6/49 false logic

[GillesD]

LT's comment

In your posting on 07-11-02, you made the following comment: "lotto tip - the number combo 1,2,3,4,5,6 is never going to happen". On what basis can you make that comment?

If 01-02-03-04-05-06 has no chances, then it mus be true also for 05-10-15-20-25-30. Then the combinations 01-06-09-16-27-36 and 01-04-11-15-16-44 and 02-08-22-30-32-39 are also to be excluded, since they follow a regular pattern (although not easy to spot). But why accept 02-04-23-36-44-49 or 15-25-30-33-45-46 or would you reject them also?

Let's not forget that we are dealing with a very immature lotery with only about 0.014% of the possible 6-number combinations having come out.

 

[winhunter]

Re: LT's comment

In your posting on 07-11-02, you made the following comment: "lotto tip - the number combo 1,2,3,4,5,6 is never going to happen". On what basis can you make that comment?

Because it is at the extreme edge of all of the statistical models, with exception to the lotto odds themselves. It is true, it does have the same chances as any other sequence. But we are not talking ODDS! PRobability is a totally different subject, and that is usually where most people get the two confused.

For example, why would you not play last weeks drawing this week? True, the odds are the same for that number to come out. But, what is the probability that it WILL come out again? IF the probability was the same, then I would say YES, play it again! But the probability (chances) that it will happen again are reduced. Think of it this way, the balls are loaded into the machine in order 1,2,3,4,5,6... The ODDS are the same, but what is the probability that you will draw them back out? Better still, if I put your hand into a running blender, what are the odds that you will loose a finger? Hmmm, that would be 1 in 5 odds. What is the probability? Hmmm, not sure about that one. Maybe if I push hard on your hand, I would get 25% chance of lobbing a finger off... Hehehe

Actually, What does your weather man mean when he says 50% chance of rain? Does that mean it will rain over 50% of the area he is forecasting? Nope. It simply means that conditions are favorable for it to happen.

With all that in mind, would you play 1,2,3,4,5,6? Better still, everyone plays it, so the jackpot will be very low if/when it ever comes in. So, why waste your $$ on it?




Andrew

P.S. - If the difference between Odds and probability dont convince you, then nevermind my posts, because I tinker in the probability relm.

 

[Brad]

Andrew,

yes I am confused. How do you define ODDs? Are they not "probability of a specified outcome" ? Or do you mean ODDs are theoretical whereas Probability is based on empirical data?

 

[winhunter]

Odds vs Probability

Andrew,

yes I am confused. How do you define ODDs? Are they not "probability of a specified outcome" ? Or do you mean ODDs are theoretical whereas Probability is based on empirical data?

Odds are specific. In the case of a 6/49 lottery you have certain odds that one combination will come out via the 6/49. Odds do not change. The only way to improve YOUR odds, is to eliminate numbers from your selection set, but you are only improving your odds if you can guarentee your remaining numbers will still contain the jackpot for the drawing you are predicting. BY purchasing more tickets, you dont increase your odds, you increase your probability. Got it?

It all goes back to the old coin toss. You have 1in2 odds at a coin toss. if you flip the coin 5 times, and you get heads every time, the odds are still the same for the sixth subsequent toss, 1in2. But, the probability increases for a tails. Why? because you have a 50-50 chance, which means that overall the coin toss will be roughly 50% heads, 50% tails. As you continue to flip, each subsequent outcome of heads only increases the probability of a tails coming up, until it actually does.

Dont ask me to explain it better than that, I can't...



Andrew

 

[Dennis Bassboss]

It seems that we do not have all the same definition of odds versus Probability....to me
better odds= high probability
less odds= low probability
I'm having a difficult time to seperate the 2...as for the 01-02-03-04-05-06 set I have to agree with Gilles It could come in any given draws but for these kinds of draws happening frequently or repetitively is highly improbable so with much lesser odds... but in an extreme scenario it could also happen that it repeats but I for one would not and never gamble on such small odds ...just by looking back at the entire history so far we have seen extreme draws at times I recall a draw having 38 as the lowest regular number ..highly improbable but it did happen ....but it is certainly not happening frequently when compared to the whole history...so the odds and the probability of that happening is relatively low...getting better odds is the aim we are all looking for in that sense Andrew is right and Winhunter is good for that but remember that a human is feeding it and human mistakes or lack of datas from humans...could always hinder its results...or should I say its odds of success...

 

[GillesD]

Odds vs Probability

I agree, Brad, a rather confusing issue. To me, it is the same.

To get 01-02-03-04-05-06, the odds are 1 against 13,983,815 while the probability is 0.0000072% or 1/13983816. This ends up the same.

But then why limit ourselves to the 01 to 06 combination, any 6 consecutive numbers should be considered. Actually (without considering the bonus number), somebody having bet in each draw the 44 possible combinations of 6 consecutive numbers would have done just slightly better than the standard odds (about 1/57 for 3 WN, 1/1032 for 4 WN. etc.).

Why not 6 consecutive numbers? 5 consecutive numbers have already appeared once (draw #1748) in only 1938 draws. I do not have any problem considering 6 consecutive numbers coming out in the next 5000 draws (about 50 years) or maybe next Wednesday.

And the same logic is applicable for numbers of the last draw. Already it happenned once that 4 numbers (without the bonus) came out in consecutive draws (draws #1464/65), so give it time and it may happen. But then everybody will jump up and shout FIX FIX FIX.

 

[Brad]

Scot Piel (LottoMan) was mentioned today in a post somewhere, I'd like to post his quote on probability. He explains it much better than I could.
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Calculating Probability

When we speak of probability, we are really saying how often, on average, we might expect to see a particular result. For example, if we fill a bag with ten balls numbered from zero to nine, we can expect to see the one ball drawn in roughly one out of every ten trials. That is, we can guess at the result and expect to be right one out of every ten draws. Thus, we would say: "The probability of correctly guessing the result of a draw would be one in ten.".
Alternatively, we can talk about the odds of something in particular taking place. We might ask: "What are the odds of reaching in the bag and pulling out the one ball?". In this example, there is one right answer and nine wrong answers. Thus, the odds of drawing the one are 1:9 and we say "The odds of drawing a one ball are one to nine.".
The distinction between the two terms is subtle, but important all the same. Note that when we talk about probability we use the word IN and when we talk about odds we use the word TO. The probability of correctly guessing drawing a one is one in ten and the odds of drawing a one are one to nine. The significant difference is that probability tells us how often things should happen "on average" while odds tells how likely a particular outcome is.
Probability gives us a tool we can use to make some assumptions about what we might expect. It is import to understand that what we expect to happen "on average" (the definition of probability) is not likely to be what actually happens. In other words, if we were to draw balls from the bag twenty times, we can not say we will have seen the one ball exactly twice. Nor can we be certain that after drawing a one it will be exactly ten draws before we see it again. Rather, what we can say is that if we drew 1000 balls, we should expect to see the one ball roughly 100 times and that the average number of trials (drawings) between drawing the one ball will be around ten.
In reality, we are not likely to see the one ball exactly 100 times. Rather, we will most likely see a few more or less than 100. Likewise, we might pull the one ball out twice in a row and then not see it again for another twenty, or so, trials. However, if we divide the number of trials by the number of times the one ball appeared, the result will typically be close to ten. The one ball appeared "on average" in one out of every ten draws.
Odds, on the other hand, are very rigid and unchanging. Each time we reach in the bag to pull out a ball, the odds of any particular outcome will be exactly 1:9. Regardless of how many times we reach in the bag, they never change. Thus, the odds of a particular outcome gives us a clear view of what is stacked against us and a measure of how reasonable our expectations are.
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This is basicaly what Andrew said except in more detail, hope it helps.

 

[Dennis Bassboss]

Interesting but I still see it differently....because at the moment you pick the ball there is always 10 balls in the bag and I think that the definition of the word odds given in that example here is in itself debatable ...the way that I see it I usually define the word odds as follow...
odds= chances of being drawn
And by that definition I only can say that the odds would be 1/10 And the other value of 1:9 does not exist by that definition...
Now always by that definition..after many many draws knowing that every ball should tend to equalize on the numbers of times each of them would be drawn I would say at a certain point that some balls perhaps including the 01 in this case would have better odds of coming out of the bag then the other balls simply because of the laws of probability...
Therefore I do think that gamblers have reinvented the word odds
forgetting the big laws of the numbers...but I'll dig deeper into dictionnary and litterature to find the real meaning of the word odds ...it is interesting because that could be the source of the confusion around that word...but I am not a native speaker of English and I might be mistaken ....What I mean is that it is basically saying the same thing but it plays with words such as overall average, overall probability, overall results, overall chances...

 

[Beaker]

Brad, do you have the Lottoman software??

 

[winhunter]

Scot Piel (LottoMan) was mentioned today in a post somewhere, I'd like to post his quote on probability. He explains it much better than I could.
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Calculating Probability......[snip]

Glad somebody could explain it!


Andrew

 

[Dennis Bassboss]

I have done a rough search into some dictionnary for the meaning of the word odds...
Here's what the Robert and Collins French /English dictionnary tells us...
It has in fact two definitions one for betting and one for probability( Mathematical language) for

a) Betting french word would be ''cote'' like example which is making more clear for me...He gave them 5 to 1 that it is going to come...This one is like Scott Peal's definition..

b)For the definition probability's .. it is using the term ''balance of advantage'' or ''Chances of'' and ''Chances for' and ''Chances againsts''

But what it tells is that the two are very much alike and I do not see between these a ''subtle distinction'' The second definition in its essence is much more complete and in itself it does include the first one ...it is only a matter of terminology it becomes obvious to me that you would have 9 chances of picking the wrong ball and one chance of picking the correct ball but overall you always have one chance out of 10 because there are 10 balls in the bag...Furthermore the second definition is not limited as ''average'' and I would say that it is the one that include ''subtle stuff''(including the entire first definition) so I will still say that the two are very much alike...but which gambler would rather like 1 in 10 against 1:9...The same logic is seen when an object or item is sold at
1.99$ instead of 2$...and believe it or not it helps greatly to sell much more than if it would be 2$....following the same logic It tells that the house (Casino) always uses the betting definition because it makes it appear as the house have one less chance to be winner so the gamblers are more eager to play and bet more...It has a lot to do with a psychological approach..Noam Chomsky a genius of our era ..
a great linguist, psychologist, mathematician...once said '' When things become too subtle it often hides simplicity''

 

[Brad]

Dennis,

I don't know if we'll ever put this to bed, the semantics make it hard to allow for that, but we can agree to disagree . I lean towards the description put forward by Scott with TO for odds and IN for prob. It's my memory trigger for the distinction if there is one.

 

[winhunter]

I agree

The definitions do appear to be roughly the same, and so do the numbers the point to. I guess where I always see the difference, is that the ODDS never change. It is the probabilty that fluctuates depending on what you are looking at. I mean, if you are asking what is the probability of drawing the same ball 10 times in a row? If I understand the calculations correctly, because you can draw 10 different balls for each draw, then you have 1in10000000000 chances (might have one extra zero in there) because 10*10*10*10*10..., etc.

The way I understand it, is based on the math for the above problem, the probabilty of 1 being draw 10 times in a row comes out so high, that basically it will never happen. ODDS suggest that it can happen, but probability says that it won't happen.

ANYWAY! I think Im done racking my head with this one....



Andrew

 

[GillesD]

Odds vs probability

The odds of getting one specific ball in a set of 10 is always 1 in 10 while the odds of getting the same ball 10 times in a row would be 1 in 10x10x10x10x10x10x10x10x10x10. This is also your definition of probability.

And actually, when you say "drawing the same ball 10 times in a row", the real probability is 1 in 10x10x10x10x10x10x10x10x10 (yes 9 time 10). This is so because as it is stated, it is immaterial whether the first ball is a 1, a 2, a 3 ... The only concern is to get the same ball that you got in the first trial 9 other times.

The probability of 1 in 10x10x10x10x10x10x10x10x10x10 (10 times 10) would apply to "to get ball number 7 in a set ot 10 balls, 10 times in a row".

This is the same as getting a pair with two consecutives cards in a deck of 52 cards. It is 3 / 51 because whatever is the first one, there is 3 good cards in the remaining 51 to make the pair. A good difference from getting a pair of aces.

 

[winhunter]

reasoning

My reasoning for probability gives a firm foundation as to why I would not play last week's drawing this week. Simply put, the probability is simply too high that the same sequence would come out again. This is the same reason why we only see balls repeat only so many times. There is a limit (probability-wise) that almost appears to dictate what the outcome will/won't be.

Take WINHunter, and goto the Scan history form. Plugin the entire History (1-XX) and click scan. Look at the Index values. You will notice that there are gaps between the index numbers. I find that rather interesting....



Andrew

 

[GillesD]

Reasoning by Winhunter

Following your reasoning, it would mean that:
- flipping a coin, you will not bet on the side that came out for the next flip, just because it came out (still 1 out of 2 chances);
- rolling a dice, you will not bet on the number that came out in the last roll, because it just came out (still 1 of of 6 chances);
- dealing a card from a deck of cards, you will not bet on the card that you just dealt, because it just came out (still 1 of of 52 chances).

So how is it different with a 6-number combination? The last combination still has one chance in 13,983,816. Each draw is independant. Unless you can tell me how balls coming out on a Saturday can influence the outcome 4 days later (but the interval will be 3 days for the Wednesday draw), the set of balls may or may not be different, the machine used may or may not be different, the time the balls bounce around is certainly different, etc.

 

[winhunter]

Re: Reasoning by Winhunter

Originally posted by Duncan Smith (newsgroup)
<snip>0 (zero) if you don't put it back . (1/10)^10 if you do, assuming independent
drawings. ie. 0.0000000001

The above is the probability of drawing the 1 ball 10 times in a row. So, given any lottery, odds for any single draw, and raise it to the power of the repeats that you want it to repeat. See? 1:13,983,816 is 0.00000000715112384201851626194166170378672 .....

Originally posted by GillesD
<snip>So how is it different with a 6-number combination? The last combination still has one chance in 13,983,816. Each draw is independant. Unless you can tell me how balls coming out on a Saturday can influence the outcome 4 days later (but the interval will be 3 days for the Wednesday draw), the set of balls may or may not be different, the machine used may or may not be different, the time the balls bounce around is certainly different, etc.

And for the same draw to occur 2 times in a row is 1:13983816 * 1:13983816 which equals 0.00000000000000051138572203885665128471811298051 !!!

Yes, the odds are the SAME each draw. But the probabilty is the odds of each draw multiplide by the odds of the next draw. We are no longer talking about an individual draw any more, we are talking about the probability of the same sequence to occur a second time in a row. To put it simply, the combination 1,2,3,4,5,6 was just drawn on Sunday, what is the probability of the same sequence (1,2,3,4,5,6) to occur again on the next draw? Yes, the odds are still 1:13983816. But the probability is
1:195,547,109,921,856! That is because there are that many possible combinations between the two Draws! Count 'em up. IN other words, if you matched the sequence 1,2,3,4,5,6 on the first draw, there are a possible 13983816 combinations for the second draw still. This means that for EACH possible combination in the first draw, there are still 13983816 possible combinations in the second draw. Sou you must multiply the chances for each combination in the first draw by the number of chances of all the possible combinations in the second draw. Thus, you obtain the above probabilty.



Take the coin toss.
The odds are 1:2 (0.5).
For two flips it is 1:4 (0.5 * 0.5 = 0.25).
Three flips it is 1:8 (0.5 * 0.5 * 0.5 = .125).
Eight Flips it is 1:256 (0.5 ^ 8 = .00390625).

So, the probability of a repeat each successive flip is 1:Number of possible sequences ^ Number of consecutive repeats
(^ means to the power of). This is why you will only have so many runs of heads, or so many runs of tails in the coin toss before the coin eventually flips to the opposite side. Yes, after the fisrt coin toss, the probability is good that the coin will again flip onto the same side. But after a run of say 5-6 same sided flips, the probability is beginning to stack against another repeat.

BTW, this is not my definition of probabilty. This is the mathematics of the probability. Consequently, the probability of predicting any two unique combinations for consecutive draws has the same probabilty. It is because you are dealing with the odds of both games in dealing with probabilty. Odds only deals with the one individual Draw. Probability deals with the chances of whatever you specify. You see, the odds for ANY unique combination in a 6/49 are 1:13983816. That is the odds of you matching the same unique combination for THAT draw. Now, the probability that you would match two individual unique combinations for two consecutive draws is 1:195,547,109,921,856... Im not making this up, that is just how the cookie crumbles for probabilty. So, for a coint toss, the odds are 1:2 for the next flip to be the same side. That is a easy bet, you only have two choices. But for the lottery, the odds are 1:13983816 (6/49), you have 13,983,816 possible outcomes. Now, would you play the same combination this week, that also occured last week? As for me, I would not. Probability dictates that the chances of that happening are slim to none. And if it ever did happen (which I seriously doubt), Would you then play it a third time also?




Andrew

P.S.- ok Im done.

 

[GillesD]

Same draw

I think you are missing something. And it is mainly related to either you know are before a possible 2 events or there is one event that has already occured and one has not happened.

Let's start with a simple example, the flip of a coin. What are the odds of getting the same side twice in a row?

You say:
Take the coin toss.
The odds are 1:2 (0.5).
For two flips it is 1:4 (0.5 * 0.5 = 0.25).

I say: "It is 50%"

In the real world, what are all possibilities (with H=head and T=tail)? I think the answer is: H-T, H-H, T-H and T-T. And if I count right, I see 4 possible outputs with 2 of them (H-H and T-T) meeting the condition "getting the same side twice in a row", so 2 on 4 is 50%. But I agree that the odds of getting H-H is 1 in 4 and T-T is also 1 in 4. But since it was not specified what side you wanted, the odds of getting the same side twice is 1 in 2.

Let's go to a 6/49 lottery. I agree with you that if you say "What are the probability of getting twice in a row the 1-2-3-4-5-6 combination (or any other given combination) before any draw has occured?", then it is 0.00000000000000051138572203885665128471811298051 and I assume your calculation is right.

But this is not the question. The question is "What is the probability of getting the same combination as the last (and past) draw?". Last Wednesday, the winning combination was 4-7-13-25-37-45. What are the odds of getting the same combination next Saturday? Well, for next Saturday, there are 13,983,816 possibilities from 1-2-3-4-5-6 to 44-45-46-47-48-49 (or at least this is what I think). Out of these, there is only one good one (4-7-13-25-37-45) and 13,983,815 bad ones (which I do not intend to list). So the probability of getting next Saturday the combination that won last Wednesday is 1 in 13,983,816.

And this also happens to be the probability of getting any specific past wiinning combination (let's say from draw #1327). But the probability of getting anyone of the past winning combinations is now 1938 in 13,983,816. Not very likely I agree, but there is still a possibility.

To a lighter side but now with a rather sad twist, this reminded me of a old joke. You know the best way to prevent getting on an airplane where there is someone else with a bomb? Bring yourself your own bomb because the odds of getting two persons not knowing each other on the same flight and each having a bomb is very, very, very low. (Or at least it was true in the good old times).

 

[winhunter]

Re: Same draw

the odds of getting two persons not knowing each other on the same flight and each having a bomb is very, very, very low.

I see we agree!


Hehehe

 

[dwoods99]

Probabilities and statistics... what a headache
No wonder I only took one year of it at school