Can anyone calculate the odds of this for me...?

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helipilot

Well-known member
Jan 20, 2010
337
Damn I need to get a life... to continue. Those were the odds of 2 and 11 on 3 lines. For any 2 numbers repeated on 3 lines it is:

1st line.. 1.000 as any 6 numbers come out
2nd line .... has to be one of 6 numbers so 6/50 + 6/49 + 6/48 + 6/47 + 6/46 + 6/45 for the first number = 0.75887 then has to be one of 5 numbers so multiply by 5/49 + 5/48 + 5/47 + 5/46 + 5/45 = total chance of 2 numbers the same on line 2 = 0.4040
3rd line... has to be one of 2 numbers now so 2/50 + 2/49 +2/48 + 2/47 + 2/46 + 2/45 = 0.25296 and 1 of 1 number 1/49 +1/48 +1/47 +1/46 + 1/45 = 0.1065 total for line 3 = 0.002868

For all 3 lines to have the same 2 numbers = 1.000 x 0.4040 x 0.002868 = 0.0011587 or 1 in 863 chance

Night all zzzzzzzzzzzzzzzzzzzzzzzzz
 




DTES

Well-known member
Jul 7, 2003
6,022
London
Damn I need to get a life... to continue. Those were the odds of 2 and 11 on 3 lines. For any 2 numbers repeated on 3 lines it is:

1st line.. 1.000 as any 6 numbers come out
2nd line .... has to be one of 6 numbers so 6/50 + 6/49 + 6/48 + 6/47 + 6/46 + 6/45 for the first number = 0.75887 then has to be one of 5 numbers so multiply by 5/49 + 5/48 + 5/47 + 5/46 + 5/45 = total chance of 2 numbers the same on line 2 = 0.4040
3rd line... has to be one of 2 numbers now so 2/50 + 2/49 +2/48 + 2/47 + 2/46 + 2/45 = 0.25296 and 1 of 1 number 1/49 +1/48 +1/47 +1/46 + 1/45 = 0.1065 total for line 3 = 0.002868

For all 3 lines to have the same 2 numbers = 1.000 x 0.4040 x 0.002868 = 0.0011587 or 1 in 863 chance

Night all zzzzzzzzzzzzzzzzzzzzzzzzz

But that's just that any two numbers are on there isn't it? We need them specifically to be the first two numbers when written in numerical order... which I think is a whole another thing?
 


Tom Hark Preston Park

Will Post For Cash
Jul 6, 2003
72,380
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?

Tonight's numbers were... 11, 20, 30, 35, 45

https://www.national-lottery.co.uk/player/euromillions/results/prizeBreakdown.ftl

Get checking those tickets! :lol:
 




helipilot

Well-known member
Jan 20, 2010
337
But that's just that any two numbers are on there isn't it? We need them specifically to be the first two numbers when written in numerical order... which I think is a whole another thing?
Thats very true.... somewhere between the 2 is all I'm going to say now!!
 




Baldseagull

Well-known member
Jan 26, 2012
11,839
Crawley
The machine that generates lucky dip tickets does not generate truly random number sets, it is not uncommon for two numbers to appear on multiple lines purchased at the same time.
 


Stat Brother

Well-known member
NSC Patron
Jul 11, 2003
73,888
West west west Sussex
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.
Not my stat but it came up recently with the big US prize.

But I don't have to justify it, as you have in your second paragraph :facepalm:
 


beorhthelm

A. Virgo, Football Genius
Jul 21, 2003
36,029
How do you justify that? Makes no sense that randomly chosen numbers are more likely to win than numbers chosen by any other means.

agree it doesnt sound right, but ive been thinking about it and it could be due to the random number generator picking consequtive numbers or more numbers higher than 30 (people often use their date of birth), or patterns of numbers that we dont pick as we think thats less likly to win.
 




helipilot

Well-known member
Jan 20, 2010
337
Ok for the first 2 numbers being the same on all three:

Line 1: Any 6 numbers and there will always be two lowest! Therefore 1.0000
Line 2: 6 chances to hit the first number and six chances to hit the second number. Therefore (1/50 + 1/49 + 1/48 + 1/47 + 1/46 + 1/45)squared = 0.015997
Line 3: Must be the same at line 2 = 0.015997

Total probability = 1.00 x 0.015997 x 0.015997 = 0.0002559 or 1 in 3907 chance.
 


8ace

Banned
Jul 21, 2003
23,811
Brighton
But that's just that any two numbers are on there isn't it? We need them specifically to be the first two numbers when written in numerical order... which I think is a whole another thing?

You also need to check all the other numbers are >11.
 


8ace

Banned
Jul 21, 2003
23,811
Brighton
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?

This is true, they probably do a qsort on them before they print them out.
 




Mr Smggles

Well-known member
May 11, 2009
2,671
Winchester
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.

Thats completely wrong. The chances of 1,2,3,4,5,6 to come up are exactly the same as 6 random numbers. You may win less because I would imagine quite a few people play 1,2,3,4,5,6 as their numbers but the odds of them coming up, are just the same.
 


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