Hello, I need a little checkup and some help on what is probably some basic stuff relative to other questions on here.

Say it is given that you hold 234XY where X and Y are two cards 9-A that you are discarding.

What is the probability that the 234 is unsuited?, is it (24/64) = 37.5%.

Denominator is 4*4*4=64; Numerator = 4*3*2*1

What if we hold 2345X where X is a card 9-A we are discarding.

Is this one simply 24/256 = 9.4%

In the case of the 2345X, what is the probability of having exactly 3 suits?

Thanks

First off, all questions are welcome here. There is no such thing as a "too easy" or "too hard" question.

Second, there are multiple ways to answer your questions. I will post how I would do them. Others will likely have other approaches (hopefully, all will lead to the same answers).

(1) What is prob that three random cards of different ranks will have three different suits (be unsuited)?

There are 4 different suits. For the three cards to be unsuited, three of the four suits must appear and the order is important. That is, shd would be different than hds.

Permutations are best here. P(4,3)=24. Permutations are typically used when the order of items selected is important (such as if a selection of pq would be counted as distinct from a selection of qp).

Formula for P(N,K) = N! / (N-K)! where the ! denotes the factorial function.

There are a total of 4^3 (=64) possibilities for the suits of three different ranked cards.

So prob is 24/64.

(2) What is prob that four random cards of different ranks will have four different suits (be unsuited)?

There are 4 different suits. For the four cards to be unsuited, all four suits must appear and the order is important. That is, shdc would be different than hcds.

Permutations are best here. P(4,4)=24.

Formula for P(N,K) = N! / (N-K)! where the ! denotes the factorial function. (Clearly, P(N,N) = N! since by definition 0!=1.)

There are a total of 4^4 (=256) possibilities for the suits of four different ranked cards.

So prob is 24/256.

(3) What is prob that four random cards of different ranks will have exactly three different suits?

Clearly, one suit must appear twice (two cards of the four will be of that suit) and two suits must appear once (two cards of the four will each be of one of the two suits).

There are C(4,1) ways to choose the "double" suit where C(X,Y) is the combination function. C(X,Y) = X! / [Y! * (X-Y)!]. Combinations should be used (instead of permutations) when the order of items is irrelevant (where a selection of pq would not be counted as distinct from a selection of qp).

There are C(4,2) ways to choose the two ranks (cards) to have that chosen "double" suit.

There are P(3,2) ways to choose the suits of the other two cards from the remaining other three other suits, where the order of the suits (cards) matters.

So total ways is C(4,1)*C(4,2)*P(3,2) = 4*6*6 = 144.

There are a total of 4^4 (=256) possibilities for the suits of four different ranked cards.

So prob is 144/256.

To show an alternative way, I’ll use probabilities.

(1) What is prob that three random cards of different ranks will have three different suits (be unsuited)?

4/4 * 3/4 * 2/4 = 3/8

(2) What is prob that four random cards of different ranks will have four different suits (be unsuited)?

4/4 * 3/4 * 2/4 * 1/4 = 3/32

(3) What is prob that four random cards of different ranks will have exactly three different suits?

6*(4/4 * 1/4 * 3/4 * 2/4) = 36/64 = 9/16

The 6 is the number of ways to select the 2 of 4 ranks for having the same suit.

Good post.

To further the conversation, some people swear by the "combinatorial" approach while others swear by the "direct probabilities" approach. Obviously, both approaches are valid and, if done correctly, lead to the same answer.

I don't think which approach you use really matters in the grand scheme of things. Use which ever method you are most comfortable with.

Thanks!

Just to circle back abit, what is the probability of being dealt any 234 out of a five card hand to begin with? (All other cards pair the 234 or are 9 or higher)

I come up with 25,200 ways

17,664 (other cards are A-9)

6,192 (one card is paired, other is A-9)

432 (two pairs)

192 (one rank has trips)

So all in all a .97% of being dealt this

Looks correct to me.

Thanks, so appreciated

What is the probability of being dealt just the 23? (All other cards are either pairs or A-9)

For the other three cards being A-9 I think there are 32,384

With one pair and the other two being A-9 it looks like 19,872

How many more are there? I'm just getting confused in my mind right now for whatever reason.

My approach is to carefully list all the possible cases (mutually exclusive and collectively exhaustive). It is typically straightforward to tally each case at that point.

For your question, my list of cases would be:

1. {2,3} are both singletons

2. {2,3} is one pair and one singleton

3. {2,3} are both pairs

4. {2,3} is one trips and one singleton

5. {2,3} is one trips and one pair

6. {2,3} is one quads and one singleton

I tallied these cases and got a total that I'd be happy to share if you want.

yes, please!

So I guess I did #1 and #2...

1. C(4,1)*C(4,1)*C(24,3) = 32,384

2. C(2,1)*C(4,2)*C(4,1)*C(24,2) = 13,248

3. C(4,2)*C(4,2)*C(24,1) = 864

4. C(2,1)*C(4,3)*C(4,1)*C(24,1) = 768

5. C(2,1)*C(4,3)*C(4,2) = 48

6. C(2,1)*C(4,4)*C(4,1) = 8

TOTAL = 47,320

Thanks!

Only when both methods are about equally efficient. But that is not always the case. For instance, to figure the chances of being dealt two pairs, combinations are much better. To figure the chances of being dealt a badugi, fractions are.

Hey, I'm having some trouble computing what I thought she be fairly straightforward odds.

Suppose you have 2c3h7d and discard KdKh

There are 4 cards that will give you a badugi, the 4s, 5s, 6s, 8s

12 more cards will help you with the low, the above cards in the suits you already have. Thus in terms of making some sort of 8, there are 31 cards that will not help.

I tried using fractions to compute the following:

Make 8 badugi and 8 low 3.3%
Make 8 badugi only 13.69%
Make 8 low, no badugi 5.00%
Make 4 card 8, no badugi 39.96%

This sums up to 62%

To help check, I calc the probability of no improvement as (31/47)*(30/46) = 43%

Problem is that it sums to 105% so I did something wrong somewhere....

Of course there are C(47,2)=1081 possibilities for your next two cards. I broke the five card hands that result into the following buckets. (There are undoubtedly easier ways to derive these figures.) Tallies are as follows (probs are these numbers divided by 1081 and are not shown).

Bucket 1: 8 or better badugi & 5-card 8 or better low
Two cases to consider
(i) exactly one spade: C(4,1)*C(3,1)*C(3,1) = 36
(ii) exactly two spades: C(4,2) = 6
Total for bucket = 36+6 = 42

Bucket 2: 8 or better badugi & 4-card 8 or better low
Four cases to consider
(i) no pair: C(4,1)*[24-2] = 4*22 = 88
(ii) one pair, 4568 spades and another 4568: 4*3 = 12
(iii) one pair, 4568 spades and pair is 237: 4*3*3 = 36
(iv) one pair, 237 spades and 4568 of paired suit: 3*4 = 12
Total for bucket = 88+12+36+12 = 148

Bucket 3: 8 or better badugi & 3-card 8 or better low
Clearly this is impossible
Total for bucket = 0

Bucket 4: No 8 or better badugi & 5-card 8 or better low
Two cases to consider
(i) suits are [3,1,1]: C(3,1)*C(4,2) = 3*6 = 18
(ii) suits are [2,2,1]: C(4,2)*P(3,2) = 6*6 = 36
Total for bucket = 18+36 = 54

Bucket 5: No 8 or better badugi & 4-card 8 or better low
Four cases to consider
(i) no pair (one non-spade 4568 and a high card): C(4,1)*C(3,1)*[24-2] = 4*3*22 = 264
(ii) one pair, two non-spade 4568: 4*3 = 12
(iii) one pair, one non-spade 4568 and pair is 237 that is not of same suit as 4568: 4*3*2*3 = 72
(iv) one pair, one non-spade 4568 and pair is 237 that is same suit as 4568: 4*3*1*2 = 24
Total for bucket = 264+12+72+24 = 372

Bucket 6: No 8 or better badugi & 3-card 8 or better low
Clearly this requires two cards higher than an 8 or 237: C(31,2) = 465
Total for bucket = 465

Total for all six buckets = 42+148+0+54+372+465 = 1081.

Thanks! Looks like I got buckets 1 and 5 wrong in my probability calcs. Will review and see if I can figure it out.