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The game is played with a deck of $40$ cards. Three of them are dealt to each of two players, so my hand, the cards that were given to me, is formed by three cards of the deck.

At the same time, four type of cards exist: gold, swords, chalice, lumber. The deck of $40$ cards is formed by ten of each of these types.

I was requested to answer what is more likely: to be dealt a hand that contains the 1 of swords (the strongest card), or a hand that contains a $6$ and a $7$ of the same type, whatever that type is.

I have advanced this further, but I'm not really sure if my reasoning is okay.

Let $A = \{1, 2, 3, ..., 40\}$ all the cards.

Let $S = \{1, 2, 3, ...,10\}$, $G = \{11, 12, 13, ..., 20\}$, $C = \{21, 22, 23, ..., 30\}$, $L = \{31, 32, 33, ..., 40\}$ be the subsets of $A$ formed by cards of type swords, gold, chalice and lumber, respectively.

I know there are $\binom{40}{3} = 9880$ subsets of three elements of $A$, i.e $9880$ possible hands. Of those subsets, which of them contains the element $1$? In other words, which of those hands contains the 1 of swords?

Well, those that do not contain $1$ are subsets of $A'=A-\{1\}$, and there are $\binom{39}{3}=9139$ subsets of $A'$. All subset of $A'$ is also a subset of $A$, therefore there are $9139$ subsets of $A$ that do not contain the element $1$. From this it follows that there are $9880-9139=741$ subsets of $A$ that do contain $1$, or rather $741$ possible hands with the 1 of swords. From all the hands, this $741$ are just $\frac{741}{9880}*100=7.5$ percent. So there's a $7.5\%$ chances of getting a hand with the 1 of swords.

I have to main problems:

a) I'm not sure my reasoning is completely right about the chances of getting a hand with the one of swords.

b) How do I find the chances of getting a hand with a $6$ and a $7$ of the same type, whatever that type is? This is the same as asking for the chances of getting a subset of $A$ with $6, 7$ or $16, 17$ or $26, 27$ or $36, 37$ in it.

EDIT: Please note that, in the answer of how many hands would contain $1$, hands that include $1$ and pair of values $6, 7$, $16, 17$, etc., are included. In the answer of how many hands contain the pairs $6, 7$ or $16, 17$, etc., can and should contain those that also contain $1$.

I'm very sorry for the long post, it was either this or not showing what I've tried already. Thank you!

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  • $\begingroup$ Note that the $741$ is correct if you consider the four hands $\{1, 6, 7\}$, $\{1, 16, 17\}$, $\{1, 26, 27\}$, and $\{1, 36, 37\}$ to be included in the same group. Is that what you want? Similarly, you need to clarify if the "hand with $6$ and $7$ of same type" should exclude these. $\endgroup$ – Lee David Chung Lin Apr 14 at 3:52
  • $\begingroup$ You're right, I should. I will edit this. And yes, the hands that contain the $1$ and the desired pair of numbers do not exclude each other, and should be contained in both answers. $\endgroup$ – Lafinur Apr 14 at 4:30
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A hand that contains the 1 of swords must contain $2$ of the remaining $39$ cards. Hence, there are $$\binom{1}{1}\binom{39}{2} = 741$$ such hands as you found. The probability of obtaining a hand containing a 1 of swords is $$\frac{\dbinom{1}{1}\dbinom{39}{2}}{\dbinom{40}{3}}$$

A hand that contains both a 6 and 7 of the same type must contain one of the remaining $38$ cards in the deck. There are four ways to choose the suit and $38$ ways to choose the remaining card, so there are $$\binom{4}{1}\binom{2}{2}\binom{38}{1} = 152$$ such hands. The probability of obtaining a hand that contain both a 6 and 7 of the same suit is $$\frac{\dbinom{4}{1}\dbinom{2}{2}\dbinom{38}{1}}{\dbinom{40}{3}}$$

Hence, it is more likely that a hand contains a 1 of swords than a 6 and 7 of the same suit.

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  • $\begingroup$ This is actually true. As always one feels a bit silly when one's question is asked so flawlesly and simply, haha! Thanks! $\endgroup$ – Lafinur Apr 14 at 14:27

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