City-Data Forum What are the chances that TWO people win the lottary? (power, player)
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01-08-2011, 11:32 AM
 11 posts, read 7,704 times Reputation: 10

if the odds of winning the mega-million is 1 in 135Millions (ie. 0.0000000074074074).

Isn't the odds of TWO people winning the same lotto equal to (1/135M)*(1/135M)? (which equals: 0.0000000000000001)

is this right?

01-08-2011, 11:38 AM
 Location: Las Flores, Orange County, CA 964 posts, read 2,287,030 times Reputation: 572
Sounds right, but it's early - still not sure what frequency my nerves are on.

OK, here's another one. Would you bet even money that two people in a group of 25 would have the same birthday?

01-08-2011, 11:57 AM
 11 posts, read 7,704 times Reputation: 10
well, the number above, is 1 in 10 quadrillion! (1 quadrillion is 1000 billion!)
so the chances of two people winning the same mega million lotto is 1 in 10 quadrillion!!!!!!!!!!!!!

now what are the chances of 3 or 4 people winning the same lotto? (it HAS happened before!) based on my calculations, it is 0.0000000000000000000000000000000030 which is 1 in about 10 undecillion!!!!!!!!

is it me or are those odds a bit impossible without a bit of rigging?

01-08-2011, 01:47 PM
 Location: Victoria TX 42,668 posts, read 71,590,043 times Reputation: 35874
No, the odds of two people winning is 270-million. Assume that with 135-million tickets, each ticked has a different number, and every number is chosen once. One wins. If 270-million tickets are sold, there will be two with each possible number, including the winning number.

Putting it another way, the odds of two people having the winning number is the same as the odds of zero people having the winning number. However, the odds that you and I are the two winners are, as you say, 135-million squared.

270-million tickets does not mean 270-million people buying tickets. It could mean 27-million buying an average of ten each, which is not at all improbable.

01-08-2011, 02:04 PM
 11 posts, read 7,704 times Reputation: 10
so you're saying this:

1 - odds of two random people having the winning numbers is equal to 1/(135M*2) = 1/270M?
2 - odds of two "specific" people having the winning numbers is (1/135M)^2

i can understand (2) but can you explain (1)?

01-08-2011, 11:11 PM
 Location: Maryland not Murlin 8,187 posts, read 21,752,379 times Reputation: 6116
Quote:
 Originally Posted by ih8you if the odds of winning the mega-million is 1 in 135Millions (ie. 0.0000000074074074). Isn't the odds of TWO people winning the same lotto equal to (1/135M)*(1/135M)? (which equals: 0.0000000000000001) is this right?
No. The odds would be the same since they are based on one winning combination out of all possible combinations. If you bought 10,000 tickets, you would still have the same odds of winning if you bought one ticket since each individual ticket has the same odds.

What you are thinking of is probability, which is different. So, yes, the probability of two people having the winning numbers would decrease.

01-09-2011, 09:02 AM
 11 posts, read 7,704 times Reputation: 10
Quote:
 Originally Posted by K-Luv No. The odds would be the same since they are based on one winning combination out of all possible combinations. If you bought 10,000 tickets, you would still have the same odds of winning if you bought one ticket since each individual ticket has the same odds. What you are thinking of is probability, which is different. So, yes, the probability of two people having the winning numbers would decrease.

not entirely true. if you buy 10k tickets, "EACH ticket" has the same odds of winning....you're correct......BUT YOU would still NOT have the same odds as one ticket. YOUR odds of winning are not 1 in 135M.....your odds are 10K in 135M (which is still very low....but higher than the 1 in 135M)

the original question still stands i guess.....what is the probability of two people (random people) having the winning numbers?

01-09-2011, 09:26 AM
 Location: Somewhere in northern Alabama 16,852 posts, read 51,350,636 times Reputation: 27730
Perhaps the best way to understand the concepts is to look up "factorals" and see how they are used.

01-09-2011, 09:32 AM
 Location: Victoria TX 42,668 posts, read 71,590,043 times Reputation: 35874
Simplify it to a shoe of ten decks of cards shuffled together. Deal 52 cards out of the shoe to 52 players, and probably, one person will get the ace of spades. Now deal 104 cards out to 104 players. Two people will get an ace of spaces. Two winners. Maybe one, maybe three, bur probably two, because when you double the number of players (same odds), you double the number of winners.

If there is one winning card out of 52, and 52 players, you get one winner. If one out of 52 cards is a winner and 104 players, you get two winners. Deal all 520 cards out of the shoe to 520 players, there wlll be ten winners getting an ace of spades.

01-09-2011, 09:47 AM
 11 posts, read 7,704 times Reputation: 10
Quote:
 Originally Posted by jtur88 Simplify it to a shoe of ten decks of cards shuffled together. Deal 52 cards out of the shoe to 52 players, and probably, one person will get the ace of spades. Now deal 104 cards out to 104 players. Two people will get an ace of spaces. Two winners. Maybe one, maybe three, bur probably two, because when you double the number of players (same odds), you double the number of winners. If there is one winning card out of 52, and 52 players, you get one winner. If one out of 52 cards is a winner and 104 players, you get two winners. Deal all 520 cards out of the shoe to 520 players, there wlll be ten winners getting an ace of spades.

first of all, there are 135M different combination for the power-ball........the odds don't change with more or less people playing. (not necessarily 135M people are gonna play......and even if they did, there isn't a guarantee that someone would pick the winning numbers)

your example above is completely different than the powerBall situation.....in the deck of card example you provided, the outcomes are known and your players are getting unique cards. in PowerBall, people are not necessarily getting unique number combinations.
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