Lotto Pens: Improbability Engines of Today?

[c4ts]

I've seen pens sold at drug stores with a numbered ball filled chamber at the end. You are supposed to shake the pen until 5 balls slide into a tube as a way to predict your lottery numbers. But I think they actually decrease you chances of winning the lottery. For example, in a 5 number game, you must choose 5 numbers and have them all match up. Assuming you have a 1 in 10 chance for each digit so your chances of winning are about 1 in 100,000. Now, if the lottery pen has 50 of those little numbered balls in it (5 for each digit, 0-9), you have about equal odds of winning the lottery. So, what are the chances that the numbers that come up in the pen are the same as the numbers that come up on television? For example, you have a 5 in 50 chance that the number at the bottom of your pen will be "1," and a 1 in 10 chance that the first number on TV comes up is "1," so you have a 1 in 100 chance of coincidence for each digit. I may have calculated probability in the wrong way, but it looks like you have a 1 in 10,000,000,000 chance of winning the 5 digit lottery using your pen numbers. Yet is that any different from just choosing a bunch of numbers off the top of your head and writing them down, or picking the digits in your birthdate or anniversary?

[Iconoclast]

You guy have a few different lottery formats, so I'm assuming this is the type where you pick a bunch of numbers between 0 and 9 inclusive, then stick them all together to make one big number, correct?

Quote:

Originally posted by c4ts
I've seen pens sold at drug stores with a numbered ball filled chamber at the end. You are supposed to shake the pen until 5 balls slide into a tube as a way to predict your lottery numbers. But I think they actually decrease you chances of winning the lottery. For example, in a 5 number game, you must choose 5 numbers and have them all match up. Assuming you have a 1 in 10 chance for each digit so your chances of winning are about 1 in 100,000.

Let's just check that, 00000 to 99999, looks like about 100,000 different combinations to me.

Quote:

Originally posted by c4ts
Now, if the lottery pen has 50 of those little numbered balls in it (5 for each digit, 0-9), you have about equal odds of winning the lottery.

Equal to what?

Quote:

Originally posted by c4ts
So, what are the chances that the numbers that come up in the pen are the same as the numbers that come up on television?

Believe it or not, that's also 1 in 100,000. For any (arbitrary) combination of balls in your little pen, there is a 1 in 100,000 chance that the lottery numbers will be the same.

Quote:

Originally posted by c4ts
For example, you have a 5 in 50 chance that the number at the bottom of your pen will be "1," and a 1 in 10 chance that the first number on TV comes up is "1," so you have a 1 in 100 chance of coincidence for each digit.

That's right, but you also need to remember that there's also a 1 in 100 chance that both your pen and the lottery have a 2 as the first number, and a 1 in 100 chance that both your pen and the lottery have a 3, and a ..... blah, blah, blah. Add them all together and you've got a 1 in 10 chance that the pen and the lottery show the same number in that position. Remember, we don't care what that digit is, only that the pen and the lottery digits are identical.

Quote:

Originally posted by c4ts
I may have calculated probability in the wrong way

Your train of thought has led you to the preposterous conclusion that a ball point pen, a goddamn biro, a freakin' Bic, can somehow influence your chance of winning the lottery. By the time you got to here you should have been backtracking to try and find the flaw in your logic.

[roger]

Quote:

Originally posted by c4ts
For example, you have a 5 in 50 chance that the number at the bottom of your pen will be "1," and a 1 in 10 chance that the first number on TV comes up is "1," so you have a 1 in 100 chance of coincidence for each digit. I may have calculated probability in the wrong way, but it looks like you have a 1 in 10,000,000,000 chance of winning the 5 digit lottery using your pen numbers.

The first sentence is correct until you say "for each digit". You just calculated the odds for matching "1". Note the beginning of the sentence - "for example". You need 9 more examples to calulate the odds of matching the first digit.

pen = '1' tv = '1' p=1/100
pen='2' tv = '2' p =1/100
...
pen='0' tv ='0' p=1/100

sum the probabilities and p = 1/10.

But all that is irrelevant.

I'll give you a different pen, with 5 balls in it. Every ball has '1' on it. Have your odds of winning been lowered in any way? (hint: no).

It doesn't matter what balls are in the pen. As long as it generates a number, any number, that is valid, your chances are equivalent to choosing a # in a different way.

[c4ts]

Quote:

Originally posted by roger


The first sentence is correct until you say "for each digit". You just calculated the odds for matching "1". Note the beginning of the sentence - "for example". You need 9 more examples to calulate the odds of matching the first digit.

pen = '1' tv = '1' p=1/100
pen='2' tv = '2' p =1/100
...
pen='0' tv ='0' p=1/100

sum the probabilities and p = 1/10.

But all that is irrelevant.

I'll give you a different pen, with 5 balls in it. Every ball has '1' on it. Have your odds of winning been lowered in any way? (hint: no).

It doesn't matter what balls are in the pen. As long as it generates a number, any number, that is valid, your chances are equivalent to choosing a # in a different way.

So one ball in showing up on TV does not lower the possibility that the next ball will also be that digit?

I thought probability worked like this:

Possible outcomes for digit 1: 0,1,2,3,4,5,6,7,8,9
Possible outcomes for pen digit 1: 0,1,2,3,4,5,6,7,8,9
Possible outcomes for pen digit 1, followed by digit 1: 0-0,0-1,0-2,0-3,0-4,0-5,0-6,0-7,0-8,0-9,1-0,1-1,1-2,1-3,1-4,1-5,1-6,1-7,1-8,1-9,2-0,2-1,2-2,2-3,2-4,2-5...9-9
Possible outcomes for digit 2: 0,1,2,3,4,5,6,7,8,9
Possible outcomes for pen digit 2: 0,1,2,3,4,5,6,7,8,9
Possible outcomes for pen digit 2, followed by digit 2: [b]0-0[/b,0-1,0-2,0-3,0-4,0-5,0-6,0-7,0-8,0-9,1-0,1-1,1-2,1-3,1-4,1-5,1-6,1-7,1-8,1-9,2-0,2-1,2-2,2-3,2-4,2-5...9-9
Possible outcomes for pen digits 1 & 2, followed by digits 1 & 2: 00-00, 00-01, 00-02, 00-03, 00-04, 00-05, 00-06, 00-07, 00-08, 00-09, 00-10, 00-11, 00-12, 00-13... 99-99

With your pen, the outcome is still the same for individual digits:

Possible outcomes for digit 1: 0,1,2,3,4,5,6,7,8,9
Possible outcomes for pen digit 1: 1
Possible outcomes for pen digit 1, followed by digit 1: 1-0, 1-1, 1-2, 1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 1-9

(Too busy to write this all down, will get back and edit in more...)

[roger]

Quote:

Originally posted by c4ts
[b]So one ball in showing up on TV does not lower the possibility that the next ball will also be that digit?

I don't know how the draws are done on tv.

Assuming: to pick 5 #s between 0-9, there are 5 separate ball tumblers, each with 10 balls in them. Each tumbler chooses 1 of the 5 digits. Then no, the draws are independent, and the second ball is not influenced by the first ball.

Or did you mean by that question that the ball being drawn on tv affects the probability of the ball drawn by the pen? In that case, my probability calculation is correct. Your odds of matching the first ball is 1/10.

Quote:

I thought probability worked like this: (snip #s)

It's hard for me to follow what you are saying. But for my pen that delivers 11111 every time, by odds of winning are 1:/100,000.
That's because on TV, there is a 1/10 chance of matching the first digit, 1/10 of the 2nd, etc. So (1/10)^5 = 1:100,000.

If my pen delivered 15783, or any other random number, the odds are still calculated as above, giving 1:100,000.

If this is not clear ask some more questions...

[Skeptical Greg]

The odds of guessing any possible outcome are wholly dependant on the combination of outcomes available.

Those odds will not be affected by any mechanism that is employed in guessing.


The possibility of getting a correct guess will increase with the number of guesses made, providing the guess is a valid possible outcome.


The simplest example being: If you are allowed three guesses on the outcome of one toss of a coin, the possibility of
you being correct is 100%.

The odds of it being ' heads ' is still 1/3..

Even if it comes up heads 1,000 times in a row, the odds it will be heads on the 1,001th toss is still 1/3..

[Cecil]

Quote:

Originally posted by Diogenes
The simplest example being: If you are allowed three guesses on the outcome of one toss of a coin, the possibility of
you being correct is 100%.

The odds of it being ' heads ' is still 1/3..

Even if it comes up heads 1,000 times in a row, the odds it will be heads on the 1,001th toss is still 1/3..

You must be talking about those newfangled 3-sided coins.

Or, maybe, you're counting the coin landing on edge as an outcome? That would make the probability for each outcome 1/3, right?

[Soapy Sam]

You know about the guy who always takes a bomb on the plane when he flies ?

[xouper]

bump