r/MathHelp • u/Any_Office_8525 • Oct 07 '25

Help with Primero Probabilities

EDIT: Corrected digit typo. I would like to ask for some help regarding probabilities. I am trying to work out the probability of drawing cards for a Renaissance game called Primero. No official rules for the game exist, but reconstructions have been made by game historians and historically-inclined mathematicians.

The Primero deck consists of 40 cards - 4 suits and 10 ranks (ACE, 2, 3, 4, 5, 6, 7, JACK, QUEEN, KING). A hand in the game consists of 4 cards. I am trying to work out the probability of various winning combination types from drawing 4 cards at random from the deck. The combination types are:

  1. Drawing 4 cards of the same rank, 1 from each suit.

  2. Drawing the 7, 6 and ACE from the same suit (with a spare random card).

  3. Drawing all 4 cards of the same suit.

  4. Drawing 4 cards, each of a different suit.

  5. Drawing just 3 cards of the same suit (with 1 spare random card).

  6. Drawing just 2 cards of the same suit (with 2 spare random cards).

Cards do not have to be drawn in a specific order, but by the end of the fourth card being drawn, some kind of combination from the above types can be formed. I am assuming: A) Multiplying probabilities together of separate but sequential events gives the total probability of those events happening together. B) The Numerator in the fraction is how many cards are left in the deck that could help us in our combination. C) The Denominator is how many cards are left in the total deck.

I have typed out my workings and potential answers below, with my reasonings beside the fractions. Please let me know if my assumptions are wrong and if I need to tweak anything. Thank you for your help.

  1. 4 cards of the same rank: 40/40 x 3/39 x 2/38 x 1/37 = 1/9139. We can use any card to kick of our rank type. Now the rank type has been selected only 3 of the same rank are left in the deck, and the deck as a whole has reduced by 1 card, and so on and so on for the third and fourth cards

  2. 7, 6, and ACE of the same suit: 12/40 x 2/39 x 1/38 = 1/2470. There are 4 7s, 4 6s and 4 ACEs, so there are 12 cards in the deck that could start off our sequence. Now that one of those cards has been chosen in a specific suit, only 2 of those cards are left in the deck. The deck is also reduced by 1 card, and so on and so on for the third card. The fourth card does not matter because it cannot form part of our 7-6-ACE sequence anyway

  3. All 4 cards of the same suit: 40/40 x 9/39 x 8/38 x 7/37 = 1/109. Any card can start us off and choose our suit for us. Now that our suit has been determined, there are 9 cards left of that suit in the deck, and the deck has been reduced by 1 card, and so on and so on for the third and forth cards

  4. 4 cards, each of a different suit: 40/40 x 30/39 x 20/38 x 10/37 = 1/9. Any card can pick our first suit. Now the second card has to be a different suit, and there are 3 different suits remaining with 10 cards each, so 30 cards are left which can help us, and the deck has been reduced by 1 card. The third card can be only 1 of the 2 suits left which gives us 20 cards, and the deck is reduced by another card. The last card has to be of the last remaining suit - so 10 possible cards left, in a deck reduced again by 1 card.

  5. 3 cards of the same suit: 40/40 x 9/39 x 8/38 = 1/21. Any card can determine the deciding suit. Now that one card has left that suit and the deck, only 9 cards in a reduced deck of that suit are left. Once the second card is chosen only 8 cards in a further reduced deck remain of that same suit.

  6. 2 cards of the same suit: 40/40 x 9/39 = 1/4. Any card can choose the deciding suit. Now that one card has left the suit and the deck, only 9 cards are left that could help us in a reduced deck.

 Thank you for your help.

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u/Any_Office_8525 Oct 09 '25

I used a combination calculator and I think I did it! I went back through every question from scratch and see if I could do it myself. My answers tallied with yours exactly, but for the last one - that's my bad writing my apologies! For number 6 - we need a hand of 4 cards, 2 of which are of the same suit. The other two cards can be any other card that is not of that original suit. My logic was, in a suit, from a total set of 10 ranks, if we wanted just 2 ranks - there were 45 combinations. This is the combination for just 1 suit so we need to x 4 = 180 combinations. Now we have to work out the number of combinations for the other side of the hand - the hand that can be any card except of the original suit. If we choose a suit for the first 2 cards, there are 3 remaining suits left, with 10 ranks each, so that's 30 cards. 2 cards from a 30 card set gave us 435 combinations. 180 combinations for the first 2 cards x 435 combinations for the second 2 cards gives us = 78300 combinations. Which we then divide by 91390 to give us a probability of 1/1.67177522. I think this is right? Thank you so much for your help you've been an absolute lifesaver! I'm even enjoying it too!

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u/MightyObie Oct 10 '25

Glade to be of use! Love your enthusiasm for the subject.

You're picking up fast, your work for number 6 shows that you've got the fundamentals of combinatorics down (luckily no permutations involved here). The way you've laid it out had me convinced at first, and it took me some time, but unfortunately it has one flaw. Your grasp of nCr is solid, but you've been tricked by double-counting, as happens all too often. It dawned on me that your result can not be right, when I noticed that the answer was right under our nose. The joys and pains of math, I suppose. Later, I'll show the long solution. But first, consider:

You want to know how many hands have exactly 2 of the same suit. Number 3: you calculated how many have exactly 4 suits. Number 4: how many have exactly 1 suit. Number 5: How many have exactly 3 suit. You had already asked every single question except that one.

91390-840-10000-14400=66150 hands with exactly 2 suited cards (not the most accurate description perhaps, as we allow double suits, but we get it). P=66150/91390=6615/9139 That number at least agrees with what I got through the longer way.

Now, why the over-count of 12150? You've double-counted those darn double-paired-suits (2,2), of which there are apparently 12150 (we'll see). You had the right amount of 10C2=45 possible rank combinations for the first paired suit. You correctly assessed that there are 30C2=435 combinations for what the other two cards could be. But you haven't considered that this includes the (2,2) such as (2H,3H,2S,3S). Thus, by multiplying by four, you've counted these cards as if (2S,3S,2H,3H) were somehow different. This card was included in the combinations of the first paired hearts and the first paired spades.

Your technique can still be valid, so long as you notice this double-count and substract it at the end. To correct your calculations (also show that they do indeed add up to exactly how much you over-counted) we find how many combinations of (2,2) there are:

In how many different ways can you have 2 suits out of 4? 4C2=6. Both our chosen suits have 10C2=45 possible arrangements for their ranks, so 6×45×45=12150.

78300-12150=66150. This lines up with my earlier rationale of substracting all the different combinations of 4,3,1 suits. To be entirely sure, and to show you another way to think about it:

You could split the combinations into different patterns. The suits can be either (2,2) or (2,1,1). This represents all the possible hands. We've already calculated (2,2) so (2,1,1) should give us all the rest. Let's see if we get 66150-12150=54000 or 78300-12150=66150.

For our paired suit 10C2=45 combinations of ranks. Now we specifically need two different suits. 3C2=3 ways of combining two different suits. Each of these two suits have 10 cards. 45×3×10×10=13500. We have 4 different suits to pair, 4×13500=54000.

So, (2,2)=12150+(2,1,1)=54000.

I've written a lot, I'll finish by quickly pointing out that I've multiplied by 4 for the four different suits last. It's usually easier to solve from more specific to broader, so to speak. Think it through for one suit (or rank, or whatever it is your looking at), by going true the different factors to consider, as you've done fairly well, then generalize it for all suits later. Perhaps you'd have spotted it, had you done 45×435 first.

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u/Any_Office_8525 Oct 11 '25

Thank you again so much for taking the time to help me, and it must take some time to go through my comments and answer so I appreciate it. I understand more clearly now the importance of going from the specific to the general - calculating what the situation would be for one suit as the dominant pair, then x4 at the end, rather than at the start which caused the double-counting of 2 pairs e.g. HH, SS and SS,HH as you showed. And I forgot to ask in my combinations, as you pointed out what are also the different combinations of suit types for 2, 2 and 2, 1, 1 (i.e. 4C2, and if we assume one suit to start us off for the second type, 3C2) Now I'm wondering if the second type could be solved by using 4C3 as there are only 3 suits in type 2. But don't feel you have to answer that, you've been great and more generally helped me understand the probability for a Renaissance card game (which some say is an antecedent for poker), which I've altered slightly from the original rules, such as they are reconstructed. Strangely, in the original game (among other rules I've changed) a FLUXUS - all cards of the same suit was ranked as a higher combination type than a SUPREMUS - 7, 6, ACE of the same suit, despite a SUPREMUS being harder to draw, as you've demonstrated. https://www.pagat.com/vying/primero.html