I'm confused... "Evening Up" and "Due"

[12AX7]

In a thread elsewhere master_helmuth was trying to tell everyone he could "beat" roulette.

I think we all know that because each bet is -EV, in the long run, this is false. An number of negatives do not add to a positive.

However, he did make a point that has me tossing and turning a little bit.

It was something along the lines of, (in my own words), "If things didn't even up, probability wouldn't work, therefore it's reasonable to expect that you are due to hit eventually if you are behind."

Sounds compelling. Taking the proverbial coin flip, if my total of heads is less than expected, seems reasonable, if things even up over the long haul to expect a "correction" and thus be "due" some heads.

Can someone help me out here? I've read discussions of Gambler's Fallacy, yet the idea of evening up over the long run almost seems to support it?

As an aside, even if you are "due" if you are getting paid -EV... you bankroll won't even up. Even if the events do even up. LOL!

In any event, I clearly understand why the next flip is still 50/50. (Independent random events) And yet... there's this evening up concept out there.

Anyway, I'm having a hard time logicing this one out and reconciling the two concepts. Any help appreciated.

[timex]

By evening up in the long run, it doesn't mean that you will have an equal number of heads and tails, it means that the difference between the number of heads and tails divided by the total number of flips approaches zero. If you flip the coin 1 trillion times, you could easily have 1 million more heads than tails, it is just a small number relative to 1 trillion. So in the long run, you don't actually become due, your results just become similar. A good way to look at this is if I flip a coin n times, before flipping the coin against you, our future results will not be altered by whether or not you know the results of the previous n flips.

[12AX7]

I'm not sure I follow this statement, but it seems to still imply an evening up.

Do you mean:

(# H's / Flips) - (# T's /Flips) -> 0

If so this would still seem to imply that if one side or the other was showing a "deficit" that it would be "due"?

[ QUOTE ]
... it means that the difference between the number of heads and tails divided by the total number of flips approaches zero. ...

[/ QUOTE ]

[drbst]

The weak law of large numbers says that the frequencies approach the probabilities, it does not say that numbers "catch up".

Example: You flip a fair coin a hundred times and may get 53 heads and 47 tails. You flip the coin 1000 times and may get 510 heads and 490 tails. The frequencies approach the probabilities while the total difference between both events get larger.

Here is another system for roulette: You go into a casino and bet $1000 on red. If you lose, you leave and come back next day. If you win, you bet another $1000 on red. You do this until you lose. This way you can never loose more than $1000 a day but you can win much more until the first time red does not come. Since losing days and winning days are equally likely, you can make a lot of money with this system.

[ThinkQuick]

lets say you are about to begin betting on heads, $1 per coinflip.

You understand that you are now (before we start playing) even.
You understand that in the future, after some large number of flips from now, you expect there to be no gain or loss and therefore you will be even.


Now lets say you play for an hour and are down $20. This is within the realm of possibility.
Right now, while you are down $20: in the future, after some large number of flips.. you expect there to be no gain or loss and therefore you can expect to be still down $20.

The big fallacy here that pertains to gamblers is that people try to decide themselves when to start/stop counting.

To analyze my analogy in another way: Lets say you're playing for two hours total.
After an hour of flipping, there were many possible outcomes, most of which put you close to zero. Unfortunately, you ended up following a less likely outcome where you ended up down $20.
You're right that from the beginning of the game, you were probably going to come out even, but now something unlikely has happened and things have changed. You are down $20 and have an hour left to play:
At this point, it is equally likely that you will win the 20 back as you will lose an additional 20. The most probable outcome is that you will come out approximately even during that hour (-$20 overall in two hours).

[ChicagoVince]

There are major flaws in his reasoning. First, things will never 'even up' in roulette because you are not getting even odds. The house makes money because they are laying 1/38 when the wager is really 1/39 or 1/40. If you continue to bet red you will eventually go broke, not because of the blacks, but because of the greens.

Second, each spin of the roulette wheel is an independent event. Like others have said, eventually the proportion will approach the proper odds (19/40 for red or black on a 0 and 00 roulette wheel), but there is no such thing as 'due'.

[Mike Haven]

regrettably, master_hellmuth was a troll poster, and, after a warning to cease went unheeded, he had to be banned from twoplustwo

[AaronBrown]

I assume this is a joke system, posted to show the flaws in such reasoning.

In this case, the flaw is that winning and losing days are not equally likely. If betting red on roulette were a 50% bet, you would lose $1,000 half the time, break even one quarter of the time (win the first spin, lose the second) and win an average of $2,000 the other quarter of the time. That comes out to a zero expected value.

The actual numbers for American roulette (two zeros and no held bets) are 53% chance of losing $1,000, 25% chance of breaking even and 22% chance of winning an average of $1,900. Your expectation from this scheme is -$100 per day.

[AaronBrown]

To restate timex's point, averages even up, counts do not.

You expect to win 47% of the time betting red or black in American roulette. If you have won 40 of 100 bets, you expect to win 47 of the next 100, the same as if you had won 60 or any other number of bets in the first 100. It doesn't matter whether you switch back an forth between red and black using any system; or if you stick with one color.

If you have won 40 of your first 100 bets in roulette, your winning percentage is 40%. After 200 spins, you expect your winning percentage to be up to 43.5%. It could be higher or lower, but with average luck, you'll win 47 of the next 100 spins, for 87 wins out of 200, for a 43.5% winning percentage. Note that although your expected winning percentage went up from 40% to 43.5%, the number of wins you are behind expectation remained constant at 7 (40 versus an expected 47 in the first 100, 87 versus an expected 94 over the first 200). The casino pays off for wins, not for percentages.

Probability works because averages move toward their expected values. Roulette systems don't work because counts meander randomly with no memory of the past.

[Thremp]

[ QUOTE ]
I assume this is a joke system, posted to show the flaws in such reasoning.

In this case, the flaw is that winning and losing days are not equally likely. If betting red on roulette were a 50% bet, you would lose $1,000 half the time, break even one quarter of the time (win the first spin, lose the second) and win an average of $2,000 the other quarter of the time. That comes out to a zero expected value.

The actual numbers for American roulette (two zeros and no held bets) are 53% chance of losing $1,000, 25% chance of breaking even and 22% chance of winning an average of $1,900. Your expectation from this scheme is -$100 per day.

[/ QUOTE ]

You are not accounting for the times you continue to win. He's got a percentage chance to win 3+ bets everytime he sets in there. But will never lose more than one bet.

I'm not saying you'll win money like this. But your analysis is oversimplified.