Yahoo Answers: Answers and Comments for Odds i can't work out? [Gambling]
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From xboxNeRd
enAU
Mon, 16 Sep 2019 13:21:53 +0000
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Yahoo Answers: Answers and Comments for Odds i can't work out? [Gambling]
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https://au.answers.yahoo.com/question/index?qid=20190916132153AAdWq7n
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From zman492: You have 120 different balls.
The odds that t...
https://au.answers.yahoo.com/question/index?qid=20190916132153AAdWq7n
https://au.answers.yahoo.com/question/index?qid=20190916132153AAdWq7n
Mon, 16 Sep 2019 19:59:14 +0000
You have 120 different balls.
The odds that the first ball picked will be one of the three is 3/120. The odds that the second ball picked will be one of the three is 2/119. The odds that the third ball picked is 1/118.
That makes the odds (3/120)x(2/119)x(1/118) = 1 in 280,840

From Coffee Drinker: If there are different colors and numbers, the...
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Mon, 16 Sep 2019 21:40:56 +0000
If there are different colors and numbers, then no 2 balls are identical. You essentially have 120 different balls. You'd have the same scenario if you just had 120 balls with the numbers 1120 on them.
You don't mention if you are trying to guess them in the correct order, or just guess which 3 will be drawn. In a standard lotto drawing the order doesn't matter.
If order does NOT matter, then the draw goes like this: you pick 3 color/number balls.
When the first ball is drawn there are 3 that match your pick out of 120, so your odds of getting the first ball right are 3/120
IF the first draw was one of your picks, then your odds are 2/119 for the 2nd draw because 2 of the remaining 119 balls will match your picks.
IF you match the 2nd ball, then you have a 1/118 chance of getting the 3rd and final ball correct.
so your odds are:
(3/120) x (2/119) x (1/118) = 6/1,685,040 or 1/280,840 or 0.00036%
Now, if you MUST get the balls in the correct order, the odds are even worse. In this scenario you have 1/120 chance to guess the correct ball on the first draw, 1/119 chance on the 2nd draw, and 1/118 chance on the 3rd draw.
So the calculation is:
(1/120) x (1/119) x (1/118) = 1/1,685,040 which is 1 chance out of 1.68 million. I'm not even going to try to figure out how many zeros you need to put that into a percentage.
Both of those options assume that you do NOT put a ball back into the mix once it has been drawn. If you put the balls back (in which case its possible to draw the same ball twice or even 3 times), then your odds are altered slightly since there are a full 120 possibilities for the 2nd and 3rd draws. You'd use the same calculations as above with either 321 if order does not matter, or 111 if order does matter, but all fractions would have 120 on the bottom.