(ignoring the rounding issues)
The number on the back is either

26.50, yeilding a payoff of 26.50-5.30 = 21.20
or
106, yielding a payoff of 106-5.30 = 100.70

If you keep the card, you get 53. You can risk 53-21.20 = 31.80 to get 100.70.

a. 53 – 100% chance, if chosen, effectively putting 53 in pocket
b. 21.20 – 50% chance, if chosen, cost \$31.80
c. 100.70 – 50% chance, if chosen, profit \$47.70
range of b and c = 79.50

1:2 odds to triple what is risked. Also, 31.80 is less than half of the 79.50 range. I would do it.

3. Andrew Says:

I’ll pay the money. The odds are with me:

50% chance the number on the back is 53/2 or \$ 26.50
50% chance the number on the back is 53*2 or \$106.00

Average the two outcomes and you get an expected outcome of \$66.25, less the \$5.30 you paid, for a total expected payoff for gambling of \$60.95.

This is better than the \$53 sure thing.

I’m not going to consider the implications of rounding in this case because even if the number on the back is \$1 off, it still won’t make a substantive difference to the result I got (\$6.05 difference).

4. Andrew Says:

Algebraic proof that gambling is a better choice for any number pulled (ignoring rounding errors):

Let x be the number pulled.
Expected outcome is
x/2 + 2x
———- – .1x
2

This simplifies to an expected payout from gambling of 1.15x, in this case 1.15 * \$53 or \$60.95

5. Phil Steinmeyer Says:

This is where it’s a bit of an enigma (perhaps riddle was the wrong term).

Yes, mathematically the payoff expectation is higher if you flip.

However, logically, this doesn’t make sense, since all you’re doing is effectively toggling between two outcomes, which you could in fact do without paying any premium a moment earlier (when your hand was in the bag). Merely because you’ve pulled your hand out now and seen a number, why does it now become worthwhile to pay a premium?

Or, to put it another way, you do the same action by feel alone, behind a screen (i.e. you can feel, but not see, the card). You’re free to flip over the card as many times as you want. You do so a few times. Then the screen is removed and you see the number is X. You can pay 1/10 of X for one last flip. Should you do it?

6. Scott Says:

In response to Dan’s question (otherwise known as the Monty Hall problem).

You should always take the door offered by Monty. Your first door selection is actually dividing up the 3 doors into a set of 1, and a set of 2. So the first set has a 33% chance of picking the prize, the second set has a 66% of picking the prize. Since the host will ALWAYS show you the nothing door of the two (which there will always be one of), this does not change the odds of the sets.

7. Mark Says:

I kind of wandered if 53 was the required outcome in your question or if is X and it required X * .10 to flip the card.

8. Phil Steinmeyer Says:

The Monty Hall problem is not a very good one in my mind, because the answer is entirely dependent upon whether the host will ALWAYS or merely SOMETIMES make you the switch offer (and usually, the question doesn’t state this either way). If ALWAYS offered, then yes, you should take it, for the reasons Scott mentions. If only SOMETIMES offered, then it’s much more variable (i.e. what if Monty only makes you the offer when you’ve already chosen the right door).

That’s why I tried to put all the suppositions to my problem clearly (including that you will always get the flip offer).

\$53 is just a random number, chosen to give the question some definiteness.

Look at it this way. Assuming there are 5 cards in the bag, but they all have the same number printed on them (X on the front, 2X on the back). Assume the experiment is conducted a large number of times (to even out randomness). If you never flip, then your expected outcome is 1.5X. If you always flip, then your expected outcome is 1.35X (1.5X -10%).

See? It’s trickier than it first appears…

9. Andrew Says:

A-ha! I think the error I (and others) made was treating the number pulled as if it were completely random / arbitrary. In reality, it is a random pull from a fixed set.

False assumption yields false result.

10. Phil Steinmeyer Says:

And yet, if you don’t know what the set is, it’s effectively random, thus the enigma.

Of course, if I tell you values of the cards initially, or even just that the X values are evenly distributed from, say, 1 to 100, then that makes the problem readily solvable.

It seems to be some sort of observer problem, because, when phrased with definite knowledge about the number you see (\$53), it seems pretty easy, even when you don’t know the set of possibilities that it came from. But thinking about it from a set perspective (even if the numbers are just represented as X), then it’s fairly easy to see that flipping is a bad choice.

11. Scott Says:

“What if Monty only makes you the offer when youâ€™ve already chosen the right door”

Well, assuming that you don’t know what the pattern Monty is following, and you are given a one shot guess, you should always pick the 2 door set over the 1 door set. Unless he decides to show you the winner in the one door or both losers in the two door. Whether he shows you a loser in the 2 door set or show you nothing at all, it doesn’t matter.

However, I do agree that your question is trickier. However, I’d argue that once you start assuming what the deck is made up of (all five cards are the same), that changes the original question, because now you actually HAVE some knowledge of the deck’s contents. So, clearly, in a deck of cards where they are all the same, you would NOT flip. But without that information, you still should flip.

12. Scott Says:

Bah. What you said.

13. Phil Steinmeyer Says:

Whether the cards are the same or not (or even whether there is 1 card, 5 cards, or 100 cards), SEEMS to make a difference, but I’d argue that it doesn’t, it only makes it easier to express the questions.

If there are 5 different cards, take the front side of each (the lower value), add it up and take the mean. Call that mean value ‘X’, and then the logic still holds (i.e. never flipping gives you an expected value of 1.5X, always flipping gives you 1.5X-10% = 1.35X.) The same holds whether there is 1 card, 5, or 100…

14. Dan Says:

The Monty Hall problem should correctly be stated as always having the offer. It’s simply meant to be a problem in probability that has a definite solution that a lot of people get wrong, there’s no trick to it.

In the problem you gave, you draw a card with 53 on it and have a 50/50 chance of either \$100.70 or \$21.20 by flipping. What does the set I drew it from matter? The 53 is part of the knowns of the problem — if I did the experiment an infiinite number of times, I am to assume that every time I see the 53. The only unknown for the experiment being regenerated is whether the number on the back is half that or twice that. Better result by flipping.

15. Phil Steinmeyer Says:

And yet, if you populated a large number of bags with identical cards, put 100 students in one room with the order to flip no matter what, and put another 100 students in a different room with orders to NOT flip, the non-flippers would produce a superior aggregate result.

16. Dan Says:

Another way of stating it:

It’s not about the X and 2X in the bag with what I know. It’s about the X in my hand, and the 50% probability of .5X and the 50% probability of 2X by flipping. (Ignore the 10%, it doesn’t matter).

Again, if I do this an infinite number of times, it’s not that sometimes I get 53 and sometimes I get 106 (or if the 53 was 2X sometimes I get 26.50). I was told I got 53. Don’t know what’s on the back. Flip it for a better average outcome.

17. Rohan Verghes Says:

Your problem is a variant of the Two Envelope Problem (http://en.wikipedia.org/wiki/Two_envelopes_problem).

The math is generally beyond me, but the wikipedia page implies that the paradox has not yet been resolved by mathematicians, and that there is a difference when dealing with an infinite number of possibilities for X, or a fixed number.

18. Phil Steinmeyer Says:

(BTW, I’m not fully convinced either way – just playing devil’s advocate if anybody comes down hard on one side or the other…)

This problem is a rephrasing of a problem I saw here: http://www.wiskit.com/marilyn/flipping.html

Here’s another enigma from that 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! “

19. Dan Says:

And I don’t mean to be argumentative, I just feel that once you’ve got your card in hand and have to flip or not, it’s just a “take a 50/50 shot at doubling or halving your money” question. If faced with this conundrum in reality, I’d flip because (A) in either case I’d make money and (B) better to take an even shot at making an extra 50 bucks in exchange for making 26 less. (If my number is 530k, I just take the money…)

About the wallet thing, which I haven’t fully thought through… Let’s say I always have 50 bucks in my wallet. (I know, now we’re qualifying things with assumptions!) If I have a 50/50 chance of winning, then I will win half the time. BUT — the amount that I win is less than 50 dollars. When I lose, I lose 50 dollars. SO — is it really advantageous for me to take the bet?!?!?

20. Dan Says:

Oh, geeze, I read the problem wrong, sorry.

21. Phil Steinmeyer Says:

I think the solution to the (original) problem is that the set of potential values on the card is not infinite, just very very large (i.e. let’s say it’s bounded on the upper end by, say \$1 billion). Thus at the upper end of this threshold, clearly a flipping strategy has a negative expected value (even if you exclude the 10% premium). The negative expected value at the upper end balances out the positive expected value at the lower end. The problem is determining the potential range of values.

If it’s truly infinite, then yes, you’ve got a real paradox.

But then again, when dealing with infinite numbers/ranges, you find all sorts of strange behaviour/paradoxes.

22. David Dunham Says:

Wow, this one is tricky. The correct answer is to thank the billionaire for the \$50 (\$53 but I’m simplifying the math) and walk away.

One of two things happen if you flip. You get \$95, or you get \$20. It’s not actually correct to use these for the expected value, however. Depending on what the original card was, there is actually either a 100% chance of getting \$95, or a 100% chance of getting \$20.

The card was labelled \$25/\$50, and simply flipping it over doesn’t somehow generate a \$100 card. There are not in fact always 3 available numbers (x showing, and either 2x or x/2 on the other side). The expected value of the \$25/\$50 card you drew is always \$37.50. And having to pay to flip reduces your expected value.

23. David Dunham Says:

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.

24. Tom Cain Says:

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.

25. Thomas Warfield Says:

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.

26. Tuck Says:

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

27. PhilSteinmeyer.com » Blog Archive » Riddle #2 Says:

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

28. Esteban Says:

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.

29. PJayTycy Says:

(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.

30. PJayTycy Says:

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 ?”

31. Dominik Says:

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?

32. PhilSteinmeyer.com » Blog Archive » Riddle Answers Says:

[...] 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. [...]