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I'm located in Florida and Play Florida Fantasy5, which has a 5/36 matrix. Total number of combo's for a 5/36 matrix is 376,992.
Of those 376,992 combo's, how many combo's have the same last two digits? (In the case of a single digit number, the number itself is the last digit)
If possible, I'd also like to know how many combo's have same three last digits, four last same digits, and how many combo's have no last digits the same.
I'd like to know the answer to those questions, but I have zero coding skills, and no idea how to mathematically calculate the answers.
Any/all help would be greatly appreciated! Thanks in advance. G5
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Quote: Originally posted by GiveFive on Sep 30, 2021
I'm located in Florida and Play Florida Fantasy5, which has a 5/36 matrix. Total number of combo's for a 5/36 matrix is 376,992.
Of those 376,992 combo's, how many combo's have the same last two digits? (In the case of a single digit number, the number itself is the last digit)
If possible, I'd also like to know how many combo's have same three last digits, four last same digits, and how many combo's have no last digits the same.
I'd like to know the answer to those questions, but I have zero coding skills, and no idea how to mathematically calculate the answers.
Any/all help would be greatly appreciated! Thanks in advance. G5
0 combos with all the same last digit (for obv. reasons)
1380 combos with last digits same 2 numbers
42096 combos with last digits same 3 numbers
184308 combos with last digits same 4 numbers
149208 combos with all different last digits
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I took a different approach, and assumed you're asking about numbers rather than digits.
The last four numbers all the same is the easiest, because if only the first number can change then there are as many possibilities as there are numbers to choose from, which is 36. That's assuming that we're working with the numbers in the order in which they're drawn. If you put them in numerical order then the answer is variable and depends on what the last 4 numbers are, since the first number has to be lower than any of the other 4. If the last 4 numbers are 33,34,35,36 then any of the other 32 numbers could be the first. OTOH, if the last 4 are 2,3,4,5 then 1 is the only possibility for the first number.
Next easiest is having a specific number in 5th place (what I assume you really meant by "same last two digits". In that case the first 4 numbers all have to be combinations made of the other 35 numbers, so C(35,4) or 52,360. The same thing follows for same last 2 or 3 numbers: C(34,3) = 5984 and C(33,2) = 528.
"how many combo's have no last digits the same"
I don't know what the question really is. 11,22, and 33 are the only possibilities for the last number that have the same digits. Of the other 33 possible numbers ten are single digit numbers, so may or may not fit the criteria of "no last digits the same" depending on what you really mean.
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Quote: Originally posted by KY Floyd on Oct 1, 2021
I took a different approach, and assumed you're asking about numbers rather than digits.
The last four numbers all the same is the easiest, because if only the first number can change then there are as many possibilities as there are numbers to choose from, which is 36. That's assuming that we're working with the numbers in the order in which they're drawn. If you put them in numerical order then the answer is variable and depends on what the last 4 numbers are, since the first number has to be lower than any of the other 4. If the last 4 numbers are 33,34,35,36 then any of the other 32 numbers could be the first. OTOH, if the last 4 are 2,3,4,5 then 1 is the only possibility for the first number.
Next easiest is having a specific number in 5th place (what I assume you really meant by "same last two digits". In that case the first 4 numbers all have to be combinations made of the other 35 numbers, so C(35,4) or 52,360. The same thing follows for same last 2 or 3 numbers: C(34,3) = 5984 and C(33,2) = 528.
"how many combo's have no last digits the same"
I don't know what the question really is. 11,22, and 33 are the only possibilities for the last number that have the same digits. Of the other 33 possible numbers ten are single digit numbers, so may or may not fit the criteria of "no last digits the same" depending on what you really mean.
"I took a different approach, and assumed you're asking about numbers rather than digits."
His question is about the last digits of the numbers in a combo, aka the 'ones place' as we usually learn it in grade school. For example, the combo 3-14-24-25-33 has last digits 3-4-4-5-3, which falls under G's category of "same three last digits."
Florida - West Coast United States
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Quote: Originally posted by GiveFive on Sep 30, 2021
I'm located in Florida and Play Florida Fantasy5, which has a 5/36 matrix. Total number of combo's for a 5/36 matrix is 376,992.
Of those 376,992 combo's, how many combo's have the same last two digits? (In the case of a single digit number, the number itself is the last digit)
If possible, I'd also like to know how many combo's have same three last digits, four last same digits, and how many combo's have no last digits the same.
I'd like to know the answer to those questions, but I have zero coding skills, and no idea how to mathematically calculate the answers.
Any/all help would be greatly appreciated! Thanks in advance. G5
I did a such poor job of articulating what I'm looking for please allow me to try again. This time I'll use some examples and maybe I'll be able to get across what I'm looking for!
The last "digit" is the number/integer in the units position of each of Fantasy5's 36 numbers. For instance in the number 16, 6 is the "last digit". For #2, 2 is the last digit.
In this example there are no digits that are identical in the units position ====> 3 12 19 24 35. How many of the total 376,992 combos that make up a 5/36 matrix look like 3 12 19 24 35 and have no numbers in the units position that are identical?
This combo has two numbers in the units position that are the same ====> 10 13 17 20 28 How many combo are in a 5/36 matrix that look like that?
This combo has three numbers in the units position that are the same ====> 5 15 18 27 35 How many combo's are in a 5/36 matrix that look like that?
This combo has four numbers in the units position that are the same ====> 4 14 24 29 34 How many combo's are in a 5/36 matrix that look like that?
I know that combos that have two identical numbers in the units ("last") position are drawn very frequently. Combo's that have four identical numbers in the units position has happened just six times in the 7,382 Fantasy5 drawings that have been held since July 16, 2001.
On June 22, 2016, this combo was drawn in Florida Fantasy5 ===> 1 11 21 31 36. According to The Florida Lottery's Top Prize Winners Report (available as a PDF file for downloading on the Fantasy5 page of their website) there were 17 jackpot winners on 6/22/2016, none of the 17 was a QP, and each winner won $13,001.96. Why am I not surprised by that??? BTW - this was the last time/most recent date it happened! Over 5 years ago!
My ultimate goal is to identify the odds of winning a jackpot by playing the types of combos included in each of the examples above. The overall odds of winning a Fantasy5 jackpot are 1 in 376,992. What are the odds of winning a jackpot by playing only combo's with zero, two, three or four last "digits" being the same? I want a breakout for each type.
THANKS TO ALL OF YOU for your very kind efforts! G5
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Quote: Originally posted by cottoneyedjoe on Oct 1, 2021
0 combos with all the same last digit (for obv. reasons)
1380 combos with last digits same 2 numbers
42096 combos with last digits same 3 numbers
184308 combos with last digits same 4 numbers
149208 combos with all different last digits
This is close to what I'd like to know...except I'm thinking it's backwards.
I would think that there are 1,380 combo's with 4 last digits being the same. Four last digits being the same has happened just six times in 7,382 FF5 dawings. (It's rare)
Combo's with 2 last digits being identical are drawn a lot. (Happens all the time) Therefore 184,308 combo's makes me think that cant be the correct number for 4 last digits being the same. I think it's for two last digits being the same.
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Quote: Originally posted by GiveFive on Oct 1, 2021
This is close to what I'd like to know...except I'm thinking it's backwards.
I would think that there are 1,380 combo's with 4 last digits being the same. Four last digits being the same has happened just six times in 7,382 FF5 dawings. (It's rare)
Combo's with 2 last digits being identical are drawn a lot. (Happens all the time) Therefore 184,308 combo's makes me think that cant be the correct number for 4 last digits being the same. I think it's for two last digits being the same.
Make sense? or no? G5
If I'm correct in my post above,
Then if I play a combo that has two last digits being the same, the odds of winning a jackpot are 1 in 184,308. (Provided a combo is drawn with two last digits being the same)
That's down from 1 in 376,992 or a 48.9% reduction in the odds. G5
Delaware United States
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Quote: Originally posted by GiveFive on Oct 1, 2021
I did a such poor job of articulating what I'm looking for please allow me to try again. This time I'll use some examples and maybe I'll be able to get across what I'm looking for!
The last "digit" is the number/integer in the units position of each of Fantasy5's 36 numbers. For instance in the number 16, 6 is the "last digit". For #2, 2 is the last digit.
In this example there are no digits that are identical in the units position ====> 3 12 19 24 35. How many of the total 376,992 combos that make up a 5/36 matrix look like 3 12 19 24 35 and have no numbers in the units position that are identical?
This combo has two numbers in the units position that are the same ====> 10 13 17 20 28 How many combo are in a 5/36 matrix that look like that?
This combo has three numbers in the units position that are the same ====> 5 15 18 27 35 How many combo's are in a 5/36 matrix that look like that?
This combo has four numbers in the units position that are the same ====> 4 14 24 29 34 How many combo's are in a 5/36 matrix that look like that?
I know that combos that have two identical numbers in the units ("last") position are drawn very frequently. Combo's that have four identical numbers in the units position has happened just six times in the 7,382 Fantasy5 drawings that have been held since July 16, 2001.
On June 22, 2016, this combo was drawn in Florida Fantasy5 ===> 1 11 21 31 36. According to The Florida Lottery's Top Prize Winners Report (available as a PDF file for downloading on the Fantasy5 page of their website) there were 17 jackpot winners on 6/22/2016, none of the 17 was a QP, and each winner won $13,001.96. Why am I not surprised by that??? BTW - this was the last time/most recent date it happened! Over 5 years ago!
My ultimate goal is to identify the odds of winning a jackpot by playing the types of combos included in each of the examples above. The overall odds of winning a Fantasy5 jackpot are 1 in 376,992. What are the odds of winning a jackpot by playing only combo's with zero, two, three or four last "digits" being the same? I want a breakout for each type.
THANKS TO ALL OF YOU for your very kind efforts! G5
The winning combination of 1-11-21-31-36 with so many winners was due to the fact that people pick card patterns to play. They played through the diagonal of the card. There were probably a lot of 4/5 winners who went the other way also using #6. Similar large amounts of winners has happened in other states smaller jackpot games.
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Quote: Originally posted by lottobrain on Oct 1, 2021
The winning combination of 1-11-21-31-36 with so many winners was due to the fact that people pick card patterns to play. They played through the diagonal of the card. There were probably a lot of 4/5 winners who went the other way also using #6. Similar large amounts of winners has happened in other states smaller jackpot games.
Your explanation makes sense to me! What doesn't make sense to me is players actually doing something like that. Given how hard it is to win a jackpot and then to be rewarded with just $13,000 for doing it. Crazy.
Interestingly enough, on May 16, 2015 this combo was drawn ====> 2 12 22 27 32. Just four jackpot winners of $64,652.98, none being a QP.
On March 17, 2013 1 6 11 21 31 was drawn, but this time there were two QP winners among the seven (total) player picked winners.
August 18, 2015 saw 2 5 15 25 35 be drawn. One QP winner was among the nine player picked jackpot winners that day. G5
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Quote: Originally posted by cottoneyedjoe on Oct 1, 2021
0 combos with all the same last digit (for obv. reasons)
1380 combos with last digits same 2 numbers
42096 combos with last digits same 3 numbers
184308 combos with last digits same 4 numbers
149208 combos with all different last digits
I'm guessing that this would be the correct number of combo's for each situation;
1,380 combo's for 4 last digits being the same.
42,096 combo's for 3 last digits being the same.
149,208 combo's for 2 last digits being the same.
184,308 combo's for zero last digits being the same.
The numbers of combo's for each situation totals 376,992, (that's the correct number of combo's for a 5/36 matrix) so it's just a matter of which number of combo's is assigned to which number of last digits being the same. G5
bgonçalves Brasil
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hello give five, let's try this approach the initial digits
ex= draw= 01 11 21 31 36= initial digit
gave = 1,1,2,3,3
five do you know how many formations you need for the 4 formations the 5th position will be chosen at random from
28 to 36
know how many in the first 4 positions at 100% only 35 formations drawings ascending order
I repeat the 5th or last position will be chosen from 28 to 36 in the random
here are the 35 formations of 4 positions at 100% check it out
you will see a delay to play 5% to 10% of these 35
of course after the work will be the finishing or last digit from 0 to 9, when you win don't forget the dr
hey the only 35 gold
0 0 0 0
0 0 0 1
0 0 0 2
0 0 0 3
0 0 1 1
0 0 1 2
0 0 1 3
0 0 2 2
0 0 2 3
0 0 3 3
0 1 1 1
0 1 1 2
0 1 1 3
0 1 2 2
0 1 2 3
0 1 3 3
0 2 2 2
0 2 2 3
0 2 3 3
0 3 3 3
1 1 1 1
1 1 1 2
1 1 1 3
1 1 2 2
1 1 2 3
1 1 3 3
1 2 2 2
1 2 2 3
1 2 3 3
1 3 3 3
2 2 2 2
2 2 2 3
2 2 3 3
2 3 3 3
3 3 3 3
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Quote: Originally posted by GiveFive on Oct 1, 2021
I'm guessing that this would be the correct number of combo's for each situation;
1,380 combo's for 4 last digits being the same.
42,096 combo's for 3 last digits being the same.
149,208 combo's for 2 last digits being the same.
184,308 combo's for zero last digits being the same.
The numbers of combo's for each situation totals 376,992, (that's the correct number of combo's for a 5/36 matrix) so it's just a matter of which number of combo's is assigned to which number of last digits being the same. G5
Unfortunately, the computation I did for a slightly different question cannot be salvaged to answer your actual question. You're also missing a few edge cases, namely, what do you consider it when the last digits split into two pairs and a fifth different digit, and what do you consider it when the last digits split into a triple and a pair. There are actually 6 mutually exclusive cases for last digits:
A B C D E <- none the same
A A B C D <- one pair
A A B B C <- two pairs
A A A B C <- one triple
A A A B B <- full house
A A A A B <- one quadruple
It's not much different than writing a script to evaluate poker hands. I can do it later when I have more time if nobody else gives you the answer first. Feel free to PM me.
bgonçalves Brasil
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hello five, of course within the 35 formations of the inicail digit of the 5/36
there is a formation that will never be drawn ex = 0000 or 3333 very difficult to be drawn you will be able to reduce it even further to 15 to 20 formations! wonderful!! the last or 5th number is random from 28 to 36, well we have the hard work of the last digit from 0 to 9 in each position
