Probability and the lotto

[Dark Warrior]

Calculating The Odds explained from www.howstuffworks.com
Let's take a look at how to calculate the odds of picking the right number for a typical Lotto game. In order to win our example game, you have to pick the correct six numbers from 50 possible balls. The order in which the numbers are picked is not important; you just have to pick the correct six numbers.

The odds of picking a single correct number depend on how many balls have been chosen already. For instance, let's say none of the six numbers had been picked yet and you had to guess just one number correctly. Since there are 50 numbers to chose from, and since six balls are going to be picked, you have six tries at picking the number correctly. The odds of picking one number correctly are 50/6 = 8.33:1.

Using a similar calculation, we can determine the odds of picking another number correctly after one number has already been drawn. We know there are 49 balls left, and that five more balls will be drawn. So the odds of picking a number correctly after one has been drawn are 49/5 = 9.8:1.

Now let's say five numbers have been picked and you have to guess what the last number is going to be. There are only 45 balls left to choose from, but you only get one shot at it, so your odds are only 45:1.

In a similar manner, we can calculate the odds of picking the right number when two, three, four and five balls have been drawn. You know the odds of a coin toss resulting in heads are 1/2 = 2:1. The odds of two consecutive tosses both resulting in heads are 1/2 x 1/2 = 4:1. The odds of three consecutive tosses all resulting in heads are 1/2 x 1/2 x 1/2 = 8:1. The odds of picking all six lottery numbers are calculated the same way -- by multiplying together the odds of each individual event. In this case:

50/6 x 49/5 x 48/4 x 47/3 x 46/2 x 45/1 = 15,890,700:1

Probably as clear as mud to some, but one thing that is correct and cannot be questioned is that the odds of your number being picked out the first ball is 1 in 49

 

[Robbo]

I'm afraid if the lottery don't want duplicate numbers chosen then a diminishing count of 1 per round is the way the probability needs to be calculated.
As already stated, first number is 1 in 49 (48 to 1), second number 1 in 48 (47 to 1) and so on.
Once the function is established, it's just a matter of simple maths to calculate the overall odds.

 

[spangley_special]

I'm afraid if the lottery don't want duplicate numbers chosen then a diminishing count of 1 per round is the way the probability needs to be calculated.
.

Now maybe its me being daft, but i just cannot see that being right unless you predict the third ball after the first and second have been drawn.
Because if you were to predict it before any numbers are drawn you'd still have the chance that the number you think will be drawn third actually comes out first or second.

I can only see it as being like drawing straws, where regardless of the numbers of straws or when you draw yours, the odds are always even because of the possibility the short one could've already been taken.

 

[spangley_special]

Calculating The Odds explained from www.howstuffworks.com
Let's take a look at how to calculate the odds of picking the right number for a typical Lotto game. In order to win our example game, you have to pick the correct six numbers from 50 possible balls. The order in which the numbers are picked is not important; you just have to pick the correct six numbers.

For us the order is important as i'm am talking about predicting a certain ball (for example the third one drawn). And predicting it prior to any of the balls being drawn)

The odds of picking a single correct number depend on how many balls have been chosen already. For instance, let's say none of the six numbers had been picked yet and you had to guess just one number correctly. Since there are 50 numbers to chose from, and since six balls are going to be picked, you have six tries at picking the number correctly. The odds of picking one number correctly are 50/6 = 8.33:1.

We're talking about the chances of a single ball, which to be honest makes that passage pretty irrelevant.

The chances of a single ball (lets say 37) being drawn first is 1/50 and lets say it makes it, getting drawn first.

Now the chance of a single ball being drawn 2nd, based on a prediction made prior to the first ball being drawn, cant be 1/49 because I could've chosen 37 which had a 49/50 chance of still being in the draw but in fact isn't.

But all the balls had an equal chance of being drawn as the first ball, so i was not at any dissadvantage picking 37 because at the time that i chose it, it had the same odds as any other ball.

Using a similar calculation, we can determine the odds of picking another number correctly after one number has already been drawn. We know there are 49 balls left, and that five more balls will be drawn. So the odds of picking a number correctly after one has been drawn are 49/5 = 9.8:1.

Now let's say five numbers have been picked and you have to guess what the last number is going to be. There are only 45 balls left to choose from, but you only get one shot at it, so your odds are only 45:1.

In a similar manner, we can calculate the odds of picking the right number when two, three, four and five balls have been drawn. You know the odds of a coin toss resulting in heads are 1/2 = 2:1. The odds of two consecutive tosses both resulting in heads are 1/2 x 1/2 = 4:1. The odds of three consecutive tosses all resulting in heads are 1/2 x 1/2 x 1/2 = 8:1. The odds of picking all six lottery numbers are calculated the same way -- by multiplying together the odds of each individual event. In this case:

50/6 x 49/5 x 48/4 x 47/3 x 46/2 x 45/1 = 15,890,700:1

Now all these passages are refereing to making predictions after the previous ball has been drawn, making them too pretty much irrelevant.

However this is were my conviction begins to waver slightly, i still do not agree with what most people seem to believe.
If any thing based on 50 balls (easier maths ) I see the probaility possibley being something like this
49/50 x 48/49 x 1/48
But again I'm not convinced...

I still believe that all balls have an equal chance of being drawn 3rd, as while 2 balls would have already been drawn, there are equal odds that those balls drawn could be any numbers meaning there is still the full range of possible outcomes (50 for the sake of easy maths)

Probably as clear as mud to some, but one thing that is correct and cannot be questioned is that the odds of your number being picked out the first ball is 1 in 49

Yep that is true, but not what i am trying to disscuss, nor are the rules.

 

[Gran]

Jack - Go buy some paint for training, I'll make sure you get atleast 1 bonus ball without buying a ticket.

 

[Piper]

I think you should all get out more and stop wasting bandwidth/forum space with intelligent questions and go back to posting drivel!

 

[Dark Warrior]

Pot calling kettle

 

[paintballJett]

Its all a bit petty. At the beginning of the raffle everyone has an even chance, its like saying after the first number everyone after that all has a 0/48 chance of winning, and thats unfair.

using number 16 as an example (cause thats going to win it)

Probability is measured by saying if we done the draw 49 times 16 would come out once, so if you dont the draw 49 times the probability of it being number 3 the probability would be 1 in 49. Its the same chances. Thats my thoughts anyways.

 

[Robbo]

Now maybe its me being daft, but i just cannot see that being right unless you predict the third ball after the first and second have been drawn.
Because if you were to predict it before any numbers are drawn you'd still have the chance that the number you think will be drawn third actually comes out first or second.

I can only see it as being like drawing straws, where regardless of the numbers of straws or when you draw yours, the odds are always even because of the possibility the short one could've already been taken.

Aha, I see the point you're making .....apologies, it's me having a blonde moment in that the answer I gave was when a sequential choice was made and in the case of the lottery, the choice of 6 numbers is done all at once.

Stats was never my strong suit when teaching but after some reflection, I'm not so sure anything changes and the fact all 6 numbers are drawn at once cannot influence the overall probability.
What I am saying here is, the probability in winning the lottery when numbers are drawn sequentially must be the same as when all the numbers are drawn in parallel and you just sit back and await for six numbers to be pulled out.

My reasoning goes like this:- if we assume you have selected an array of 6 numbers and then await the ball selection.
The important thing to note is (I think) it doesn't matter which order you chose them, or in fact which order the balls will be selected because as soon as the first ball comes down that pipe, it either corresponds to one of your selections or it does not.
If it does, then the probability for the next ball to be the same number as any one of the five you have left must have increased due to the fact there is one less ball.
This sequence of events and probabilities continues until all 6 balls have been selected.
This is one of those little problems that seem to me to be counter-intuitive at first but I think the resultant probabilities are still the same.

But...I'm not 100% sure I got to the right answer in the right way and look forward to other rationales and ideas.

 

[Sid Sidgwick]

Ok im going to have a go but I will keep this in lamens terms as hell they are easy to talk in. The chance of the 1st ball is 1 in 49 as we are all aware, in this lotto style of raffle the number ball removed from machine is not relevant and has no effect on the probability of the person winning or not, every one in the raffle has a 1 in 49 chance of winning no matter which ball is selected to be the winning ball.

If however people has say two tickets for the draw, and we made it into a multiple number draw, then there odds of winning would change as number were picked out.

Now back to the classroom me thinks