What are the odds of guessing at least one number of 3 played in a 10/20 loto

Say you have a lotto game 10/20, which means that 10 balls are drawn from 20 which are in a cylinder.

Now, lets say I bet on one exact number to come out (for example number 6), I have the 50% chance of doing that, right?

Ok, now please tell me if I'm correct by stating that the chance of guessing three numbers (in example: i say that from 10 numbers which are drawn there will be 2, 5, 8) is this: 10/20 * 9/19 * 8/18 = 0,10 => 10%

Ok, now further more (and this one I don't know how to approach): what is the probability of me guessing exactly one number, or exactly 2 numbers, or none of them (of those three which I chose). Even further what is the probability to guess AT LEAST one number, or AT LEAST two numbers.

I will be grateful for any help, or any reading material reference.

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You are right about the probability of guessing $3$. Your method can be adapted to solve the other problems. But my preferred way of doing it involves the binomial coefficients, which in general count the number of ways of choosing $r$ objects from $n$ objects, order irrelevant. Maybe also look up combinations. Note that if you can handle exactly $0$, $1$, $2$, $3$ you can handle the "at least" stuff. For example the probability of at least $2$ is prob. of exactly $2$ plus prob. of exactly $3$. – André Nicolas Jun 2 '11 at 11:12
@user6312: Thank you for your comment I now understand (I better say you've reminded me) about the "at least" stuff. Please take a look at the answer provided by Henry and my response to that answer and see if you can maybe shed some light on my problem. – Nikola Jun 2 '11 at 12:34
You need to be clear whether order matters. Henry's answer assumes that guessing (2,5,8) and getting (2,8,5) is three correct, not one. – Ross Millikan Jun 2 '11 at 13:08
@Ross: yes, he is right, so if I say that numbers 2,5,8 will be n that 10 numbers that are drawn then the order of them actually "coming" out is not important. – Nikola Jun 2 '11 at 13:16

If 10 out of 20 are drawn and you guess 3, then the probabilities are

• Three correct: $\dfrac{10 \times 9 \times 8}{20 \times 19 \times 18} = \dfrac{2 }{19} \approx 0.10526$
• Two correct: $\dfrac{3 \times 10 \times 9 \times 10}{20 \times 19 \times 18} = \dfrac{15}{38} \approx 0.39474$
• One correct: $\dfrac{3 \times 10 \times 10 \times 9}{20 \times 19 \times 18} = \dfrac{15}{38} \approx 0.39474$
• Zero correct: $\dfrac{10 \times 9 \times 8}{20 \times 19 \times 18} = \dfrac{4 }{38} \approx 0.10526$

As a check, those numbers add up to $1$. The factors of 3 come from the possibility of reordering correct and incorrect answers

• At least two: $\dfrac{2 }{19} + \dfrac{15}{38} = \dfrac{1}{2} = 0.5$
• At least one: $\dfrac{2 }{19} + \dfrac{15}{38} + \dfrac{15}{38} = \dfrac{17}{19} \approx 0.89474$
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@Henry: thank you for your answer, though I must admit I don't understand how can the possibility of guessing two numbers be the same as guessing one? Somehow I'm thinking that it should be easier (bigger number) to guess one number than two. Can you please elaborate? – Nikola Jun 2 '11 at 12:27
@Nikola: with 10/20, half the numbers are right and half wrong, so the probability of getting 2 right and 1 wrong is the same as 2 wrong and 1 right. – Henry Jun 2 '11 at 12:58
@Nikola: It is indeed easier to guess 1 number correctly (probability of being right is 10/20 = 50% if you guess only 1 number, as you calculated yourself) than to guess 2 numbers correctly (i.e., if you guess two numbers, probability of both of them being right is $\displaystyle\frac{10\times9}{20\times19} \approx 23\%$, indeed less than 50%). What you have here is that if you guess 3 numbers, then the probability of [1 right, 2 wrong] is same as [2 right, 1 wrong]. But perhaps you want to compare the probability of [at least 1 right] with [at least 2 right], which are 34/38 and 19/38. – ShreevatsaR Jun 2 '11 at 13:26
@all: ok, yes you made your point though I really must admit this just doesn't cut in my logic. Ah, always was jealous to those colleagues that understood combinations and stuff like that at college :) – Nikola Jun 2 '11 at 14:03
@Nikola: Maybe this will help: whether you say "exactly 1 right and exactly 2 wrong" or you say "exactly 2 right and exactly 1 wrong", you're forcing the entire set of 3 results. Because there are the same number (10) of both "right" and "wrong", the labels "right" and "wrong" are irrelevant. This is also why the probability of "exactly 0 right and exactly 3 wrong" is the same as "exactly 0 wrong and exactly 3 right". – ShreevatsaR Jun 2 '11 at 16:52