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devonwoody

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We had a lottery win last night. :D :D :D :D

Something very unusual happened though :twisted:

We normally have the same numbers every week (and for the last 10years)

Yesterday the wife ticked one of the numbers 47 in mistake for 37.

47 was a winning number .

Isn't that odd!!!!

Regret only a ten pound win.

Still we share 50/50 so I might get some sandpaper.
 
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Anonymous

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In a bored moment I once decided to calculate individual odds for the lottery balls

my calcs gave the following odds for your 6 to appear

first ball 8.16 to 1
secodn ball 9.6 to 1
third ball 11.7 to 1
forth ball 15.3 to 1
fifth ball 22.5 to 1
sixth ball 44 to 1

So, the odds of getting all 6 are 9.6*11.7*15.3*22.5*44

= 13983816 to 1 !!!!!!

Just to cheer you up on this cold and damp sunday morning :lol:
 

devonwoody

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Thanks tony, odds look promising :roll: :roll: :roll:

But could you do another calculation. I don't think a week hardly goes by on the Saturday lottery without there being consecutive pair of numbers.
My mental calculations think this is unusual? So what logistic can you provide on this phenomia <(grammar or spelling problem?)

Not a gambler by the way just a £1 per week.
 

StevieB

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DW,

It might seem unlikely, but just think about it for a moment. After drawing ball 1, ball 2 then has two chances to be consecutive - one less than or one more than ball 1. For ball 3, there are now 4 chances of being consecutive to either ball 1 or ball 2. By the time you get to ball 6, there are now 10 chances of being consecutive to balls 1 to 5.

there are 49 balls in total, minus the 5 already drawn means that 10 of a possible 44 balls left will be consecutive to one already drawn. This is better than a 1 in 5 chance.

Oh, and for the pedantic, I know 1 and 49 cannot have a ball both sides of them drawn, this is a rough calculation. :roll: :wink:

Steve.
 
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Anonymous

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StevieB

not sure I understand your logic here - sorry if I'm being a bit dim :oops:

Probability is usually not intuitive and most people get the wrong answer when attempting to assess odds from an intuiitive standpoint

I am no expert on probability but this is my take on the problem

Lottery balls do not know either of the flollowing

A) what they are (which number)
B) what number appeared before them

This is a bit like tossing a coin. Heads and tails will even out to 50:50 over time for an evenly balanced coin as the coin does not know one side from the other and does not know which way up it landed last time.


I would say that the chances of consecutive balls appearing are the same as the chances of your balls on the ticket appearing. We multiply probabilities to get the final probability and so my calcs for the sequential balls are as follows: (I will assume that the second ball is higher than the first in this case as it is simpler than considering higher an dlower where 1 and 49 are problematic)

for the second ball to be numerically sequential to first 8.16* 9.6 = 78.336 to 1
third ball being numerically to second ball 9.6 * 11.7 = 112.32 to 1
forth ball being numerically to third ball 11.7* 15.3 = 179.01 to 1


etc.
 

StevieB

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Tony,

yup, confusing isnt it :? you have got me wondering if I have got it right now :oops:

My greater than 1 in 5 chance was actually the chance of pulling a ball that was consecutive to one of the 5 previously drawn balls, assuming those previous 5 balls were not themselves consecutive (and assuming they were neither number 1 or 49). I did not add to this the probability of the previous balls being consecutive.

The way I tackled this is as follows:

Ball 1 is drawn.

Probability of drawing a consecutive ball on ball 2 is now 2/48 (assuming ball 1 was not 1 or 49). 2/48 = 0.04 or 4%.

If ball 2 is not consecutive, the chance of pulling consecutive ball for ball 3 is now 4/47 (with the same 1 and 49 caveat). 4/47 = 0.085 or 8.5%.

If 1 2 and 3 are not consecutive, the chance of pulling a consecutive ball on the 4th draw is now 6/46. 6/46 = 0.13 or 13%.

and so on until ball 6 is drawn. If balls 1 to 5 do not make 2 consecutive numbers, then 10 balls of the remaining 44 balls can be drawn to make two consecutive numbers. Thus 10/44 = .227 or 22.7%.

Thus if balls 1 to 5 are not consecutive, there is a greater than 1 in 5 chance that the 6th ball drawn will be consecutive to one of the previously drawn 5 balls.

I think our discrepancy arises because I have presented the chance of the last ball only, while you have been better at maths and presented the probability of all options :?:

I also am not an expert in probability :D, if I was I would be in Vegas, or possibly Blackpool waiting for the mega-casino :twisted:

Steve.
 

devonwoody

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(I'm not playing poker with you lot)

So is there something funny going on?

At a guess I would state that over the last 52 weeks, 40 weeks have had at least one set of consecutive numbers and a few have had even more.

So where does that leave probability?
 
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Anonymous

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DW

No balls have any knowledge of how many times they have come out of the machine nor the order. They are balls - no past memory and so the chances ofthe same ball coming out every week are the same every week, even if that ball appeared the previous 40 weeks

The lottery is truly fair as it is truly random every week.

well, fair if you fancy your 1 in 14,000,000 chance :wink:
 
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Anonymous

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Stevie

tough game eh?

I think an error is creeping in here:

Probability of drawing a consecutive ball on ball 2 is now 2/48 (assuming ball 1 was not 1 or 49). 2/48 = 0.04 or 4%
First ball is 1 in 49, but second ball is ONE in 48, not 2 in 48 :?

3rd ball is 1 in 47 etc.

balls have no memory or knowledge of what occured before :wink:
 

Johnboy

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Tony wrote
First ball is 1 in 49, but second ball is ONE in 48, not 2 in 48
It is 2 in 48 as long as the first one was not 1 or 49. The ball can be the one higher or the one lower.

John
 

Alf

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I can't believe you chaps could seriously give time and effort to this on a Monday morning. :shock: :roll: :lol:

Cheers, Alf
 

StevieB

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First ball is 1 in 49, but second ball is ONE in 48, not 2 in 48

It is 2 in 48 as long as the first one was not 1 or 49. The ball can be the one higher or the one lower.
Yes, Johnboy has it right, it can be one higher or one lower. If the first ball drawn is number 7, then either number 6 or number 8 would make two consecutive numbers. Hence 10/44 by the time ball 6 is drawn if the previous 5 were not consecutive in themselves.

Steve.
 

mudman

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Interesting.
I hadn't considered the probabilities of consecutive balls occuring before but it does make sense.
<Ramble Mode>
This seems to imply that to get a better chance of winning, you don't want an even spread of numbers and that two of your numbers should be consecutive. It would appear at first glance that this would be better, but, as soon as you choose your numbers, the probabilities still remain at the 1:14,000,000 that all your numbers come up.
</Ramble Mode>
 

StevieB

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ROFL MudMan :lol:

Two different problems I am afraid. The probability of any specific 6 numbers coming up are always the same, be they consecutive or otherwise.

The only thing you can do is try to minimise your chance of sharing a jackpot with other winners. This is because people have patterns in the numbers they pick. Many people pick birthdays for example. By definition these can only range from 1-31 so picking numbers higher than this in your set of 6 increases the chance someone else does not have the same numbers as you. It doesnt mean your set of 6 is any more likely to come up, merely that if they do, you are more likely to be the only person to have those 6 numbers.

the most popular set of numbers is apparently 1 2 3 4 5 6. These are no more or less likely than any others to be picked, but if they are then the jackpot is going to be split between an awful lot of people :shock:

People also dont pick numbers truly randomly, it would be more accurate to say they pick them with an even distribution. The best way to pick random numbers is to pay for a lucky dip. How often have you filled out a ticket with no numbers in mind only to say to yourself 'hmm, got a few low ones, lets pick a high one to even it up'. This is therefore not random but even distribution. The brain loves symmetry and will always try to smooth out an apparently lopsided selection. :roll:


Having said all this I dont actually play the lottery :!: :wink:

Steve.
 

devonwoody

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Go back to sleep Alf.


AH........................Yes but the balls don't come out consecutively, they are arranged con. after they have all been selected. Somehow it shouldn't make any difference?
 

mudman

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It does make me chuckle when everyone has their own little system for picking the numbers. Birthdays, lucky dips, counting the number of times a ball has come up. But it does come down to the 1:a-tad-less-than-14,000,000 is an absolute and the only way it may change is if the balls are not identical (of course they aren't but not enough to matter much) and the differences favoured certain numbers.
I don't play the lottery, the main reason is that the odds are pretty much stacked against me. But also the fact that the sort of luck I get is where even if the odds of myself winning a £10million jackpot approached unity, then there would be either a) another 9,999,999 other people that had a share in the jackpot or b) I have an equally amazing stoke of opposite bad luck that lands me with a bill for £10,000,001 pounds and so actually end up out of pocket. :roll:

Cheers,
Barry.

Who really does think that if not the Universe, then all retailers, really are out to get him.
 

DaveL

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Maybe I should quote a well known fact:

One in a million chances happen nine times out of ten :shock:

I am quite sure lots of you will know where that comes from so no prizes :roll:
 

StevieB

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Ooh, Ooh I know, its from Guards Guards by TP, said by Colon, Nobby and Carrot on top of the Opera house while trying to shoot a dragon with a bow and arrow :D :D

And the anorak award goes to me! I believe there was mention of a Trend T5 going spare...... :wink:

Steve.
 
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