Calculating a percentage

KnuckleHead's avatar - box

Morning all,

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 30 drawings. My formula would look like; 6/30. The answer would be 20%.

What I'm wondering is, should the be calculation be: (6/30)/5 ?  The "/5" being the number of chances that the number 3 ball call show up in a single draw? If this calculation is correct, than the answer would be 4%.

Would someone clear this calculation up for me? Thanks for any insights...

In response to KnuckleHead


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.


KnuckleHead's avatar - box
In response to phileight

Evening phileight,

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 realistic percentage figure?

I don't know...maybe something like:

((specific drawn number count/total number of draws)/(1/5))

Would something like this be closer to a realistic percentage?

Who knows, my original formula is right and I'm using it, but in an attempt to get at a more realistic percentage, am I just over thinking it to much...

Any insite into this would be greatly appreciated, and thanks too for the patience while looking into this...


I'm not sure about the format for Powerball but isnt it the case a number can only be drawn once per draw?

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.

KnuckleHead's avatar - box
In response to Newb

Morning Newb,

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 much experience in these types of calculations. I'm still new at this and attempting to figure it out.

Thank you for asking.


I think I see where you are coming from but I suspect it comes down to what really looking for.

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 drawn the second number has a 1/49 chance of being drawn because there are 49 balls left to draw. The third number 1/48 etc...

KnuckleHead's avatar - box


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 circle my mind is going through asking questions... Now, keeping in mind that I could go out and purchase a program that would work very well, but, I like a challange in figuring out how to get things working and creating it myself...

Grandpa once told me that the only DUMB question was the question you didn't I apologize if my questions seem strange or dumb.

And thanks again for the insights. You may not think so, but answers are helping me to comprehend this better.

johnph77's avatar - avatar

In MegaMillions, any particular number has a 5 in 56 chance of being drawn. It doesn't make any difference in what order it is drawn, as long as it is drawn. The only thing this sort of thing will accomplish is to add suspense to the draw, if you're into that sort of thing.

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 particular number being drawn as the drawing is conducted.

But keep this in mind - when those five numbers are drawn, they could also have been drawn in 120 different orders. For instance, if the drawn numbers were, in order, 01, 02, 03, 04 and 05, the order of the draw could also have been 05, 04, 03, 02 and 01 without changing the final result.

Incidentally, there are 3,819,816 possibilities in a 5/56 matrix, and each number from 1 to 56 will appear exactly 341,005 times when all possibilities are listed.

KnuckleHead's avatar - box
In response to johnph77

Hello johnph77,

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.

johnph77's avatar - avatar
In response to KnuckleHead

Not really. Knowing which number is drawn first isn't going to have any effect when the order of drawn numbers isn't of importance.

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 matrix to a 4/55 matrix. And that's true no matter if the first ball was a number you selected or not.

If the order draw were to became a factor in MegaMillions, then the odds of matching the five balls drawn in the main draw would increase from 1::3,819,816 to 1::458,377,920, a factor of 120 - and the bonus ball draw hasn't been brought into consideration yet.

There aren't many lotteries conducted anywhere at the present where, when balls are drawn from a single pool, the order of appearance of the numbers is a factor.

pick4hawk's avatar - Trek HAND3.gif

Mega Millions tickets cost $1.00 per play.

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 equal to all the cash in the Mega Millions jackpot prize pool.

In addition to the jackpot, there are other prizes ranging from $2 to $250,000*.

KnuckleHead's avatar - box
In response to johnph77

Hello againjohnph77,

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 at this, is that, this equals 100%.  5 divided by 100 equals 20%. Wouldn't each number of the 5 balls/numbers drawn then have a 20% chance of being drawn from 100%? Therefore each ball starts with a 20% chance and when all 5 are drawn, you would then have the 100%. Would adding in the formula to divide the 5 white ball draws into 20% for each, then continuing with the standard formula, give a more accurate final figure?

Wouldn't this final figure, while being smaller, actually give a larger percentage when compared with the other drawn numbers, therefore giving you a better idea of which number may show up in the next draw?

Since the PowerBall is 1 of 46, you have only 1 chance that that number would show up. Therefore the standard formula works for this, the compeletly seperate second bin draw.

I do appreaciate your interest and patience, Thank you.

johnph77's avatar - avatar
In response to KnuckleHead

The fallacy in this argument is that, if one assumes there will be five numbers drawn, that would equal 100% of the numbers. That would only be true if the five numbers to be drawn were already known or predetermined, or if there were only five balls to be drawn. Then your argument is valid, and the odds for drawing a specific number from that draw would be:

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 ball
1::1, or 100%, on the fifth ball.

Also if the draw was predetermined, then, of course, you have a 100% chance of drawing any of the five numbers on each and every draw.

But we know that that is not true. There are 56 numbers from which to draw.

The odds of drawing one of the five numbers you selected on the first draw is 5::56.

If none of your numbers appeared in the first draw, then the odds of one of your numbers appearing in the second draw is now 5::55, since one of the 51 numbers you didn't select appeared in the first draw. But - if one of your five picks did appear, then the odds of one of your four remaining picks appearing in the second draw is 4::55. In other words, by being successful in your first pick your odds for success in the second draw have increased by 20% and the percentage of success, thin as it may have been to start with, decreased by the same amount.

This chain goes on. I won't hog bandwidth and time by listing all the odds in the subsequent draws with the varying degrees of previous successes possible, but the figures should be apparent.



chattanoogadog's avatar - disney28

I suspect the people that win these things have nothing to do with web site.... "Just play"

KnuckleHead's avatar - box

Thank you johnph77, I believe I understand.

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