A Simple Math Question?

Quote: EvenBob

Math questions for sure, nobody here got it right.

expletive deleted
I'm going to track down that other forum and analyse your wondrous contribution to it. I might even bow to your genius there and to the geniuses that advise on that forum....
... or I might just p155 myself laughing at you all.

Quote: OnceDear

expletive deleted
I'm going to track down that other forum.

It's invitation only, the best ones always
are. I'm on several. (I should have said
nothing and let you look).

Quote: EvenBob

Math questions for sure, nobody here got
it right.

Everyone got it right. Except the person thrashing around right now with some copied off wikipedia explanation he clearly doesn't understand about empirical data.

It's 75%.

Quote: thecesspit

doesn't understand about empirical data.

If it's not empirical, what's the answer then. If each
dozen is worth 50% what is the math to determine
how often you will get one right right when you bet
2 dozen. (the door question isn't the same, as somebody
pointed out.)

50% + 50% is 100%. What's the next step.

You have to determine the probability of being wrong both times (25%) and then subtract that from 1.

So the answer is you will get one of them right 75% of the time.

Quote: EvenBob

Quote: OnceDear

expletive deleted
I'm going to track down that other forum.

It's invitation only, the best ones always
are. I'm on several. (I should have said
nothing and let you look).

Ummm....dafuq?

Oh to hell with it

Quote: Thermos

You have to determine the probability of being wrong both times (25%) and then subtract that from 1.
.

50% chance of being right +
50% chance of being right =
25% chance of being wrong?

What's the equation for that?

Quote: OnceDear

You could never have had "a 50% of getting one dozen right."

You could if you were a theoretical good guesser.
But there is no math for that, so it's falls under
empirical probability where you would use the
data from actual trials to determine the probability.

Thanks for clearing that up.

Quote: EvenBob

Bet on two dozens where each has
a 50% chance of being correct. What are the chances
of getting one correct?

Guess I misunderstood this theoretical scenario. If you bet on two dozens, each with a 50% chance of winning, you will get one correct 100% of the time.

I got a coin....50% chance to land on heads, 50% chance to land on tails. What I don't understand is, what's the chance that it won't land on either?

If EvenBob spends 50% of his life upstairs in a building and 50% of his life downstairs in a building and 50% of his life outside in the sunshine, how long would he live?
Who cares?

Bob, I have tried to give you some straight answers. You did reply to my first post in this thread with a question -- I would have replied to that, but someone else gave the correct answer before I read your question.

However, I never got a reply to my question about your empirical data approach.

Quote: Doc

Bob, does your empirical probability approach suggest that you can collect data on real roulette rolls and summarize the outcomes to find that 50% of the results are in the 1st dozen, 50% are in the 2nd dozen, and 50% are in the 3rd dozen, as suggested in your earlier post?

I'm still trying to understand what it is that you are trying to describe that all of the rest of us here are having difficulty interpreting.

Quote: Doc

Bob, does your empirical probability approach suggest that you can collect data on real roulette rolls and summarize the outcomes to find that 50% of the results are in the 1st dozen, 50% are in the 2nd dozen, and 50% are in the 3rd dozen, as suggested in your earlier post?

I never saw this, I don't know why.

If you could guess 50% on each dozen, yes,
you would have sessions on a real wheel and
analyze the data.

Quote: RS

I got a coin....50% chance to land on heads, 50% chance to land on tails. What I don't understand is, what's the chance that it won't land on either?

Using EB's logic and always assuming you're a good guesser, it will be

Heads 1 in 3 times
Tails 1 in 3 times
Heads and tails 1 in 3 times
Neither heads or tails, about 50% of the time.

Empirically speaking of course.

Quote: Jeepster

Heads 1 in 3 times
Tails 1 in 3 times

That's not the same as 3 outcomes, you
only have 2 when flipping a coin.

Quote: EvenBob

That's not the same as 3 outcomes, you
only have 2 when flipping a coin.

So what if you are a bad guesser ( is tosser the appropriate word?) and each time you guess the outcome but you are right with any guess 33% of the time. But then, to edge your bets, you stake a one dollar bet on heads and a one dollar bet on tails. Somehow you would still manage to lose money. The scenario is not too different.
Ask your buddies on that other forum about bad tossers guessers.

* In the UK, we don't flip coins, we toss them.

EvenBob has a sense of humour, it's very dry.

Quote: EvenBob

That's not the same as 3 outcomes, you
only have 2 when flipping a coin.

You must've never seen the movie Mouse Hunt.

You only seem to be satisfied when you hear what you want to hear.

Quote: Dalex64

You only seem to be satisfied when you hear what you want to hear.

I'm satisfied when I hear the correct answer.
Aren't you? Are you satisfied with wrong
answers? That would be odd.

Bob,

Are you sticking by your claim that on a 36-number roulette wheel, for one spin, you are assigning the following probabilities:

first dozen: 50 percent
second dozen: 50 percent
third dozen: 50 percent

?

Do you not see how absurd this is?

Go draw a circle. Divide it in half with a line. Each portion of the circle is 50 percent:



Do you see room for another 50 percent in there? Are we extending to dimensions that don't exist in space or time?

Your options can't add up to more than 100 percent. There isn't room.

If you can't grasp this concept you need to go repeat 5th grade. I'm serious.

Quote: sodawater

Your options can't add up to more than 100 percent. There isn't room.
.

You really need to catch up. We've already
established that if you guess better than
average, math cannot be used to determine
probability here. You need calculate it by
gathering data from actual sessions and
doing it that way. Empirical probability.

I’m going to try once again to give Bob some straight answers. I know I have used just a tiny bit of sarcasm in this thread, but nothing like most of the other comments he has received.

I don’t really think that Bob is deliberately trolling the membership; I think he is describing things in different terms and a different manner than some of us would use, with the result being that (to us) it sounds as if he doesn’t understand arithmetic. I don’t think that’s the case, so I’m going to try again.

Here is one possible scenario that Bob might be describing, but I’ll leave it to him to say whether this is what he means or not:

If everything is normal and random about a roulette game, except that this wheel has no zeroes, then each dozen has a probability of 1/3 being the winner on each roll. Suppose, though, that you are a good guesser, or that you have mind-over-matter influence on the ball, or that the game is partially rigged somehow. Suppose that under that scenario, when you bet on a particular dozen you are able to win at a 50% rate. (It would be like a gold mine in terms of probable win rate and payout rate on the wins, except that it would require a lot less labor than a gold mine.)

Suppose you can do that on any dozen that you like, and whichever dozen you pick for a given roll will hit at a 50% frequency. Yes, the 1st, 2nd, and 3rd dozen all will hit at a 50% rate, if that is the dozen that you have bet on for that roll and if you have used whatever influence, game rigging, or “good guessing” you have at your disposal to make that the 50% dozen. Note that this does not mean or imply that any particular dozen (or every dozen) hits 50% of all rolls – just the rolls on which you have picked that dozen.

If that is the hypothetical (non-real-world) situation that Bob has attempted to describe, then I think his question goes something like this: Suppose you used that “good guessing”, influence, game-rigging, whatever on two of the dozens and bet those two dozens for the same roll. What would be the probability of winning one of those bets or the other, accepting that it is not possible to win both of them on the same roll?

Bob, is that anything like what you have tried to ask? If so, perhaps someone can suggest an appropriate analysis. Here is one possible analysis:

You chose two of the dozens to wager on, call them X and Y. We are going to consider the results in terms of dozens A and B, where A might be either X or Y, with B being the other one. Your wager on dozen A has a 50% chance of winning. If it loses (50% likely), then you still have a 50% chance that dozen B will be the winner. In that case, there is an overall 25% chance that B is the winner because you must have both that A loses and that B wins. Combining these gives 50% for A and 25% for B for a total of 75%. This is more commonly calculated as was shown in a previous post:

P = 1 – (0.5 * 0.5) = 1 – 0.25 = 0.75

Note: If you use this two-dozens strategy, you will have converted your gold mine into a silver mine. For the same total wager per roll, you will have a reduced net expected win. For example, with a $20 wager per spin/roll, the single-dozen strategy would provide an expected net win of $10 per spin/roll (50% chance of losing $20 and 50% chance of winning $40). If instead you split that $20 wager between two of the dozens and use your influence on both, you will have a 25% chance of losing the full $20 and a 75% chance of losing one of the $10 wagers and getting paid $20 in winnings on the other, for a net expected win of:

E = 0.25 * (-$20) + 0.75 * (-$10 + $20) = -$5 + $7.50 = +$2.50

Better to stick with just one of the dozens.

There is a completely different analysis of this hypothetical situation in which you decide that you are going to bet on the same dozens X and Y every time and that your guessing, influence, whatever must get each of those bets as winners 50% of the time. Under that analysis, you get a combined 50% + 50% = 100% likelihood of one or the other winning on every roll. If you split your $20 wager per spin/roll under that scenario, then on every turn you would lose one $10 wager and win $20 on the other one for a net win of $10 each time, the same as the expected average win for the single-dozen strategy but without the variance.

However, I don’t think that is what you have in mind, since you have said that the other (third) dozen will still win some of the time.

Let’s start with whether the hypothetical scenario I described is what you had in mind, Bob, and then maybe members can offer better analyses.

Quote: EvenBob

You really need to catch up. We've already established that if you guess better than average, math cannot be used to determine probability here. You need calculate it by gathering data from actual sessions and doing it that way. Empirical probability.

OK. This is absolutely my last attempt.
Kudos to Doc for his reply with all the calculations and kindness, but I'm going to not use too much math. I'm going to use Empirical Probability and here is the challenge to EvenBob: Create a set of outcomes which give Empirical probability that each guess has a 50% success rate and populate a little list or spreadsheet with those outcomes and results. The set of outcomes can be made up in any way that EB likes. He can observe and jot down a million real spins or he can just fake up three or more results as though he had seen them himself. From that list of outcomes we will calculate all the Empirical probabilities. He should then be able to demonstrate to himself why none of this works. Here's a spreadsheet from where I had a go. He can use that as a template.

www.oncedear.com/EB_EmpiricalNonsense.xlsx

First Let's define 'Empirical Probability' http://en.wikipedia.org/wiki/Empirical_probability
I'll paraphrase as 'It's our best estimate of probability based on the observation of a real set of events'

Here is a picture of my made up results. I'm damned if I can Empirically observe any set of outcomes which satisfy the question and which don't get 100% as the probability of winning every time.

Quote: Doc

....Combining these gives 50% for A and 25% for B for a total of 75%. This is more commonly calculated as was shown in a previous post:

P = 1 – (0.5 * 0.5) = 1 – 0.25 = 0.75

Doc, A few of us have done that calculation, but here's where it breaks down. The two bets are not independent of eachother. It's all one spin. The option of both bets being the same is excluded by the condition of the test

Consider instead that our good guessers have no talent and that each guess has a 33¹/₃ % probability of being right. That flips to a 66²/₃ % probability that any guess will be wrong.
These are average quality guessers in the real world. Let's plug those numbers into your calculation ( and the calculation of previous posters.)

Stay out of this EB. You are not qualified!

Probability of at least one success = (probability of any outcome)-(Probability of both bets losing)
Probability of at least one success = (probability of any outcome)-((Probability of bet1 losing)*(Probability of bet2 losing)) (Flawed)
Probability of at least one success = (1)-((²/₃)*(²/₃))
Probability of at least one success = (1)-(⁴/₉)
Probability of at least one success = ⁵/₉

Probability of at least one success = 55.5555555555555555556%

Which we know is not true, because the probability of both bets losing is actually 33¹/₃%

It does not matter what you say, your logic, nor your reasoning. EB is simply trolling. He couldn't be doing anything other than trolling.


Seriously, think about it. He's basically spelled it out to us -- you really think that this "invite only" gambling forum has invited EB to join them?


Now, perhaps there is a chance that EB is actually a super pro AP who is doing all the best plays, knows the best APs and has the best information, he's on these "private invite only" forums, etc. etc....and he decides to waste his time trolling this peasant-like forums where anyone can join (WOV). That is definitely a possibility.




Either EB is one of two things:

1) He is a super-pro AP who decides to troll public forums. If this is the case, a response is not warranted.

2) He's just a typical degenerate ploppy/troll who thinks he can beat some game with his excellent guessing work, kinda like jybr0dyz [or whoever the f*ck is making threads about betting systems].





edit: Plus, the thread is "a simple math question".....with an absolutely f***** retarded question.

Quote: RS

It does not matter what you say, your logic, nor your reasoning. EB is simply trolling. He couldn't be doing anything other than trolling.


Seriously, think about it. He's basically spelled it out to us -- you really think that this "invite only" gambling forum has invited EB to join them?

I fear you are right. Far too many of us have expended far too much time on this. No good turn goes unpunished.
EvenBob has been ignorant and rude and does not deserve any help or attention. Sadly such behaviour as Bob's can only serve to make helpful people less inclined to be helpful in future. I'm blocking EB and will never reply to him again in any form. I'll no longer contribute to any thread that he starts. I expect to look in on this forum less often as a direct consequence of EB. That might very very slightly reduce the viability of this forum. Let idiots, trolls and non-contributors keep the forum busy. Let's see how the forum owners business model prospers under those circumstances.

Quote: OnceDear

I fear you are right. Far too many of us have expended far too much time on this. No good turn goes unpunished.
EvenBob has been ignorant and rude and does not deserve any help or attention. Sadly such behaviour as Bob's can only serve to make helpful people less inclined to be helpful in future. I'm blocking EB and will never reply to him again in any form. I'll no longer contribute to any thread that he starts. I expect to look in on this forum less often as a direct consequence of EB. That might very very slightly reduce the viability of this forum. Let idiots, trolls and non-contributors keep the forum busy. Let's see how the forum owners business model prospers under those circumstances.

I thought the same -- don't respond to these (IMO idiotic) trolls -- let the forum become a wasteland of EvenBobs and their "systems". However, I don't think that is productive to AP's or serious/smart gamblers in general. Instead of avoiding these people -- we should work towards educating gamblers and APs (or future APs.

Don't block EvenBob so you can't see what he writes -- rather, every time he says something (stupid), fight back -- eventually [and theoretically], when one person is spuring nonsense BS - that person will eventually get ousted -- either he will leave the forum in shame of being wrong, or he'll be suspended (rightfully so). IMO, someone who is supporting betting systems or other degenerate-style gambling should not be a member of this forum, as that is exactly the opposite of Wizard's philsophy.

Of course, some will come on here with questions and asking for answers, typically something like "if I double my bet after a loss....and whenever I win I go back to 1 unit....I will definitely win right?" [martingale]. I've thought the same before -- a friend in college came up with the idea and told me to bet that way. [I didn't bet that way, and explained to him, that essentially you're giving up X% to the House Edge]. Those who don't know any better -- sure, ask all the questions and advice and discuss all they want. But those who know better [ie: EvenBob] should not be given the same leniency. Players supporting betting systems or other bullsh** shouldn't be posting on here -- it just confuses people who don't know any better.

After all -- the purpose of this forum is to EDUCATE gamblers into making better bets.



TL;DR: Don't block EvenBob -- rather, respond to his posts and prove why he is (almost always) wrong. Eventually, other members will realize how full of sh** he is -- and the mods won't put up with this nonsense anymore.

EvenBob has been wrong, he has bragged about a winning Roulette method with no details shared, and he does not always have the best demeanor, but my bet is on someone else getting suspended in any confrontation with him because he is very good at measuring his words to stay inside of the lines.

If you decide to dispute his claims and enter into a tussle with him, think before pushing the post button...it may save you some time in the cooler.

Quote: EvenBob

I'm satisfied when I hear the correct answer.
Aren't you? Are you satisfied with wrong
answers? That would be odd.

Hearing what you want to hear, and being satisfied by the answer, does not imply or prove that the answer is correct.

The opposite is also true. Not liking an answer doesn't make the answer wrong.

Forum Rule 1: Politely attack the writing, not the writer.

Quote: OnceDear

Doc, A few of us have done that calculation, but here's where it breaks down. The two bets are not independent of each other. It's all one spin. The option of both bets being the same is excluded by the condition of the test.

(snip)

There may be lots of reasons that the calculation I presented may not be the correct one. The purpose of my post was to try to clarify what it is that EvenBob has been trying to describe. If (and maybe that’s a tremendous “if”) we can do that, then maybe someone can figure out the proper analysis method and calculations.

I suggested that (maybe) the question is about a non-real-world condition in which a player has the Voo-Doo to influence (or guess fairly reliably) the outcome. If they use that influence on just one of the three dozens, they are successful 50% of the time. No, this scenario might not fit the usual calculations for a random event very well – it’s specifically not a typical random event since we have the influence of that Voo-Doo.

We have accepted similar non-real-world assumptions for discussion in a variety of threads in the past – remember the Newcomb Being’s paradox thread? If you will accept this current assumption that there is a 50%-successful Voo-Doo that can be applied to roulette, then the question (I think) becomes “What happens if you use your Voo-Doo on two of the dozens at the same time, perhaps working partially against yourself? What would happen then?”

I wanted to find out first whether this is really the scenario that Bob is trying to describe and the question that he is trying to ask. He has not yet responded. I suggested that if this is indeed the scenario and the question, then perhaps someone can present a good analysis technique. I then presented two possible analysis techniques that are clearly opposed to each other. I think that before we debate the techniques, we need to establish what scenario Bob is really talking about and what the question is that he would like analyzed.

And I have tried to do that without attacking EvenBob.

Quote: EvenBob

50% of getting one dozen right. If you add 50+50 you get 100

(I assume the payout of any Dozen is the traditional 2-to-1.)

On every spin:

• You risk 2 units.
• One of your units loses. (-1)
• The "other" unit wins. (+2 [won] plus +1 [original-wagered unit returned])
• You drag 3 units to your stack.
• You win 50% of your combined bet-pair, every spin.

This is like 50% interest, compounded as fast as you can shovel chips onto felt.

Better yet, only wager 1 unit on the magical 50/50 Dozen:

• 50% of the time, you lose your wagered unit (-1).
• 50% of the time, you win two units and retain your wagered unit (+3).
• You lose your wagered-unit half the time, and triple it half the time.
• Expected value: (.5 * -1) + (.5 * +3) = 1. That's 100% return, average per spin.

You're doubling your wager (on average), every spin. Play this game!

Just for the fun of it, lets assume like EB you could pick a winning dozen with 50% accuracy.
The other 2 dozens would stand at 25% each of being the winner.
So by combining your first and second choice you get a 75% chance of winning every spin.
As indignant99 says just forget the second dozen and bet the 50% dozen
A veritable gold mine.

Error

Quote: indignant99

Better yet, only wager 1 unit on the magical 50/50 Dozen:

• 50% of the time, you lose your wagered unit (-1).
• 50% of the time, you win two units and retain your wagered unit (+3).
• You lose your wagered-unit half the time, and triple it half the time.
• Expected value: (.5 * -1) + (.5 * +3) = 1. That's 100% return, average per spin.

You're doubling your wager (on average), every spin. Play this game!

I already described the advantage of sticking with the single 50% dozen. However, I thought I should point out the error in your EV calculation:

Quote: indignant99

Expected value: (.5 * -1) + (.5 * +3) = 1. That's 100% return, average per spin.

Are you trying to calculate the expected value of the net winnings per spin or the expected value of how much you have in hand afterward, including whatever remains of your original bet? If you want net winnings per spin, you need to replace that 3 with a 2 and find that the net expected winnings per spin is 0.5 for a 50% return, not 100%. If you want the expected amount that you have left after resolving the bet, then that -1 should be replaced by the zero you will have in the event of a loss, giving a calculated remaining funds amount of 1.5 units from your 1 unit wager -- again a return of 50%.

In my example, I showed a $20 wager leading to an expected win of $10 per spin on average.

The probability of winning any wager on a Roulette layout is 50%. Either you will win it, or you will not win it.

/amidoinitrite?

Quote: Deucekies

The probability of winning any wager on a Roulette layout is 50%. Either you will win it, or you will not win it.

/amidoinitrite?

I suppose the same is true at craps. Just keep betting heavily on the 12. Either you will win it or you will not win it, so the probability must be 50%, right?

Not.

I now understand that Bob's idea of this problem is less one of the mind-over-matter control of the ball that I suggested earlier and more of the "good guesser" or crystal ball source of better-than-usual information about the future. Since I strongly suspect that many on here view good guessers and crystal balls with a high degree of skepticism, I have prepared a different description of the problem that may be of interest to some. Perhaps there is someone who might be able to present a sound approach to analyzing the question at the end.



Consider this scenario: There is an on-line casino that offers roulette. They actually generate the outcomes with a standard roulette wheel and broadcast the full, continuous video of the wheel and ball. Unfortunately, they don’t run the game quite the way they should. Instead, the continuous broadcast is delayed, which presents an opportunity for past-posting if someone could find out the results between the time the video was recorded and the time that it is played for wagering.

The casino has a disgruntled employee who has access to that information, and he decides to share it with some of the players. He knows that if he gives complete info, it will be over-used and the management will detect it quickly. Instead, he decides to give semi-accurate information to just a few players – enough for them to make some money and enough to cause some pain for the employers/managers that he dislikes.

He knows that you have a fondness for the first dozen, 1 through 12. He has agreed to provide you with a signal that will help you win more often than usual when you bet on this dozen. First, he will tell you occasionally not to bet at all. He will do this every time that the result is 0 or 00 to protect you from the green numbers. He also tells you not to bet on some other rolls that are not 0 or 00 just to discourage you from betting 0-00, since he thinks the management would pick up on that.

On many but not all other rolls, he will tell you to go ahead and bet that first dozen. Rather than make it a sure thing and give you a perfect record that management will detect, he gives you the “bet” sign roughly half the time when the first dozen will win and half the time when it will lose. He figures that this will give you a gold mine to exploit, but it will not be completely obvious to the “suits”. Your man on the inside does not know when you are at the table and using the info that he is providing; you play a lot but not constantly.

Note: the fact that you don’t place a bet every time that you get the bet sign might mean that your exact win percentage is different from the percentage of winning info that is being provided by your source – he might say “bet” for a roll when the 1st dozen is the winner, but you might be on break or something, so you miss out on that opportunity, and the opposite thing when he advises you to bet on a loser. However, you collect data on your results, and you find that you are winning almost exactly half of the bets you place on the first dozen when following his advice.

After you have exploited this for some time and have found his information to be reliable at the 50% level, you learn that he is also providing similar info to Sam, a buddy of yours who has a fondness for the second dozen, 13 through 24. The disgruntled employee does not realize that you and Sam know each other, but Sam has found that the info that he has been receiving about the 2nd dozen is also reliable at the 50% level.

You and Sam begin to ponder different ways that you might combine the info that you have been receiving. One of the strategies you have considered is for both of you to bet your favorite dozen any time that you both get the signal to bet. This may not be the best strategy, but it is one that you are currently considering.

The question you would like answered is this: What can you expect as the overall outcome for your action and Sam’s combined if you were to follow this "partnership" strategy? Assume that you and Sam wager equal amounts on the same spins of the wheel/rolls of the ball.


Who can offer a good analysis of this proposed strategy with the inside information? What problems can you see for your own analysis? I can see one complication that I do not know how to analyze.

Quote: Doc

Consider this scenario...

This whole scenario you have outlined involves multiple, repeated spins.

Bob insists that each dozen has a 50 percent chance to win on a SINGLE spin, thus revealing a staggering chasm of mathematical and logical ignorance that should shock anyone reading the thread.

Quote: sodawater

Bob insists that each dozen has a 50 percent chance to win on a SINGLE spin,.

So does Doc's scenario. Each bettor of a single
dozen has a 50% chance on each spin.

I think the scenario that I described matches Bob's original description and at least most of his subsequent posts. You have repeatedly posted that these conditions are not possible. How about analyzing the problem I presented, if you can, and predict the expected results of the "partnership" strategy.

Quote: Doc

Quote: indignant99

Expected value: (.5 * -1) + (.5 * +3) = 1. That's 100% return...

... you need to replace that 3 with a 2...

Thanks for correcting my flawed EV calc. But, both of our analyses were wrong.

Betting one Dozen:

Probability	Dozen 1	Net
Fraction Decimal
1/2 ________ .500	Lose (-1)	-1
1/2 ________ .500	Win (+2)	+2
		+1

So, we get an expected return of +1 unit (over 2 events), 2 units wagered.
(1 unit / 2 units) = .5 = 50%

Betting two Dozens:

Probability	Dozen 1	Dozen 2	Net
Fraction Decimal
1/4 ________ .250	Lose (-1)	Lose (-1)	-2
1/4 ________ .250	Lose (-1)	Win (+2)	+1
1/4 ________ .250	Win (+2)	Lose (-1)	+1
1/4 ________ .250	Win (+2)	Win (+2)	+4
			+4

So, we get an expected return of +4 units (over 4 events), 8 units wagered.
(4 units / 8 units) = .5 = 50%

Betting three Dozens:

Probability	Dozen 1	Dozen 2	Dozen 3	Net
Fraction Decimal
1/8 ________ .125	Lose (-1)	Lose (-1)	Lose (-1)	-3
1/8 ________ .125	Lose (-1)	Lose (-1)	Win (+2)	0
1/8 ________ .125	Lose (-1)	Win (+2)	Lose (-1)	0
1/8 ________ .125	Lose (-1)	Win (+2)	Win (+2)	+3
1/8 ________ .125	Win (+2)	Lose (-1)	Lose (-1)	0
1/8 ________ .125	Win (+2)	Lose (-1)	Win (+2)	+3
1/8 ________ .125	Win (+2)	Win (+2)	Lose (-1)	+3
1/8 ________ .125	Win (+2)	Win (+2)	Win (+2)	+6
				+12

So, we get an expected return of +12 units (over 8 events), 24 units wagered.
(12 units / 24 units) = .5 = 50%

Shocker! Counter-intuitive! Perhaps. I had no firm intuition; more like a Mersenne-Twisted mind.

Quote: indignant99

Shocker! Counter-intuitive!

You have both dozens winning at the
same time on several spins. You realize
that's not possible, right? Only 1 dozen
can ever win on a single spin.

Quote: EvenBob

You have both dozens winning at the
same time on several spins. You realize
that's not possible, right? Only 1 dozen
can ever win on a single spin.

You have a fantasy scenario in the first place, right?
You said that any dozen enjoys a 50/50 chance of winning, right?
There's nothing in your problem, as presented, to preclude simultaneous wins.
Because if you enforce that proviso, you must drop back to 1/3, 1/3, 1/3 reality.
There's nothing in your problem to preclude 3 simultaneous losses, either.

Is there a specific Dozen than enjoys this 50% probability? If so, then all the pockets in this Blessed Dozen are double-wide, occupying half the wheel. The other 24 pockets (the Damned Two Dozen) have a strict average width of half as wide; all 24 of them in aggregate occupying the "other half" of the wheel.

Quote: indignant99

There's nothing in your problem, as presented, to preclude simultaneous wins.
.

Each dozen has 12 unique numbers, how could
two dozens win at the same time.

Quote: EvenBob

Each dozen has 12 unique numbers, how could
two dozens win at the same time.

It happens in the same universe where the dozens come up 50% of the time.