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A game consists of 32 cards (A, K, Q, J, 10, 9, 8, 7) in four suits, and each player gets 8 cards.

I need to find the probability that I am being dealt a 10 of one suit, and have all my other cards be of a different suit. I know the chance of having a 10 is 1/8, but I get stuck on this.

The card game is a Dutch card game called 'klaverjassen'. There are 4 suits, just like a normal card game. Each player gets 8 cards out of the 32 cards. Now I need the probability that I get a 10 of one suit, and all of my 7 other cards of different suits. For example if my 10 is a diamond, the other 7 cards either need to be clubs, spades or hearts. It doesn't matter which one it is. So I have 24 cards left where I need to pick 7 cards out of.

There are 4 10's in the game. The goal I have is that I have at least 1 10 in my hand, with all other 7 cards being of a different suit than that 10. I can have multiple 10's, as long as one of them is 'unique', meaning that the other 7 cards are not of the 10's suit. So I can have a 10 of hearts, 10 of clubs, 10 of spades and 10 of diamonds in my hand and when all other cards are also hearts, it is still good

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  • $\begingroup$ This isn't clear. I assume there are four suits, each with these $8$ ranks? Do the other $7$ cards have to have the same suit? $\endgroup$
    – lulu
    Mar 23, 2018 at 16:42
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    $\begingroup$ Note: the chance of having a $10$ in eight cards is not $\frac 18$. It's $1-\binom {28}8/\binom {32}8\approx 0.705$. $\endgroup$
    – lulu
    Mar 23, 2018 at 16:43
  • $\begingroup$ You could add that this is the popular card game 29. $\endgroup$ Mar 23, 2018 at 16:55
  • $\begingroup$ The card game is a Dutch card game called 'klaverjassen'. There are 4 suits, just like a normal card game. Each player gets 8 cards out of the 32 cards. Now I need the probability that I get a 10 of one suit, and all of my 7 other cards of different suits. For example if my 10 is a diamond, the other 7 cards either need to be clubs, spades or hearts. It doesn't matter which one it is. So I have 24 cards left where I need to pick 7 cards out of. $\endgroup$ Mar 23, 2018 at 17:06
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    $\begingroup$ Still don't think it's clear. What if there are multiple $10's$ or is this excluded? $\endgroup$
    – lulu
    Mar 23, 2018 at 17:15

2 Answers 2

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Note: the final answer I arrive at seems too high to me, so I suspect that some arithmetic errors have been made. The core methodology should be sound but the calculation should be checked carefully.

Assuming that multiple $10's$ are allowed. We will work by the number of $10's$ in the hand. The case where only one $10$ is allowed is Case I.

Case I: exactly one $10$. Then there are $4$ suits for the $10$, after which there are $21$ allowable cards in the other suits so: $$\boxed {4\times \binom {21}7\Big/\binom {32}8\approx 0.044220074}$$

Case II: exactly two $10's$ There are $\binom 42=6$ ways to choose the ranks, fix one choice. Then if one of the ranks (of the two $10's$) is to be a singleton, there are $21$ acceptable cards of the other suits so the probability that the other six cards are acceptable is $\binom {21}6/\binom {32}8\approx 0.005159009$ Similarly, the probability that both $10's$ are singletons is $\binom {14}6/\binom {32} 8\approx 0.000285502$ It follows that the answer in this case is $$\boxed {6\times \left(2 \times 0.005159009-0.000285502\right)\approx 0.060195089}$$

Case III: exactly three $10's$ There are $\binom 43=4$ ways to choose the ranks, fix one choice. The probability that a specified rank is a singleton is $\binom {21}5/\binom {32}8\approx 0.001934628$. The probability that two specified ranks are both singletons is $\binom {14}5/\binom {32}8 \approx 0.000190335$ and the probability that all three are singletons is $\binom 75/\binom {32}8\approx 1.99652E-06$. Thus, by Inclusion Exclusion, the answer in this case is $$\boxed {4\times \left(3\times 0.001934628-3\times 0.000190335+1.99652E-06\right)\approx 0.020939505}$$

Case IV: four $10's$. The probability that a specified rank is a singleton is $\binom {21}3/\binom {32}8\approx 0.000126446$. The probability that two specified ranks are both singletons is $\binom {14}3/\binom {32}8\approx 3.46064E-05$ and the probability that three are singletons is $\binom 73/\binom {32}8\approx 3.32753E-06$ whence, again by Inclusion Exclusion, the probability of this case is approximately $$\boxed {0.000311457}$$

The final answer is then, barring arithmetic error (highly probable!): $$.044220074+.060195089+.020939505+000311457= \boxed{ 0.125666125}$$

Note: this seems too high to me, so I would strongly recommend checking the steps carefully.

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  • $\begingroup$ I got about $0.127$ (as with you, via a method I'm not entirely confident in), so I suspect the answer is somewhere in that vicinity. $\endgroup$
    – Brian Tung
    Mar 23, 2018 at 18:11
  • $\begingroup$ @BrianTung Thanks. I can't see anything conceptually wrong with it, but my intuition is unhappy. I suppose I ought to just simulate it. $\endgroup$
    – lulu
    Mar 23, 2018 at 18:13
  • $\begingroup$ For comparison's sake, my expression was $[4C(24, 7)-6C(16, 6)+4C(8, 5)]/C(32, 8)$. $\endgroup$
    – Brian Tung
    Mar 23, 2018 at 18:37
  • $\begingroup$ The way you answer is correct I think, but the answer @BrianTung had, was correct in the program I had to enter it in. Thank you all very much!!! $\endgroup$ Mar 23, 2018 at 18:59
  • $\begingroup$ @BrianTung I found $0,1271$ as answer, via a method that is trustworthy. $\endgroup$
    – drhab
    Mar 23, 2018 at 19:30
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Let $S$ denote the event of getting the $10$ of spades and no other spades.

Let $D$ denote the event of getting the $10$ of diamonds and no other diamonds.

Let $C$ denote the event of getting the $10$ of clubs and no other clubs.

Let $H$ denote the event of getting the $10$ of hearts and no other hearts.

With inclusion/exclusion and symmetry we find:

$$\begin{aligned}\mathsf P\left(S\cup D\cup C\cup H\right) & =\binom{4}{1}\mathsf P\left(S\right)-\binom{4}{2}\mathsf P\left(S\cap D\right)+\binom{4}{3}\mathsf P\left(S\cap D\cap C\right)\\ & =\binom{32}{8}^{-1}\left[4\cdot\binom{24}{7}-6\cdot\binom{16}{6}+4\cdot\binom{8}{5}\right]\\ & =\frac{1336592}{10518300}\\ & \simeq0,127073006 \end{aligned} $$

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  • $\begingroup$ This is exactly the way I reasoned to the answer, and it is exactly the value I obtained. It seems conceptually simpler to me than the other way, but they should both be valid in principle. lulu must be overcounting or undercounting something minor somewhere. $\endgroup$
    – Brian Tung
    Mar 23, 2018 at 22:09

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