 Posers and Puzzles

1. 08 May '16 04:02
I have an infinite number of cards each of which has a number on each side.

On one side the cards are labelled \$1, \$2, \$3, \$4, ...
The other side has exactly double the amount.

At random you are given a card.
Either STICK and I pay you that amount.
Or TWIST and I pay you the amount on the reverse.

Your card is an even amount.
Do you STICK or TWIST and why?
2. 08 May '16 05:31
So there's a 1:2, 2:4, 3:6 etc, one of each card going to x:2x as x approaches infinity...

There are twice as many even numbers on the "bigger" side than the "smaller" side, since only half the "smaller" sides are even.

So if I see an even number, 2/3 chance that the other side is smaller. STICK.
3. 08 May '16 21:39
Originally posted by AThousandYoung
So there's a 1:2, 2:4, 3:6 etc, one of each card going to x:2x as x approaches infinity...

There are twice as many even numbers on the "bigger" side than the "smaller" side, since only half the "smaller" sides are even.

So if I see an even number, 2/3 chance that the other side is smaller. STICK.
I'll give that answer 4/10 as it is incomplete.
4. 09 May '16 04:33
Originally posted by wolfgang59
I have an infinite number of cards each of which has a number on each side.

On one side the cards are labelled \$1, \$2, \$3, \$4, ...
The other side has exactly double the amount.

At random you are given a card.
Either STICK and I pay you that amount.
Or TWIST and I pay you the amount on the reverse.

Your card is an even amount.
Do you STICK or TWIST and why?
I would stick with an even numbered card, because if you double any number (odd or even) the doubled number will be an even number.

If the card was odd I would automatically twist, because it would mean the other side is double.
5. 09 May '16 05:28
Originally posted by lemon lime
If the card was odd I would automatically twist, because it would mean the other side is double.
The problem is what do you do with an EVEN number.
You say STICK. Why?
6. 09 May '16 06:154 edits
Originally posted by wolfgang59
The problem is what do you do with an EVEN number.
You say STICK. Why?
Because of the possibility of it being either a doubled odd number or a doubled even number. An odd number cannot be a doubled number, but an even number can be a doubled number regardless of whether the other side is even or odd.

Or to put it another way, half of an even number can be an odd or even number. But an odd number must be half of what is on the other side. ATYoung pointed out there would be more even numbers than odd when considering both sides of all the cards... so putting this all together, there's probably a better than 50/50 chance of an even number being double the number on the other side.

Twisting on an even number means the possibility of getting 1) double that number 2) half that number as an even number or 3) half that number as an odd number. Sticking on an even number means the possibility of 1) it being half the number on the other side or 2) it being double the number on the other side. There appears to be less that can go wrong with sticking and more that can go wrong with twisting.
7. 12 May '16 19:23
Originally posted by lemon lime
Because of the possibility of it being either a doubled odd number or a doubled even number. An odd number cannot be a doubled number, but an even number can be a doubled number regardless of whether the other side is even or odd.

Or to put it another way, half of an even number can be an odd or even number. But an odd number must be half of what is on ...[text shortened]... ere appears to be less that can go wrong with sticking and more that can go wrong with twisting.
You have not taken into account the risk.
For instance if you are showing \$4 the other side is either \$2 or \$8.
So if you twist you are risking losing \$2 against gaining \$4.
8. 12 May '16 21:16
Originally posted by wolfgang59
You have not taken into account the risk.
For instance if you are showing \$4 the other side is either \$2 or \$8.
So if you twist you are risking losing \$2 against gaining \$4.
Sounds risky. But okay, why not, either way I get paid... right?
9. 12 May '16 23:56
Does not matter. Your expected return from sticking is the same as your expected return from twisting.

3 out of 4 of the numbers on the cards are even. Of those even numbers, only 1 out of 3 are low numbers (good for twisting) while 2 out of 3 are high numbers (bad for twisting). So, if we twist all cards, 1/3 are good twists and 2/3 are bad twists.

If we know nothing about the odds and evens, just that twisting (I prefer "flipping"😉 either pays twice as much or half a much as is shown on the face, then the expected return from twisting is ((2 * facevalue) + (1/2 * facevalue)) / 2 = (1 + 1/4) * facevalue = 5/4 of facevalue. So, over time, you will get 25% more from always flipping the card if we know nothing else.

But we know that the likelihood of a good twist is 1/3 and the likelihood of a bad twist is 2/3 so our expected return from always flipping is 1/3 * 2 * facevalue + 2/3 * 1/2 * facevalue = 2/3 * facevalue + 1/3 * facevalue = facevalue.
10. 13 May '16 02:41
Originally posted by ParShooter
Does not matter. Your expected return from sticking is the same as your expected return from twisting.

.
Correct.
11. 13 May '16 19:14
Can someone please re-explain this for me?

12. 14 May '16 20:45