Probability and the lotto

[spangley_special]

Ok, so in one of he raffle threads the subject of probabilty came up and confused me slightly.

So as not to take that thread off topic i've started one here.

original thread here ---->
http://www.p8ntballer-forums.com/vb/showthread.php?t=111914&page=2

Not quite correct

47 numbers will have the same chance
The 48th number will have a less chance but will have more chance than the 49th number

All 49 will have a chance to win but not the same chance
The same reasoning is applied when you have a selection of prizes which dictates that the top prize has to be drawn first (unless numbers are re-entered)

Let me state that I'm not attempting to question the rules here.

Now I've not studied probability in like about forever, but here is my thoughts. If I am mistaken could some one explain it for me

So we have the numbers 1-49 picked at random, and then removed from the draw.

Obviously the first number could be any of them with an equal chance (1/49)

Now my intial thoughts are that the odds of the second draw being a certain number (prior to the first being drawn) would still be equal for all numbers and therefore still be 1/49.
As although there is indeed a chance the number may have been drawn first, there is an equal chance that it was any of the other numbers.

...so am i just confusing myself here or am I along the right lines?

 

[Paintball_Mad]

The first number to be choosen has the smallest chance of getting picked (1/49)
The second number to be choosen has the 2nd smallest chance of getting pickes (1/48) - because you only have one number thats all you have, the 1, and there are 48 numbers left as 1 has already been taken away from the total 48...
this works the whole way down, obviousaly, you have 1 number out of 2 possible numbers then you have a (1/2) chance...

EDIT: after a re-read of what you said, I think you are right.

 

[Cetanu]

I was under the impression that each subesquent number had an increase in probability, so number 19 has a 1in 49 chance the first time, then a 1/48, 1 in 47 and so on. So the longer you keep drawing the higher the probability. If the ball is put back into the draw then the probability of any number is still 1 in 49.
If the number is put back in it's equally as likely to be drawn again as any number is to be drawn (1/49 * 1/49).

All chances are equal...as far as i know.

 

[Impulse raider]

Yikes

Yikes a debate , lol .. buying tickets is a good probability to me ,, the more you buy the better odds you have in winning ..

So as you all debate , bring on the purchases so i can get my new gat

 

[dave t]

Now I've not studied probability in like about forever, but here is my thoughts. If I am mistaken could some one explain it for me

So we have the numbers 1-49 picked at random, and then removed from the draw.

Obviously the first number could be any of them with an equal chance (1/49)

Now my intial thoughts are that the odds of the second draw being a certain number (prior to the first being drawn) would still be equal for all numbers and therefore still be 1/49.
As although there is indeed a chance the number may have been drawn first, there is an equal chance that it was any of the other numbers.

...so am i just confusing myself here or am I along the right lines?

The first number drawn has a probability of 1/49 of being drawn as stated, however, there are only 48 balls remaining, meaning only 48 possible outcomes of a certain number being drawn, ie a probability of 1/48. Then there are only 47 balls left, or 47 possible outcomes remaining, etc, etc.

 

[Paintball_Mad]

confused now?..... thought not

 

[spangley_special]

The first number drawn has a probability of 1/49 of being drawn as stated, however, there are only 48 balls remaining, meaning only 48 possible outcomes of a certain number being drawn, ie a probability of 1/48. Then there are only 47 balls left, or 47 possible outcomes remaining, etc, etc.

yes, but you are over looking the chance that it couldve already been drawn as the first number.

 

[dave t]

yes, but you are over looking the chance that it couldve already been drawn as the first number.

Not really...none of the numbers can be drawn more than once

 

[spangley_special]

Not really...none of the numbers can be drawn more than once

yes that is true, but i dont think its relevant. bare in mind that all the predictions are done prior to any numbers being drawn. so the number i predict will be drawn third could in reality be drawn first, and therefore not even be in the draw by the time the third number gets picked.

 

[dave t]

I know what you mean but the number of outcomes to each event are fixed and diminishing but, most importantly, unaffected by your predictions
I, personally, think that whatever number out of the hat, a persons chances of winning the gat, and thats all we really care about, yeah, are the same, ie 1/49 but thats not to say the odds of winning the gat are the same as as the odds of a certain ball arriving at a certain time. As Cetanu rightly pointed out, as the number of balls decreases, each ball then has a marginally higher probability of being next but at the outset, they all have the same probability of being third or first or whatever
Of course there's always the possibility ( much higher than 1/49 ) that I could be talking complete bollocks this late on a Sunday