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I came up with a betting system that seems to defy logic... explain why its wrong?

Here's the "logic" behind it

Rule: If you flip a coin enough times(x) the number of heads(H) and tails(T) will be equal to each other (law of large numbers?)

H + T = x  AND 1/2H = 1/2T AND 1/2H / 1/2T = 1

This could happen at after flipping HT or THTTHH or HHTHTHTT... extra

And Now The System!!!

The first bet(more like lack of bet) you make is for $0, you have 1/2H / (1/2H + 1/2T) chance of getting heads.

The coin flips tails. Therefor x is now x-1 and T is now T-1 and H is still the same. So now you have 1/2H / (1/2H + 1/2T - 1) chance of flipping heads. Because 1/2H / (1/2H + 1/2T) = 50% chance of flipping heads and 1/2H / (1/2H + 1/2T - 1) is not equal to 1/2H / (1/2H + 1/2T) then 1/2H / (1/2H + 1/2T - 1) is not equal to a 50% chance of flipping heads. How can this be?

PLEASE do not say every time the next flip will be heads or tails, therefor you have a 50% chance of flipping heads... of course I know this!!! I want to know what is wrong with my logic not that it is wrong, I already know that it is wrong...

On a side note this system would never work in a real life senario because no casino offers 1:1 odds which this system needs.

The system itself is

  1. Watch first flip
  2. Bet on opposite of flip amount of money you want to profit
  3. Continue making bet until heads and tail flips are equal
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Another note, I ran this through a homemade program $1,000,000 to start and it took only about 10 billion flips on average to make 1000 dollars –  java May 11 '13 at 3:51
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The Law of Large Numbers does not say what you think it does, but in fact in this case with probability $1$ the number of heads and tails will equalize, no matter how far (say) tails are ahead at any particular time. Your "system" is correct. But, as your experiment suggests, it is not a get rich quick scheme. You need to have deep pockets, and persistence. –  André Nicolas May 11 '13 at 4:04
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It doesn't matter what you bet. The step that's doing all the work is 3., which is equivalent to "continue betting until you make money." Of course you will make money if you don't stop until you make money, unless you go bankrupt... –  Qiaochu Yuan May 11 '13 at 4:13
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Your system isn't wrong at all and it does not defy logic. You can make money with this system. Problems you should try to answer are: suppose you start with n dollars and bet one dollar on each flip. Work out the probability that you go bankrupt using your scheme. Now work out the expected amount of time (again, starting with n dollars) it takes you to win that dollar in the cases you do win. –  Eric Lippert May 11 '13 at 7:46

5 Answers 5

You are taking a One-Dimensional Random Walk on the integers, starting at 0. With probability 1, you will visit every integer infinitely often, including the starting point, but also the point where you're broke. Curiously, if the random walk is expanded to two dimensions, the same result holds, but in three dimensions it's only a 34% chance to return back to the origin.

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I believe the problem is your interpretation of the law of large numbers. Wikipedia says:

"According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed."

I don't believe there is anything that says that says the number of heads flipped must equal the number of tails flipped. Given enough flips, they almost certainly will be equal at some point, but they don't have to.

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The law of large numbers is irrelevant, I believe they have to become equal at some point given an inf. amount of time, would you disagree?? –  java May 11 '13 at 4:24
    
But you can't go on betting for an infinite amount of time. –  Qiaochu Yuan May 11 '13 at 4:28
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If you take a random walk on the line, one step forward with probability $1/2$, one step back, then if you start at say $(1,0)$, with probability $1$ you will return to the origin. This is a known result, not very easy to prove. A similar remark holds for a random walk in the plane (up, dow, left, right each with probability $1/4$). In $3$-D space, the result does not hold. –  André Nicolas May 11 '13 at 4:32
    
@QiaochuYuan this is purely theoretical, it is not a realistic way to make money by any means –  java May 11 '13 at 4:33

The result is correct, albeit the reasoning is not. As I mentioned in a comment, your interpretation of the Law of Large Numbers is not right. And any argument that uses the word "infinity" is almost automatically at least incomplete.

However, given that the first toss was a tail, with probability $1$ the number of heads and tails will equalize. Also with probability $1$, if the first toss is a tail, then at some time tails will be $17$ ahead. These facts are not implied by the Law of Large Numbers, but they are true.

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Forget the betting. I think what you're confused about is the tension between the following two observations:

  • If the first flip is heads, all of the subsequent flips are independent, so there's no reason to expect more tails than heads in the future.
  • By the law of large numbers, the number of heads and the number of tails must be about equal eventually.

These two observations seem to contradict each other, but they don't. The reason is that the law of large numbers is a probabilistic statement about what you should expect to happen before you see any of the coin flips. After you see the first coin flip, you need to condition on it, and when you do the law of large numbers now says that the number of heads and the number of tails after the first coin flip must be about equal eventually.

This might be clearer if, instead of seeing that the first flip is heads, you saw that the first 1,000 flips were heads. This is very unlikely, but conditioned on it happening, you have no reason to expect the universe to magically force 1,000 extra flips to be tails in the future to compensate; that's the gambler's fallacy.

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Imagine this... you can see into the future and find out when the number of heads and tails will equal each other, assuming you agree that eventually heads will = tails then you will be able to see into the future to this point. Say this happens on flip 100, and your first 50 flips are tails. Well then you can assume that every flip from then on will be heads because (50-0)/(100-50) = 1 or 100%. This future point can also be called x. Just because you don't know when this point is does not mean it can change just as the number of tooth picks in a tooth pick jar will not change simply bc u –  java May 11 '13 at 4:44
    
don't know how many tooth picks are in the jar –  java May 11 '13 at 4:44
    
@java: you're conditioning on a whole bunch of extra information there. –  Qiaochu Yuan May 11 '13 at 5:24

Given any amount of money you want to make, there is a (computable) way to do so on any infinite sequence of coin flips. However, you can't do this uniformly; i.e. almost surely (i.e. with probability 1) you cannot use a single strategy to win arbitrarily much.

Your strategy is not wrong, it's just not uniform. Almost surely there will indeed be a point with the same number of heads as tails. In fact infinitely many of them. If you know that point ahead of time, then you could indeed exploit that to make money in the way you described. But without that, all you might be able to do is guesse correctly a place before which there will be the same number of heads as tails. But this isn't so useful.

BTW, The way to interperet almost surely is that it would be foolish to think anything but that will happen, even though it theoretically could.

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