Calculating the Expected Value of the Double-On-Coin-Flip Paradox

[Pds3.14]

Suppose that you given a $1 and flip a coin, if heads, you double it and flip again, if tails, you keep it and nothing else happens.

How much, on average, would this opportunity be worth? In other words, how much, on average, would you make.

Hint: the answer is most-certainly not $2.

 

[crunchynet]

Very interesting question. I would probably approach it from this end:

1) Figure out how many individual people (each with a single coin) it would take to have them all flip their coin at the same time and count the results, such that each time they do this the result is very nearly 50/50 tails and heads respectively. Let's call this number P.

2) So (1/2)*P = the number of people who have 2$ and another shot. Let's call them P2

3) The odds get really shaky starting here however because now only HALF of the people we established to be a good number for getting 50/50 are present to flip again, which means that the variance spreads out quite a bit... So when these people (P2) go for the flip again, it's no longer guaranteed so well that half will get heads and half will get tails. It's more likely than before to get say 60% and 40%, or 35% and 65%... or any range of values in between. I expect a statistician would be able to answer this part with some crazy equation I haven't seen yet :P, probably something that relates variance to average number of heads. I'm no statistician haha.

So now going backwards and figuring out how many people out of the original P make 1$, 2$, 4$, 8$ and so on... would give you the odds and average you are looking for. Anyway, I would love to see the answer, though I can already see the answer is based on what you define to be a happy amount of 'P'.

 

[D H]

The original post is poorly worded. Fix that and this is an oldie but goodie. The expected outcome is infinite.

Alternatively, some physicists might say that, thanks to the magic of zeta function regularization, the expected outcome is -1/2.

 

[crunchynet]

D H you're not going to attempt to explain it for us less wise beings lol? I'm still curious!

Also what in the world is the average age of users on this forum? Are most people here grad students or professors??

 

[D H]

D H you're not going to attempt to explain it for us less wise beings lol? I'm still curious!

I made a mistake in my first reply. Some physicists would say the expected value is -25 cents, not -50 cents.

Here's the infinite answer:
Suppose you get tails on the first flip. You collect $1. The probability of this event is 1/2, so the expected gain from this event is 50 cents. The other possibility is getting heads on the first flip, which is also has a probability of 1/2, but this doubles the amount in the pot. Suppose you get heads on the first flip, tails on the second. You collect $2. This probability of this event is (1/2)², or 1/4, so the expected gain from this event is also 50 cents. If you get heads on the first two flips the amount doubles yet again to $4. If you get tails on the next flip you collect that $4. The expected gain from this event: 50 cents. Keep on going, ad infinitum, and the expected gain is $\sum_{n=1}^{\infty} \$0.50$, which is obviously a divergent series.

Here's the physicist's answer:
As everyone knows, 1+1+1+1+… = -½. (Not so snarky: See http://en.wikipedia.org/wiki/1_+_1_+_1_+_1_+_⋯.) Therefore, ½+½+½+½+… = -¼.

 

[strangerep]

Suppose that you given a $1 and flip a coin, if heads, you double it and flip again, if tails, you keep it and nothing else happens.

How much, on average, would this opportunity be worth? In other words, how much, on average, would you make.

Hint: the answer is most-certainly not $2.

If you mean: "how much would you make at a real casino?", the answer is "probably nothing". They have upper betting limits, and the technique you describe is one reason why.

I actually knew a gambler who played roulette by this principle, but he used a "3 state" system rather than the "2 state" red/black. He would bet on 2-out-3, hence much more likely to win on each spin, and he could generally stay under the casino's betting limit. OTOH, he also knew when to walk away...

 

[1MileCrash]

There is no limit to the possible prize, so the average outcome could not possibly have a limit, either.

Or at least, that's how I see it.

 

[MathJakob]

Suppose that you given a $1 and flip a coin, if heads, you double it and flip again, if tails, you keep it and nothing else happens.

How much, on average, would this opportunity be worth? In other words, how much, on average, would you make.

Hint: the answer is most-certainly not $2.

The potential return will be infinite. I forget exactly where I saw this question, I think it was on Vsauce or Numberphile but basically the return on such a gamble could spiral to infinity so for a casino to host such a game, they would need infinite money before taking customers. I will do an absolute thorough google search if you simply must have the link.

 

[D H]

It's been discussed right here on this forum before.

 

[PAllen]

The way I look at it is that chances less than 1 in million are meaningless over the scale of my lifetime. The expectation based on events with at least 1 in a million chance of occurring is $10. That is the most I'd be willing to spend for this opportunity. Also, note that there isn't enough money in the world to make good on this scheme - out to the largest payout that could be covered, the expected value is still peanuts.