Are number 01 to 09 at a disadvantage in this lotter draw

magpie · 2011-05-15 04:21:24

Lotterty numbers are 1 to 83

Two balls are drawn from two seperate pots of balls 0 to 9.

If the 1st ball is 1 and the second ball is 7 number 17 wins.

However if ball 0 is drawn and then 0 is drawn again there is a complete redraw of both balls.

Is therefore numbers 01 to 09 at a statistical disadvantage of winning.

Thanks
Colin

bobbym · 2011-05-15 04:34:40

Hi;

I am not following you. You have not said what wins and what loses.

gAr · 2011-05-15 04:39:43

Hi all,

I think he means the person having lottery number 17 wins.

But what happens for numbers > 83?

magpie · 2011-05-15 04:40:41

The first ball is drawn it it is a 1, 2, 3, 4, 5, 6, 7, or an 8 then there is no chance of a redraw and whatever the 2nd ball is that number wins.
However if ball 0 is drawn first then drawn again hence number 00 then both balls are redrawn.

Does this put numbers 01, 02, 03, 04, 05, 06, 07, 08, 09 at a statistical disadvantage?

Thanks

Last edited by magpie (2011-05-15 04:41:40)

bobbym · 2011-05-15 05:33:43

Hi;

The fact that (0,0) resets the game does not effect the odds at all. Each number (0,0) to (9,9) has the same odds of being drawn. One in a hundred. So I would say there is no advantage here.

magpie · 2011-05-15 05:51:36

But ) is the only ball that can trigger a redraw

If ball 1 to 8 is drawn the winning ball can only take 2 balls.

If 00 is drawn there is a total redraw so

If 00 is drawn the winning number say 07 could take 4 balls to win.

bobbym · 2011-05-15 06:22:12

I am assuming that after a (0,0) the two zeroes are put back in the pots as you call them. So they can be drawn again.

magpie wrote:

If 00 is drawn the winning number say 07 could take 4 balls to win.

You are forgetting that you can also get a ( 1, 8 ) in 4 draws too. And the odds of that are the same as ( 0,7 ) after ( 0, 0 ).

To get (0,7) is 1 / 10 on the first and 1 / 10 on the second that equals 1 / 100.

To get ( 1,8 ) is 1 / 10 on the first throw and 1 / 10 on the second that equals 1 / 100.

If you drew a ( 0, 0 ) the chance of now getting a ( 0, 7 ) is still 1 / 100 and the chance of getting ( 1, 8 ) after the ( 0,0 ) is still 1 / 100.
Odds have not changed. It is not harder to get a (0,7) after a ( 0,0 ) than a (1,8) after a ( 0, 0 )

magpie · 2011-05-15 07:09:11

But drawing 7 means only 70, 71, 72, 73, 74, 75, 76, 77, 78 & 79 can win.

If you draw 0 first ball there is still a chance that all numbers can still win if the 2nd ball is also 0.

Bob · 2011-05-15 10:41:21

hi magpie,

You raise an interesting point. There is a small chance that the draw could go on for a long time.

I have made a tree diagram (see below) showing four initial outcomes, one of which leads to a redraw, leading to four outcomes, one of which leads to a redraw, and so on ...........................

Then I analysed the probablility of getting a win for 01, 02, 03, ...09 in one group
and 10, 20, 30, 40, 50, 60, 70, 80, 90 in a second
and for the rest.

Then I split the probabilities in each group to get the individual probabilities for all 99 potential winners. (Within each group each number has an equal chance)

Here's that analysis:

P(win for number in range 01 to 09) =0.09 + 0.01x0.09 + 0.01x0.01x0.09 + 0.01x0.01x0.01x0.09 + ......

= 0.09 x (1 + 0.01 + 0.01x0.01 + 0.01x0.01x0.01 + ......) = 0.09 / 0.99 = 1/11

There are 9 numbers in this category.

So P(any of these) = 1/11 x 1/9 = 1/99

P(win for any of 10,20,30,40,50,60,70,80,90) = 0.09 + 0.01x0.09 + 0.01x0.01x0.09 + 0.01x0.01x0.01x0.09 + ......

= 0.09 (1 + 0.01 + 0.01x0.01 + 0.01x0.01x0.01 + ......) = 0.09 / 0.99 = 1/11

There are 9 numbers in this category.

So P(any of these) = 1/11 x 1/9 = 1/99

P(win for any number 11 or over not ending in zero) = 0.81 + 0.01x0.81 + 0.01x0.01x0.81 +.........

= 0.81 x (1 + 0.01 + 0.01x0.01 + 0.01x0.01x0.01 + ......) = 0.81 / 0.99 = 9 / 11

There are 81 numbers in this category.

So P(any of these) = 9/11 x 1/81 = 1/99

So all are equally likely.

This analysis ignores the additional problem of what to do when the number drawn is over 83.

The diagram gets horribly complicated if you try to allow redraws for these too, but I think the final result is the same.

Bob

gAr · 2011-05-15 16:17:25

Hi magpie,

I tried using an absorbing markov chain and I get the same answer as Bob's.

Below are the probabilities of getting into the absorbing states:

P(Number is in between 01 and 09, inclusive) = 9/99
P(Number is in between 10 and 79, inclusive) = 70/99
P(Number is in between 80 and 83, inclusive) = 4/99
P(Number is in between 84 and 89, inclusive) = 6/99
P(Number is in between 90 and 99, inclusive) = 10/99

Hence probability of getting any number is equally likely, 1/99

If we want the balls to be redrawn for numbers>83, with slight modification of the markov chain, the probability becomes 1/83

Math Is Fun Forum

#1 2011-05-15 04:21:24

Are number 01 to 09 at a disadvantage in this lotter draw

#2 2011-05-15 04:34:40

Re: Are number 01 to 09 at a disadvantage in this lotter draw

#3 2011-05-15 04:39:43

Re: Are number 01 to 09 at a disadvantage in this lotter draw

#4 2011-05-15 04:40:41

Re: Are number 01 to 09 at a disadvantage in this lotter draw

#5 2011-05-15 05:33:43

Re: Are number 01 to 09 at a disadvantage in this lotter draw

#6 2011-05-15 05:51:36

Re: Are number 01 to 09 at a disadvantage in this lotter draw

#7 2011-05-15 06:22:12

Re: Are number 01 to 09 at a disadvantage in this lotter draw

#8 2011-05-15 07:09:11

Re: Are number 01 to 09 at a disadvantage in this lotter draw

#9 2011-05-15 10:41:21

Re: Are number 01 to 09 at a disadvantage in this lotter draw

#10 2011-05-15 16:17:25

Re: Are number 01 to 09 at a disadvantage in this lotter draw

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