A Simple Math Question?

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March 29th, 2015 at 11:54:42 PM permalink

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

Actually, Doc explained it quite well. Much
more interesting than how I presented it.

"It's not called gambling if the math is on your side."

michael99000

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March 29th, 2015 at 11:55:35 PM permalink

Quote: EvenBob
all THREE dozens
win FIFTY PERCENT of the time. 50-50-50.

I realize it's only March, but here's an early candidate for 2015 WOV Post of The Year

EvenBob

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March 30th, 2015 at 12:00:43 AM permalink

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

Ah, you have the answer then.

Waiting. (sound of crickets)

"It's not called gambling if the math is on your side."

EvenBob

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March 30th, 2015 at 12:03:25 AM permalink

Quote: michael99000
I realize it's only March, but here's an early candidate for 2015 WOV Post of The Year

Exactly. In the scenario I presented they do. In
the casino they do not. The question is hypothetical
in nature, a variation on reality, as it were..

"It's not called gambling if the math is on your side."

Thermos

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March 30th, 2015 at 12:05:49 AM permalink

In your hypothetical, do you know which of the three dozens has a 0% chance of coming up?

EvenBob

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March 30th, 2015 at 12:20:50 AM permalink

Quote: Thermos
In your hypothetical, do you know which of the three dozens has a 0% chance of coming up?

The 4th one.

"It's not called gambling if the math is on your side."

Thermos

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March 30th, 2015 at 12:22:26 AM permalink

The fourth what?

Face
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Face

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March 30th, 2015 at 12:49:59 AM permalink

I'm just posting to subscribe to the thread. I figure I may need an example to use in the future of the perfect way to handle a post initially viewed as absurd.

Good on ya, Doc.

The opinions of this moderator are for entertainment purposes only.

indignant99

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March 30th, 2015 at 12:55:16 AM permalink

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

That's my question to you.

If the 3 distinct dozens are mutually exclusive (don't overlap),
and each of the three can happen,
and no other possibility whatsoever can occur,
and (this is a key word) "cover" all possibilities...

Then, your problem is incoherent - it doesn't hang together. It's a festering pile of garbage.

Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

RS

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March 30th, 2015 at 2:18:42 AM permalink

Quote: EvenBob
Ah, you have the answer then.

Waiting. (sound of crickets)

I have a super easy question that anyone should be able to answer -- it's so easy -- I asked on another forum and everyone knew it.

If EvenBob is a troll....and most trolls live under a bridge.......how many pancakes can my dog eat?

Another question, but this one's 10x easier: If I can roll a 12 every single roll in craps....what's the chance I can roll an 8? I know the answer isn't 0%, because 0% + 100% does not equal 100% in this fantasyland I live in. AnY help PLEAZE?

sodawater

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March 30th, 2015 at 2:30:25 AM permalink

Quote: Face
I'm just posting to subscribe to the thread. I figure I may need an example to use in the future of the perfect way to handle a post initially viewed as absurd.

Good on ya, Doc.

A long block of text doesn't mean anything in the post is actually coherent.

Doc's post is disappointing in that it attempts to make sense out of something that is obviously wrong.

Just because you can try to muddle the wrongness with a big block of words doesn't make it less wrong, or even less obviously wrong.

RS

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March 30th, 2015 at 3:03:06 AM permalink

Quote: Doc
Doc's long post:
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.

OK. So there are three dozens, let's call them A, B, and C. I like dozen A while Sam likes dozen B. Without being in cahoots with each other (just me + disgruntled employee), I win 50% of my bets. Sam wins 50% of his bets when he only gets info from employee.

If we decided, "Hey, instead of always betting when we get the signal, let's ONLY bet when we both get the signal!" But....not so easy -- this is impossible, if we both have a 50% chance of success. [I'm taking into assumption that, "disgruntled employee" will also be signalling both of us to bet even if column C is going to hit.....after all, that's what EvenBob's question is about -- right?]

If I get the signal, this means there's a 50% chance A will hit, 25% chance B will hit, and 25% chance C will hit.
If Sam gets the signal, then 25% chance for A, 50% for B, and 25% for C.

Excluding 0/00:
There is a 66% chance that I get the signal because 33% of the time A will hit [I always get signal when A hits] and a 66% chance B or C will hit [and I get signaled half the time B or C hits].
There is a 66% chance Sam gets the signal because 33% of the time B will hit [Sam always gets it when B hits] and a 66% chance A or C will hit [and Sam gets signaled half the time A or C hits].

Let's look at a 9-spin cycle, where every combination of signal/no-signal is possible.

1-A hits: I get signal, Sam gets signal.
2-A hits: I get signal, Sam doesn't get signal.
3-A hits: I get signal, Sam doesn't get signal.

4-B hits: I get signal, Sam gets signal.
5-B hits: I don't get signal, Sam gets signal.
6-B hits: I don't get signal, Sam gets signal.

7-C hits: I get signal, Sam gets signal.
8-C hits: I don't get signal, Sam gets signal.
9-C hits: I get signal, Sam doesn't get signal.

I got the signal for spins 1,2,3,4,7,9. That's 6 total spins I got the signal. Out of 9 spins. That's 6/9 = 66%
Sam got the signal for spins 1,4,5,6,7,8. That's also 6 signals out of 9 spins. Equals 66%.

So far, so good.

But, we only care about the times when we BOTH get the signal. So we get rid of all the times when either I didn't get the signal, Sam didn't get the signal, or neither of us got the signal. Which leaves us with:

1-A
4-B
7-C

This would mean that, whenever we both get the signal, we each have a 50% chance of winning, right? Well, considering that 1-A, 4-B, and 7-C each occur the same amount of times with the same probability, this means that

1-A hits 33% of the time
4-B hits 33% of the time
7-C hits 33% of the time

This leads us to an impossible situation -- our premise, that I win 50% of the time and Sam wins 50% of the time cannot be true [if we only bet when we both get the signal], because we will each end up winning 33% of the time [or either of us will win 66% of the time].

Your turn, EvenBob/Doc.

If Sam and I were smart, I would bet whenever I'd get the signal and Sam would bet whenever he got the signal, even if I'd be betting and Sam wouldn't be betting, or vice versa.

charliepatrick

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March 30th, 2015 at 3:32:14 AM permalink

I may be falling for the goat syndrome in the above, but how does this sound :-

Dozen   I     S  (I'm informed, Sam's Informed) Probability
01) A-- I  is S  is 1/3 1/1 1/2 = 1/6  (If it's A, I'm always told, and Sam is 50/50 to be told)
02) A-- I  is S not 1/3 1/1 1/2 = 1/6
04) -B- I  is S  is 1/3 1/2 1/1 = 1/6  (If it's B, I'm told 50/50, Sam is always told)
05) -B- I not S  is 1/3 1/1 1/1 = 1/6
07) --C I  is S  is 1/3 1/2 1/2 = 1/12 (If it's C, I'm told 50/50, Sam is told 50/50 ; hence four possible scenarios)
08) --C I  is S not 1/3 1/2 1/2 = 1/12
09) --C I not S  is 1/3 1/2 1/2 = 1/12
10) --C I not S not 1/3 1/2 1/2 = 1/12

Thus 1 4 7 have probabilities of 1/6 1/6 1/12, so each person's chances are 2/5.
Obviously A's best chance is to bet when he's told and Sam isn't (2 and 8), his chances are then 2/3.

Doc

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March 30th, 2015 at 7:24:32 AM permalink

Quote: Face
I'm just posting to subscribe to the thread. I figure I may need an example to use in the future of the perfect way to handle a post initially viewed as absurd.

Good on ya, Doc.

Thanks. Unfortunately, some posters are so focused on jumping on EvenBob that they won't even read my post close enough to realize that it presents a scenario where EvenBob's question is completely valid.

Quote: RS
...

Your turn, EvenBob/Doc.

If Sam and I were smart, I would bet whenever I'd get the signal and Sam would bet whenever he got the signal, even if I'd be betting and Sam wouldn't be betting, or vice versa.

First of all, RS, thanks for actually reading my post and presenting an analysis of it. I think I need to re-read your analysis and ponder a while in order to understand it fully, and I will do that. Here are some preliminary comments:

I agree that it is not the best strategy for Sam and his associate to both bet if and only if they both get the "bet" sign. I mentioned that possibility in my "long" post. I agree with your spoiler, but I would go even further: I think if I were Sam's partner and we were smart, then I would bet when I get the "bet" sign but Sam does not get it, and that Sam should bet when he gets the "bet" sign and I do not. However, the question that corresponds to EvenBob's original post is "What happens if they DO follow that bet-at-the-same-time strategy?"

I think that with the inside info as I described, it is quite possible that if I bet the first dozen every time I get the "bet" sign from the inside source then I will win 50% of the time. It is also possible for Sam to win 50% of his wagers while betting the second dozen when advised to. And sometimes the third dozen will win. That corresponds to what EvenBob has described and is what some people keep claiming cannot be possible.

Here is where I think the problem arises, and it may be what RS has tried to present -- I still need to look at his post closer. This is the part that I said I didn't yet know how to analyze.

(Edit: I also have not studied charliepatrick's post, and he may have covered what I describe below.)

Assuming that each partner was betting each time he got the bet sign and each was consistently winning at a 50% rate, we still do not have any info on (1) how often both partners get the bet sign for the same roll or (2) how often either partner wins when the other partner gets the bet sign. I think this is related to co-variance, but that term may not be appropriate, since we may not have conventional random events.

I suspect that it is a bit of a challenge -- probably impossible -- for the man on the inside to give both of the players 50%-reliable info on their respective dozens on every roll. I suspect that if Sam gets the bet sign for the second dozen, then the first dozen is less than 50% likely to hit, and I would be less than 50% likely to win even if I have been advised to bet that time. I make up for those shortfalls by winning at a higher rate than 50% when I am advised to bet the first dozen and Sam is NOT advised to bet the second dozen. Sam also loses more than 50% of the time when I am told to bet the first dozen, but he wins more than 50% when I am advised not to bet. This is perhaps comparable to saying that your crystal ball cannot maintain its 50% accuracy when it starts working against itself -- or at least trying to predict multiple, mutually-exclusive events at the same time.

Overall, Sam and his partner each can win at a 50% rate when betting independently, but if they were each to bet their favorite dozen if and only if they both got the bet sign for that spin, then they would both wind up winning less than 50% of the time. I just have no idea how to calculate the percentage, given that they are both working with inside information. EvenBob has provided me with some info on what he thinks the answer is, but I don't really follow the reasoning behind his calculations, and I think he has been jumped on without adequate reason too many times already and doesn't think its worth the aggravation of trying to present it in the thread.

OnceDear

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March 30th, 2015 at 8:33:45 AM permalink

Quote: Doc
Thanks. Unfortunately, some posters are so focused on jumping on EvenBob that they won't even read my post close enough to realize that it presents a scenario where EvenBob's question is completely valid.....

I think that with the inside info as I described, it is quite possible that if I bet the first dozen every time I get the "bet" sign from the inside source then I will win 50% of the time. It is also possible for Sam to win 50% of his wagers while betting the second dozen when advised to

Doc, I think that you are very kind to EvenBob and that you don't believe that he is being totally mischievous. I fell into that trap too.

There is strong Empirical evidence that EvenBob has raised this thread with the intention of wasting the time and effort of those who would intelligently reply. I suggest that you save your time and your energy and simply ignore him. This is not a personal insult, but is a personal observation and suggestion.

Note the style of all EvenBob's follow-up contributions. Each consists of a few lines that tell us that we are wrong and telling us that some-one else has given the correct answer and re-challenging us to give him the correct answer that he already, supposedly knows. Do you see Evenbob answering any of our... any of your, follow up questions in any meaningful way? Did he, or you manage to fill in a faked set of observations in the 'Empirical nonsense' workbook at https://wizardofvegas.com/forum/gambling/tables/21514-a-simple-math-question/8/#post444708 such that the modeled scenario could ever exist?

It's all mischief, pure and simple. EB is trolling... and quite successfully.

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.

Kerkebet

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March 30th, 2015 at 10:20:06 AM permalink

Quote: OnceDear
It's all mischief, pure and simple. EB is trolling... and quite successfully.

I had been thinking, April Fools' is about due. However, this was a fun diversion when taken "seriously" nonetheless. Eases the mind along.

I guess the saying should read, "You can't embarrass an (apparently) ignorant man, either."

Supplementary note: "Planet Earth to Bob, planet Earth to Bob... anyone EVEN there?"

Nonsense is a very hard thing to keep up. Just ask the Wizard and company.

RS

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March 30th, 2015 at 2:46:14 PM permalink

Doc, yes pretty much.

The problem is that when we both get the bet sign, we are no longer at 50/50. Sometimes we are at 66% winrate and other times we are at 33% winrate.

It's sort of like blackjack. You win 43% of the hands, right? Well, what if we're only looking at scenarios where the first card is an ace? We can no longer say that the winrate is 43% when the first card is an ace.

That's what's going on here. The events are related.

Adding on to me + Sam problem, if we had a third person, we could each bet with 100% accuracy. I bet when I get the bet sign and the other 2 do not get the bet sign. Likewise I can also figure out my % to win given "complete info". If Sam and I both get sign, while Suzy (column C) gets no sign, me and Sam have a 50% chance. If we all get sign, we each have 33% chance to win. If no one gets the sign, it'll be 0/00. For perfect play, just wait until you get the sign and the other two don't (you will always win).

Doc

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March 30th, 2015 at 4:04:19 PM permalink

Quote: RS
The problem is that when we both get the bet sign, we are no longer at 50/50. Sometimes we are at 66% winrate and other times we are at 33% winrate.
...

Adding on to me + Sam problem, if we had a third person, we could each bet with 100% accuracy. I bet when I get the bet sign and the other 2 do not get the bet sign. Likewise I can also figure out my % to win given "complete info". If Sam and I both get sign, while Suzy (column C) gets no sign, me and Sam have a 50% chance. If we all get sign, we each have 33% chance to win. If no one gets the sign, it'll be 0/00. For perfect play, just wait until you get the sign and the other two don't (you will always win).

I don't think we have enough info to calculate those percentages, not the 66%, the 33%, or the 100%.

If you get the bet sign and the other two do not, then no, I don't think you are assured of a win. That might just be one of the times that the man on the inside is expecting you to bet a loser and keep the managers from catching on that something is up -- after all, your source gives you "bad" advice half of the time. At the same time, he could be telling the other players not to bet because this is part of his plan for disguising (for them) the conditions for his saying "don't bet." He doesn't want them to start believing that "don't bet" implies 0 or 00 is coming or for that matter that a different dozen is coming, so he tells Sam not to bet sometimes when a 15 is coming and tells Suzy not to bet sometimes when the 30 is coming up. Otherwise, if they might think "don't bet" means 0 or 00, and they might start betting green numbers and attracting attention (if he really did that only for 0 and 00).

I don't see any way to calculate the expected results of collaboration with just the information that we have. We are not dealing with random results -- we are dealing with the results that the disgruntled employee wants to lead us to with imperfect advice. All we really know quantitatively is that a single player following the advice given to him will win very close to half of his wagers, even though he won't get to bet on every roll/spin.

indignant99

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March 31st, 2015 at 12:40:06 PM permalink

Quote: Doc
Consider this scenario:
...broadcast is delayed...
...disgruntled employee...
...give semi-accurate information...
...provide you with a signal...

Here's a simpler scenario, that works the same way. You bring a Clairvoyant with you to the roulette table. A "clairvoyant" is also called a "savant," but this guy is an Idiot-Savant. He can foresee the future accurately, but not completely. We'll call him "Savvy" for short.

Savvy can tell (with perfect accuracy) one of the Dozens that's going to fail. But his powers don't extend to telling both Dozens that are going to fail. Whichever Dozen he nixes, you bet the other two Dozens.

Every spin you'll get one loser (-1) and one winner (+2), for a net of +1 unit, on 2 wagered. You win +1 unit, for your 2 units wagered, every spin. 50% Player's Edge, with absolutely no variance.

You're in heaven, or at least Rainman land, with Tom Cruise and Dustin Hoffman.

Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

Doc

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March 31st, 2015 at 1:28:34 PM permalink

First, I deliberately avoided such concepts as clairvoyants, even though EvenBob mentioned good guessers and crystal balls, since I expected skepticism from many of the members here. I thought they might better accept the idea of an online casino with a disgruntled employee.

Second, you have described a different problem with a different expected outcome than the one I described and the one EvenBob described. Knowing that the third dozen will not win is quite different from knowing that the first dozen and the second dozen each have a 50% probability of winning on the spins they have been recommend..

RS

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March 31st, 2015 at 1:32:24 PM permalink

Quote: Doc
I don't think we have enough info to calculate those percentages, not the 66%, the 33%, or the 100%.

If you get the bet sign and the other two do not, then no, I don't think you are assured of a win. That might just be one of the times that the man on the inside is expecting you to bet a loser and keep the managers from catching on that something is up -- after all, your source gives you "bad" advice half of the time. At the same time, he could be telling the other players not to bet because this is part of his plan for disguising (for them) the conditions for his saying "don't bet." He doesn't want them to start believing that "don't bet" implies 0 or 00 is coming or for that matter that a different dozen is coming, so he tells Sam not to bet sometimes when a 15 is coming and tells Suzy not to bet sometimes when the 30 is coming up. Otherwise, if they might think "don't bet" means 0 or 00, and they might start betting green numbers and attracting attention (if he really did that only for 0 and 00).

I don't see any way to calculate the expected results of collaboration with just the information that we have. We are not dealing with random results -- we are dealing with the results that the disgruntled employee wants to lead us to with imperfect advice. All we really know quantitatively is that a single player following the advice given to him will win very close to half of his wagers, even though he won't get to bet on every roll/spin.

I was under the premis that if my dozen will hit, I will definitely get the signal.

indignant99

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March 31st, 2015 at 1:36:38 PM permalink

Quote: Doc
Knowing that the third dozen will not win is quite different from knowing that the first dozen and the second dozen each have a 50% probability of winning on the spins they have been recommend..

I'm too dense to see the difference. My scenario has one gambler; yours has two - merge them.

Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

OnceDear

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March 31st, 2015 at 1:42:57 PM permalink

Quote: Doc
I think that with the inside info as I described, it is quite possible that if I bet the first dozen every time I get the "bet" sign from the inside source then I will win 50% of the time. It is also possible for Sam to win 50% of his wagers while betting the second dozen when advised to. And sometimes the third dozen will win. That corresponds to what EvenBob has described and is what some people keep claiming cannot be possible.

Nope. No can do, unless the third dozen only wins when neither has made a wager. It can do that as often as it wants and won't afftect win rate. Maybe the guys are out having a beer and not answering their phones.

Quote: Doc
Knowing that the third dozen will not win is quite different from knowing that the first dozen and the second dozen each have a 50% probability of winning on the spins they have been recommend..

If you mean

Quote:
Knowing that the third dozen will not win (on the spins they have been recommend) is quite different from knowing that the first dozen and the second dozen each have a 50% probability of winning on the spins they have been recommend.

OK. It's a teensy bit different in that the individual percentages of 1st dozen against 2nd dozen might not be exactly 50:50, but if the sum of those two probabilities is to be 100% then any 3rd dozen wins will have to be cancelled out by adding a second ball to the spin or using an anti-matter ball occasionally.*

There is no scenario where that can be true. Never was, never can be. If the third dozen can win, even once, then one of the 50%s must sacrifice that win: To get back up to 50% then the third dozen will need to hit a few negative times, or the first two dozens must occasionally both win at the same time. Clairvoyant scenario or disgruntled employee camaflaging his cheating from his boss in cahoots with any number of related or inrelated accomplices: None of it makes any difference.

* That's no more absurd than the original premise, nor did anyone say the game wasn't rigged using extra balls or antimatter.

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.

indignant99

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March 31st, 2015 at 2:10:46 PM permalink

Quote: OnceDear
If the third dozen can win, even once,

Now, Now, OnceDear, the trick is that Doc's 2 gamblers don't bet every spin.

Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

OnceDear

OnceDear

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March 31st, 2015 at 2:19:03 PM permalink

Quote: indignant99
Now, Now, OnceDear, the trick is that Doc's 2 gamblers don't bet every spin.

That'll be the other 50% :o) The 50% where the gambler(s) are out having a beer with Bob.

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.

RS

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March 31st, 2015 at 2:21:48 PM permalink

Is Bob even still here? (See what I did there!) He hasn't posted in a while. Maybe he'll go on that big hiatus again. Ah! The dawn of a new era.

indignant99

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March 31st, 2015 at 2:23:51 PM permalink

Quote: RS
I was under the premise (fixed that for ya) that if my dozen will hit, I will definitely get the signal.

And, you'll get some signals even when you're not gonna hit.

Hmm, maybe that was wrong. Maybe every signal does guarantee a win - just that sometimes you don't get the signal, but your Dozen is a winner anyway.

I gotta go back and re-read Doc's scenario. Yuck.

Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

RS

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March 31st, 2015 at 2:39:11 PM permalink

Of course I'd get signal sometimes when it won't hit. That's not what I'm saying.

I'm saying if it is going to hit, I will get the signal.

In other words, there would never be a situation where I do not get the signal and my dozen hits.

I believe I wrote this as part of my premisE on my long post earlier.

indignant99

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March 31st, 2015 at 2:48:59 PM permalink

Quote: RS
Of course I'd get signal sometimes when it won't hit...
...if it is going to hit, I will get the signal.

Agreed. (I did review Doc's scenario, and your premise.)

Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

indignant99

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March 31st, 2015 at 3:02:18 PM permalink

Quote: OnceDear
unless the third dozen only wins when neither has made a wager.

Not true. One of the gamblers can make the wager (a losing wager). Just not both of them at the same time. Because if that happened, the 2 guys jointly cannot ever get to 50%. Why? Because that two-losers spin is gone forever, and we're below 100% to be divvied up between 'em.

Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

Doc

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March 31st, 2015 at 3:15:16 PM permalink

Quote: indignant99
Quote: RS
Of course I'd get signal sometimes when it won't hit...
...if it is going to hit, I will get the signal.

Agreed. (I did review Doc's scenario, and your premise.)

I didn't realize I had made the inside-info concept so complex. I'll try to clear up some of the things that have arisen lately, including that the above agreement ("if it is going to hit, I will get the signal") is in error.

The disgruntled employee is providing info independently to you and to Sam. He doesn't even realize that the two of you know each other, much less think that you might be collaborating. He doesn't provide info to either one of you for every spin of the wheel. He might only provide info to you less than half of the time, and half of the times that he tells you to bet your favorite dozen, it will lose.

We have no info that he will tell you to bet every time that your favorite dozen will win, and I have allowed that there may be potential wins that he doesn't bother to tell you about. In fact, it could turn out that he never provides info to both you and Sam on the same spin, but we have no info about that.

What we do know is that when he advises you to bet the first dozen, the first dozen will hit 50% of the time, and when he tells Sam to bet the second dozen, that dozen wins 50% of the time. With this limited information, we are trying to determine whether we can decide the expected average outcome if you both bet your favorite dozen every time that you are both advised to do so (if it turns out that this ever happens.) I speculated some things about the results of this strategy, assuming that the both-get-the-bet-sign situation occurs fairly often, but I decided that I cannot conclude anything quantitatively, so I asked for others to offer their analyses.

I think that the inside-info scenario I described presents the same basic constraints and objectives that EvenBob described, but mine does not rely on good guessing, a crystal ball, a clairvoyant, or anything but a disgruntled employee who is willing to share some semi-accurate info about past-posting opportunities, something I felt was possible to encounter in the real world of skeptics.

Doc

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March 31st, 2015 at 3:29:43 PM permalink

Quote: indignant99
Quote: OnceDear
unless the third dozen only wins when neither has made a wager.

Not true. One of the gamblers can make the wager (a losing wager). Just not both of them at the same time. Because if that happened, the 2 guys jointly cannot ever get to 50%. Why? Because that two-losers spin is gone forever, and we're below 100% to be divvied up between 'em.

The above was posted while I was typing my previous post, so I didn't address it there.

It is possible for the third dozen to win on a spin when both you and Sam had been advised to bet. When that happens, you would both be behind and would have to catch up on other spins. That catch-up is possible if there is a first-dozen spin where you are told to bet and Sam is told not to bet, followed by a second-dozen win that Sam is told to bet on and you are told not to bet. Understand? You don't both stay at 50% constantly, which would be impossible for a series of wins and losses, but overall you each tend to win 50% of the time so long as you follow the advice you receive.

indignant99

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March 31st, 2015 at 3:43:00 PM permalink

Quote: Doc
both be behind and would have to catch up on other spins. That catch-up is possible

I actually thought about that, and hesitated a lo-o-o-ng time before I hit "post."

Quote: Doc
Understand?

Yes.

Quote: Doc
You don't both stay at 50% constantly

I never thought that. Nor said it.

Your catchup scenario is, of course, possible. Would you agree it's not very likely, though?

Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

Doc

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March 31st, 2015 at 3:51:49 PM permalink

Quote: indignant99
Would you agree it's not very likely, though?

I think (1) it might be very common, and (2) it's one of the many things about this problem for which we don't have quantitative info.

I commented in a much earlier post that there is an issue about "co-variance" (which might not be a proper term) and the fact that we have no info at all about the values of these co-variances. We have no info about the patterns by which the source decides to share info, only the accuracy percentages of the advice he provides independently to you and to Sam.

RS

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March 31st, 2015 at 4:07:42 PM permalink

Ugh.....now you're making the math more complex, Doc! Haven't looked at it yet, but, I'm fairly certain we will end up with the same result -- where we end with either an impossibility being the case OR the original premise cannot be true. But the premise and conclusion cannot both be true (at least I'm fairly sure it cannot be true).

Doc

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March 31st, 2015 at 5:46:37 PM permalink

I didn't mean for my previous two posts to make the math more complex or even to change anything of substance. I was trying to clarify things that some folks seemed to misunderstand.

I think that a key source of confusion was that people read my post (or maybe not) and then added in a bunch of assumptions that would apply to a typical roulette probability problem, even though those assumptions didn't apply to the inside-info scenario. The same was true for a lot of the responses to Bob's original posts. I concede that many of his (brief as always) posts weren't completely clear, but that doesn't mean they were wrong to the extent a lot of people mocked them.

For example, perhaps now people can see that with the inside info, you (betting first dozen), Sam (betting second dozen), and even Suzy (?) who likes the third dozen could all win on 50% of the wagers each of them placed as recommended. That doesn't mean that the "50-50-50" expression that Bob used should mean that there would be 1.5 winning dozens per spin. He even specifically pointed out that 50% winning for bets on the first dozen and 50% winning for bets on the second dozen does not imply that the third dozen never hits. He may not have explained it clearly -- just as I may not have -- but it didn't justify people being so adamant that he was posting foolishness. They just needed to look a little harder as to how his claims could be true without imposing their own additional assumptions.

I'm not sure that I would agree that "we end with either an impossibility being the case OR the original premise cannot be true." As far as "the premise and conclusion cannot both be true", I don't even know what the "conclusion" is just yet.

What I suspect is that (assuming you and Sam both get the bet sign on a reasonable number of spins) we still don't have enough information to calculate the expected result of the bet-at-the-same-time strategy specifically. Thus far, most everyone seems to have tended to add in their own assumptions based on the usual roulette odds, as if the 38 slots determine the likelihood of winning your wager.

EvenBob

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March 31st, 2015 at 6:04:25 PM permalink

Quote: Doc
I don't even know what the "conclusion" is just yet.
.

That's why I posted the problem, it's a mind
bender. I asked the math prof daughter and
she immediately said she doesn't teach
probability math and only took one class in
it 20 years ago. That was that..

"It's not called gambling if the math is on your side."

RS

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March 31st, 2015 at 8:09:29 PM permalink

Premise is "with 50% win rate on dozen A and 50% on dozen B" while conclusion is "dozen C winrate is not equal to 0%".

If the premise is true, then the conclusion is false. If the conclusion is true, then the premise cannot be true.

I'll take a stab at this tonight. Maybe.

EvenBob

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March 31st, 2015 at 8:28:11 PM permalink

Quote: RS
Premise is "with 50% win rate on dozen A and 50% on dozen B" while conclusion is "dozen C winrate is not equal to 0%".
.

Yes. Find win rate on dozen C and you will
have the answer for A and B.

"It's not called gambling if the math is on your side."

Doc

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March 31st, 2015 at 8:42:43 PM permalink

Sorry that I did not understand. I was thinking that "conclusion" referred to what the win percentage would be if you and Sam bet only when both received the bet sign. That was the question that EvenBob and I had asked.

When you take a stab at this, remember that you and Sam are receiving the "bet" sign separately. When you each verified and exploited the 50% win rates, your wagers on the first dozen were not always on the same spins as Sam's wagers on the second dozen.

The history (and future expectation) of 50% win rates is for when you are following the advice given to only you, and it might not work out that way if you choose some particular subset of those recommended bets. Since your dozen is expected to lose half of the time that you are advised to bet, it is even possible that with some very bad decision making (or bad luck) about which recommended bets you actually put money on, you could even have a win rate of 0% -- choose only the half of the recommended bets that are losers and sit idle for the bets that are winners. (Yes, that would be incredibly bad results, but it is possible even when given good, 50% advice.)

I have already said that if you are betting the first dozen if and only if Sam is betting the second dozen (because you have both been given the bet sign for that spin), then I expect at least one of you will average less than 50% wins. I think you can only maintain that 50% win rate because each of you tends to win more than 50% when the other player has been advised not to bet. That's why it's my opinion that betting together is not a good strategy; it just happens to be the one that you are trying to evaluate quantitatively right now.

indignant99

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March 31st, 2015 at 9:02:50 PM permalink

Quote: EvenBob
That's why I posted the problem, it's a mind bender.

The betting strategy boils down to:

Refrain from betting, when conditions are unfavorable. (Even though a wager could win under these conditions, it's unlikely.)
Refrain from betting, when conditions are unknown.
Proceed with betting, when conditions are favorable. (Even though a wager could lose under these conditions, it's a 50/50 proposition.)
Figure out some way to detect when conditions are unfavorable versus favorable. This method would have to be both plausible and quantifiable.

Indignant's "idiot-savant" method is already quantified; but not plausible.
Doc's "disgruntled insider" method is sorta plausible, but not yet quantified. Doc says that with a few more factors ("co-variance" and the insider's "bet-signal strategy") clarified/revealed/discovered/concocted, the "insider" method can be quantified.
Have at it.

Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

RS

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March 31st, 2015 at 9:34:32 PM permalink

Quote: Doc
Sorry that I did not understand. I was thinking that "conclusion" referred to what the win percentage would be if you and Sam bet only when both received the bet sign. That was the question that EvenBob and I had asked.

When you take a stab at this, remember that you and Sam are receiving the "bet" sign separately. When you each verified and exploited the 50% win rates, your wagers on the first dozen were not always on the same spins as Sam's wagers on the second dozen.

The history (and future expectation) of 50% win rates is for when you are following the advice given to only you, and it might not work out that way if you choose some particular subset of those recommended bets. Since your dozen is expected to lose half of the time that you are advised to bet, it is even possible that with some very bad decision making (or bad luck) about which recommended bets you actually put money on, you could even have a win rate of 0% -- choose only the half of the recommended bets that are losers and sit idle for the bets that are winners. (Yes, that would be incredibly bad results, but it is possible even when given good, 50% advice.)

I have already said that if you are betting the first dozen if and only if Sam is betting the second dozen (because you have both been given the bet sign for that spin), then I expect at least one of you will average less than 50% wins. I think you can only maintain that 50% win rate because each of you tends to win more than 50% when the other player has been advised not to bet. That's why it's my opinion that betting together is not a good strategy; it just happens to be the one that you are trying to evaluate quantitatively right now.

I would only looking for the results when both Sam and I get the bet signal.

Also, if there is the chance that my dozen will hit but I am not given the bet sign....I believe I would need the frequency this happens, although I'm not sure...yet.

indignant99

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April 1st, 2015 at 12:16:30 AM permalink

Quote: RS
if there is the chance that my dozen will hit but I am not given the bet sign....I believe I would need the frequency this happens, although I'm not sure...yet.

If you don't get the signal, refrain. You'd have no clue, and are left with ¹/₃, ¹/₃, ¹/₃ natural probabilities. So if you bet the non-signal, all such bets in aggregate will be a push. Only bet when you get the signal, and the corresponding artificial ¹/₂, ¹/₂ probabilities - one of those halves being in your favor. Oh... you can refrain even with the signal, sometimes. It can't hurt you to skip one/several/lots of 50/50 shots. Repeat... Only bet when you get the signal, but it needn't be always.

Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

RS

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April 1st, 2015 at 2:27:56 AM permalink

Quote: indignant99
If you don't get the signal, refrain. You'd have no clue, and are left with ¹/₃, ¹/₃, ¹/₃ natural probabilities. So if you bet the non-signal, all such bets in aggregate will be a push. Only bet when you get the signal, and the corresponding artificial ¹/₂, ¹/₂ probabilities - one of those halves being in your favor. Oh... you can refrain even with the signal, sometimes. It can't hurt you to skip one/several/lots of 50/50 shots. Repeat... Only bet when you get the signal, but it needn't be always.

Right. That's not what my question was about, though. It was how frequently do I not get the signal when my dozen hits? I am not asking what the chance is that when I do get the signal that my dozen will hit.

After all, we are dealing with a problem where we only care about the situation where both Sam and I get the signal, because EvenBob's claim is that we both have a 50% chance of winning. The problem is trying to figure out the % of dozen C being hit when both dozen A and dozen B get the signal. In solving for the C% chance, we need to know the frequency that A will hit and I don't get the signal and the frequency B hits and Sam doesn't get the signal.

But, either way, if I get the signal and Sam gets the signal, then using that information, I know that I can no longer have a 50% chance of being successful, and he knows this as well.

It would be like playing Mississippi Stud. I get dealt a J and K. Normally you'd play (1x) according to basic strategy. But I realize the guy to my left has J,J and the guy to my right as K,K. Using this information, I know that I don't have the same chances of making a high pair anymore. There is only 1 remaining Jack and 1 remaining King in the deck.

The plays are correlated.

charliepatrick

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April 1st, 2015 at 4:42:52 AM permalink

I do not understand some of the pontificating going on - it's more easily understood if one considers we had a friend who gave us reliable tips on the horses. As it happens it may give tips to other people, including (say) my brother, but we'll assume these actions don't affect the odds being offered by the bookies.

Quite simply we should each bet when we receive a tip (assuming in the first instance brothers keep their betting secret from each other).

If I knew my brother had received a tip for another horse in the same race, we might both be less inclined to make our wagers.

It makes little difference to our strategy how often we receive tips as long as they are reliable.
(i) no tips, no bet.
(ii) a tip - don't know about brother - bet.
(iii) two tips for same race - probably bet less (technically insufficient info), but if we do bet, back both horses.
(iv) a tip, know the brother hasn't had a tip - bet (technically equal or slightly better odds than (ii)).

You can do the maths only if you have ALL the facts. I did this earlier where we always received tips based on the dozens in roulette.

indignant99

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April 1st, 2015 at 1:02:25 PM permalink

Quote: charliepatrick
tips on the horses

We want more than just the tip. We want to peel back the skin, to see the infrastructure and workings that went into generating the tip.

Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant

beachbumbabs

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April 1st, 2015 at 3:06:05 PM permalink

Quote: RS
So EvenBob's starting sh*t again. Obvious troll is obvious :) ......who wants to bet $$$ BeachBanhammerBabs will step in and defend EB?

Actually I'm just kidding.

I don't want to bet moneyz.

On a side note, something I just realized, a member on another forum I used to participate on, was very EvenBob-like, and his initials were also EB. Except the other EB was like 13 years old or something.

Boy, have you got me figured wrong. What forum have you been reading? Not this one.

If the House lost every hand, they wouldn't deal the game.

beachbumbabs

beachbumbabs

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April 1st, 2015 at 3:26:19 PM permalink

Quote: EvenBob
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.

I haven't read all the way through this thread yet. However,

I think what EB's trying to say is:

If you isolate the guessing of whether the ball will land in the first dozen, the guesser has a track record of a correct prediction 50% of the time (whether the guess was that it would or wouldn't - per the hypothetical question).

Then isolate the guessing, same way, for dozen 2, where the guesser has a 50% correct guessing record. Right or wrong. There could be overlap; if he correctly guessed it would land on dozen 1, and so would not land on dozen 2, on the same spin, and it does land on dozen 1, there are 2 correct guesses. If it lands on dozen 2, there are 2 incorrect guesses. If it lands in dozen 3, there is still 1 correct guess.

Which means that if the guesser says it will not land in dozen 1 or dozen 2, those would also be correct guesses if the ball landed in dozen 3. (One would hope that the guesser refrained from betting on either dozen if he thought it would land in dozen 3 and he believed in his system.) In this scenario, no matter where the ball landed, the guesser would have at least 1 right guess.

It's not about whether the guesser actually bet on winners/losers; it's about whether the guesser correctly predicted it would or would not land in the specified dozen.

If the House lost every hand, they wouldn't deal the game.

RS

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April 1st, 2015 at 4:05:48 PM permalink

Just havin' a little fun at your expense, BBB. :)

The problem with the 50%'s thing is, they have a 50% chance of winning independently of each other. If I make the bet without knowing what Sam is doing, I'll have a 50% chance. But if I know what Sam is doing, then my chance of winning is no longer 50%.

Imagine a different scenario -- I'm playing a coin-toss game with a psychic/future-teller. This person, David, can predict with 100% accuracy what the next coin toss is going to be (heads or tails). If I make a guess (50/50) that the toss will be heads, I have a 50% chance of being correct. However, if I guess it'll be heads and David predicts it will be heads, I no longer have a 50% chance of being correct. If I guess heads and David predicts tails, again, I no longer have a 50% chance of being correct.

"Guess" = random guess, no information.
"Predict" = accurate (100%) correct.

In a way it's like Shrodinger's Cat.

indignant99