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I am not good in math, but it would be very helpful if you could help me understand something.

Let's say we have ten numbers: 1,2,3,4,5,6,7,8,9,10

Chances I will guess correct number are 1 vs 10.

However what are the chances I will guess the number from one to ten in second, third round as well ? AND what are the chances I will not guess correct number in first, second, third round? How can I calculate this? Thank you.

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    Not sure this is clear. Is each round independent of the first? So knowing that the correct number was, say, $7$ the first time tells us nothing at all about the next round? Or did you mean something else? – 2017-01-14
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    it can be repeated, thank you. – 2017-01-14
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    I suppose, if you randomly guess '2' on the first round, then on the second round you will randomly guess from among the nine numbers 1, 3, 4, 5, 6, 7, 8, 9, 10. And so on, for the third. You need to clarify whether this is true. – 2017-01-14

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If numbers can be repeated

Probability of guessing correct number in first round $=\dfrac{1}{10}$

Probability of guessing correct number in first round and second round $=\left(\dfrac{1}{10}\right)^2$

Probability of guessing correct number in first round and second round and third round $=\left(\dfrac{1}{10}\right)^3$

Probability of not guessing correct number in first round and second round and third round $=\left(\dfrac{9}{10}\right)^3$

If numbers cannot be repeated

Probability of guessing correct number in first round $=\dfrac{1}{10}$

Probability of guessing correct number in first round and second round $=\left(\dfrac{1}{10}\right)\left(\dfrac{1}{9}\right)$

Probability of guessing correct number in first round and second round and third round $=\left(\dfrac{1}{10}\right)\left(\dfrac{1}{9}\right)\left(\dfrac{1}{8}\right)$

Probability of not guessing correct number in first round and second round and third round $=\left(\dfrac{9}{10}\right)\left(\dfrac{8}{9}\right)\left(\dfrac{7}{8}\right)$

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    It may depend on whether the numbers can be repeated – 2017-01-14
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    Hi Henry, yes it can be – 2017-01-14
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    edited my answer – 2017-01-14
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    Thank you Kiran, I have to wait for 6 more minutes to mark your answer as solution... new member i am:) thank you. – 2017-01-14
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    welcome. glad that I could help – 2017-01-14
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    @Kiran I've noticed you're specialty is along the lines of combinatorics and probability ;) – 2017-01-14
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    @TripleA, i am learning these two topics through this site and hence focusing on the same now. good observation. :) – 2017-01-14