Since you already know 5 numbers are drawn, I really don't see the difference in expressing it as being draw... [ More ]]]>

A person waiting to be struck by lightning isnt going to go out into the middle of a desert. They know theres a better chance of being hit by going to the right place at the right time - raining season.]]>

So while several people try to convince you it's just luck.........keep doing what your doing! I'd be willing to bet they've never looked at past draws. Everyone I know buys QPs o... [ More ]]]>

I suspect that your right.

Several people have told me that one cannot pick winning numbers since all games are random. It's just luck they tell me.

That to me is a challange.

So I'd like to learn as much as possible to be able to have better odds of the numbers I choose hitting instead of just spending money. I look at it as an investment in my future. That my sound odd, but then I've been told worse.

Hope your luck is beter than mine...]]>

1::5, or 20%, on the first ball.

1::4, or 25%, on the second ball.

1::3, or 33%, on the third ball.

1::2, or 50%, on the fourth... [ More ]]]>

I'm follow your explanation so far. I understand the formula:

Specific number drawn so many times [Divided by] Total number of draws. (this is a standard formula)

This is where my mind is spinning; it takes 5 of 56 and 1 of 46 which is the complete draw. But, the complete draw comes from 2 different bins, 56 white and 46 red, seperate systems. (with me so far?) The first bin needs a 5 ball/number draw to complete that stage of the draw.

The way mind is looking a... [ More ]]]>

Players may pick six numbers from two separate pools of numbers - five different numbers from 1 to 56 and one number from 1 to 46 - or select Easy Pick. You win the jackpot by matching all six winning numbers in a drawing.

What if you win the jackpot?

Annuity option: Provides 26 annual payments. For every $1,000,000 in the jackpot, you will receive approximately $38,500 per year before taxes.

Cash option: A one-time, lump-sum payment that is... [ More ]]]>

To continue with the 5/56 example let us assume (yeah, I know the joke) that a number you didn't select was drawn first. The odds of your number being drawn on the very next ball draw have now been reduced from 1::56 to 1::55. But that's also true for any other undrawn number. What has been done, though, is to change the matrix for the remaining balls from a 5/56... [ More ]]]>

I understand what you're describing and the traditional way of figuring the percantage, but does my question make any sense? Would calculating the 1/5 of a chanced pick provide a better or more realistic percentage?

Thanks for the intrest.]]>

But, yes, any given number has a 1 in 56 chance of being drawn first. If that number isn't drawn, it has a 1 in 55 chance of being drawn second, and so on. The odds are lower and the percentages are higher on any particula... [ More ]]]>

I understand that concept, but, wouldn't a percentage also go down according to the drawn postion, ie 1/50, 1/49, 1/48, etc. (100%, 80%, 60%, etc.), or does each number have a 100% chance of 5/50 of being drawn?

Or would the percentage chance of being drawn go up from 20%, 40%, 60%, etc? For instance, the first draw would be 20% of the total draw, 2nd draw would be 40%, etc., since there are 5 numbers drawn to complete the 5/50. 5 drawn numbers would be 100%.

Anyway, that's the ci... [ More ]]]>

We know a number can be drawn 0 to 1 times per draw. And over N number of draws that number can occur 0 to N times. That's the way I look at numbers over range N number of draws. Some numbers will be picked more than others.

But then looking at numbers for each draw, say for eg a 5/50 draw, the theory is that each number has an equal chance of being drawn ie 1/50. After the first number has been... [ More ]]]>

Your correct, each of the 5 white balls can only be chosen once per draw.

The counts are the number of times that the white ball has been chosen, divided by the number of daws that have occured. That's why I started thinking that a correct percentage would be to calculate the 1/5 percentage for each drawn ball per drawing.

That's why I'm wondering if I'm over thinking the calculation/formula or, would this type of calculation actually be more helpful. I don't have very muc... [ More ]]]>

So if #3 was drawn 6 times out of 30 draws that's it - there is no potential for it to be drawn 7, 8, 9 etc times unless the ball was drawn in different draws. Therefore /5 doesn't seem neccessary.]]>

I realize that that formula is correct, I'm using it now.

I've begun to wonder about this though. If you start off with 100%, then take 5 from that, doesn't that make each 1 of the 5 drawn balls 20% of the whole draw? (You couldn't include the MegaBall or PowerBall in this calculation since they are drawn from different bills). So each number drawn isn't 100% in a calculation, but, actually only 20% of the whole draw. Wouldn't a calculation reflecting this give a more re... [ More ]]]>

I don't understand what it is you are trying to do in your first statement.

as far as calculating a percentage. a percentage is simply the ratio of two numbers expressed as parts of 100, and your first method is correct.

the probability that a specific number will come up is totally different in the way that it is calculated.

p8]]>

I hope someone is able to answer this question for me.

I'm attempting to calculate the percentage of times that a number shows up in a drawing verses the number of times the same number has showed up in the history of the draws.

For instance: Power Ball Game Example: 5 white balls and 1 red ball.

I'm counting the number of times each number ball is drawn, then dividing that number by the toal number of draws. For instance, the number 3 white ball shows up 6 times out of]]>