Silent Acorns

Suppose I were to offer you the following opportunity:

I hand you a non-biased two sided one-dollar coin to flip. If it comes up heads, I give you another dollar and you can flip again. Each time the coin comes up heads, I double the money I've given you so far. Once tails comes up, the game stops. You get to keep the flipping coin.

For example:

If you flip:
H H H T

I give you:
$1 (the flipping coin, for you to keep)
$1 (for the first heads)
$2 (for the second)
$4 (for the third heads)

for a total of $8

The questions:

1) How much would you be willing to pay (up front) for this opportunity?
2) How would your answer change if you were only allowed one chance (i.e. once you flip tails, the game is closed to you forever)?

Baloo

Fun experiment.

If limited to zero flips, the expected value of this bet is $1. If limited to one flip, it is $2, and if 3 flips, $3, etc. Since there is no limitation on the number of flips, the expected value of the entire bet is infinite - thus, in a perfect mathematical world, I'd bet everything I had to bet.

However, we live in the real world. My first question is: how much money does the "bank" have? If you are the bank, and you have $3,000 in savings, the bet is limited to 12 flips (you conuldn't cover any winnings beyond that). I'd pay you $10 for that.

If you're backed by a bank, or Bill Gates, and can cover $1 billion (after which no more money can be given), then the bet is implicitly limited to 28 flips. Thus, I'd pay about 26 dollars or so to play (but I'd want the $1 billion contract in writing).

To make it more interesting, let's say all of the governments of the world - in fact, all of humanity - has agreed to honor my payout check no matter what it's. The net dollar value of all of the combined purchasable assets of the earth (land, property, minerals, resources, etc.) probably falls just shy of $10,000 trillion. Any money I won beyond that would be worthless (there'd be nothing else to buy with it, as I'd already own EVERYTHING that was for sale).

So even with the backing of every person on the face of the planet, I'd say the bet becomes worthless (and is thus limited) at about 52 flips - which corresponds to an expected value of $52 - so I'd cough up about $50 for that (but I'd ask for everyone's signature, up front).

How about you?

Baloo

p.s....it goes without saying that my answer does not depend on the number of times I'm allowed to bet.

Silent Acorns

Baloo, your approach is the same as I would take. But what if I change it slightly:

Instead of giving you $1 and then doubling it with each "heads", what if I give you $1000 and double it with each "heads"?

If I have the entire Earth backing me up, and we assume that it's wourth $10 000 trillion, this works out to about 43 flips. Would you be willing to pay $43,000 for the chance?

Bumble Bee Tuna

Well first I tried to solve this without complex math.

If you tried this bet 2^11 (2048) times, for instance, and everything followed the perfect statistical bell curve at a price of x dollars, you would win the following amount:

1024*(1-x)
+512*(1-x)
+256*(1-x)
+128*(1-x)
+64*(1-x)
+32*(1-x)
+16*(1-x)
+8*(1-x)
+4*(1-x)
+2*(1-x)
+1*(1-x)
+2048-x
which translates to
(2^10)*12 - (2^11)*x

and it seems that this would extend to any number of bets. I would keep it at powers of two for easy calculations:

2^n bets will yield winnings of

(2^(n-1))*(n+1) - (2^n)*x
which simplifies to (2^(n-1))*((n+1)-(2*x))
So this means if (n+1) is greater than (2*x) then you will make a profit of 2^(n-1).

So if I had infinite funding (through loans) and could change the mechanism to some sort of computer algorithm, I would bet any amount of money.

If I only got one chance, that makes it a bit more interesting.

A full half of all bets will earn $2 or more, so $2 seems pretty safe.
at $2,
1 in 8 wins at least $6.
1 in 16 wins at least $14
1 in 32 wins at least $30
1 in 1024 wins at least $1022
1 in1 in 850000 wins at least $849998

The odds are a hell of a lot better than the lottery, but I still would not bet very much because it would most likely just be tossing my money down the shitter if I bet much more. I could maybe be convinced to bet something up to $10 but anything past that would probably be pretty silly to a poor fella like me.

-B

Baloo

This is the mathematical equivalent of asking me on a single coin toss worth $44,000 for "heads", would I wager $43,000? For simplicity, that is how I will treat it.

If I were a millionaire, I'd take your bet in a heartbeat (as soon as you had a contract with all of the the signatures ).

But now we get to the heart of the matter: I'm not a millionaire.

The real problem is - I can't afford to lose $43,000. This is, again, real life, not mathematical utopia, where money is an abstraction for what we're really talking about: the happiness/security/improved living associated increased levels of wealth.

Let's say the happiness associated with various levels of wealth is measured in units of "grogs". Right now, say I have 5000 grogs (as a combination of wealth and debt association with home, job, car, savings, credit cards, etc.)

A $44,000 windfall might be worth another 500 grogs to my life (I could pay off my debts, then pay a good chunk off my house, maybe invest something for my future childrens' educations, and maybe buy a few toys on the side). It would be nice - but it wouldn't be earth-shattering. I'd be at 5500 grogs after such a win.

However, losing $43,000 right now would cost me 4500 grogs - it would wipe me out. My house, car, savings - gone. Investments - gone. I'd be buried in high interest debt. And I won't even start with my wife's reaction to the news. My life would most likely be ruined for the next 5 years.

So when we make the real-life comparison, the wager is laughable - I'd wouldn't wager $4500 for $500, nor would I wager 4500 grogs for 500 grogs.

I'd say a 500 grog gain would be equivalent to a financial loss of maybe $2,500 right now (which would hurt, but not cripple me). Thus, I would personally wager about $2,500 on your latest scenario.

If we reduce the starting numbers to, say, $10 - which would require 48 flips or so to reach $10,000 trillion, well, I might consider laying $200 (-$200=-90 grogs) on the line for the expected $480 (+$480=+100 grogs).

And I will change what I said at the beginning. At $1 a dollar start-off, I would only put up $30 (2 grogs) for the $52 (3 grog) payoff...

Jesse

Here's another expected-returns problem. Suppose I show you two envelopes, and tell you that one envelope contains twice the amount of money as the other. You pick one, and I open it, revealing that it contains 100 dollars. But then I make an offer--if you like, you can return this envelope to me and take the other envelope, pocketing how ever much money is in that one. You calculate the average expected return from taking me up on the offer: (1/2)(50) + (1/2)(200) = 125, which is more than what you'd get it you stuck with your initial choice. Therefore, you reason, you should take the other envelope.

But there's a problem with this reasoning. Even if I hadn't shown you how much was in the first envelope before offering you the second one, you could still calculate the expected return, assuming there were X dollars in the first envelope and therefore finding that your expected return for taking the second would be (1/2)(0.5X) + (1/2)(2X) = 1.25X, so you should take the second. But this can't be right, since both envelopes are identical, so there can't be any basis for thinking one is a better choice than the other when neither has been opened! So the question is, what's the flaw in the reasoning?

Lobstrosity

Could the flaw be that you're taking an average where no average is allowed? Let's take the first case. Suppose we are given that envelope A will always have $100. Next suppose we know that 50% of the time envelope B will be twice what is in envelope A ($200). We also know that if envelope B is not twice envelope A, it must be half what is in envelope A ($50). Given this setup, we expect that if we choose envelope B N times, where N >> 1, we will have approximately N x ($125). Had we chosen envelope A every time, we would have exactly N x ($100), so clearly we benefitted from our knowledge that the value on envelope A was fixed over multiple trials. After all, that's where the expected returns actually makes you money, isn't it?

Now, if you're only given one trial, you have a 50% chance of getting the envelope with the most money. Simply by symmetry one can argue that swapping will not increase your odds of getting the maximal dollar amount for that trial. Multiple trials where you know nothing about the contents of either envelope can just be considered a series of single trials where each time your parameter X has a new value, which voids the idea of averaging the trial results to produce 1.25X. Because these are single trials, I would argue that your consideration of average returns (i.e. 1.25X) is not valid.

Is this anywhere close to the right track?

Bumble Bee Tuna

Wow. Jesse, you've definitely stumped me for the moment. I mean, it's obvious that the logic is wrong, but I can't see an easy flaw in it. This requires more thought. I really like this problem. What Lobstrosity said makes sense, as always. but to me it doesn't really provide a good answer to the question 'what's wrong with the logic'.

-B

SULPHUR

Is there any reason why heads could not turn up an infinite amount of times