A Simple Math Question?
you have the 3 dozen bets. 33 1/3% chance
on each, bet two of them and add 33 1/3 and
33 1/3 and you have a 66 2/3 chance of getting
1 dozen correct if you get rid of the zeros.
Lets say theoretically you have a 50% of getting
one dozen right. If you add 50+50 you get 100,
and I know that's not correct for the chance of
getting one right. So how do you do the math
to figure the chance of getting one right.
Quote: EvenBobI can't wrap my head around this. In roulette,
you have the 3 dozen bets. 33 1/3% chance
on each, bet two of them and add 33 1/3 and
33 1/3 and you have a 66 2/3 chance of getting
1 dozen correct if you get rid of the zeros.
Lets say theoretically you have a 50% of getting
one dozen right. If you add 50+50 you get 100,
and I know that's not correct for the chance of
getting one right. So how do you do the math
to figure the chance of getting one right.
lol, what?
OK, so imagine a roulette wheel with 36 numbers and no zeroes.
You want to know the chances of getting at least one dozen right in two spins?
This would be simply: 1 - (2/3 * 2/3) = 5/9
OK, so on the same 36-number roulette wheel, you want to know what are the chances of getting red at least once in two spins. This is more accurate for your question of 50% for one spin.
1 - (1/2 * 1/2) = 3/4
Quote: sodawater
You want to know the chances of getting at least one dozen right in two spins?
No. Bet on two dozens for one spin. 1/3 + 1/3 = 2/3 chance
of getting one right.
Quote: sodawateryou want to know what are the chances of getting red at least once in two spins.
No, not at all. Bet on two dozens where each has
a 50% chance of being correct. What are the chances
of getting one correct?
Quote: EvenBobNo. Bet on two dozens for one spin. 1/3 + 1/3 = 2/3 chance
of getting one right.
Yes, that is correct.
Quote: EvenBob. Bet on two dozens where each has
a 50% chance of being correct. What are the chances
of getting one correct?
This question is silly by definition. If a 36-number wheel is being divided into three dozens, and you stipulate that two of those dozens each has a 50% chance of being correct, then the answer getting one out of those two is by definition 100%. If you can truly "eliminate" one of the dozens such that the other two dozens each has a 50% chance, then of course the answer is 100%.
This is the same as asking on a 0-free wheel, what are the chances of getting one correct betting both red and black. It's 100%.
Now say in the same scenario, you bet on ONE of the dozens that you stipulate somehow has a 50% chance of winning. Then the chances of getting at least one right in two spins is 1 - (1/2 * 1/2) = 3/4
Quote: EvenBobLets say theoretically you have a 50% of getting one dozen right.
"50% chance" from what? Do you mean, if you bet on two dozens and win, each has a 50% chance of being the winning dozen?
You can do it two ways:
1. "Obviously" each dozen has the same chance of winning, so the chance of a particular dozen winning is 1/3.
2. You have a 2/3 chance of it being one of the two dozens you bet on; if you win, each of the two has a 1/2 chance of being the right one; 2/3 x 1/2 = 1/3.
This is simply 50% + 50% = 100%
This is the exact same thing as asking, on a 36-number wheel, what are the chances of getting red OR black. 100%.
Quote: sodawaterNo, Bob is saying that each of two dozens has a 50% chance of winning. He wants to know what are the chances of winning if you bet on both of them.
This is simply 50% + 50% = 100%
This is the exact same thing as asking, on a 36-number wheel, what are the chances of getting red OR black. 100%.
So the third dozen is teflon or something? There are only 24 numbers that can be hit?
Quote: DeucekiesSo the third dozen is teflon or something? There are only 24 numbers that can be hit?
That is his stipulation.
Quote: sodawater
This is simply 50% + 50% = 100%
That is not correct. The dozen numbers that
you didn't bet on are still there and will win
sometimes, so you can't win 100% of the
time.
The question is simple. Each dozen has a 50%
chance of winning. No zeros. If you bet on
dozen 1 and dozen 2, what's the probability
of one of them winning.
Quote: EvenBobThat is not correct. The dozen numbers that
you didn't bet on are still there and will win
sometimes, so you can't win 100% of the
time.
The question is simple. Each dozen has a 50%
chance of winning. No zeros. If you bet on
dozen 1 and dozen 2, what's the probability
of one of them winning.
OMG -- this has got to be a troll.
If each dozen has a 50% chance of winning, then the two of them are 100%.
If "the dozen numbers that you didn't bet on are still there and will win sometimes" -- then BY DEFINITION the first two dozen don't have a 50% chance each.
You can't have it both ways.
Quote: sodawaterThat is his stipulation.
Where did I say that??? Of course they can
be hit, there are 36 numbers and you are
betting 24 of them. So your answer of 100%
is wrong.
Quote: sodawater
If "the dozen numbers that you didn't bet on are still there and will win sometimes" -- then BY DEFINITION the first two dozen don't have a 50% chance each.
Sigh. It's theoretical, as I said in the first post.
Lets make this real easy to understand. It's
rigged. You know how to get 50% right on
the first and second dozens. You can't win
100% of the time because the 3rd dozen
will still win part of the time. So what's
the probability of getting the first or
second dozen right. Maybe a better way
of asking is how often will the 3rd dozen
win?
Quote: EvenBobSigh. It's theoretical, as I said in the first post.
Lets make this real easy to understand. It's
rigged. You know how to get 50% right on
the first and second dozens. You can't win
100% of the time because the 3rd dozen
will still win part of the time. So what's
the probability of getting the first or
second dozen right. Maybe a better way
of asking is how often will the 3rd dozen
win?
If the third dozen will still win part of the time, then the probability for each of the first two cannot be 50%. You need to adjust the probability of the first two dozens down to account for the third dozen winning part of the time.
Without knowing how often the third dozen will win, you cannot answer your question.
Quote: sodawater
Without knowing how often the third dozen will win, you cannot answer your question.
As stated in the first post, all THREE dozens
win FIFTY PERCENT of the time. 50-50-50.
Quote: EvenBobAs stated in the first post, all THREE dozens
win FIFTY PERCENT of the time. 50-50-50.
This is hilarious. You can't have probabilities that add up to more than 1. Therefore all three dozens cannot each have a 50% chance of winning. Do you know what 50% means?
If all three dozens have an EQUAL chance of winning (also known as 1/3), then the chance of you betting on two of them and winning on one spin is simply 2/3.
Quote: EvenBobAs stated in the first post, all THREE dozens
win FIFTY PERCENT of the time. 50-50-50.
Those three things are each equally likely to occur. And, one and only one of those things must occur on each spin. The sum of the probabilities of these events occurring cannot be anything other than 100%. 50% + 50% + 50% is not 100%. So you're doing something wrong in there somewhere.
Quote: sodawaterThis is hilarious. You can't have probabilities that add up to more than 1. Therefore all three dozens cannot each have a 50% chance of winning.
Let me explain how it works. Read carefully.
You bet on 1 and 2 and each wins 50%
of the time. But on this spin the first
dozen loses because it only wins half
the time. The second bet loses because
it only wins half the time. The third
dozen wins. So yes, all 3 can have
theoretical chance of being 50%.
So how often will the first and second
dozen win?
Quote: EvenBobLet me explain how it works. Read carefully.
You bet on 1 and 2 and each wins 50%
of the time. But on this spin the first
dozen loses because it only wins half
the time. The second bet loses because
it only wins half the time. The third
dozen wins. So yes, all 3 can have
theoretical chance of being 50%.
So how often will the first and second
dozen win?
I am sorry but that is simply not how math works.
The question is impossible to answer because your assumptions are wrong.
If there is an event with three options, and one of them must happen, then all three options cannot each be 50%.
The percentages of all possible options have to add up to 100.
Quote: sodawaterI am sorry but that is simply not how math works.
.
If you have a 50% chance of getting 12 out
of 36 numbers correct that does not mean
you will win every time. If you a have 50% chance
on each of two dozens, you still won't win
every time. How often will the dozen not
bet on show up? Keep in mind you're
obviously not betting the same 2 dozens
every time.
Quote: EvenBobIf you a 50% chance
on each of two dozens, you still won't win
every time.
By definition, if you have a 50% chance on each of two dozens on the same wheel on the same spin, you will win 100% of the time.
If you have a 50% chance on each of two dozens on different wheels and/or different spins, you will win 3/4 of the time.
has a car or a donkey behind it. With each
door you have a 50% chance of winning a
car. You can pick two doors at once.
If one of them has the car, you win. A
coin was flipped for each door to determine
car or donkey.
You pick door 1 and 2. Open door 1 and
there is a donkey. Door 2 still has a 50%
chance of having a car. What is the probability
of picking the right two doors with a
car behind one of them?
Kind of like Monty Hall, but here you get
to pick 2 doors. No you can't win 2 cars,
if the first door has a car, the second door
is not opened.
Quote: EvenBobTry it this way. You have 3 doors. Each door
has a car or a donkey behind it. With each
door you have a 50% chance of winning a
car. You can pick two doors at once.
If one of them has the car, you win. A
coin was flipped for each door to determine
car or donkey.
You pick door 1 and 2. Open door 1 and
there is a donkey. Door 2 still has a 50%
chance of having a car. What is the probability
of picking the right two doors with a
car behind one of them?
Kind of like Monty Hall, but here you get
to pick 2 doors. No you can't win 2 cars,
if the first door has a car, the second door
is not opened.
Bob, I interpret your latest description – the three-doors version – to mean that each of the doors has a 50-50 chance of having a car vs. a donkey, and that these chances are completely independent of what happens to be behind either of the other two doors. Is that correct? And you want to know the probability of winning at least one car if you can choose two doors. Correct? Simple enough – you have a 75 percent chance of getting at least one car.
However, this is quite different from the way most anyone would interpret your description of the roulette problem. While there might be a car behind all three doors (12.5% probability), the roulette ball cannot possibly stop in all three of the dozens on the same roll. I think that’s where there is disagreement between what you seem to be trying to describe and what people get when reading your description.
Quote: EvenBobTry it this way. You have 3 doors. Each door
has a car or a donkey behind it. With each
door you have a 50% chance of winning a
car. You can pick two doors at once.
If one of them has the car, you win. A
coin was flipped for each door to determine
car or donkey.
You pick door 1 and 2. Open door 1 and
there is a donkey. Door 2 still has a 50%
chance of having a car. What is the probability
of picking the right two doors with a
car behind one of them?
Kind of like Monty Hall, but here you get
to pick 2 doors. No you can't win 2 cars,
if the first door has a car, the second door
is not opened.
This model does not fit your question of a single spin on a roulette wheel.
But I will try to be helpful and answer this question as you have stated it. But please remember it absolutely cannot and doesn't apply to a single spin at roulette.
Given: There are three doors, each with a 50% chance of a car behind them. You get to pick two doors. If you win a car on your first door, you don't open the second door.
The chance you will win a car is 1 - (1/2 * 1/2) = 3/4.
This is exactly the same answer as TWO spins in roulette, having a 50% chance to win each time. Each door = 1 spin.
The 3/4 answer absolutely cannot apply to a single spin of roulette.
Quote: Doc– you have a 75 percent chance of getting at least one car.
.
What's the math for that conclusion?
Quote: EvenBobWhat's the math for that conclusion?
1 - (1/2 * 1/2) = 3/4.
Quote: sodawater1 - (1/2 * 1/2) = 3/4.
.50 times .50 is .25, not .75
Quote: EvenBob.50 times .50 is .25, not .75
Jesus Christ.
But... oh forget it.
Quote: RSwas like 13 years old or something.
Who would have guessed you hung out
on forums with 13 year old's. Makes
sense when you think about it..
Or rather, for any specific event, you can add up all the bets you've made, and figure out the chance of you winning any given bet, simply by adding up the %'s. If I'm playing craps, and I have the 4,5,6,8,9,10 covered (24 ways to roll one of those numbers), and I have a $5 horn and a $6 seven [the amounts on each number doesn't matter]....then I have a 100% chance of winning one of the bets [not saying that I will have a NET win, but simply, that I am guaranteed for one of the bets to pay out].
The math gets different when you're talking about separate events. For example, I have 5 numbers covered -- what is the chance that one [or at least one?] of the numbers is hit within 3 rolls? You can't just add up each of the %'s and figure that you have a 135% chance of winning [or w/e it is, I didn't do the math].
And it is impossible to have 3 separate bets that have a 50% chance of winning. You can easily figure this out. Let's say that you believe there are 3 bets that each have a 50% chance of winning (3 dozens, 1-12, 13-24, 25-36). So, if you run a simulation, either on a computer or in real life by spinning a ball, you would expect for each dozen to show up 50% of the time [or close to 50% of the time]. And the more spins or whatever you do, the closer each of the numbers should be getting to 50%.
given: The total sum of the %'s has to be 100%.
if 3 bets at 50% chance each sums to 150%.
then 150% != 100%
:. EB is trolling
And it was a gaming forum. Think I joined it like 5 or 6 years ago. I didn't play a lot [well, I played a decent amount], but I worked a lot on writing programs/scripts/maintenance as well as forum/game moderation. But yeah, there were people of all ages.
Edit: EB, you make yourself seem all hot n mighty.....but you can't even figure this out?
Quote: EvenBobI can't wrap my head around this. In roulette,
you have the 3 dozen bets. 33 1/3% chance
on each, bet two of them and add 33 1/3 and
33 1/3 and you have a 66 2/3 chance of getting
1 dozen correct if you get rid of the zeros.
Lets say theoretically you have a 50% of getting
one dozen right. If you add 50+50 you get 100,
and I know that's not correct for the chance of
getting one right. So how do you do the math
to figure the chance of getting one right.
I'm sorry if this has been said before, but here is how I interpret the question:
There is a zero-zero biased roulette wheel, with these probabilities:
Column 1 = 1/2.
Column 2 = 1/2.
Column 3 = 0.
The probability of the ball landing in column 1 or 2 = 1/2 + 1/2 = 1.
Surely I must be misunderstanding the question as this seems too obvious.
When the ball landed within the first two dozen, there is a 50% chance that it landed in the first dozen, and a 50% chance it landed in the second dozen.
If you don't know which dozen the ball landed in, there is a 1/3 probability for each dozen.
Since I did not see this answered,
1 - ( 1/2 x 1/2 ) = 3/4
Yes, a half times a half is a quarter. Subtract that from one to get 0.75.
Quote: Wizardbiased roulette wheel, with these probabilities:
Column 1 = 1/2.
Column 2 = 1/2.
Column 3 = 0.
The probability of the ball landing in column 1 or 2 = 1/2 + 1/2 = 1.
Surely I must be misunderstanding the question as this seems too obvious.
Pay attention Mike. Evenbob clearly said that the three options each had a 50% probability.
Set 1 = 1/2
Set 2 = 1/2
Set 3 = 1/2
And Mike, don't confuse columns with dozens, because everyone knows the laws of physics are different for rows and columns :p
Quote: Evenbob
As stated in the first post, all THREE dozens
win FIFTY PERCENT of the time. 50-50-50.
So, as quite clearly stated by EB, half the time the ball lands in the first twelve: The other half the time it lands in the second twelve and the other, other half of the time it lands in the third twelve. Simple ! The table is divided into three separate and equal halves. How hard can that be? LOL!
Or, if I'm confusing time with space (area of the table) then I apologise: For any sixty minutes of time, the ball will spend thirty minutes landing in the first twelve: It will spend thirty minutes landing in the second twelve and it will spend thirty minutes landing in the third twelve. It will spend ninety minutes existing in the universe. Oh! I'm being so stupid! It will spend some time actually spinning!
Evenbob is showing an abject failure to grasp the most simple of mathematical concepts. But of course, he did say it was theoretical.
Quote: sodawater1 - (1/2 * 1/2) = 3/4.
FFS. that made me spit my coffee out. EB cannot be serious !!!Quote: evenbob.50 times .50 is .25, not .75
Huh?Quote: CrystalMathI'm pretty sure there is a chance of winning up to three times in a single spin.
yeah, I guess you could double, or is it triple, your money if you place 1/3 of bankroll on the first 12, 1/3 on second 12 and 1/3 on third 12 and that one spin landed on a number from each of those sets at the same time. EB did say it was rigged: Maybe a special kind of ball?
Quote: EvenBobI can't wrap my head around this. In roulette, you have the 3 dozen bets. 33 1/3% chance on each, bet two of them and add 33 1/3 and 33 1/3 and you have a 66 2/3 chance of getting 1 dozen correct if you get rid of the zeros.
I assume this means:
If you bet on two of them, there is a 66 2/3% chance that of the two will win.
This is correct.
Quote: EvenBobLets say theoretically you have a 50% of getting one dozen right.
This is the part that is confusing. You have only a 33 1/3% chance of getting one of the three dozens right; you said so in your first paragraph.
Where do you get 50% from?
Maybe using a specific example would make things clearer.
Quote: OnceDearHuh?
yeah, I guess you could double, or is it triple, your money if you place 1/3 of bankroll on the first 12, 1/3 on second 12 and 1/3 on third 12 and that one spin landed on a number from each of those sets at the same time. EB did say it was rigged: Maybe a special kind of ball?
I was just being a little sarcastic. EB said it was like having three doors and each door has a 50% probability of winning. Using that example, he would win on all 3 doors, 12.5% of the time. Of course, this is impossible in roulette, and you cannot win on more than one section at a time.
Quote: EvenBobAs stated in the first post, all THREE dozens
win FIFTY PERCENT of the time. 50-50-50.
You all seem to forget that we're not dealing with a mere mortal here.
Given EB's unique ability with random numbers and his outstanding prowess with roulette nothing is amiss here.
All of the three dozens winning 50% of the time while impossible (even theoretically) in the real world, isn't in EB's.
Quote: JeepsterIn the real world, isn't in EB's.
EB will just jump through two one-sided doors between motel rooms of imaginary or one-sided time, and into the arms of one of those "celibate hookers" to smooth all of this out before it collapses into the singularity and takes us all with him/her.
if you're a good guesser of the next outcome
you could have a 50% chance on each dozen.
However, the probability has to be figured
out empirically, and not theoretically. Problem
solved.
"Empirical Probability of an event is an "estimate" that the
event will happen based on how often the event occurs after
collecting data."
"Empirical probability, also known as relative frequency, or
experimental probability, is the ratio of the number of
outcomes in which a specified event occurs to the total
number of trials, not in a theoretical sample space but
in an actual experiment."
I suggest you keep a close watch on your accountant.
Quote: EvenBobI've been told on another forum that indeed,
if you're a good guesser of the next outcome
you could have a 50% chance on each dozen.
However, the probability has to be figured
out empirically, and not theoretically. Problem
solved.
"Empirical Probability of an event is ...."
EvenBob, You need to stop listening to the man on the internet. You need to stop using the word 'probability' (because you are not qualified). You need to stop digging yourself into a big embarrassing hole and most of all you need to stop believing and sprouting complete nonsense.
If someone is a lucky guesser of the next outcome, then he has a 32.42% chance on each dozen and might observe a 50% success rate. That does not change the FACT that the probability (chance) of those outcomes was 32.42% all along. You can continue to observe such good luck for an arbitrary amount of time and that could easily be a significant amount of time without being at all statistically unusual. There is no edge or advantage in luck. There is no such thing as a 'good guesser' there is only 'lucky guesser' or a guesser with a spell of 'good luck.' I have good luck in Blackjack. I have won more than I've lost even without counting. It's luck matey coupled with care. I'm not a good guesser.
Now, once and for all...
Draw three big circles of about 15cm diameter on a large piece of paper.
Take an item, any item will do, it can be an apple, the entire contents of your wallet, a pack of 20 cigarettes, or something intangible as the count of hours in a day. Or the probability of getting laid.
Put 50% of that item in the first circle. If it's the intangible hours, just write it in.
Put 50% of the item (otherwise known as the remaining 50%) in the second circle.
Now..... Put the other 50% in the third circle.
Do you see the issue? Do you?
I just know you are going to come back and say that the 50% only applies to the one that your good guesser likes at that instant in time. Well whoopy doo. Give 100% of your worldly wealth to him. He won't need it because, being a good guesser, he will already be as rich as Croesus.
'Good guessers' tell all their friends and anyone who will listen about all the times that their skill gave them a winning trip. They don't tell you about the trips that didn't go so well. Even I don't like to talk about those days. It's human nature. Get real. Use the empirical evidence of your own eyes and your own brain. And then knock off 17% for your own Confirmation bias
EvenBob just outed himself as having a sub-middle-school understanding of mathematics.
Quote: OnceDearI'm not a good guesser.
Obviously. I stated in the original question
that this theoretical, not real. IF a person
could guess the next outcome more often
than not, he could indeed get 50% right
on each dozen. But it's not a fact, just
theory.
The interesting thing is, when I asked the
same question on a gambling forum, nobody
was confused, nobody laughed or mocked,
nobody 'set me straight'. I had the correct
answer in no time.
"Empirical probability is the most accurate scientific "guess"
based on the results of experiments to collect data
about an event."
Perfect.
Quote: EvenBobObviously. I stated in the original question
that this theoretical, not real. IF a person
could guess the next outcome more often
than not, he could indeed get 50% right
on each dozen. But it's not a fact, just
theory.
The interesting thing is, when I asked the
same question on a gambling forum, nobody
was confused, nobody laughed or mocked,
nobody 'set me straight'. I had the correct
answer in no time.
"Empirical probability is the most accurate scientific "guess"
based on the results of experiments to collect data
about an event."
Perfect.
Every time you have the urge to post here, you should post it on your "gambling forum" instead.
Quote: RSEvery time you have the urge to post here, you should post it on your "gambling forum" instead.
Math questions for sure, nobody here got
it right.
