Past numbers have *NOTHING* to do with future events?
Lets say its the numbers 5, 17 and 34. I am looking for a YES or a NO to this question......try not to DRIFT from the question. Didn't those three numbers have to have two hits on them (within 37/38) BEFORE it had 3 hits? The number 5 is not going to magically jump from having ONE hit on it, then out of no where, it has 3 hits on it. My point being, if PAST numbers mean nothing, then HOW is it that we can gage that there will 2-4 numbers with 3 hits on them?
An example of a little contest we can have >> We'll track some numbers. Of course we know CHEATING may be involved with that, as usual. lol When we have 25 numbers recorded, I will pick 3 numbers that I think will have 3 hits on it by 38 (or 37) spins. *BUT* I also get to pick ANY 3 numbers for you, you dont get to choose your numbers. At the end of 38 numbers, we'll see who has the most numbers with 3 hits.
Who knows, I might not have any. Also, we'll do this for 30 groups of 38 numbers. At the end of the 30 groups, according to the slide ruler guys, it should be roughly 15/15. Maybe not exact but it should be close. Why? Because "the ball has no memory. All numbers are independant from one another, PAST NUMBERS MEAN NOTHING". Ken
In those 25 spins, you will have numbers that have hit once, twice, and maybe even three+ times. And there is a guarantee that there will be at least 13 numbers (on a 00 wheel) that have zero hits, which you will obviously give to the challenger.
Past spins don't matter to what will be hit next, but if you pick a numbers with previous hits, your question CHANGES from "Pick 3 numbers that will have 3 hits in 38 spins, and I'll pick 3 for you.", to "Pick 3 numbers that will have 0, 1 or 2 hits in the next 13 spins, and you can have 3 numbers that still need 3 hits in the next 13 spins."
By taking those past events into account, you are no longer asking the same question.
Quote: mrjjjYes, I know EACH spin is independant of each other, blah blah blah, I dont need the speach. My point being, the number of times you have read regarding past numbers meaning NOTHING. This misconception is usually from AP guys but there are sometimes others. In 37/38 spins, on average, there are 3 numbers with 3 hits on it. Sometimes 2 numbers, sometimes 4 numbers, whatever.
Lets say its the numbers 5, 17 and 34. I am looking for a YES or a NO to this question......try not to DRIFT from the question. Didn't those three numbers have to have two hits on them (within 37/38) BEFORE it had 3 hits? The number 5 is not going to magically jump from having ONE hit on it, then out of no where, it has 3 hits on it. My point being, if PAST numbers mean nothing, then HOW is it that we can gage that there will 2-4 numbers with 3 hits on them?
An example of a little contest we can have >> We'll track some numbers. Of course we know CHEATING may be involved with that, as usual. lol When we have 25 numbers recorded, I will pick 3 numbers that I think will have 3 hits on it by 38 (or 37) spins. *BUT* I also get to pick ANY 3 numbers for you, you dont get to choose your numbers. At the end of 38 numbers, we'll see who has the most numbers with 3 hits.
Who knows, I might not have any. Also, we'll do this for 30 groups of 38 numbers. At the end of the 30 groups, according to the slide ruler guys, it should be roughly 15/15. Maybe not exact but it should be close. Why? Because "the ball has no memory. All numbers are independant from one another, PAST NUMBERS MEAN NOTHING". Ken
So, what you're saying is basically "the number 5 must have hit two times before it can hit a third time. And hitting 3 times makes the number special!" And what everyone else in the whole world is saying is "I don't care. The fact that it's hit twice does nothing to increase the probability that it will hit again. What difference does it make if the next hit is the 1st, 3rd, or 7th hit in 25 spins? The odds of that number hitting on the next spin are still 1/38."
Quote: mrjjjDidn't those three numbers have to have two hits on them (within 37/38) BEFORE it had 3 hits?
Not necessarily. If the number of times a number hits is a quantum state, then it can jump from 1 to 3 without ever being 2.
The rest of your rant appears to have a question in it, but it's too incoherent to amke out. Please try again. This time try not to interrupt your exposition >>I SAY NOT with comments which are not related to your question like THE WHEEL HAS NO MEMORY.
Now>> if we put a cat in a closed box along with a poison GAS container connected to a roulette wheel which...
MrJjj,
What would you like to hear from us today?
Mark
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17.....So you are saying, its possible there are THREE 17s within this group? Thats usually a LAST resort, trying to be tricky/creative with an answer. Ken
Quote: mrjjj@marksolberg & Dween >> Let me ask this. Following my rules....I can pick my 3 numbers and I also pick your 3 numbers. Would you say I have an *ADVANTAGE* over you? If yes, how can that be? Everything is EQUAL in roulette. Nothing is DUE etc., all numbers are independant from one another so it would be IMPOSSIBLE for me to have any type of an advantage for this little challenge, correct? Ken
I think people think you would already have the results before your test. How do you propose to do this challenge on an online forum?
Anyways, you MADE MY POINT and dont even realize it. You said, the fact that it hit twice does not increase the probability that it'll hit again. Thats cool, I agree but you just silenced a few in the 'crew'. If you are correct, you would then agree? >>> So after the 30 groups of 38 numbers, the RESULTS should be around 15/15 (even 13/17 is close enough). Afterall, I have no EXTRA advantage of choosing numbers with two hits on them, correct? Ken
Quote: mrjjj
An example of a little contest we can have >> We'll track some numbers. Of course we know CHEATING may be involved with that, as usual. lol When we have 25 numbers recorded, I will pick 3 numbers that I think will have 3 hits on it by 38 (or 37) spins. *BUT* I also get to pick ANY 3 numbers for you, you dont get to choose your numbers. At the end of 38 numbers, we'll see who has the most numbers with 3 hits.
So you're going to start a race right before the finish line, while your opponent starts way back at the beginning, and you're wondering who's going to win? Seems obvious to me.
But since you posted this in the math forum, here's the interesting question: what odds would need to be offered to make this a fair bet? Alternately, assuming the optimal picking strategy (pick 3 numbers for yourself that require the fewest additional repeats, and pick numbers that have not occurred for your opponent), what are the chances that you will win the contest? It's not 100%, but it's very high...
I calculated that if you play Martingale, the probability is about 2:1 that you will go bankrupt before you double the value of your bankroll (assuming your bankroll is the minimum required to cover 5,6,7,... losses in a row). That calculation is without a house edge (i.e. a fair coin). A real life casino game with a house edge will result in probabilities worse than 2:1.
Quote: mrjjj"I think people think you would already have the results before your test" >>> That was my concern. lol You see, I dont do what some others (mainly into AP) do. They will lie, cheat, steal and sell their own grandma, if it means one more soul might be interested in AP. 'They' will do ANYTHING to get their point of view across, I dont. Thats in response to your question regarding me KNOWING the results ahead of time. Ken
Of course, the other thing about the APs is that the actually understand the math well enough to know that they have an advantage. You blindly ignore math and common sense in favor of utter lunacy.
Quote: mrjjjNO NO NO, past spins are NOT SUPPOSE TO MEAN ANYTHING for future events in *ANY* MANNER.
This is simply not true. Some events are dependent, some are independent.
Past spins mean a lot for all future events that depend on them, but do not have any effect on future events, that are independent (like other spins).
Some number that was hit twice before has a higher probability to be hit three times by the next spin, than some other number, that has not yet bit hit. However, the fact that it was hit before does not change the probability of it being hit on the next spin - it's still 1/38.
Quote: mrjjjLets say its the numbers 5, 17 and 34. I am looking for a YES or a NO to this question......try not to DRIFT from the question. Didn't those three numbers have to have two hits on them (within 37/38) BEFORE it had 3 hits?
Yes.
Quote: mrjjjThe number 5 is not going to magically jump from having ONE hit on it, then out of no where, it has 3 hits on it. My point being, if PAST numbers mean nothing, then HOW is it that we can gage that there will 2-4 numbers with 3 hits on them?
By using simple statistics. However, we don't know what the 2 - 4 numbers will be (in advance of the full 38 spins), and there is no guarantee that this will even happen. It's just more likely.
Quote: mrjjjAn example of a little contest we can have >> We'll track some numbers. Of course we know CHEATING may be involved with that, as usual. lol When we have 25 numbers recorded, I will pick 3 numbers that I think will have 3 hits on it by 38 (or 37) spins. *BUT* I also get to pick ANY 3 numbers for you, you dont get to choose your numbers. At the end of 38 numbers, we'll see who has the most numbers with 3 hits.
Who knows, I might not have any. Also, we'll do this for 30 groups of 38 numbers. At the end of the 30 groups, according to the slide ruler guys, it should be roughly 15/15. Maybe not exact but it should be close. Why? Because "the ball has no memory. All numbers are independant from one another, PAST NUMBERS MEAN NOTHING". Ken
I'm not sure what the rules of your contest really are.
You say you will record 25 numbers, and then pick the three numbers that have the highest probability of having 3 hits by 38 spins. Do the 38 spins you are talking about include the 25 you have already recorded? If so, the contest has no merit. Of course numbers that have already hit twice by the 25th spin have a better chance of hitting 3 times by the 38th spin if you look back on all 38 spins. In this case, the correct question is: Will the three numbers you pick have a better chance of hitting ONCE in the next 13 spins than the random three numbers you choose for me? The answer is, no. They won't.
But, if you are saying that your three numbers have a better chance than any other three numbers over the NEXT 38 spins, then I can see no basis for this claim.
Please clarify the rules of your contest.
Quote: mrjjj"It's not 100%, but it's very high" >>> It should not be high, it should be around 15/15 in the end. You guys dont get to have it both ways. You preach and preach and preach that past numbers mean NOTHING for future events. Then I offer a question and I get...."Umm, umm hang on now Ken! This isn't fair at all. You cant use PAST numbers to help you in the challenge". lol Also, I was not sure if to post this in the math section, I took a shot. Ken
The problem you posed is actually related to one of the oldest problems in probability theory, one which actually started the field: "The problem of points". The question is this: in a race to 10 points, the game is halted when the score is 6 to 4. How should you divide the pot? Fermat and Pascal spent many weeks discussing this ... in the mid-1600s. The answer is not trivial in the context of what was known in the 1600s.
Your question is similar: in a race to 9 points, where you start with somewhere between 3 and 9 points and I start with zero, what are my chances of winning? It's an interesting question, but it's not a "fair" race. You have misapplied your understanding (or your understanding is incorrect) of the independence of events in roulette.
Quote: scotty81Will the three numbers you pick have a better chance of hitting ONCE in the next 13 spins than the random three numbers you choose for me? The answer is, no. They won't.
No, I think the question really boils down to "are the chances that his three numbers will hit 0, 1 or 2 times (as the case may be) in the next 13 spins greater than the chances that my three numbers will hit 3 times." That's obviously yes.
"However, the fact that it was hit before does not change the probability of it being hit on the next spin - it's still 1/38" >>> BINGO, very true. So I should NOT have any advantage. The results should end up around 15/15. Ken
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No, I think the question really boils down to "are the chances that his three numbers will hit 0, 1 or 2 times (as the case may be) in the next 13 spins greater than the chances that my three numbers will hit 3 times." That's obviously yes. <<<<< LOL, so here we have two different answers from two other posters. Ken
So, if he picks 5, 17 & 34 because they have already hit twice, and chooses 2, 25 & 16 for me because they have not come in at all, he would win if either 5, 17 or 34 comes in before any one of my numbers comes in three times in the next 13 spins.
Quite a mouthfull, but that's how I interpreted his contest. If that is what he meant, it's certainly not a fair contest.
Mrjjj will need to clarify exactly what his contest is.
Quote: mrjjjWill the three numbers you pick have a better chance of hitting ONCE in the next 13 spins than the random three numbers you choose for me? The answer is, no. They won't.
No, I think the question really boils down to "are the chances that his three numbers will hit 0, 1 or 2 times (as the case may be) in the next 13 spins greater than the chances that my three numbers will hit 3 times." That's obviously yes. <<<<< LOL, so here we have two different answers from two other posters. Ken
Two different answers, but to two different interpretations of an ambiguous contest.
Please clarify what the contest really is.
Quote: mrjjj"Some events are dependent, some are independent" >>> Ohhh, so now its SOME events are dependant. lol If I had a nickel for every post I ever read that said the opposite of that. :)
Are you talking about posts on this board? Can you provide a quote and a link to one of those posts if so?
Not that I don't trust you, but I have reasons to believe that you have misinterpreted those earlier posts. If not, I promise to pay you a nickel for every link you provide that really says the opposite of my statement above.
Quote:
"However, the fact that it was hit before does not change the probability of it being hit on the next spin - it's still 1/38" >>> BINGO, very true. So I should NOT have any advantage. The results should end up around 15/15. Ken
You would not have any advantage if your challenge was about whether or not a given number hits next time.
If you want to stick to your original challenge, than the first part of my statement applies, the one that you (I am sure, accidentally) neglected to quote - the probability of a number hitting N times by some point in the future obviously depends on the number you start counting from. If you have read differently somewhere, then either that place was wrong, or you just misunderstood what it was saying.
Now that we are in agreement, that some events are independent, and some are not, would you like to move on to the point you wanted to make, or is this already it?
30 groups of 38 numbers. We look at the first 25 numbers that are recorded. At that point, I will pick 3 numbers I feel will have 3+ hits on them by the 38th spin. I will also pick YOUR 3 numbers, to see if any of them have 3 hits on them by the 38th spin. At the end of the 38 spins (or RNG, whatever) we'll add it up to see who has more. It might even be a tie, 2-2. We do this challenge for 30 groups. It should finish around 15/15, correct? (I'll give you a hint. I've already been doing this for 10 days, I have 280 groups already recorded) Ken
"the probability of a number hitting N times by some point in the future obviously depends on the number you start counting from" >>> Question for you in regards to definition. 'You start counting from'....is that also the SAME definition as 'past'? Ken
Just Wow.
The outcomes of a series of roulette wheel spins are i.i.d. r.v.s (independently identically distributed random variables). The summary statistic for which outcomes are observed more often over a given number of trials relies on posterior probabilities. That is, the posterior probability of seeing one number appear 3 times in 38 spins given that you have seen it X times in 25 spins DEPENDS ON X.
Using Bayesian probability on independent trials does not invalidate the notion of independence.
Take a stats course.
Wow.
Quote: mrjjj"Do the 38 spins you are talking about include the 25 you have already recorded?" >>> Yes.
Good. We got that out of the way.
Quote: mrjjj"Of course numbers that have already hit twice by the 25th spin have a better chance of hitting 3 times by the 38th spin" >>> Hold on, hold on. lol "Have a BETTER chance of hitting 3 times by the 38th spin (says scotty81). Remember, you cant have it both ways
How on earth could anyone interpret this as having it both ways?
Quote: mrjjjIf this is your view, then you would agree (yes or no) that PAST numbers do/could have an affect on future events? Ken
No.
They have a better chance hitting three times in 38 spins because THEY HAVE ALREADY HIT TWICE IN 25 SPINS.
This in no way implies that they have a better chance of hitting ONCE in the next 13 spins - which is what really counts in the casino. The next 13 spins are the only FUTURE events in this contest. The first 25 are PAST events.
I've answered your questions. Now, please answer mine. A YES or NO, please:
Starting with the 26th spin, betting your three numbers for 13 spins vs. my three numbers for 13 spins. Do you have a better chance of winning than I do?
Please - just YES or NO.
"The next 13 spins are the only FUTURE events in this contest. The first 25 are PAST events" >>>
Yes correct but it should make NO difference to you because PAST events (numbers) mean nothing towards future event!
"Starting with the 26th spin, betting your three numbers for 13 spins vs. my three numbers for 13 spins. Do you have a better chance of winning than I do?" >>> I dont understand your question? We are not betting during those 13 spins, where did you get that from? Ken
The fact that you don't understand my question speaks volumes about your understanding of statistics.
You are including past information in your requirements for a future bet.
Here is another YES or NO question:
Do you agree that the probability of all of your three numbers hitting TWICE in 38 spins is 100%?
If so, how is this possible? How can ANY gaming event have a probability of 100%?
Quote: mrjjj"Can you provide a quote and a link to one of those posts if so?" >>> I'm on 8 boards for over 6 years, with around 7K posts of my own and another 30K posts read, no I cant. My point? Its funny how I ask the tough questions.
Your questions aren't tough at all, in fact, they are very basic. The problem is not with your questions, it's that you are not listening to the answers.
Quote:You see, in the past, ALL events were independant of one another.
Nope. That never was the case. If somebody told you that in the past, either they were wrong, or you misunderstood. Since you don't have the link, we'll never know now. In any event, recanting that incorrect statement now does not serve any purpose, since everybody is in agreement, that it is wrong.
Quote:Now, SOME events dependant. lol
Yes, and I am not sure why you find it funny. It is not science, just common sense.
Quote:
"You would not have any advantage if your challenge was about whether or not a given number hits next time" >>> You mean the next spin, right? I agree with that BUT thats not my question. My question is in regards to 38 total spins.
Exactly.
Quote:
"the probability of a number hitting N times by some point in the future obviously depends on the number you start counting from" >>> Question for you in regards to definition. 'You start counting from'....is that also the SAME definition as 'past'?
You can have the number you start counting from defined by any means you want. It could come from past spins, a divine epiphany, a random number generator or anywhere else. It does not matter where it came from, what matters is to understand that there is nothing new in realization that if you start counting from 2, you will reach 3 faster, than if you started from 0. Everybody agrees with you on that. Let's move on.
Did you actually have a point you wanted to make besides this trivial (and obviously correct) statement?
