I don't often do the lottery so this maybe not so unusual but I thought if I bought 3 lucky dip euro-millions lines the numbers are completly random. I bought a ticket with 3 lines on it all lucky dip and on all 3 lines the first two (ie lowest numbers were 02 and 11. I know there are only 5 numbers but this appears to be extremely unlikely. Any mathematicians out there?
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Can anyone calculate the odds of this for me...?
- Thread starter Gangsta
- Start date
You are more likely to get lottery winners with a lucky dip facility.
Which is why our lotto went for the best part of 2 years without one.
More than enough time for everyone to select their numbers, and have a few big roll overs, too.
Which is why our lotto went for the best part of 2 years without one.
More than enough time for everyone to select their numbers, and have a few big roll overs, too.
- Thread starter
- #4
Yeah thats how they came out. I guess its probably not that weird but it just looked weird. Did a double take and walked on. Just wondered what the odds were?
Yeah so what's the range of numbers. 1-49?
- Thread starter
- #6
Yes 1-49. 5 digits a line
I think the Euromillions goes 1-50. UK lottery is 1-49?
- Thread starter
- #9
Yes sorry you maybe right I dont do it that often and I had 49 on one line so thought it may be the same. Looked on website it does indeed go to 50
There are roughly 1,175 possible different combinations of the first two numbers (1,225 ways of picking 2 numbers from 50, but some aren't possible as they don't leave enough numbers left to fill the remaining three, given that they're in numerical order). On that basis, the chance of three lines all having the same first two numbers is 1 in 1,175^2 = 1,380,625
The complication is that each of those 1,175 combinations has a different probability, some being thousands of times more likely than others... so it isn't anywhere near that simple.
The complication is that each of those 1,175 combinations has a different probability, some being thousands of times more likely than others... so it isn't anywhere near that simple.
- Thread starter
- #13
DTES, this looks like great stuff. So are you saying that the answer is 1,380,625 or is it not?
No, sadly not that simple. In practice probably slightly more likely than that as this includes combinations like 40 41 X X X which are far less likely than 1 2 X X X - this completely skews the answer
Another example - if you take the first line as drawn with 2 11 X X X (like your example) then the chance of getting another two of them is only in 53,748. (9,139 of the 2,118,760 combinations begin with 2 and 11, then square it). Still bloody small, but nowhere near as small as the number above. Each combination of 2 numbers will have a different probability...
Bugger, how to generalise this to get the overall chance...... I really don't know, sorry
- Thread starter
- #15
No, sadly not that simple. In practice probably slightly more likely than that as this includes combinations like 40 41 X X X which are far less likely than 1 2 X X X - this completely skews the answer
Another example - if you take the first line as drawn with 2 11 X X X (like your example) then the chance of getting another two of them is only in 53,748. (9,139 of the 2,118,760 combinations begin with 2 and 11, then square it). Still bloody small, but nowhere near as small as the number above. Each combination of 2 numbers will have a different probability...
Bugger, how to generalise this to get the overall chance...... I really don't know, sorry
How can certain strings of numbers be less likely than others - they all have the same chance of coming out? I think you're right about 53,748 - that sounds about right. But then what are the odds of them being the first two numbers in each line as on my ticket? That is a different question.
You need to get exactly a 2 on one line out of 6 numbers so the odds are:
1/50 + 1/49 + 1/48 + 1/47 + 1/46 + 1/45 = 0.2648
You then need a 11 too so need to square this... giving 0.0160
To get this on 3 lines you need to cube this number giving 0.000004093 or 1 in 244,275 chance I reckon.
1/50 + 1/49 + 1/48 + 1/47 + 1/46 + 1/45 = 0.2648
You then need a 11 too so need to square this... giving 0.0160
To get this on 3 lines you need to cube this number giving 0.000004093 or 1 in 244,275 chance I reckon.
- Thread starter
- #17
I am in no position to argue with that. There are plenty of numbers in your calculation and fractions so thats good enough for me. Im off to bed.
Dirk Gently
New member
- Dec 27, 2011
- 273
How do you justify that? Makes no sense that randomly chosen numbers are more likely to win than numbers chosen by any other means.
In fact, you're more likely to have a big win with lucky dip numbers, as you're less likely to share it with others since people choosing numbers tend to pick from a smaller range - but you're just as likely to win regardless of how the numbers are picked.
Notters' Dad
Invicta
Although they may not have been originally, randomly drawn in that sequence. Just printed in the logical sequence once drawn. Which may(?) add a further layer of complexity?
Every combination of 5 numbers has exactly the same chance of coming out - but when you write the numbers on the ticket, they are in numerical order. To give the extreme cases as an example - there are more combinations of 1 2 X X X (any other numbers can fill the Xs) than there are of 46 47 X X X (there is only one combination here - 48, 49 & 50).