Testing my accuracy as I learn combinatorics.

Ok, guys, this will be simple for you, but it's all new to me. :-)

Four cards.

You see the first card.

What are the chances of getting a pair, trips, or quads.

Now, first, the pair.

I figured this out by saying how many ways can I NOT get a pair - (48/51)*(47/50)*(46/49).
That gives me 83.05% chance of not getting a pair, which is 16.95% chance that I do.
(I don't think I am fully done here, though, because I still have to subtract trips and quads.)

Next, trips.

I know that once I see the first card, there are only 3 more cards to make a trips, 3 more cards to reveal out of the 4 drawn, and only 3 ways to get two of those three cards to be duplicates of the first one: xxo, xox, and oxx. Armed with that info, I can compare that to the total number of combinations of possible cards from the deck. I figure that by (51*50*49), or 124,950. So, the chances of getting trips are 3/124,950, or .0024%.

Lastly, quads.

Only one way to get 3 more duplicates of the first card, so it's 1/124,950, or .0008%.

So, overall, my chances of getting a pair, trips, or quads are still 16.95%, but .0016% of that is trips, and .0008% of that is quads. So therefore, my final conclusion is:

Pair: 16.71%
Trips: .0016%
Quads: .0008%

Dang it.

Pair: 16.9476%
Trips: 00.0016%
Quads: 00.0008%

Do your pairs, trips, and quads have to match the first card? Or would a first card of ace of spades and the next 3 cards all fours be trips?

Ah, you are right.

Same with the pairs.

Ok.

So, for pairs:

1*(48/51)*(44/50)*(40/49) = .676110 = 67.611% chance of not getting a pair.

Therefore, 32.389% chance of getting a pair in 4 cards.


For trips:

Previous calculation is correct, EXCEPT I need to add in three of a kind NOT of the original card.

I can calculate that by figuring out in 3 cards the number of ways to get trips, x 12.

So, 4 choose 3 yields 4 combinations, multiplied by 12, is 48.

That is 48 more ways.

The proper answer is 51/124950, or .000408, or .0408%

Quote: 21Flip

Ok.

So, for pairs:

1*(48/51)*(44/50)*(40/49) = .676110 = 67.611% chance of not getting a pair.

Therefore, 32.389% chance of getting a pair in 4 cards.

That is the probability for pair + 2-pair + trips + quads. Is that what you wanted? If you want just pairs, you can do those directly instead of subtracting from 1.

52/52 * 3/51 * 48/50 * 44/49 * 6 ≈ 30.425%.

We multiply by 6 at the end because there are C(4,2) = 6 ways to choose the 2 cards that can be paired. If we left off the 6, we would be counting only cases where the first 2 cards are paired. You can also do this with combinations:

13*C(4,2)*C(12,2)*4^2 / C(52,4) ≈ 30.425%.

That is, 13 choices for the rank that makes the pair, times C(4,2) = 6 pairs of that rank, times C(12,2) = 12*11/2 ways to choose the other 2 ranks, times 4 ways to choose a card from each of those ranks, all divided by C(52,4) combinations of 4 cards which is 52*51*50*49/(4*3*2*1). It's usually cleaner to use combinations like this instead of fractions once you are used to them, even though in this case it might look more complicated. When you use fractions, you have to worry about the order that cards come in or "permutations", and you have to multiply by the number of permutations, in this case 6 as above. Using fractions is equivalent to using permutations instead of combinations.

Quote: 21Flip

For trips:

Previous calculation is correct, EXCEPT I need to add in three of a kind NOT of the original card.

I can calculate that by figuring out in 3 cards the number of ways to get trips, x 12.

So, 4 choose 3 yields 4 combinations, multiplied by 12, is 48.

That is 48 more ways.

The proper answer is 51/124950, or .000408, or .0408%

You would actually need to multiply what you had before by 2*C(4,3)*48, not just add 48. Before you had 3/(51*50*49), but that isn't the probability of making trips with the first card for 2 reasons. First, it ignores the 48 choices for the card that doesn't make trips. Second, the 3 needs to be a 6 because your denominator 51*50*49 is the number of permutations of 3 cards out of 51 or P(52,3). It is not the number of combinations C(52,3) as you stated. That is 6 times the number of combinations since each combination has 3! = 3*2*1 = 6 permutations. If you count permutations in the denominator, then you must count permutations in the numerator as well. So each of your xxo, xox, oxx is actually 2 different permutations each, or 6 total. Doing it your way would give

6*48/(51*50*49) * C(4,3) ≈ 0.00922%.

But if you are going to use fractions, I would recommend instead

52/52 * 3/51 * 2/50 * 48/49 * 4 ≈ 0.00922%.

That is, first we pretend that the first 3 cards are going to be the trips, so that's 3/51 * 2/50 * 48/49. Then we multiply by 4 since there are really C(4,3) = 4 ways to choose the 3 trip cards, or equivalently, 4 ways to choose the non-trip card. Using combinations you can do

13*C(4,3)*48 / C(52,4) ≈ 0.00922%.

Quote: 21Flip

Lastly, quads.

Only one way to get 3 more duplicates of the first card, so it's 1/124,950, or .0008%.

Again, it would have to be 6/124950 ≈ 0.00480% since your denominator is permutations, not combinations. You would use 1 if you had used combinations. For all 4 cards you can do

13/C(52,4) ≈ 0.00480%

since there are 13 ranks that can make quads out of C(52,4) total combinations of 4 cards.


If you want 2-pairs, you can do

C(13,2)*C(4,2)*C(4,2) / C(52,4) ≈ 1.037%.


Now we can check that if we add pair + 2-pair + trips + quads, we get your number 1 - 48/51 * 44/50 * 40/49 ≈ 32.389%.

30.425% + 1.037% + 0.00922% + 0.00480% ≈ 32.389%.

Thank you for that very complete explanation.

I still struggle with knowing whether to use combinations or permutations, and with if I have to worry about the repeating or not repeating aspect.

I understand the principles of each concept, I struggle to know when to apply it.

Any hints or tips on a way to get my mind right?

Quote: 21Flip

Any hints or tips on a way to get my mind right?

practice
practice
practice

some have to (must) do more (like me) and some can do less (like Bruce)


the challenge in the art of counting, imo,
is there are more than just 1 way to count correctly all the possible ways


here is from Brian Alspach

http://people.math.sfu.ca/~alspach/comp17/

"In forming a 3-of-a-kind hand,
there are 13 choices for the rank of the triple, there are 4 choices for 3 cards of a given rank, and there are 48 choices for the remaining card. Altogether, there are 13*4*48 = 2,496 3-of-a-kind hands."


BruceZ showed this
"Using combinations you can do (and NOT "no can do" - Hall and Oats)

13*C(4,3)*48 / C(52,4) ≈ 0.00922%."