What is "luck"?

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19 years 8 months ago #13187 by kcody
Reply from Kevin Cody was created by kcody
No, that's not luck. It's chance. Big difference.

Chance arises when you make multiple attempts at an unlikely outcome. In this case, millions of people picked numbers, leaving about a 1/1500 chance that someone would be right. Decent odds actually.

I was raised by an Irish-American family. "Luck" is easily the most common thing to get shot by the proverbial breeze. "Good Luck" is events of chance <i>occurring exactly when you need them</i>, and "Bad Luck" is when such events happen at an inconvenient time.

In other words, "luck" is entirely circumstantial and perceptual.

-Kevin

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19 years 8 months ago #13188 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by shando</i>
<br />In as much as the "magic" of creation of something from nothing is prohibited in the MM, I would expect that "luck" is probably disallowed too.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">There is nothing magical about luck. It is merely the fulfillment of random chance required by the laws of mathematics.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">So can anyone explain this:<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Yes, you calculated the probabilities incorrectly. The probability of a match to the first number drawn is not 1 in 49, as you assumed, but is 6 in 49 because it only needs to match any one of your six chosen numbers. Likewise, the other probabilities are 5 in 48, 4 in 47, 3 in 46, 2 in 45, and 1 in 44, making the real probability of winning 1 in 14 million.

The reason why you had 50 winners is no doubt because, when there is no winner and the jackpot rises, more than the average number of entries are made in the next lottery, so the chances of a winner rise above the roughly 1 in 2 probability that would exist with only 7 million entries. -|Tom|-

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19 years 8 months ago #13189 by shando
Replied by shando on topic Reply from Jim Shand
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by kcody</i>
<br />
In other words, "luck" is entirely circumstantial and perceptual.

-Kevin
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Hmmm ... perceptual ... just like the rest of the universe [;)].

Thanks Kevin

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19 years 8 months ago #13214 by shando
Replied by shando on topic Reply from Jim Shand
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
<br />
Yes, you calculated the probabilities incorrectly.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
<br />
Ah, 10,068,347,520 / (6*5*4*3*2) = 13,983,816

Thank you Tom.

Just one little quibble - the 6.77 million entry number sets were selected <u>before</u> the winning set was determined. So what is the probability of the selected winning set matching one of the 6.77 x 10^6 entries?

The probability of the winning set matching one of the 6.77 million entry sets would appear to be 1 in 6.77 x 10^6. The number of possible sets is 10 billion. Voila - 10 x 10^9 / 6.77 x 10^6 = ~1500 to 1.

Evidently this does not make a difference.

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19 years 8 months ago #13269 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by shando</i>
<br />So what is the probability of the selected winning set matching one of the 6.77 x 10^6 entries?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Close to 50% (6.77 million in 13.98 million).

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The probability of the winning set matching one of the 6.77 million entry sets would appear to be 1 in 6.77 x 10^6. The number of possible sets is 10 billion. Voila - 10 x 10^9 / 6.77 x 10^6 = ~1500 to 1. Evidently this does not make a difference.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Your calculation would only be relevant if one had to select the winning numbers in the correct order. Then the probability of getting the first number correct would be 1 in 49. But because order is not important, that probability is actually 6 in 49. -|Tom|-

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19 years 8 months ago #13216 by EBTX
Replied by EBTX on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The odds against there being a winner were about 1,487 to 1.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
These are the odds of winning anything at all in your lotto. 2 bucks, 5 bucks or a million bucks (or whatever you call a buck up there ;o). They mean that they are paying some amount of cash on 1 out of 1487 tickets in that particular lottery.

Next time ... do what I do. Take your dollar and flush it down the toilet. There is a finite chance that the toilet will back up and send you a hoard of drug loot that someone was trying to get rid of ... quick. It's not likely ... but the odds are better than the lotto.

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