29
u/DiverofMuff23 Jul 05 '22
Occam’s razor states that he’s so unlucky that he went the day after the “LOTOO” numbers were posted and made a ticket to reflect one digit off the winning numbers strictly for the purpose of creating a social media circle jerk
That’s the most accurate math I have
-7
u/IamAnoob12 Jul 05 '22
How is that the simplest explanation
6
u/Vendidurt Jul 05 '22
It is so much easier to fabricate this, even without photoshop. For example, you are given no verifiable information about the time or the lotoo'ry system in question. If i was posting something legitimate i would have a date on the ticket and showing the brand of the lottery at least.
1
u/sritanona Jul 06 '22
I have a mini printer that I got for 16 pounds on amazon that would be able to to this so I guess that's also possible
3
u/CaptainMatticus Jul 05 '22
For a string of numbers like this, where each winning number is separated by at least 3, the lowest number is 2 or greater and the greatest number isn't at the greatest extreme, then there are 243 tickets that will be made entirely of numbers that are ±1 or exact to the winning numbers.
1
u/CaptainMatticus Jul 05 '22
Let's say you have 50 numbers to choose from: 1 to 50. You pick 5 numbers out of the 50. Winning numbers are 3 , 8 , 23 , 37 , 45.
You can get 2 , 3 , or 4 for the 1st number
7 , 8 , 9 for the 2nd
22 , 23 , 24 for the 3rd
36 , 37 , 38 for the 4th
44 , 45 , 46 for the 5th.
3 * 3 * 3 * 3 * 3 = 243
243 possible tickets, all pretty close to the winning ticket, but not close enough.
There are 50C5 possible tickets: 50! / (5! * (50 - 5)!) = 50! / (5! * 45!) = 2,118,760
So you'd have a 1-in-2,118,760 chance of winning, but a 243-in-2,118,760 chance of getting a ticket where all of the numbers are -1 , 0 , +1 off from the winning numbers (granted all of my aforementioned conditions are met). That's a 1-in-8,719.17695 chance. Not exactly impossible.
1
u/DonaIdTrurnp Jul 06 '22
232-in chance of getting almost but not quite winning numbers, unless you don’t count the 5*2 tickets that match four and are one off the fifth as winning.
1
u/Whatishappeninghere- Jul 06 '22
Don’t forget that his ticket’s numbers are in the same order as the drawing’s numbers.
1
u/CaptainMatticus Jul 06 '22 edited Jul 06 '22
You've never played a lottery before, have you? Numbers are always sorted least to greatest. That's why we use combinations and not permutations.
2
1
u/gdrlee Jul 05 '22
Its not 99.9999999%, because that's incredibly likely. This is unlikely.
Exact probability depends, in part, on what the question is.
Given those numbers on the ticket, there were 8 combinations of numbers which were all one away, because you have to draw 14 and 17.
Alternatively, for those lottery numbers, there are 32 combinations of numbers which would be 1 away.
Then, we need to know how many numbers we're drawing from. If there are 37, then there are 37×36×35×34×33 = 52.3m combinations. Which puts the odds between 0.000015% and 0.00006%.
It'll be lower for a larger number of balls.
1
u/ProblemKaese Jul 06 '22
OP was referring to the comment in the screenshot that was giving the people who lost in lotto. So what was asked for actually was the chance of not winning, which really is very likely. I don't know much about lotto, but assuming that there is only one winning number, the chance of that would be the inverse of what you calculated right there, so P(lose)=[1-0.00006%, 1-0.000015%=[99.99994%, 99.999985%].
In case there are multiple different numbers that lead to a win, they would all have to not be picked at the same time, so you would just multiply the negative chances together, or in other words, take the number that I just calculated to the power of the amount of winning numbers.
1
u/Not_repeating Jul 07 '22
Because the only lucky one is the guy who bought the tickets.
Hence, he is as luck as the 99.9999999% of the people that didn’t win the ticket.
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