Probability of drawing at least one repeated cards on multiple draws?

Question

Those are the mechanics:

• You always draw from 2 to 10 cards.
• You only use from A to 10. So 40 cards.

What I want to obtain:

• The % of times you gonna obtain a number repeated.

I want to make a graphic with how much % of at least one repeat you obtain with every number of draws.

I'm really really ignorant on this field, so please, teach me.

On the 2 draw I know I have to do 4/40*3/39 if I am not wrong. I do that because you have 4 possibilities to draw a certain number on the 40 cards and then 3 in 39 of the rest to repeat that number. It's that correct?

But I got lost in the 3 draw, because you can have the repeat in the first and second draw or in the second and third one.

How can I calculate this and the rest from 3 to 10 draws? Please, be as simple as possible so I can understand.

Answer 1

It's easier to calculate first the probability of getting no repetitions. Let $N_k$ be the event of no repetitions with $k$ draws, you have: $$P(N_k) = \frac{36}{39}\cdots\frac{40 - 4(k - 1)}{40 - k + 1} = \prod_{i = 1}^{k - 1} \frac{40 - 4i}{40 - i}$$ You can think of the process as follows: at each draw, you are "burning" one value. Therefore you have to discount all the other cards with the same value and different suit. In total they are four (the drawn one plus three in the deck).

It follows that the sought probability is $$p_k = 1 - P(N_k) = 1 - \prod_{i = 1}^{k - 1}\frac{40 - 4i}{40 - i}$$

This is how the "distribution" looks like:

Answer 2

On the 2 draw I know I have to do 4/40*3/39 if I am not wrong.

You are wrong.

The probability for drawing at least one repeated value, when taking two draws without replacement from a deck of four suits of ten values each, is:$$\require{cancel}1~-\color{silver}{\cancel{\dfrac{40}{40}}\cdotp}\dfrac{36}{39}$$

Calculated by considering that the complement event is : drawing all distinct values.

So the probability for drawing at least one repeated value when taking $n$ draws without replacement from a deck of four suits of ten values each is:

You can do it if you try.   The complementary event is selecting $n$ distinct values from ten, each from one of four suits, when selecting $n$ cards from forty.