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.
Starting with Riddle 2-A. In brief:
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, youâ€™ll gain more money than when you lose.
I think this is the easiest to address:
Clearly, no new money is being generated in the interaction, so there can be no overall positive expectation for playing the game (assuming equal information between the players).
If in fact, this game DID generate positive total expecations, then I would encourage all married individuals to go right now to their spouses and play this game – the two of you will come out ahead, in the aggregate.
Or not. Anyways, I could run a simulation to address this problem, but I think it’s simple enough that it’s unnecessary. I won’t present math to try to analyze WHY this apparent paradox exists, but follow the Wikipedia link at the end to those who do.
2-B is basically the same as 2-A, except that rather than their being an unknown difference between the two player’s wallets (envelopes), it’s known that one player has twice the other player’s money. I’ll skip it, because I think I address the case better in analyzing Riddle 1, below.
Riddle 1. In brief:
An eccentric billionaire invites 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. You draw one, and see that it shows 53. By prior agreement, 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?
The key here is that, although it APPEARS that the amount of money on the other side of the card has a 50/50 chance of being twice or half what’s on the side you see, it’s not quite that simple. In fact, the cases where the amount on the other side is higher are disproportionately drawn from the set of card face values with low values on them. Confusing? Sure, but it’s easy to simulate:
Here’s the code to test this. It sets up a simulation where 10 cards receive a random value from 1 to 100, and the back sides receive double that amount. Then I draw a card and record the results. I repeat the experiment 1,000,000 times (with a new, random set of cards each time), and here are the results:
Flip mean: 68.35
No-Flip mean: 75.53
Flip advantage (penalty) -9.51%
Flip winners percent: 50.34%
Flip winners mean (starting): 50.51
No-Flip winners mean (starting): 100.89
By flipping, you end up with a 9.51% overall penalty – very close to the 10% penalty you’d expect (i.e. the cost of the flip.)
Note that the mean result from not flipping is a bit over 75. That’s the average of the front side (1..100, average = 50.5), and the back (2..200, average = 101), total average = (50.5+100)/2 = 75.25.
Note that on the 4th line, we see that in this simulation, the odds of ‘flip’ improving your position on any ONE bet were 50.34% – close to the expected value of 50%.
However, the next two lines show why the whole paradox exists: The occasions when ‘flip’ works are those where the starting value of the card (i.e. the number you see) are relatively low (mean 50.51), and the no-flip works on higher relative values (mean 100.89). So even though, on any SINGLE observation, flipping should work half the time, the problem is that those observations cluster into the group where the initial observed value is low.
I also did the same test altering one aspect of the simulation. What if, rather than the ‘flip’ option giving you whatever is ALREADY on the other side of the card, it gives you a random (50/50) chance at double or half your money. Here’s the results:
Flip mean: 87.41
No-Flip mean: 75.53
Flip advantage (penalty) 15.72%
Flip winners percent: 50.47%
Flip winners mean (starting): 75.54
No-Flip winners mean (starting): 75.52
Flipping here improves your outcome by about 15%, as expected. The key difference? The starting values of the flip winners and the no-flip winners are effectively the same (~75). So flipping is not paying off only in the low value cases, but rather, in an even spectrum across all cases.
Some might argue that in the original problem, the eccentric billionaire’s range of possible numbers for the cards was unstated, and in my simulation, I’ve arbitrarily chosen 1-100. Fair enough. Here’s the results on two other ranges.
(All using the ‘preset deck’ scenario)
Card Range 1-20,000:
Flip mean: 11636.22
No-Flip mean: 12858.08
Flip advantage (penalty) -9.50%
Flip winners percent: 50.36%
Flip winners mean (starting): 8596.28
No-Flip winners mean (starting): 17180.83
Card Range 1524-7621 (i.e. a fairly arbitrary low and high end)
Flip mean: 6006.02
No-Flip mean: 6637.75
Flip advantage (penalty) -9.52%
Flip winners percent: 50.35%
Flip winners mean (starting): 4437.15
No-Flip winners mean (starting): 8869.06
Note that I couldn’t easily test with max values above 32,768, because of limitations of my random number generator. But if you still have doubts, download the code, put in a better random number generator, and test from 1 – 1 billion. The results will be effectively the same.
Finally, I know this is only a simulation, and doesn’t include a good mathematical explanation of WHY this works in such a counter-intuitive manner. The math of this problem is beyond me, but I’d refer you to the Wikipedia article on this subject, with many links to mathematical analyses at the bottom.