In the game Badacey, you are dealt five cards, and I am trying to calculate the odds of three of the cards being 6 or lower of different ranks. So Ac2h3s but not Ah2h6c or Ah2s2c. Here is how I thought to calculate it:

[First Card, A-6, all four suits) x Second Card (5 numbers left, 3 suits) x Third Card (4 numbers left, two suits)] x number of ways to make a 3 card set when given 5 cards

(24/52) x (15/51) x (8/50) x 9 = ~19.5%

The part I'm not sure of yet is multiplying the odds of getting a perfect three card set by the number of possible three card sets given five cards. Anyone out there that can confirm or correct this for me?

combinations are the easier way to calculate this imo. why is A26 not allowed if 3 cards need to be 6 or lower while A23 is?

let me try: you have 6 ranks to choose from (A,2,3,4,5,6) and you must choose 3. from each of those ranks choose 1 out of possible 4 cards. then, you need 2 cards from all 7,8,9,T,J,Q,K (choose 2 cards from 28).

so, the number you're looking for is 6C3 * 4C1^3 * 28C2 = 483840. (exactly 3 cards from 3 different low ranks + any 2 out of higher ranks, can be paired)

total number of possible choices is 52C5 = 2598960, so probability youre looking for is 483840/2598960 = 288/1547 = 18.62%.

with your way, it would be 24/52 * 20/51 * 16/50 * 28/49 * 27/48 * 5C2, which is again the same.

It is 10=C(5,3) ways instead of 9 (a typo?). Otherwise it should be correct

EDIT: I assumed that OP is asking for 3 different suits

Here is my detailed case-by-case attempt at the correct answer of tallying how many 5-card hands have exactly three cards of ranks A-6 of different suits.

Case 3: Exactly 3 cards in Ranks A-6
C(6,3)*C(4,3)*(3!)*C(28,2) = 181,440

Case 4A: Exactly 4 cards in Ranks A-6 with exactly three different ranks in A-6
C(6,3)*C(4,2)*C(3,2)*(2!)*[C(4,2)-1]*C(28,1) = 100,800

Case 4B: Exactly 4 cards in Ranks A-6 with exactly four different ranks in A-6
C(6,4)*C(4,3)*C(4,2)*(3!)*C(28,1) = 60,480

Case 5A: Exactly 5 cards in Ranks A-6 with exactly three different ranks in A-6, with rank counts = [3,1,1]
C(6,3)*C(4,2)*C(3,2)*(2!)*C(4,3) = 2,880

Case 5B: Exactly 5 cards in Ranks A-6 with exactly three different ranks in A-6, with rank counts = [2,2,1]
C(6,3)*C(3,2)*[C(4,2)*C(4,2)*C(4,1)-C(4,2)*2] = 7,920

Case 5C: Exactly 5 cards in Ranks A-6 with exactly four different ranks in A-6, with the three singletons having different suits
C(6,4)*C(4,3)*C(4,3)*(3!)*C(3,2) = 4,320

Case 5D: Exactly 5 cards in Ranks A-6 with exactly four different ranks in A-6, with the three singletons having exactly two different suits
C(6,4)*C(4,3)*C(4,2)*C(3,2)*(2!)*[C(4,2)-1] = 10,800

Case 5E: Exactly 5 cards in Ranks A-6 with exactly five different ranks in A-6, with the suit counts = [3,1,1]
C(6,5)*C(5,2)*C(4,2)*(2!)*C(2,1) = 1,440

Case 5F: Exactly 5 cards in Ranks A-6 with exactly five different ranks in A-6, with the suit counts = [2,2,1]
C(6,5)*C(5,1)*C(4,1)*C(4,2)*C(3,2) = 2,160

GRAND TOTAL = 372,240

As a pct of all 5-card hands: 372,240 / 2,598,960 = 14.3226521%

Besides kapw's correction shown in bold, you made the mistake of counting the same possibilities multiple times. You allowed the 4th and 5th card to be anything, and some of those anythings are cards lower than 6. When you multiply it all by 10, the possibilities of 4 or 5 cards lower than 6 get counted more than once, which is why your answer is too high. Overlap is a bitch.

To avoid that, you can either split it into cases like Whosnext did, or you can use inclusion-exclusion (affectionately referred to as PIE in this forum). I'm a big fan of PIE, but I don't think it saves time for this problem, due to the fact that different possibilities are over-counted by varying magnitudes (at least, at a glance I think they are). PIE is handy when the over-counting is uniform; if it were uniform here, the problem would be reduced to 3 steps (as opposed to the 9 shown by Whosnext). But if it's not uniform, PIE can be as tedious as the direct method, sometimes more so.

Thanks everyone, this is very helpful.

heehaww, suppose I were to fix my mistake regarding change my question to: What are the chances of having 3 of 5 dealt cards be unpaired, of rank 6 or lower, and all different suits, allowing the other two cards to be anything left in the deck? Would my method be correct then, or would I still be at the effect of overlap?

the answer to that question is in whosnext post, first entry. 181440 / 52C5 = 6.98%.

24*(24-4-5)*(24-8-8) * 28*27 *5!/(2!*3!) /(52*51*50*49*48)

you have to take into account other 2 cards (28*27) and factor in number of possible permutations (5!/(2!*3!) ). you must calculate for 5 cards, in your first post you state the problem for 5 cards but calculate only with 3.

No, in fact that's how I originally interpreted your question. I guess in your original question you meant "exactly 3" as opposed to "at least 3". Either version can be done with inclusion-exclusion (but it would be crazy to use PIE for the "exactly 3" version, since it can be done in one step the way md46135 showed).

For the "at least" version, I change my mind -- I think PIE actually will be better than splitting into cases. I'll show how to do it in a bit.

I just woke up. Glad to see the thread is still going.

As we all know, we must be very precise in asking/answering these types of questions. It is very easy to get confused when "ambiguous" terms are used.

Here, what does "three" cards mean? Does "three" mean "exactly three" or "at least three"? In addition, in problems such as this there is another ambiguity.

Which of these hands does/did OP want to include in the tally?

(A) 2s 4d 5h Jc Ks

(B) 2s 4d 5h 5c Ks

(C) 2s 4d 5h 6c Ks

Everyone, presumably, thinks Hand A should be included.

Hand B is a "three-card" badecey but it has four cards in A-6 (the pair of fives are "counterfeited" or whatever the right term is).

Hand C is a "four-card" badecey, of course.

My tally in post #4 above tallied all hands like Hands A and B ("three-card" badeceys). It would not be difficult to tally how many "four-card" badecey hands there are using a similar method. (Doing quick math I think there are 15,840 of those.)

In case anyone cares or if this is of some interest in other calcs, here is my tally for the total number of "4-card badecey" hands (referenced at the bottom of the previous post).

Here is my case-by-case attempt at the correct answer of tallying how many 5-card hands have exactly four cards of ranks A-6 of different suits.

Case 4: Exactly 4 cards in Ranks A-6
C(6,4)*C(4,4)*(4!)*C(28,1) = 10,080

Case 5A: Exactly 5 cards in Ranks A-6 with exactly four different ranks in A-6
C(6,4)*C(4,1)*C(4,4)*(4!)*C(3,1) = 4,320

Case 5B: Exactly 5 cards in Ranks A-6 with exactly five different ranks in A-6
C(6,5)*C(4,4)*C(5,2)*(4!) = 1,440

GRAND TOTAL = 15,840

As a pct of all 5-card hands: 15,840 / 2,598,960 = 0.609474559%