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Back to Basics (Dice)

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Before I listen to somebody claiming they know the right way to predict lottery numbers, they have to prove to me they can predict the next roll of the dice.

 

That's right.  And I'd put a dollar on every dime Bill Gates has (as if I could...wouldn't be on this web site if I did) if there is anybody that can predict the next roll of the dice.

 

This is all I know about dice:

1. There is a 1 in 6 chance of rolling a one (for example).

2. There is a 1 in 36 chance of rolling a one twice in a row (1:6 X 1:6), etc.

 

This does not mean that rolling a one on the second roll is less likely to happen just because a one was rolled on first roll.  The odds of getting a one on the second roll are still 1:6.  Each roll is independent of the other rolls.  They cannot be related by mathematics in any way.

 

If there is anybody out there that disputes this, please point me in the direction of the approved college level text book you used to learn otherwise.

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Interesting.

I would like to see what everyone else has to say, first.

My statements to follow.

Thoth's avatar - binary

Did someone on this site specifically state that a lottery combo will not be drawn again just because it was the last number drawn or was just drawn recentently??

You are correct in stating that each throw of a die has the same odds of producing the same digit.  If you toss the die once and you wager on the five, your odds are 1 in 6.  If you wager on the five for the next throw your odds, again, are 1 in 6.  The probability of both tosses resulting in two fives is 1/6 X 1/6 = 1 in 36.

The probability to toss a five ten consecutive times in a row is 1/6^10 = 1 in 60,466,176.  The reverse of that event is tossing the die ten times in a row and not landing a five even once.  The probability of that occuring is 1/5^10 ÷ 1/6^10, which equals 16.15%.  Obviously, this means the chance of having AT LEAST one five in ten tosses is 83.85%

If you toss the die and it lands on five, there IS a 50% that another five will be thrown within the next 4 tosses.  Also, your chance to throw at least one five within 51 tosses is 99.99%

This is the mathematics that IS releated to our lottery, as well as your dice throwing game.  Analyze a series of several thousand truly random dice throws and these things will prove to be true.

 

 

 

Thoth's avatar - binary

"Each roll is independent of the other rolls.  They cannot be related by mathematics in any way"

 Also, I must add that the two rolls are mathematically related through combinatorics and the laws of probability.  The odds of guessing two consecutive die rolls (before the first takes place) is 1 in 36, because there are 36 possible outcomes...or 36 permutation of the two dice.  If you did not care about the order of the two dice within the group of two, then there would be six doubles (1-1, 2-2, 3-3, 4-4, 5-5, 6-6) and 15 "combinations" where the order does not matter, i.e. - a roll of a 4 then a 5 would be the same as a roll of 5 then a 4.

The odds of rolling 5-4 OR 4-5 would be 36/2 = 1 in 18.  In 81 trials there is a 99% chance that the 4 will follow the 5 OR the 5 will follow the 4.  As a possible system for throwing dice (if such a system exists), you could opt to only wager a 4 when a 5 is thrown, or a 5 when 4 is thrown, but only if any combination of 4 and 5 has not been thrown since the start of the 81 game trial.

Another real life test that could be done is to watch how each of the 15 dice combos repeat within 12 to 13 trials 50% of the time.  In this case, the key would be to predict two future dice rolls at once, instead of basing one throw on the last.

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Hmmm...

 

I just rolled a die 13 times.  The first 12 numbers, in order, were 1, 3, 6, 2, 6, 6, 3, 1, 4, 5, 5, and 5.

 

Using your combinatorics and probability, what was the 13th number?

 

You would impress me if you had a formula that not only worked with this set of numbers, but also the next set of numbers I have not yet thrown.

Thoth's avatar - binary

First of all, I never claimed to be able to predict numbers, let alone the exact outcome of the next roll of your die.  The mathematics of combinatorics does play a huge role in lotteries as well as dice throwing.  While it is impossible to consistently predict the next number to be drawn, their are many things that probabilty will tell you besides the "odds". 

Even in I wanted to try to predict dice rolls, I would still want a much larger history than your 12 throws.  If I was forced to choose I would go with a 1 or 4 - based on the 50% repeat chance, although that could easily be a fourth 5 as well. 

If you chose to roll the dice 12 times because I stated that "...each of the 15 dice combos repeat within 12 to 13 trials 50% of the time," then I should clarify that one trial is equivelant to 2 rolls of the dice.  So, if you was trying to guess two consecutive throws at a time and the dice was actually thrown 20 times, there would only be "10 trials" so to speak.

I am in agreement that there is no formula to tell you the next outcome of an event, whether it be lottery or dice, but probabilty does work within a measured amount of trials.

RJOh's avatar - chipmunk

The way I see it, if any one could really come close to predicting the winning combination of any lottery, they could just post a scanned image of a winning ticket.  That would convince me that they could pick a winner.

Amazing Grace's avatar - lion

3

truecritic's avatar - PirateTreasure

You would impress me...

Why is someone supposed to impress you?

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Amazing Grace, 

It wasn't 3 but any one of the next 5 people might get it right.

 

truecritic,

Did I hurt your feelings?  My family doesn't have to impress me.  My friends don't have to impress me.  My next door neighbor doesn't have to impress me.  My co-workers don't have to impress me.  But, an on-line board of anonymous lottery prognosticators MUST ABSOLUTELY impress me before I would believe anything any of them had to say.

FYI:  One of the synonyms of impress is to affect.  How could I not be affected by somebody that could predict the next roll of the dice?  Since I know that the next roll cannot be predicted, I will never be impressed.  You see, you aren't supposed to impress me because you can't.

truecritic's avatar - PirateTreasure

Quote: 

"My family doesn't have to impress me.  My friends don't have to impress me.  My next door neighbor doesn't have to impress me.  My co-workers don't have to impress me.  But, an on-line board of anonymous lottery prognosticators MUST ABSOLUTELY impress me..."

Maybe I should rephrase the question...who cares if you are or aren't impressed?

Click Resources->Search for something. Do some reading and just make up your own mind if you want to join in.

for99c's avatar - Lottery-058.jpg

4

Thoth's avatar - binary

Through combinatorics you can deduce the probability of all outcome patterns for throwing a die six times.  You could take the chart below and break it down still further and allow a letter to represent a particular digit. 


 

 DICE   BASE COMBOS    WAYS  PERMUTATIONS    Chance

AAAAAA        6                1            6        0.01%
AAAAAB      30                6          180        0.39%
AAAABB      30              15          450        0.96%
AAAABC      60              30        1,800       3.86%
AAABBB      15              20          300        0.64%
AAABBC      120              60        7,200       15.43%
AAABCD       60              120        7,200       15.43%
AABBCC      20              90        1,800       3.86%
AABBCD      90              180       16,200       34.72%
AABCDE      30              360       10,800      23.15%
ABCDEF        1              720          720        1.54%
 
                                 Sum:  46,656
                                 6^6 = 46,656

Thoth's avatar - binary

Always losing the format...grr

RJOh's avatar - chipmunk

Hmmm...

 

I just rolled a die 13 times.  The first 12 numbers, in order, were 1, 3, 6, 2, 6, 6, 3, 1, 4, 5, 5, and 5.

 

Using your combinatorics and probability, what was the 13th number?

 

You would impress me if you had a formula that not only worked with this set of numbers, but also the next set of numbers I have not yet thrown.

When I look at your previous results, I am confident that your next number will be a repeat of one of the earlier rolls. The odds of a correct guess with no formula is 1:6, however with more observations one probably could reduce those odds to 1:4 or a 50% improvement.  Over a short period of time even in a random environment, certain outcomes will appear to be favored and it's those types of treads that the system lottery players are looking for. By elimination the results that are least likely to happen, one can improve one's odds of guessing correctly what will happen.  In your dice rolls, eliminating two possible outcomes will improve the odds of a correct guess by 50%.  No system can reduce the odds of guessing correctly to 1:1 and everyone old enough to play lotteries knows it's unrealistic and foolish to expect such results.

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