Calculating a percentage

Reply #19

Stack47 — Sat, 09 May 2009 21:16:16 GMT

The number 3 can only be drawn 30 time in 30 drawings even though it has 5 chances of being drawn in each drawing. If it was drawn first in 3 of the 6 times it was drawn, there would be no chance of it being drawn in the other 4 chances because there is only one number 3. In pick-3 and pick-4 games where there are multiple number 3s, it could be drawn more than 30 times in 30 drawings.

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

Reply #18

Newb — Sat, 09 May 2009 11:19:55 GMT

That's true but its good now that even an a suppossedly random game you can pick out patterns and systems that help you pick numbers.

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.

Reply #17

ca-dreamin* — Fri, 08 May 2009 23:42:11 GMT

Not to get off topic but here goes...... I used to enter sweepstakes. For sending in one 3 by 5 piece of paper with my name, addy, phone number, etc. I won a 42 plasma TV with a surround sound DVD player and a check for 200 dollars. Yet everyone I know says that sweepstakes aren't real and nobody wins those. REALLY?

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 ]

Reply #16

ca-dreamin* — Fri, 08 May 2009 23:13:05 GMT

I know alot of people that just play and they aren't winning more than a couple dollars here and there.

Reply #15

KnuckleHead — Fri, 08 May 2009 02:50:07 GMT

Good evening chattanoogadog,

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...... [ More ]

Reply #14

KnuckleHead — Fri, 08 May 2009 02:37:40 GMT

Thank you johnph77, I believe I understand.

Reply #13

chattanoogadog — Fri, 08 May 2009 00:46:21 GMT

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

Reply #12

johnph77 — Fri, 08 May 2009 00:40:18 GMT

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... [ More ]

Reply #11

KnuckleHead — Thu, 07 May 2009 19:18:29 GMT

Hello again johnph77,

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 ]

Reply #10

pick4hawk — Thu, 07 May 2009 17:15:41 GMT

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... [ More ]

Reply #9

johnph77 — Thu, 07 May 2009 16:59:33 GMT

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... [ More ]

Reply #8

KnuckleHead — Thu, 07 May 2009 16:21:29 GMT

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.

Reply #7

johnph77 — Thu, 07 May 2009 04:53:07 GMT

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 particula... [ More ]

Reply #6

KnuckleHead — Wed, 06 May 2009 15:16:34 GMT

Newb,

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 ]

Reply #5

Newb — Wed, 06 May 2009 14:33:22 GMT

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... [ More ]

Reply #4

KnuckleHead — Wed, 06 May 2009 12:12:41 GMT

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 muc... [ More ]

Reply #3

Newb — Wed, 06 May 2009 06:58:19 GMT

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.

Reply #2

KnuckleHead — Wed, 06 May 2009 01:56:03 GMT

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 re... [ More ]

Reply #1

phileight — Tue, 05 May 2009 22:56:40 GMT

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.

p8

Calculating a percentage

KnuckleHead — Tue, 05 May 2009 12:25:01 GMT

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... [ More ]