United States
Member #130,789
July 25, 2012
80 Posts
Offline
Quote: Originally posted by Stack47 on Apr 4, 2013
"The statement is unclear to me because of typos and awkward construction. And I don't know why you are quoting it in this context. I believe Stack47 wrote it to be critical of your approach."
A group of 35 numbers creates 324,632 combinations so any group of 35 numbers has a 1 in 12 chance matching five numbers. The probability is that any group of 35 numbers should match five numbers 10 times in 120 drawings plus or minus standard devation.
"I cannot make sense of "or 9 times a year". If that is "for 9 times a year", I don't know why Stack47 would choose 9, since you talked about 17."
There are 104 drawings in a year and the group Ronnie is isolating on, matched five numbers 17 times which is much better than the probable 8 or 9 times (104/12). Nobody mentioned "exactly" 9 times in one year, but there probably are billions of groups of 35 numbers that will match five number exactly 9 times in any one year period.
"The probability of any of the full wheel combinations matching 9 times in a year is about 13.69%."
Does that include the group of 35 numbers that matched five numbers in the last 10 consecutive drawings?
(With apologies to Ronnie316 for the digression.)
Stack47 wrote: "A group of 35 numbers creates 324,632 combinations so any group of 35 numbers has a 1 in 12 chance matching five numbers".
Only if we play all such combinations(!). Since no one said anything about doing that, I think we need to stipulate that as an assumption.
Alternatively, we could play "any group of 35 numbers" by simply dividing them into 7 sets of 5. In that case, the odds are not 1 in 12, of course.
In fact, as we learned later, Ronnie316 intends to do an abbreviated wheel.
Stack47 wrote: ``Nobody mentioned "exactly" 9 times in one year``.
Well, I think you did. You wrote (here): "The probability is about 1 in 12 drawings that any 35 numbers should match five numbers or 9 times a year".
I wrote: "The probability of any of the full wheel combinations matching 9 times in a year is about 13.69%".
Stack47 wrote: "Does that include the group of 35 numbers that matched five numbers in the last 10 consecutive drawings?"
Who said anything about 10 times in a year?
If the probability of success is 1/12 for one event, the probability of success for 9 of 104 independent events is (in Excel):
(1/12)^9 * (1 - 1/12)^(104-9) * COMBIN(104,9)
or simply BINOMDIST(9,104,1/12,0).
And yes, that includes all instances of 9 consecutive successes, among others.
mid-Ohio United States
Member #9
March 24, 2001
20,272 Posts
Offline
Quote: Originally posted by jimmy4164 on Apr 5, 2013
MathHead,
"I had toyed with the idea of demonstrating that point with a simulation of Ronnie316's strategy; a fun little program. But enough's enough."
Wise decision. The people disagreeing with you here are not impressed with simulations. I know; I wrote them and posted the results. I even posted source code which wise guys edited and reposted. You would have to first convince them that computerized random number generators are valid tools, and then teach the value of Monte Carlo Techniques. I tried - even started a thread for MCT. I failed. Too many innumerates literate enough to obfuscate your posts.
Good luck. If you're really lucky, they'll wear you out before you waste as many hours as I did.
--Jimmy4164
"The people disagreeing with you here are not impressed with simulations. I know; I wrote them and posted the results."
If your simulations had showed them how to win, maybe they would have been more interested.
* you don't need to buy every combination, just the winning ones *
United States
Member #93,943
July 10, 2010
2,180 Posts
Offline
Quote: Originally posted by jimmy4164 on Apr 5, 2013
MathHead,
"I had toyed with the idea of demonstrating that point with a simulation of Ronnie316's strategy; a fun little program. But enough's enough."
Wise decision. The people disagreeing with you here are not impressed with simulations. I know; I wrote them and posted the results. I even posted source code which wise guys edited and reposted. You would have to first convince them that computerized random number generators are valid tools, and then teach the value of Monte Carlo Techniques. I tried - even started a thread for MCT. I failed. Too many innumerates literate enough to obfuscate your posts.
Good luck. If you're really lucky, they'll wear you out before you waste as many hours as I did.
--Jimmy4164
RJOh's response to my post above was, "If your simulations had showed them how to win, maybe they would have been more interested."
United States
Member #116,263
September 7, 2011
20,243 Posts
Offline
It sounds like Jammy is still ignoring the human ability to pick winning numbers. I have proven over and over it can be done (as many other LP members have also) and I will continue to prove it, Stay tuned.
Kentucky United States
Member #32,651
February 14, 2006
10,302 Posts
Offline
Quote: Originally posted by mathhead on Apr 5, 2013
(With apologies to Ronnie316 for the digression.)
Stack47 wrote: "A group of 35 numbers creates 324,632 combinations so any group of 35 numbers has a 1 in 12 chance matching five numbers".
Only if we play all such combinations(!). Since no one said anything about doing that, I think we need to stipulate that as an assumption.
Alternatively, we could play "any group of 35 numbers" by simply dividing them into 7 sets of 5. In that case, the odds are not 1 in 12, of course.
In fact, as we learned later, Ronnie316 intends to do an abbreviated wheel.
Stack47 wrote: ``Nobody mentioned "exactly" 9 times in one year``.
Well, I think you did. You wrote (here): "The probability is about 1 in 12 drawings that any 35 numbers should match five numbers or 9 times a year".
I wrote: "The probability of any of the full wheel combinations matching 9 times in a year is about 13.69%".
Stack47 wrote: "Does that include the group of 35 numbers that matched five numbers in the last 10 consecutive drawings?"
Who said anything about 10 times in a year?
If the probability of success is 1/12 for one event, the probability of success for 9 of 104 independent events is (in Excel):
(1/12)^9 * (1 - 1/12)^(104-9) * COMBIN(104,9)
or simply BINOMDIST(9,104,1/12,0).
And yes, that includes all instances of 9 consecutive successes, among others.
"You wrote"
And the reply was to Ronnie and he had no problems understanding what I meant.
"Who said anything about 10 times in a year?"
I don't know; have a problem reading my question?
I asked if the probability you posted applied to the group of 35 numbers that had a five number match in each of the last 10 consecutive drawings. FYI, if the group matches five numbers in 10, 15, or 17 drawing in a year, they matched in 9 drawings too.
"And yes, that includes all instances of 9 consecutive successes, among others."
We already discussed your math mumbo-jumbo. Do you have any suggestions about playing recently drawn numbers to win a jackpot?
Kentucky United States
Member #32,651
February 14, 2006
10,302 Posts
Offline
Quote: Originally posted by Ronnie316 on Apr 5, 2013
It sounds like Jammy is still ignoring the human ability to pick winning numbers. I have proven over and over it can be done (as many other LP members have also) and I will continue to prove it, Stay tuned.
I haven't seen any decline in the number of systems and ideas posted since July 10, 2010 so it appears whatever his message is, nobody is listening.
mid-Ohio United States
Member #9
March 24, 2001
20,272 Posts
Offline
Quote: Originally posted by Stack47 on Apr 5, 2013
"You wrote"
And the reply was to Ronnie and he had no problems understanding what I meant.
"Who said anything about 10 times in a year?"
I don't know; have a problem reading my question?
I asked if the probability you posted applied to the group of 35 numbers that had a five number match in each of the last 10 consecutive drawings. FYI, if the group matches five numbers in 10, 15, or 17 drawing in a year, they matched in 9 drawings too.
"And yes, that includes all instances of 9 consecutive successes, among others."
We already discussed your math mumbo-jumbo. Do you have any suggestions about playing recently drawn numbers to win a jackpot?
'We already discussed your math mumbo-jumbo."
I am surprise mathhead posted his math mumbo-jombo without checking it first.
* you don't need to buy every combination, just the winning ones *
Kentucky United States
Member #32,651
February 14, 2006
10,302 Posts
Offline
Quote: Originally posted by RJOh on Apr 5, 2013
'We already discussed your math mumbo-jumbo."
I am surprise mathhead posted his math mumbo-jombo without checking it first.
The same math that's being preached means there is not difference in the odds against winning whether the same amount of sets of numbers are chosen by the player, picked by an RNG, fortune cookies, or any other method. I see no reason to argue when someone says their set has better odds because they can't have worse odds.
United States
Member #116,263
September 7, 2011
20,243 Posts
Offline
Quote: Originally posted by Stack47 on Apr 6, 2013
The same math that's being preached means there is not difference in the odds against winning whether the same amount of sets of numbers are chosen by the player, picked by an RNG, fortune cookies, or any other method. I see no reason to argue when someone says their set has better odds because they can't have worse odds.
Exactly my point Stack. It's impossible to do worse than Jammy, so why not at least try to do BETTER?
mid-Ohio United States
Member #9
March 24, 2001
20,272 Posts
Offline
Quote: Originally posted by Stack47 on Apr 6, 2013
The same math that's being preached means there is not difference in the odds against winning whether the same amount of sets of numbers are chosen by the player, picked by an RNG, fortune cookies, or any other method. I see no reason to argue when someone says their set has better odds because they can't have worse odds.
Getting better odds is only important to players looking for an edge, those satisfied playing quick picks or other randomly picked numbers don't care.
* you don't need to buy every combination, just the winning ones *
United States
Member #93,943
July 10, 2010
2,180 Posts
Offline
MathHead,
Do you see what I mean?
When you said,
"These forums seem to be all about voodoo practices; some are cloaked in the language of math, but that is only an illusion for the most part. I see no basis to criticize one voodoo practice more than another."
and
"...because it is [a] hoot to see the voodoo priests argue over whose practices have more mojo...."
...you cracked me up!
I'm afraid it's a waste of precious time trying to teach math, logic and critical thinking skills to people who are either Innumerate and incapable of understanding, or have a vested interest in propagating ignorance. The Innumerate ones are just literate enough to obfuscate teaching efforts as they expound on their fallacious beliefs. The Numerate ones are probably praying that they will someday be forgiven.