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why guessing the lottery probability...

 
 
Reply Fri 29 Jul, 2011 02:11 pm
The chance for a certain lottery draw is 1 in 14 million. However the player has his own "drawing" sort of say. Why isn't the real probability of winning the lottery a bigger number. Take two dice for example, one representing the lottery the other the player. The chance both lottery and player draw the same drawing, is 1/6 x 1/6 = 1/36, not just 1/6.

Can you prove otherwise?
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Type: Question • Score: 2 • Views: 1,072 • Replies: 10
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View best answer, chosen by w0lfshad3
markr
  Selected Answer
 
  2  
Reply Fri 29 Jul, 2011 09:10 pm
@w0lfshad3,
The chance of the player matching the lottery is 6/36, not 1/36. The six matching combinations are:
1,1
2,2
3,3
4,4
5,5
6,6
cicerone imposter
 
  1  
Reply Fri 29 Jul, 2011 09:40 pm
@w0lfshad3,
How did you calculate 14 million? I wanna see your formula.
maxdancona
 
  1  
Reply Fri 29 Jul, 2011 09:43 pm
@w0lfshad3,
I predict the next time you roll a dice (or die for the pedantic) you will roll a "2".

What is the chance that I guessed correctly?
w0lfshad3
 
  1  
Reply Sat 30 Jul, 2011 03:38 am
@markr,
Actually you're right, I blame friday night Very Happy
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w0lfshad3
 
  1  
Reply Sat 30 Jul, 2011 03:39 am
@cicerone imposter,
http://en.wikipedia.org/wiki/Lottery_mathematics
w0lfshad3
 
  1  
Reply Sat 30 Jul, 2011 03:39 am
@maxdancona,
1/6 for a 6 sided dice
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w0lfshad3
 
  1  
Reply Sat 30 Jul, 2011 05:20 am
@w0lfshad3,
Truth is, the player and lottery are independent events, so the math to get the probability starts with the independent events formula, then divide the result by the number of possible outcomes. It is important for me to put it like this.
maxdancona
 
  1  
Reply Sat 30 Jul, 2011 06:43 am
@w0lfshad3,
How is me picking what number you are going to roll on a dice (or die for the pedantic) different then you picking what number is going to come up on the lottery?

You are getting it right with the dice example.
w0lfshad3
 
  1  
Reply Sat 30 Jul, 2011 07:56 am
@maxdancona,
It's not any different, but I was missing a factor, the fact that there are only as many possibilities to match as they are Smile

I'm thinking that a player picking his drawing from a bowl of numbers is an independent event from the lottery drawing, thus the two events have to be multiplied before divided to the possible number of matchups. This is important for more complicated calculations

So to preserve the metaphysics of this calculation, the player - lottery probability formula analogy to two 6 sided dice, is (1/6 x 1/6) = all combinations, divided by the number of ways they can match up, which is the entropy of one 6 sided dice (hope I hit that fancy term right:) ), thus (1/6 x 1/6)/(1/6)=(1/6 x 1/6)*(6/1)=6/36=1/6

Legend:
http://en.wikipedia.org/wiki/Entropy_%28information_theory%29
"In information theory, entropy is a measure of the uncertainty associated with a random variable."
a measure of the uncertainty = probability
0 Replies
 
cicerone imposter
 
  1  
Reply Sat 30 Jul, 2011 09:53 am
@w0lfshad3,
Thanks for that Wiki article on how the math was done. The reason I came up with a different answer is here (also from Wiki). It depends on how numbers are used to determine the winner.

Quote:
On February 16, 2002, the California Lottery's then-highest payout of $193 million was won by three tickets.

Overall chances of winning a prize are 1 in 23. Chances of winning the jackpot are 1 in 41,416,353 (41,416,353 = 47/5 x 46/4 x 45/3 x 44/2 x 43/1 x 27; if the first five numbers had to be picked in one specific order to win the jackpot, the odds would be 1 in 4,969,962,360).


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