This is a statistics question but it has a number of correct answers

>A random number can be any number.

So the mean number given will be aprox half of infinity?

It can't be any number because there is a limited amout of dollars in the world. Your envelope cannot say 999quatrillion because there aren't that many dollars available.

>B has a higher expected return. Your answer should depend on how risk averse you are.
this

I don’t think you understand what I was saying. If you assume that it’s a completely random number between 0 and infinity then the expected return and utility of both options will be infinitely large no matter what your utility function is. In other words, the question isn’t very interesting unless you define the sample space first.
No. B has a higher expected return. Your answer should depend on how risk averse you are and what the sample space is.

>What is your choice, and why.

Depends on the amount of money in the envelope.
If it is lose change then why not double or nothing. If the amount is high then it depends on whether you need it or not, for myself I would go for A if the number is high enough. Meaning at least 1k.
If the number is 2 or something then fuck the envelope.

Any number under 10k double or nothing, based on savings/current income wouldn't feel too bad about losing anything under 10k and wouldn't feel too excited about winning anything under 10k either

Anything over 10k, just keep it, especially if it's a large amount

B
but got only a brainlet explanation of 2 picks
If money = 10,
2A = 20, average =10
2B in case of split is 20(double) +5 (half) , average = 12.5

Assuming all numbers have equal chances of being pulled i'd pick A for obvious reasons.

There is a limited amout of money in the world and option B for all amounts of money greater than half of the total supply cannot be doubled. There is a 50% chance that the card has a number on it more than half the total supply. Plain and simple. Option A therfore has a higher expected return. Pick option B if you like to gamble on losing odds.

I'm good at games so I'll go with B.

Edited my post because of a typo. To the guy who deleted his comment asking me to prove that B has a higher expected retur. The expected return of option A is X. The expected return of option B is 0.5*0.5X+0.5*2X=1.25X. Where X is the number.

Its not a brainlet explanation, its called expectancy in statisics.

It really doesn't have a single correct answer though. It depends on the number and your propensity to risk. Yes, B has a higher theoretical return by 25%, but that's not everything.

Say the number is 5 million, and you know for sure that it's enough for you to never work again and support the lifestyle you want to live. If you win the flip and get 10 million, then nothing really changes for you, and the marginal gain is very small. If you lose and get 2.5 mil, then you will have to cut down on a lot of things and the marginal loss will be much greater, therefore you will not want to flip the coin.

What if it's zero and we are all JUST'd

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Isn't the expected return for both A and B identical?

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What if it’s 100 trillion dollars and we all make it, fren?

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Really it depends on the amount inside of the envelope. Without knowing, I would take the guaranteed money. But if it only contained like $50 then I would take option B.

If you all want to read more about this decision making process read this paper.
uzh.ch/cmsssl/suz/dam/jcr:00000000-64a0-5b1c-0000-00003b7ec704/10.05-kahneman-tversky-79.pdf

No, read the question again. You probably read option B as double or nothing. It’s not.

...

Might as well be a negative number for all of us.

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Actually that changes the whole question as people are risk adverse and practice risk seeking tendencies when presented with the chance to lose money,

What if its 999quatrillion and the next world war begins because you just devalued the money everyone else already owns by >90%.

B is the only right answer. It has the same expected return as option A and allows you to play the odds.

How are the odds the same on with only 1 draw?
Option A: 100% of X
Option B: 50% 2x or 50% .5x

If we are talking about a random number of all positive real numbers then you should probably fetch a math phd, who would question which option is actually better for eons.
>muh hilbert's hotel

B has higher expected returns unless the money supply is finite which it is which makes A have the higher expected returns.

The expected value if you don't take the money straight up is (((0.5)*(2x))+((0.5)*(0.5x)))/2 =

(x+0.25x)/2 = 0.625x

Since the expected value is less than x, you wouldn't want to take the chance.

A
number : 10
a= 10
b = 20 or 5
i rather have 10 as it sure ill have it

Don't divide it in 2

This is wrong. The expected return of B is 1.25X. You shouldn’t decide by 2.

Retarded question, OP is a brainlet or a troll. Next.

The question is worded retardedly and can be interpreted two ways, which is causing the confusion.

A = 100% of the number picked
B = 50% / 2 + 200% / 2

or

A = 100%
B = either 50% or 200% (basically a coin flip)

Muh monte calro simulation (excel spreadsheet) say that B is better
figures from 0-10:
return a: ~5
return b: ~6.2

Do the math on Option B. 50% * 2 + 50% * 0.5 = 1 = Option A.

You might want to put that in a calculator.

the wording is retarded for this question. Does it ask, if you choose:

A, you get 100%

B, you either get 50% or 200%

Is that the premise? Because I'd choose A as I'm risk adverse

also, is this assuming you only get to play once?

This, fpbp

YES