N.Y. woman wins lottery for second time

[SpaceAce]

[ QUOTE ]
I don't understand the odds they gave. Is that the odds of buying a single ticket for one lottery then buying a second ticket for a second lottery and winning both?

Clearly if the odds of one person winning two lotteries is ~3.6 trillion to one, its unlikely that she's the 3rd person to have done this.

[/ QUOTE ]

I'd love to see someone do the math for having three two-time winners in the New York State Lottery assuming everyone in the state (using the current population) bought exactly two tickets since the inception of the game.

SpaceAce

[elitegimp]

[ QUOTE ]

I'd love to see someone do the math for having three two-time winners in the New York State Lottery assuming everyone in the state (using the current population) bought exactly two tickets since the inception of the game.

SpaceAce

[/ QUOTE ]

Population (2005): 19,254,630 [source]

I'm guessing you actually want the probability of at least 3 2-time winners, but I'm also guessing that the difference is negligable. Just in case, I'll do both (:

1) Exactly 3 winners. Using the numbers quoted in the article, let p = Probability of winning = 1/3,669,120,000,000

The probability of exactly 3 winners is C(population, 3)*p^3*(1-p)^(population-3), or approximately 1 in 1.415*1^16 (that's 14,151,775,022,528,610). At least, according to matlab.

2) p0 = probability of no two-time winners = (1-p)^population = 0.99999475197658


p1 = probability of one two-time winner =
population*p*(1-p)^(population-1) = 5.2477 * 10^-6 (1 in 190,000)

p2 = probability of two two-time winners =
0.5*population*(population-1)*p^2*(1-p)^(population-2) = 1.37*10^-11 (1 in 72,000,000)

probability of at least 3 = 1 - p0 - p1 - p2 = ~1 in 3.5 billion

this is an indication that rounding error is messing up my calculations, and I'm not sure which to believe... I'm using Matlab for all these calculations, which I believe uses double precision. As curious as I am, I'm not gonna roll my own structure for better precision (:

[BruceZ]

[ QUOTE ]
[ QUOTE ]

I'd love to see someone do the math for having three two-time winners in the New York State Lottery assuming everyone in the state (using the current population) bought exactly two tickets since the inception of the game.

SpaceAce

[/ QUOTE ]

Population (2005): 19,254,630 [source]

I'm guessing you actually want the probability of at least 3 2-time winners, but I'm also guessing that the difference is negligable. Just in case, I'll do both (:

1) Exactly 3 winners. Using the numbers quoted in the article, let p = Probability of winning = 1/3,669,120,000,000

The probability of exactly 3 winners is C(population, 3)*p^3*(1-p)^(population-3), or approximately 1 in 1.415*1^16 (that's 14,151,775,022,528,610). At least, according to matlab.

2) p0 = probability of no two-time winners = (1-p)^population = 0.99999475197658


p1 = probability of one two-time winner =
population*p*(1-p)^(population-1) = 5.2477 * 10^-6 (1 in 190,000)

p2 = probability of two two-time winners =
0.5*population*(population-1)*p^2*(1-p)^(population-2) = 1.37*10^-11 (1 in 72,000,000)

probability of at least 3 = 1 - p0 - p1 - p2 = ~1 in 3.5 billion

this is an indication that rounding error is messing up my calculations, and I'm not sure which to believe... I'm using Matlab for all these calculations, which I believe uses double precision. As curious as I am, I'm not gonna roll my own structure for better precision (:

[/ QUOTE ]

Believe the first method, except you have an extra 1 at the beginning (1 in 4.15*10^16, not 1 in 1.415*10^16). In the second method, you are subtracting a number just less than 1 from 1 and losing significant figures. Then you continue to subtract numbers that are almost equal to each other until all the digits remaining are garbage.

The actual result for >=3 is almost the same as for exactly 3. You can get the right order of magnitude "in your head" as follows:

The population is about 20 million, and the probability of winning twice is about 1 in 4*10^12.

C(20*10^6,3)* 1/(4*10^12)^3

=~ 8*10^21/6/(6*10^37)

=~ 2/9 * 10^-16

=~ 2 * 10^-17