0
$\begingroup$

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

$\endgroup$
14
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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. $\endgroup$
    – Prasun Biswas
    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. $\endgroup$
    – Leon van den Boogaard
    Mar 23, 2018 at 17:06
  • 1
    $\begingroup$ 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

Reset to default
1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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!!! $\endgroup$
    – Leon van den Boogaard
    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
1
$\begingroup$

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:

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

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.