# odds i can't work out?

I have 12 lots of balls numbers 0-9 (120 balls in total) each lot (10 balls) are different colours.

I put them into a lottery type machine, what are the odds of guessing 3 balls to come out and the correct colour of them 3 also?

### 2 Answers

- Coffee DrinkerLv 710 months ago
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 1-120 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 3-2-1 if order does not matter, or 1-1-1 if order does matter, but all fractions would have 120 on the bottom.

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- zman492Lv 710 months ago
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

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