Powerball odds and a question of statistics...

General Questions
#1

Lets say I wanted to purchase all of the possible combinations for the lottery (any five numbers from 1 to 49 and then another number 1 to 42 for the Powerball). My odds of winning then would be 1:80,089,128 (or so says a Powerball home page).

But what if I wanted to eliminate any combinations that included three consecutive numbers (No 1-2-3-x-x-x or x-32-33-34-x-x or x-x-x-20-21-22, etc)?

How many number combinations would I be eliminating, and what would it do to the odds?

I’m hopeless at this kind of math. Can anyone help?

#2

Well, I am not going to figure out how many sets of numbers you would eliminate by removing sequential tickets. Thats a bunch of calculations, and by this point of a Friday afternoon, I am lucky I can finish this sentence.
What I WILL tell you is that the odds are exactly the same regarless. Each ticket you purchase, whether it has sequential numbers or not, has the same EXACT chance of being picked as any other ticket. So, by removing tickets that only have sequential numbers we accomplish —absolutely nothing! Say it again…huuh
So, lets say for this discussion that removing sequential numbered tickets we only have to buy 55,000,000 tickets instead of 80,000,000, we would have exactly the same chance of winning as if we simply bought 55,000,000 quick picks.
If I had a dollar for everytime I had to explain this to someone, I wouldn’t have to win the lotto…
Just remember…any set of 6 numbers has EXACTLY the same chance of being drawn as any other set of 6 numbers.
So, although it SEEMS incorrect, a ticket with 123456 has the same chance of being drawn as 2 14 19 20 42 51 or any other combination.
Think about it.

Michael

#3

You’re absolutely correct, of course. Eliminating sequentials wouldn’t affect the odds at all.

I guess my question should be: How many “tickets” could I eliminate if I removed all of the tickets with any three sequential numbers?

#4

…or, to make it worse…three or more sequential numbers?

#5

But what’s your point? Why do that? Let’s say that if you eliminate all the triples, you eliminate 37% of the tickets. OK. Now, instead of having a sure winner, you only have a 63% chance of winning. Again: What’s the point?

#6

Well, the good news is that it doesn’t make it worse to eliminate all those sets with three or more consecutive numbers, because those sets were all eliminated in the previous round of eliminating sets with three consecutive numbers.

Thank goodness!

#7

the number of normal lottery combinations is:
(494847464542)/(54321)=80,089,128 as you say.
i can calculate the number with at least 3 consecutive numbers by treating the 3 consecutive as a single wide number. there are 47 places i can put the start of the 3 consecutive.once those are placed, there remain 46 places for the 4th number and 45 for the fifth. the order of the my 3 choices (the 3consecutive group and the other 2 ) doesn’t matter so i divide by 321.
so total number of lottery combinations with 3 consecutive is:
(47464542)/(321)=681,030.
so the lottery tickets with 3 consecutive represent just 0.8% of the total.
-Luckie

#8

Ok, so let me get this straight. You all are looking for an advantage to shave a few hundreds off powerball odds? Whatever happened to good old card counting at the casino? At least there you reduce your odds to single or double digits. (Not that I’d know of course, I’d never hooked up with a crew, honest…)

Anyways, if you’re looking for a steady income, do what all us oldtimers (27, but lots of experience) do. Play 3 digit numbers. Yes, you don’t win as much, but you can play several times a week, your odds are better, and if you win a couple times a month, you might earn enough money to break even with welfare…

Wanna make real money? Start a poker game with people you don’t like anyways. Then learn a few false shuffles, and a couple easy slight of hand tricks (most people couldn’t pick a SOH if they were taught it). Keep the limit low, but entice your guests to play all night. Add in a second player (one that only plays once in a while, and rotate them), make a deal to split winnings, and deal that second all the winning hands you want. They look like they cheated, at worst, and you (not winning), can still make a buck. 'Course you have to trust them, but they want to be invited back, right?

Of course, I’ve never done this… :slight_smile:

#9

Rysdad is asking a combinatorics question. Nowhere does it say anything about trying to implement this strategy to win more at Powerball. So without any tips for superior money making schemes, here is the answer to the question…

If you buy one of each of the 80,089,128 possible tickets you will have a 100% chance of winning. If you don’t buy any tickets with at least 3 consecutive numbers (I am not counting the Powerball number as part of the sequences, is this what you had in mind?) you are eliminating 1,956,150 combinations and you would have a 97.56% chance of winning. I hope that answers the question.

The calculation by Luckie is on the right track but incorrect. That method double and triple counts some combinations, and dividing by (321) is incorrect. For example, if the non-Powerball numbers are 1,2,3,4,5, Luckie’s counting method counts it 3 times depending on which three numbers are considered as in the set of 3.

Ignore the Powerball at first (multipy by 42 later). For the 5 regular numbers, there are 45 ways to get all 5 in sequence. There are 1980 ways to get exactly 4 in sequence: 442 (44 ways to pick the out of sequence number, in the 2 cases when the sequence falls on the end of the 1-49 range) plus 4344 (43 ways to pick the out of sequence number for each of the 44 sequences of 4 that do not lie on the ends of the 1-49 range). There are 44550 tickets with exactly 3 in sequence: (4544)/(21)2 + (4443)/(21)45. The total for exactly 3, 4, or 5 is 44550+1980+45=46575. You could also start with Luckie’s approach (correcting the error from dividing by 321), and subtract for double counting the sequences of exactly 4 and triple counting the sequences of exactly 5: 47*(4645)/(21)-1980-2*45=46575. Multipy by 42 to account for the Powerball to get the number given above.

#10

Manlob you left out all the possible winning combinations that have the powerball as part of the sequence of three consecutive numbers.

Rysdad, I’m not sure exactly what you’re asking. Are you saying that you’d buy a ticket with the numbers 1-10-2-20-30 but not the ticket 1-2-3-10-20-30? If that is the case, you’d eliminate none of the 80089128 possible winning tickets. However, if you want to eliminate all the combos that have three consecutive numbers anywhere on the ticket you would be eliminating 3657972 tickets, or about 4.5% of all the tickets. Your chance of winning would then be about 95.5%. If it was me buying the tickets, I’d spend the extra 3 million bucks just to be sure.

The winning number will contain three consecutive numbers once about every twenty drawings. The numbers drawn on 11/25/00 were 37-02-39-38-42-22. That was about twenty drawings ago.

#11

This “1-10-2-20-30” should be this “1-10-2-20-3-30” in the post above.

#12

Lance Turbo, the ticket “1-10-2-20-3-30” cannot exist. The tickets are printed with your selected (or randomly picked via quick-pick) numbers in numerical order. Obviously, the results come out in random order.

Rysdad would be excluding the 11/25/00 winner you cited.

#13

I know. That is why it is mentioned in my post. Rysdad was not clear about this. I assume the Powerball is marked in a different area of the ticket than the regular numbers. So a sequence of 3 which includes the Powerball would not stand out as obviously on the ticket as one entirely in the regular numbers.

If you include the Powerball in sequences, I concur with your value of 3,657,972.

#14

I always went under the notion that odds and chances were totally different. You flip a coin 100 times and there all heads the 101 time you flip it the chances of it being heads is still 50-50. The odds though 100 to one. I’m no betting man, but I’d bet on tails. (Although maybe I should consider the odds of a coin landing on heads 100 time and there not having a form of manipulation are stacked agenst. OH, but wait, I still have a 50-50 chance) I buy a lottery ticket and my chances are one in a lot. I buy two lottery tickets and my chances are still one in a lot, but my odds are now one in a half of a lot.

#15

You’re on ace … lets just assume that you or I have spun 100 consecutive heads and save all that tedious waiting. OK, we’re waiting on the 101st spin and you want to back tails … you reckon it’s 100:1 against the head … I’ll back the head. What odds are you going to give me? :smiley: