Probability question in the card game of Cut-throat pinochle

[Willy d]

Hello, I have a question pertaining to the game of Cut-throat pinochle. Wasn’t sure where to propose this question, so I’m hoping someone here can help me w/ it…
I’ll lay out some conditions to the question, so there will be no need to understand the rules of the game.
1) There are 48 cards in the deck consisting of aces, 10s, kings, queens, jacks, & 9s. Each card has a duplicate in each suit. (2 ace of spades, 2 king of diamonds etc…)
2) it is a 3 player game, with each player being dealt 15 cards randomly (45 cards in all)
3) the last three remaining cards are kept separate in a category called the kitty. (Their value & suit unknown to the players…Face down)
4) players bid based on the value of their current hand, & the possible expectation of what’s in the kitty

My question - what are the odds that a particular card that you need is in the kitty?

-if you hold 15 of the 48 cards in the deck, 33 remain.
- Say you need the jack of diamonds. There are 2 jack of diamonds available of those 33 or (16.5-1)
-However, you are drawing at 3 of those 33 simultaneously (this is where my understanding of probabilities falls short).
*an equation would be helpful

Please let me know if there is any additional info needed that I’ve left out

 

[Steven G]

Math does not always use equations but rather thought process.
Do you have one of the jack of diamond? Having one jack of diamonds with change the probability of the other jack being in the kitty.
Let's assume that you have one of the jack of diamonds. There are 33 cards other than yours. The chance that the jack of diamonds is in the kitty is 3 out of 33 or 3/33=1/11.
What do you think will change if you don't have any j-d?

 

[Dr.Peterson]

- Say you need the jack of diamonds. There are 2 jack of diamonds available of those 33 or (16.5-1)
-However, you are drawing at 3 of those 33 simultaneously (this is where my understanding of probabilities falls short).

I'll suppose that you don't have a jack of diamonds, so there are two of them out there. (Not knowing the rules, I don't know whether you might need a given card only when you have another of the same, or only when you don't; but this assumption fits what you said better. It's also a harder question!)

Imagine laying all the other 33 cards in a row, with the 3 in the kitty first. How many ways are there for the two jacks of diamonds to be in any of the 33 places, ignoring the other cards? (This is a question about how much you know about combinations and permutations.) (And that comment is a hint about what to look up if you don't.)

How many ways are there for at least one of the two jacks of diamonds to be among the 3 in the kitty? The easiest way to answer this is to first find how many ways there are for neither of them to be in the kitty.

 

[Willy d]

I'll suppose that you don't have a jack of diamonds, so there are two of them out there. (Not knowing the rules, I don't know whether you might need a given card only when you have another of the same, or only when you don't; but this assumption fits what you said better. It's also a harder question!)

Imagine laying all the other 33 cards in a row, with the 3 in the kitty first. How many ways are there for the two jacks of diamonds to be in any of the 33 places, ignoring the other cards? (This is a question about how much you know about combinations and permutations.) (And that comment is a hint about what to look up if you don't.)

How many ways are there for at least one of the two jacks of diamonds to be among the 3 in the kitty? The easiest way to answer this is to first find how many ways there are for neither of them to be in the kitty.

Dr Peterson

Thx for the quick response. Ok, you lost me in that riddle of a reply, but maybe you can help me work through it. I’m obviously way out of my depth here.

So here we go, let’s start here..

*To clarify, I was assuming we did not have a Jack of Diamonds in our 15 card hand, & 2 were available of the 33 remaining.

If by permutation you mean the order in which the kitty lays, we don’t have to worry about that. I’m really only interested in the percentage of combinations in which the jd lies in the kitty. So, I’m guessing the total combinations possible would be 33x32x31 or 32,736, but that would include multiple combinations of same cards w/ different orders. I have no clue how to condense that to exclude duplicate combos. & even if I did, i I wouldn’t know what to do w/ it. Nor do I know how to find the # of combos in which both jd’s aren’t in the kitty. This is going to take baby steps, I apologize in advance.

 

[Willy d]

Math does not always use equations but rather thought process.
Do you have one of the jack of diamond? Having one jack of diamonds with change the probability of the other jack being in the kitty.
Let's assume that you have one of the jack of diamonds. There are 33 cards other than yours. The chance that the jack of diamonds is in the kitty is 3 out of 33 or 3/33=1/11.
What do you think will change if you don't have any j-d?

Steven G

Thx for the quick response. When proposing the question I was assuming I did not have a jack of diamonds.

But in reference to your answer, if I did possess one JD, & there are 3 cards in the kitty, w/ 33 cards remaining, & the odds are 1/11 of getting 2nd JD…I’m guessing you arrive there by taking # of cards in kitty/cards remaining? If there were 5 cards in kitty it would be 5/33?

Now assuming I don’t have a jd, that original 3* in the 3/33 wouldn’t double to 6/33 would it? If so, what would happen if there were 19 cards in the kitty? That probability when not possessing a jd would be impossible.

Sorry, I still don’t understand.

 

[Dr.Peterson]

So, I’m guessing the total combinations possible would be 33x32x31 or 32,736

No, that's the number of permutations! Search for an introduction to this topic.

You can use either permutations or combinations; and you can ignore the locations of other cards, or include them. (A lot will cancel if you do the latter.)

Think of 33 blanks, and you are choosing where to put the two JD's, leaving everything blank (to be filled in later). Note: I'm suggesting you focus on the two cards whose location you care about, not the three locations you care about (the kitty)!

(I'm in a hurry to get to work, or I'd say more.)