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I'm making this table.

You have classic cards from A to 10. 4 of every number. So 40 cards.

It's a hand, so you have one less card on every draw.

I want to calculate the repeats. (only in the numbers)

enter image description here * numbers are rounded

I'm basically ignorant on maths, even this basic, so I didn't understand how I to continue now.

I was calculating the probabilities to at least one repeat, but now I want to calculate the probabilities to obtain a double, a triple and a quadruple.

I have to take in account that now when you have a triple, you didn't have a double and the same for the quadruple and triple.

How can I calculate the rest?

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    So you mean, that if you have 8 8 8 8 (four eights), you want to count it as 1 "triple repeat" but not as a "double repeat", right? (It's not 100% clear) – 2017-01-12
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    If you have four eights, that count as a quadruple and not as a double or a triple. (Ignoring the 2 doubles here, it just "At least one double %") Sorry for the bad explanation. Feel free to edit the question. – 2017-01-12
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    Do you put the cards back in the stack at the end of a draw ? (if you do, the math will be easier) – 2017-01-12
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    No, it's a hand. You have one less card every time. – 2017-01-12
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    It think you will have no other choice than to write a huge probability tree: say you first card is `a`, then for your second draw, you have 3 other `a`, and 39 cards left. You will have $${3}\over{39}$$ probabilities to have `aa`, and $${36}\over{39}$$ to have `ab`. – 2017-01-12
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    Didn't understand from where you obtain those numbers to make the division. – 2017-01-12
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    @Malkev: You have 4 cards of each category. You have a total of 40 cards. If you get one card from category `a`, you only have 3 cards left in that category, and since you took a card, you only have 39 cards left. – 2017-01-12
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    So 3/39 * 36/39. But what I do with that? – 2017-01-12

1 Answers 1

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Suppose you have a deck of 4 Ones, 4 Twos, ... ,4 Tens.

You can write the individual situations as tupels. The tupel will denote how many Ones, Twos, .... , Tens you have after drawn K cards. E.g. $(0,0,2,0,0,0,0,0,0,1)$ denotes that you have drawn 3 cards in total, your draw consisting of 2 Threes and 1 Ten.

Now the probability for drawing such a tupel is given by the multivariate hypergeometric distribution. When you have drawn $K$ cards, consisting of $k_1$ Ones, $k_2$ Twos, ..., $k_{10}$ Tens, you have

$$ p(k_1,k_2, ... , k_{10}) = \frac{{4 \choose k_1} {4 \choose k_2}\cdots {4 \choose k_{10}} }{{40 \choose K}} $$

So for the above example, $$ p(0,0,2,0,0,0,0,0,0,1) = \frac{{4 \choose 2} {4 \choose 1}}{{40 \choose 3}} = 0.0024 $$

Now for your table, you can sum up the probabilities of all tuples which lead to the desired result.

E.g. 2 draws, 1 double (you have that already):

$$ p(2,0,0,0,0,0,0,0,0,0) = \frac{{4 \choose 2}}{{40 \choose 2}} = 0.0077 $$

You can have the 2 at ten positions, so you get

$P($ 2 draws, 1 double$) = 10 \; p(2,0,0,0,0,0,0,0,0,0) = 0.077$

Next: 3 draws, 1 double:

$$ p(2,1,0,0,0,0,0,0,0,0) = 0.0024 $$ (see above)

You can have the 2 at ten positions, and for each of these positions you can have the 1 at nine positions, so you get

$P($ 3 draws, 1 double$) = 10 \cdot 9 \cdot 0.0024 = 0.2160$

Next: 3 draws, 1 triple:

$$ p(3,0,0,0,0,0,0,0,0,0) = \frac{{4 \choose 3} }{{40 \choose 3}} = 0.000405 $$

You can have the 3 at ten positions, so you get

$P($ 3 draws, 1 triple$) = 10 \cdot 0.000405 = 0.00405$

This concludes the 3 draws.

You will have to sum up increasingly many different situations as you go to 4 draws and higher. It may be a bit tedious but it works.

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    Oh, I think I understand how to positionate the numbers, but I don't understand how to calculate it. (4/3) / (40/3) does not give me 0.000405. Sorry for being that ignorant. – 2017-01-18
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    Dear Malkev, $ {4 \choose 3} $ and the other entities are not fractions but binomial coefficients, see e.g. here (https://en.wikipedia.org/wiki/Binomial_coefficient) for computation, also in MATLAB this can be computed directly by the command nchoosek(4,3) . – 2017-01-18
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    Ok! I understand now this point. Last problem I having is with the draws more than 4. For example 7 draws and a quadruple. I have to start with (4,1,1,1,0,0,0,0,0,0) and end with (4,3,0,0,0,0,0,0,0,0)? Because you can't draw more than 4. Is this correct? Then I have to multiply it for 10*9*8*7? – 2017-01-18
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    Multiplying with 10*9*8*7 would be overcounting, since it doesn't matter if your first 1 is on position 2, say, and the second 1 is on position 3, or the other way around. If there are multiple entries with the same number, e.g. as in your example, 3 times the 1, and if you have 9 positions available for these, than the number of arrangements is ${9 \choose 3}$. So the total for one 4 and three 1's is $10 \cdot {9 \choose 3}$. – 2017-01-18
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    Before you run into trouble, here is the general case. Example: (1,1,1,2,2,0,0,0,0,0). How many cases of two doubles and 3 singles are there in total? More generally: If you have a tupel of N (here:10) and if you want to arrange D different events in this tupel (here: D = 3 events: doubles, singles and none's), and if these events occur $k_1, k_2, ... , k_d$ times (here: k1 =2 (doubles), k2 = 3 (singles), k3 = 5 (none's)), with $k_1+ k_2 + ... + k_d = N$, then the total arrangement number is the multinomial $N \choose {k_1, k_2, ... , k_d} $ = $\frac{N !}{k_1 ! k_2 ! \cdots k_d !} $ – 2017-01-18