Welcome Guest
You last visited December 11, 2016, 8:42 am
All times shown are
Eastern Time (GMT-5:00)

# Check my math, please?? :)

Topic closed. 7 replies. Last post 7 years ago by Ammothief41.

 Page 1 of 1
New Member
Duluth, MN
United States
Member #91721
May 23, 2010
6 Posts
Offline
 Posted: May 24, 2010, 4:06 pm - IP Logged

I am new to the lottery math.  On a website I was reading it stated that there was a benefit to winning a prize by playing a number from the previous numbers, or avoiding previous winning numbers.  I computed the odds and became puzzled because it didn't look like that would help at all and was wondering if someone could check the math to make sure it's correct.

It was for a particular lotto game, 5/56, and the site stated that there was a 44% chance of a number from the previous winning set being played in the next drawing.  I think that's correct because you have 56 total possibilities and five numbers you could pick from each game.

56 / 5 = 11.2

So 1 in 11.2 chances of 1 particular number being chosen, and if there are 5 numbers drawn each drawing 11.2/5 =  1 in 2.24 or 44%.  After that, I tried computing the odds to see if there was benefit to picking a number and the results didn't seem to be all that amazing...

if you have 44% chance that a number will repeat it would be represented as 100 / 44?  for 100 possibilities and 44 correct.  Then you have a 1 in 5 chance of picking which number is going to be the one that repeats so 5 / 1, five possibilities and one chance.  And finally instead of picking from five numbers, it's going to be from four because we already picked one number out.. so ( 55 / 4 ) * ( 54 / 3 ) * etc

( 100 / 44 )  * ( 5 / 1 ) * ( 55 / 4 ) * ( 54 / 3 ) * ( 53 / 2 ) * ( 52 ) = 3.8m.  So the odds of picking 1 number to play again is the exact same 3.8m to 1 odds as if you'd just taken QP's for the same game?

And what happens if you decide to play the game by disregarding any of the previous winning results...

( 100 / 56 ) * ( 51 / 5 ) * ( 50 / 4 ) * ( 49 / 3 ) * ( 48 / 2 ) * 47 = 4.1m.  So is that right?  You have worse odds than what you even started with by not selecting the numbers that won the last jp?  I'm just very confused, I was expecting there to be some kind of benefit to doing that and I can't figure out where I would be coming out ahead, or I must be doing something wrong.

Thanks,

Jason

United States
Member #75358
June 1, 2009
5345 Posts
Offline
 Posted: May 24, 2010, 11:18 pm - IP Logged

Here's where you are confused. You're confusing yourself with your own words....lol

First you say there is a 44% chance of getting a repeat number, and you prove it. But you run into trouble when you equate the 44% chance with the odds of hitting the JP. I'll show you...The following is your quote:

"So 1 in 11.2 chances of 1 particular number being chosen, and if there are 5 numbers drawn each drawing 11.2/5 =  1 in 2.24 or 44%."

You proved that it is 44%. Then at the end you're puzzled because you still have a 3.8 million to 1 odds of hitting the jackpot. As if, because you have a 44% repeat hit rate on a number/numbers out of 5, you should win the JP much much easier than 3.8 million to 1. You're assuming the benefit should reflect a better odds of hitting the JP. But the problem is that you should accept the fact that you have a 44% chance of getting a repeat number/numbers out of 5, and nothing else. By hitting a repeat number/ numbers out of 5, there would only be a slight advantage goal towards hitting the JP.

United States
Member #75358
June 1, 2009
5345 Posts
Offline
 Posted: May 24, 2010, 11:31 pm - IP Logged

I should also add that If you had a 44% edge over the total 56 numbers in the matirix, then you'd be justified in your confusion. That's nearly 1 in 2 of hitting the JP. But that's not the case here. I think that's where the confusion comes in.

New Member
Duluth, MN
United States
Member #91721
May 23, 2010
6 Posts
Offline
 Posted: May 25, 2010, 6:24 pm - IP Logged

Excellent, thank you.  That was just what I was wondering.  The "great" advice of some lottery websites are a little misleading.  Anyway, I looked at the history of drawing results for that particular lottery and found something mildly cool.  Of the last hundred drawings, a number from the previous drawing showed up in the new drawing 35% of the time instead of the expected 44%.  I did the math and that should put the odds well over QP's by picking a number from the previous results, but it also lowers the odds for the method of avoiding all the previous results.  It's not a big difference, but a little bit is better than nothing, right?  (Again, asking in case there's something I'm missing here)

QP Odds = 1 in 3819816

Here's the math, for the pick 1 previous number:

100 / 35 * 5 / 1 * (55 / 4 ) * ( 54 / 3 ) * ( 53 / 2 ) * ( 52 ) = 4,872,214

And for picking results by not including any of the immediately adjacent numbers:

100 / 65 * ( 51 / 5 ) * ( 50 / 4 ) * ( 49 / 3 ) * ( 48 / 2 ) * ( 47 ) = 3,613,938

If that's correct, by avoiding the previous draw, one would have 5.4% better odds on picking all 5.  Is that right?

mid-Ohio
United States
Member #9
March 24, 2001
19831 Posts
Offline
 Posted: May 25, 2010, 8:51 pm - IP Logged

MATCH   ODDS                     WINNING COMBOS
5/5    1 : 3819816                1
4/5    1 : 14980                    255
3/5    1 : 300                        12750
2/5    1 : 18                          208250
1/5    1 : 3                            1249500
________________________________________________
overall odds are 1 : 2.5

Why do you think matching 2/5 and 1/5 pay nothing?  It happens too often because the odds are it will.

* you don't need to buy more tickets, just buy a winning ticket *

New Member
Duluth, MN
United States
Member #91721
May 23, 2010
6 Posts
Offline
 Posted: May 26, 2010, 12:17 am - IP Logged

Thank you for the chart.  I'm not shocked by the odds being slim.  I was surprised that my awful math came out with the right numbers.  Not expecting to flip a light bulb and be able to pick JP combos, but any little extra couldn't hurt.

NY
United States
Member #23835
October 16, 2005
3475 Posts
Offline
 Posted: May 26, 2010, 3:49 am - IP Logged

"So 1 in 11.2 chances of 1 particular number being chosen,"

That's fine for whether or not a particular number will be selected in any given drawing, but you're looking for the chances that any of the 5 numbers from a drawing will repeat in the next drawing. Here's how that works.

5 of the 56 numbers were in the last drawing and 51 of 56 were not. When the first number is selected there's a 51/56 chance it will  not be any of the 5. If that happens there will be  55 numbers remaining, of which 50 were not selected in the previous drawing. That means there's a 50/55 chance of selecting anothe rnumber that's not a repeat. For the 3rd number it will be 49/54, then 48/53 for the 4th and 47/52 for the 5th number. Overall that means there's a 51/56 * 50/55 * 49/54 * 48/53 * 47/52 chance that none of the previous numbers will be repeated. Doing the math we get .615, or a 61.5% chance that none of the previous numbers will be repeated. Subtracting from 100% that leaves a 38.5% chance that 1 or more of the numbers will repeat. Note that that figure matches the history you looked at much more closely than the 44% figure.

So what does that do to help you out? Well, you could play all 5 numbers from the previous drawing and have a 38.5% chance of getting at least one number right. As a practical matter, almost all of that 38.5% is the chance of getting only 1 of the numbers right.  How much do you win for doing that?

Still, let's pretend that  there's actually an advantage to using numbers from the previous drawing. The advantage you'd get would only happen the 38.5% of the time that a number(s) does repeat. The other 61.5% of the time you'd lose that advantage. The theoretical advantage only happens some of the time, and is offset by the chance that there won't be repeats.

For a simple comparison consider playing 2 odd and 3 even numbers or 3 odd and 2 even numbers.Most of the possible combinations fit into one of those two patterns, so most of the winning combinatins fit one of those two patterns. That  makes you more likely to have the winning combination if the winning numbers fit one of those patterns. If you have the right "patern" but not the right numbers you win nothing, so there's no advantage to that. If the winning numbers aren't one of those patterns, then you can't possibly have chosen the winning combination and you've got a 0.0% chance of winning.

In short, all you're doing is recreational math to calculate the odds for when a particular what if happens. Most of the time it's a different what if that happens. The only way to figure the odds for the real world, where you don't know which what if will happen, is to use the odds that include all of the what if's. Those are the odds that RJ gave you.

New Member
Duluth, MN
United States
Member #91721
May 23, 2010
6 Posts
Offline
 Posted: May 26, 2010, 2:51 pm - IP Logged

Ok.  That is exactly what I was trying to figure out, trying to get a basic grasp of probability.  Thanks.

 Page 1 of 1