A Simple Math Question?
April 3rd, 2015 at 2:14:07 AM
permalink
Quote: indignant99Tipster does have to keep track of False-Y signals he issued to each position...
We need to add a slight nuance to this tracking... to account for different Dozens hitting at a distribution that violates the natural 1/3, 1/3, 1/3 distribution. In this table, X, Y, Z can all be different.
| Winner | Dozen-1 | Dozen-2 | Dozen-3 |
|---|---|---|---|
| Which Dozen _ Winning Count | Losing Count | Losing Count | Losing Count |
Dozen-1 ________ X | - |
~ x/2 |
~ x/2 |
Dozen-2 ________ Y | ~ y/2 |
- | ~ y/2 |
Dozen-3 ________ Z | ~ z/2 |
~ z/2 | - |
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 3rd, 2015 at 3:13:27 AM
permalink
Quote: indignant99Tipster should always give 2 Y signals:
- Y to the winning position, and
- Y to a losing position (but split equally 50/50 between them).
Major thanks to {Doc, RS} and minor thanks to {sodawater, ThatDonGuy} for helping me to spiral in to this Tipster strategy.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 3rd, 2015 at 9:13:18 PM
permalink
Quote: EvenBobEach bettor of a single dozen has a 50% chance on each spin.
Here is where you went wrong. And I do mean, by God, dead wrong. That's how you confused all of us... because that is flat f'n impossible. It cannot be true for each spin. However, it can be true for each spin that the bettor bets on. But that ain't every spin. It's a selective subset of all spins.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 3rd, 2015 at 11:47:18 PM
permalink
Quote: indignant99It cannot be true for each spin. However, it can be true for each spin that the bettor bets on..
What's the difference?
"It's not called gambling if the math is on your side."
April 4th, 2015 at 3:09:36 AM
permalink
Quote: EvenBobQuote: indignant99It cannot be true for each spin. However, it can be true for each spin that the bettor bets on..
What's the difference?
I think this explains what's going on here. EvenBob asking the probing questions for comedic effect. Nice one Bob.
https://www.youtube.com/watch?v=tzV_39xoz_E
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
April 4th, 2015 at 3:52:20 AM
permalink
Quote: indignant99Quote: EvenBobEach bettor of a single dozen has a 50% chance on each spin.
Here is where you went wrong. And I do mean, by God, dead wrong. That's how you confused all of us... because that is flat f'n impossible. It cannot be true for each spin. However, it can be true for each spin that the bettor bets on. But that ain't every spin. It's a selective subset of all spins.
Quote: EvenBobQuote: indignant99It cannot be true for each spin. However, it can be true for each spin that the bettor bets on..
What's the difference?
The difference was clearly explained in my first quoted post. Which you redacted.
As an earlier thread-post has justly and righteously exclaimed... Jesus H. Christ!
Now, I further justly and righteously exclaim... Nuclear Facepalm!
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 4th, 2015 at 11:08:34 AM
permalink
I'll repeat the original question. All 3
dozens have a 50% chance of hitting
(for whatever reason). If you bet on
two dozens at the same time, how
often will you win. Put it another way,
how often will the dozen you didn't
bet show up to make you lose.
dozens have a 50% chance of hitting
(for whatever reason). If you bet on
two dozens at the same time, how
often will you win. Put it another way,
how often will the dozen you didn't
bet show up to make you lose.
"It's not called gambling if the math is on your side."
April 4th, 2015 at 11:38:07 AM
permalink
Quote: EvenBobI'll repeat the original question.
Why ? No, seriously. WHY ???
Thanks for making the question so clear and perfectly simple to understand.Quote: EvenBobAll 3 dozens have a 50% chance of hitting (for whatever reason). If you bet on two dozens at the same time, how often will you win. Put it another way, how often will the dozen you didn't bet show up to make you lose.
I'll repeat the answer in ever more simple terms:-
In that scenario, half the time you will win. The other half of the time you will win... and the other half of the time you will lose.
(For whatever reason)
Which part of that don't you understand? Is the math of that too complex for you?
If you have a better answer from your buddies on that gambling forum, bring it here.
Waiting. (sound of crickets)
And before EvenBob says "That's why I posted the problem, it's a mind bender." No it isn't. Its abject nonsense.
There is ONE* situation where Bob's Scenario could exist. When the number of spins is ZERO. There is no other scenario in this universe that could accommodate Bob's hypothetical premise.
For any set of numbers we have
Number of wins = Probability of a win x number of spins
Therefore for any set of numbers we have
Probability of a win x number of spins = Number of wins
Substitute in the known probabilities.
For set of numbers 1-12
50% x 0 spins = 0 wins
For set of numbers 13-24
50% x 0 spins = 0 wins
For set of numbers 25-36
50% x 0 spins = 0 wins
Total number of spins = 0 so no rules of mathematics have been broken except playing fast and loose when applying probabilities to a null sample space.
Inserting into original questions as asked:-
Q ...how often will you win....
A ...You will win Zero times. Because, lets face it, the wheel spun zero times.
Q ...Put it another way, how often will the dozen you didn't bet show up to make you lose....
A ...That dozen that you didn't bet will show up to make you lose Zero times.
Simple
*
It might also work if there are an infinite number of spins where the ball lands in the first set an infinite number of times, in the second set an infinite number of times and the third set an infinite number of times. Infinity is not a number that can be manipulated, though so we would just conclude that
Q ...how often will you win....
A ...You will win an infinite number of times. Because, lets face it, the wheel spun an infinite number of times.
Q ...Put it another way, how often will the dozen you didn't bet show up to make you lose....
A ...That dozen that you didn't bet will show up to make you lose an infinite number of times,
That's a bit tricky in the known universe, because the bettor might die or the wheel might wear out.
For any set of numbers we have
Number of wins = Probability of a win x number of spins
Therefore for any set of numbers we have
Probability of a win x number of spins = Number of wins
Substitute in the known probabilities.
For set of numbers 1-12
50% x 0 spins = 0 wins
For set of numbers 13-24
50% x 0 spins = 0 wins
For set of numbers 25-36
50% x 0 spins = 0 wins
Total number of spins = 0 so no rules of mathematics have been broken except playing fast and loose when applying probabilities to a null sample space.
Inserting into original questions as asked:-
Q ...how often will you win....
A ...You will win Zero times. Because, lets face it, the wheel spun zero times.
Q ...Put it another way, how often will the dozen you didn't bet show up to make you lose....
A ...That dozen that you didn't bet will show up to make you lose Zero times.
Simple
*
It might also work if there are an infinite number of spins where the ball lands in the first set an infinite number of times, in the second set an infinite number of times and the third set an infinite number of times. Infinity is not a number that can be manipulated, though so we would just conclude that
Q ...how often will you win....
A ...You will win an infinite number of times. Because, lets face it, the wheel spun an infinite number of times.
Q ...Put it another way, how often will the dozen you didn't bet show up to make you lose....
A ...That dozen that you didn't bet will show up to make you lose an infinite number of times,
That's a bit tricky in the known universe, because the bettor might die or the wheel might wear out.
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
April 4th, 2015 at 4:34:42 PM
permalink
Each dozen can hit 50% of the time if two of the dozens cansimultaneously hit. For whatever reason.
So for each roll, the possibilities are:
Dozen 1 hits
Dozen 2 hits
Dozen 3 hits
Dozen 1 and 2 hit
Dozen 1 and 3 hit
Dozen 2 and 3 hit
Each dozen appears in 3 of the 6 possibilities, for a 50% hit rate. I'm going to claim that those are all equally probable because, why not?
So, if you bet on dozen 1 and 2,
You win one and lose one 4 ways out of 6
You win 2 one out of 6
You lose 2 one out of 6
So for each roll, the possibilities are:
Dozen 1 hits
Dozen 2 hits
Dozen 3 hits
Dozen 1 and 2 hit
Dozen 1 and 3 hit
Dozen 2 and 3 hit
Each dozen appears in 3 of the 6 possibilities, for a 50% hit rate. I'm going to claim that those are all equally probable because, why not?
So, if you bet on dozen 1 and 2,
You win one and lose one 4 ways out of 6
You win 2 one out of 6
You lose 2 one out of 6
April 5th, 2015 at 12:18:22 AM
permalink
Quote: beachbumbabsIf you isolate the guessing of whether the ball will land in the first dozen...
Then isolate the guessing, same way, for dozen 2....
One would hope that the guesser refrained from betting...
Why guess? There's no need to.
Why refrain? There's no need to.
Bob's theoretical promises 50% hit-rate, on each Dozen, each spin, EvenBab.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 5th, 2015 at 12:27:07 AM
permalink
Quote: indignant99
Bob's theoretical promises 50% hit-rate, on each Dozen, each spin
It certainly does. That's rather the point of
the question. It's the theoretical world,
not the real world.
"It's not called gambling if the math is on your side."
April 5th, 2015 at 12:44:10 AM
permalink
improved, below
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 5th, 2015 at 1:27:20 AM
permalink
Short answer: the numbers from each Dozen are labeled on half (18) of the pockets.
1/6th (6) of the pockets show only a single number, from Dozen-1.
1/6th (6) of the pockets show only a single number, from Dozen-2.
1/6th (6) of the pockets show only a single number, from Dozen-3.
1/6th (6) of the pockets show two numbers, one from Dozen-1 and one from Dozen-2.
1/6th (6) of the pockets show two numbers, one from Dozen-1 and one from Dozen-3.
1/6th (6) of the pockets show two numbers, one from Dozen-2 and one from Dozen-3.
Thus, the numbers from each/any/all Dozen(s) are shown on half (18) of the pockets.
(6 pockets by their lonesome, and 6+6 with a buddy number.)
More roulette idiocy.
1/6th (6) of the pockets show only a single number, from Dozen-1.
1/6th (6) of the pockets show only a single number, from Dozen-2.
1/6th (6) of the pockets show only a single number, from Dozen-3.
1/6th (6) of the pockets show two numbers, one from Dozen-1 and one from Dozen-2.
1/6th (6) of the pockets show two numbers, one from Dozen-1 and one from Dozen-3.
1/6th (6) of the pockets show two numbers, one from Dozen-2 and one from Dozen-3.
Thus, the numbers from each/any/all Dozen(s) are shown on half (18) of the pockets.
(6 pockets by their lonesome, and 6+6 with a buddy number.)
More roulette idiocy.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 5th, 2015 at 2:39:03 AM
permalink
Why EvenBob, Why? What's 'rather the point' EvenBob? What is it?Quote: EvenBobThat's rather the point of the question.
My answer not good enough for you?? My questions too complicated for a pathetic 3 short line reply?
https://wizardofvegas.com/forum/gambling/tables/21514-a-simple-math-question/16/#post446011
(Sound of crickets)
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
April 5th, 2015 at 4:37:13 AM
permalink
Another trick roulette wheel:
3 of the Dozen-1 pockets have an asterisk as well as the appropriate number.
3 of the Dozen-2 pockets have an asterisk as well as the appropriate number.
3 of the Dozen-3 pockets have an asterisk as well as the appropriate number.
What does the asterisk mean? It's a Wild-Card pocket; all three Dozens are winners.
Not only does each/any/all Dozen(s) enjoy its own 12 usual numbers, but also 6 more Bonus pockets lurking in the other two Dozens.
Simultaneous with the other two Dozens... 1/4 of the time, 25%.
Not at all... 1/2 of the time, 50%.
You can even add back in the Zero pocket, and the Double-Zero pocket.
One with an asterisk (all three Dozens rejoice at their wins),
and one without (all three Dozens sob at their losses).
3 of the Dozen-1 pockets have an asterisk as well as the appropriate number.
3 of the Dozen-2 pockets have an asterisk as well as the appropriate number.
3 of the Dozen-3 pockets have an asterisk as well as the appropriate number.
What does the asterisk mean? It's a Wild-Card pocket; all three Dozens are winners.
Not only does each/any/all Dozen(s) enjoy its own 12 usual numbers, but also 6 more Bonus pockets lurking in the other two Dozens.
By itself... 1/4 of the time, 25%.Quote: EvenBobMaybe a better way of asking is how often will the 3rd dozen win?
Simultaneous with the other two Dozens... 1/4 of the time, 25%.
Not at all... 1/2 of the time, 50%.
You can even add back in the Zero pocket, and the Double-Zero pocket.
One with an asterisk (all three Dozens rejoice at their wins),
and one without (all three Dozens sob at their losses).
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 5th, 2015 at 8:54:22 AM
permalink
How would empirical data collection work in this case? The hypothesis is that each dozen has an equal 50% chance of being hit. But the ball can only hit one dozen per spin. How would we prove using observed spins that the ball was equally likely to have fallen into another dozen when it didn't do so? Also, no matter what we observe in the spins, we're going to account for 100% of the outcomes. i.e. 37% fall in dozen 1, 32% fall in dozen 2, 31% fall in dozen 3. For a total of 100%. How would watching those spins "prove" that each dozen had a 50% chance of success?
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
April 5th, 2015 at 10:00:15 AM
permalink
It wouldn't. It couldn't. Maybe EvenBob would care to explain that to us. (Sound of hiding trolls)Quote: rdw4potusHow would empirical data collection work in this case?
No. It's not an hypothesis: It's a precondition, a premise: A statement of fact. EvenBob says so. It's only theoretical complete nonsense. (for whatever reason)Quote:The hypothesis is that each dozen has an equal 50% chance of being hit.
Why would we want to prove that something that EvenBob says is true when it isn't. That's his problem. His challenge, to which he steadfastly refuses to rise.Quote:How would we prove using observed spins that the ball was equally likely to have fallen into another dozen when it didn't do so?
Who said we could? What kind of idiots would we be to try?Quote:Also, no matter what we observe in the spins, we're going to account for 100% of the outcomes. i.e. 37% fall in dozen 1, 32% fall in dozen 2, 31% fall in dozen 3. For a total of 100%. How would watching those spins "prove" that each dozen had a 50% chance of success?
EvenBob, Rise to the challenge. Make sense of your own complete and utter nonsense question/premise. Put up or STHU
(sound of hiding trolls)
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
April 5th, 2015 at 6:59:18 PM
permalink
Quote: rdw4potusHow would empirical data collection work in this case?
Easy. You collect (track) six data points.
- Dozen-1, non-Bonus pocket.
- Dozen-1, Bonus pocket.
- Dozen-2, non-Bonus pocket.
- Dozen-2, Bonus pocket.
- Dozen-3, non-Bonus pocket.
- Dozen-3, Bonus pocket.
There is no stipulation that each Dozen must hit 50% of the time.
The stipulation is that 50% of the bets on each Dozen, must win.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 6th, 2015 at 4:58:14 AM
permalink
Quote: indignant99
There is no stipulation that each Dozen must hit 50% of the time.
The stipulation is that 50% of the bets on each Dozen, must win.
Well, that's a change...
Quote: EvenBobQuote: indignant99
Bob's theoretical promises 50% hit-rate, on each Dozen, each spin
It certainly does. That's rather the point of
the question. It's the theoretical world,
not the real world.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
April 6th, 2015 at 5:33:41 AM
permalink
Quote: indignant99On my trick roulette wheel...
There is no stipulation that each Dozen must hit 50% of the time.
The stipulation is that 50% of the bets on each Dozen, must win.
Very deceptive that you left out "On my trick roulette wheel..."
I'm not making the same idiotic stipulation that EvenBob is. Because his stipulation of "hits" is impossible, incoherent, nonsense, BS.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
