Need a "Statistics Guy" to Solve a Friendly Dispute

[Centropolis]

The question is:

In a 6-number lottery system with 49 numbers in total, is there less chance that I would win if I pick numbers 1,2,3,4,5,6 than if I were to pick 4,11,22,26,32,48 or 1,3,9,18,37,38?

At first, I thought that it would be the same odds. Then on second thought, I wasn't so sure.

Here is an example I am using to slightly simplify the situation.

I have 10 ping-pong/table tennis balls with numbers 1 to 10 on them....one number on each ball. Since the chances of each ball will be drawn is the same for every ball, I think it is the same chance that I would get 3 consecutive numbers (i.e. 2,1,3 or 7,5,6 or 2,3,4 - draw order doesn't matter) as I would get 1,7,8 or 10,3,2 or 4,1,8.

But does the chance of order come into play at all?

Let the discussions begin!

 

[Launch Mini]

Exact same odds of consecutive numbers as random numbers.
Any ONE of the first six has a 6/49 chance, the next 5/48, then 4/47....
I have monitored this for fun, an watch the history of prior drawings.
VERY oftern you will see 3 in a row.

 

[RocketTomato]

If order does not matter, then the chances are exactly the same. There is no difference. ANY combination of 6 numbers has the exact same probability as any other combination of 6 numbers.

 

[Tekno_Cowboy]

Now what if order did matter? What would the odds be then?

Random Observation: This is my 1425th post.

 

[Launch Mini]

Now what if order did matter? What would the odds be then?

Random Observation: This is my 1425th post.

Then the odds of ANY one of the balls for the first selection would be:
1/49
The second would be
1/48
Etc

So, to select 6 balls IN ORDER. Assuming the ball chosen is removed from the remaining balls for the next selection
Your total odds are
1/49 x 1/48 x 1/47 x 1/46 x 1/45 x 1/44 - 1 in 10,068,347,520 not good at all

 

[gswitter]

That's not quite correct.

• If you stipulate increasing order, there are only 44 possibilities for the first draw - you can't start a subset with 45 though 49.
• If you allow increasing or decreasing order, there are 49 possibilities for the first draw, but options for the second draw get complicated.

It's probably easier to just divide the number of ordered 6-item subsets (44, or 88 if you allow decreasing order) by the total number of 6-item subsets: 49! / ( ( 49 - 6 )! * 6! ). (I think that's right. It's been a while.)

 

[RocketTomato]

Let's simplify things, drawing three balls labeled A-J out of a possible ten.

The total number of possible ways to pick three balls out of ten is:

10*9*8 = 10!/7! = 720 possibilities

Since there is only one possible way to pick the balls in a specific order, for example picking the balls in order A, B, C; your total chances are 1 in 720.

However, since order does not matter, there are 3*2*1=6 possible ways to pick 3 balls in any order, i.e. ABC, ACB, BAC, BCA, CAB, and CBA. Thus your odds, if order does not matter, improve to 6 possibilities out of a total of 720 or 1 in 120.

[Using factorials your chances are 1 in 10!/7! if order matters and 1 in 10!/(3!7!) if order does not matter.]

 

[Mike Painter]

If only the "f" was left out of this link. http://tinyurl.com/39fto7
Anyway here are some of the things you are more likely to do than win the lottery.

 

[bshanahan14rulz]

I like this question:

Bag with 26 balls, labeled A-Z. What is the probability of picking the same ball 6 times in a row, assuming replacement after drawing.

 

[RocketTomato]

I like this question:

Bag with 26 balls, labeled A-Z. What is the probability of picking the same ball 6 times in a row, assuming replacement after drawing.

1 in 26^5 if it does not matter which of the 26 letters you draw first.

The odd of drawing a specific letter, "Z" let's say, are 1 in 26^6.

 

[Crenshaw]

now i know why i fell asleep during my stats class....:sick2:

But i do remember there is some formula for calculating stuff like this using the mathematical symbol Sigma or something

Crenshaw

 

[ANDREAS FERRARI]

Here's a question;

If the odds of winning 6/49 with one ticket is 14,000,000 to 1 than what are the odds of winning if you have 10 tickets?

 

[Empath]

If the odds of winning 6/49 with one ticket is 14,000,000 to 1 than what are the odds of winning if you have 10 tickets?

The "winning" isn't necessarily the same as matching the draw. Matching the draw with 10 tickets would be 1,400,000 to 1, or ten times as likely as having one ticket.

The "winning" depends on whether the draw matches any player. If the draw doesn't match any players' series, then the odds are zero; the winning would be impossible. If the draw does match a series played, then the odds against would be the number of unique series played to one.

The house concerns itself with the odds of selecting a match. The players are concerned with whether a winner is possible, and if so, the number of plays made. They aren't exactly the same. Since the number of plays can't be ascertained before hand, we entertain ourselves by playing around with the same numbers as the house.

 

[koubilaihan]

Exactly the same odds with 1,2,3,4,5,6 or 4,11,22,26,32,48 or any other combination.
The odds are 6/49*5/48*4/47*3/46*2/45*1/44= 1 chance in 13.983.816 with any given combination of numbers.

 

[ANDREAS FERRARI]

Matching the draw with 10 tickets would be 1,400,000 to 1, or ten times as likely as having one ticket.....

Sorry Empath but that is not the correct answer.

Let's see if someone else can come up with the right answer.

 

[LA OZ]

Here's a question;

If the odds of winning 6/49 with one ticket is 14,000,000 to 1 than what are the odds of winning if you have 10 tickets?

Well, assuming you don't buy the same combination in the 10 tickets than the chance will be:
1/14000000 + 1/13999999 + 1/1399998 + 1/1399997 + 1/13999996 .... 1/1399991 = 0.000000714285943878 chance of winning.

 

[ANDREAS FERRARI]

Well, assuming you don't buy the same combination in the 10 tickets than the chance will be:
1/14000000 + 1/13999999 + 1/1399998 + 1/1399997 + 1/13999996 .... 1/1399991 = 0.000000714285943878 chance of winning.

Are you saying the odds are longer if you have more tickets???? The odds you are giving are for one ticket-not ten!!!!

Wrong!!!!!

I'm sure someone will figure this out.

This was a question from first year stat/analysis class.

I'll give you a hint-koubilaihan's math is right.

I need to edit this because LA OZ is right-but I'm not sure if for the right reason????

Exactly the same odds with 1,2,3,4,5,6 or 4,11,22,26,32,48 or any other combination.
The odds are 6/49*5/48*4/47*3/46*2/45*1/44= 1 chance in 13.983.816 with any given combination of numbers.

 

[javiole]

Well, assuming you don't buy the same combination in the 10 tickets than the chance will be:
1/14000000 + 1/13999999 + 1/1399998 + 1/1399997 + 1/13999996 .... 1/1399991 = 0.000000714285943878 chance of winning.

I dont think so.

10/14000000 is your probability of getting the right combination, i.e. to win the prize (if you choose 10 different combinations of numbers).
This is 0.0000007142857142, which is lower than 0.0000007142859438 for just a bit more than 0.0000000000002296. No much difference anyway.

Although lottery is ruled by randomness, if you check statistics in the lottery history of the numbers that have already appeared, you will see that some numbers appears in a higher rate that some others. But this won't help you to calculate the wining combination!! Otherwise few clever boys will always win the lottery!! Is that a paradox? No!! again the statistics shows that the difference in the rates of appearence is not statistically significant, meaning that the more the lottery is run, the closest these rates get.

The setting of the game (number of numbers 49 and number of numbers that we pick up 6) are calculated taking into account the number of players (population), so there will be a high probability that the number of winners is very low, so the prizes are kept huge and make us play more. Simple and clever busisness!!

 

[LA OZ]

I last done statistic over two decades ago. If I was to get 10 tickets, I will be stupid to get any one of them in the same combination. My calculation indicates less odd with additional ticket and not more. I don't know how your lotto system works but the question was "If the odds of winning 6/49 with one ticket is 1 in 14,000,000".

 

[Jay R]

The chances of winning with ANY combination of numbers is exactly the same. However, if you pick a common combination such as 1,2,3,4,5,6 or lots of number under 32 ( dates ), you are much more likely to be sharing your winnings with other winners.
Best combination to chose is 6 numbers over 31. You have exactly the same chance of winning but if you do, you'll probably win more money.