Drawing three cards - combined probability

[Lenus]

In a game of Open Face Chinese poker, I encounter this problem quite often.

One draws three cards at a time and one of these cards must be discarded, so two of three drawn cards can be used. A typical problem and question in this game would be:

What is the probability of drawing one spade AND one of either 10,9 or 8 (call it Middle cards)with drawing three cards?

We could have 10,9,8 of spades but we will need either another spade or another of the same group the draw to be successful.

My approach was following -

Success scenarios:

1) Spade and Middle Card -
13/52*10/51
2) Spade and Middle Spade
10/52*3/51
3) Middle and Middle Spade
9/52*3/51
4) Middle Spade and Middle Spade
3/52*2/51

If I sum up all the results, would that bring me to the correct answer?

 

[mathman]

Major flaw. You seem to be calculating possibilities when drawing two cards, but question has you drawing three cards.
Minor flaw. The numbers don't look quite right.

 

[Lenus]

Thanks for the comment.

I realized this would have been correct if I drew only 2 cards, but couldn't figure out how to apply the fact that there are 3 cards drawn. Would that be multiplying every result by C(3,2)?

with regards to minor flow:
S,M = 13,9 (corrected)
S,Ms = 10,3
M,Ms = 9,3
Ms,Ms = 3,2

(by the way, already know the final answer, but I am trying to find out how to solve this type of problems, in general)

 

[mathman]

Your breakdown has one flaw M,Ms is part of S,M. I believe you are correct that you need to multiply by 3 (=C(3,2)).

 

[Lenus]

mathman

I realize my mistake in breakdown groups.
If the first draw is the Middle card, the second could be any of the spades, that's the catch (well, for me)...

I will re-run my calculation.

Thank you very much

 

[mathman]

The problem is a little messier than I first thought. Divide the deck into 4 parts (a) (8,9,10) not spades, (b) (8,9,10) spades, (c) other spades, (d) other cards. You are looking for 3 card sets which contain cards from (1) (a) & (b), (2) (a) & (c), (3) (b) & (c), (b) & (b). For each case you have 4 subcases depending on where the third card comes from. Furthermore some subcases repeat, so the repeats have to dropped to avoid overcounting.

Calculate the number of combinations in each subcase, making sure that permutations are properly taken into account. If the three cards come from different parts, multiply by 6. If two cards come from one part and the other is different, multiply by 3.

 

[mfb]

I would try to calculate the opposite: what is the probability that you get (a) no middle card at all, (b) no spades at all, (c) no middle card AND no spades?
All three are easy to calculate, and allow to find the result.

Note that throwing away the third card does not influence the result (compared to keeping it). There is no possible set of three cards where throwing one away would ruin the set you are looking for.

 

[Lenus]

mfb,

the result i mentioned earlier, that is known to me was achieved by the method you described, but I needed to know how to approach these problems from "this" side too ))

That probability of missing some draws are easier for me, I denote them by q:
q_any middle = C(40,3)/C(52,3)
q_any spade = C(39,3)/C(52,3)
q_any middle, any spade = C(22,3)/C(52,3)

However I was never sure what to do with these results, in spite of one gentleman from another forum kindly explaining it, still need some time to digest that info to be honest.

mathman

I am still banging my head at the method you suggest, allow me to have some time before posting my attempted results

 

[PeroK]

I would try to calculate the opposite: what is the probability that you get (a) no middle card at all, (b) no spades at all, (c) no middle card AND no spades?
All three are easy to calculate, and allow to find the result.

Note that throwing away the third card does not influence the result (compared to keeping it). There is no possible set of three cards where throwing one away would ruin the set you are looking for.

As I understand the game, if you get a middle spade then that can still be a fail. So, you would need to add (d) One spade (middle card) and no other spades or middle cards.

Then failure would be (a) + (b) - (c) + (d).

 

[mfb]

Oh, right. Well, still quite easy.

@Lenus: Those questions typically have many possible ways to solve them, but most approaches involve many different cases to consider.

 

[PeroK]

Oh, right. Well, still quite easy.

@Lenus: Those questions typically have many possible ways to solve them, but most approaches involve many different cases to consider.

I second that. For example, another simple way to look at it is to consider what card you throw away. If the card isn't a spade or a middle card, it's quite easy to see how many ways you can do that. But, you then have to consider all the cases where you throw away a spade or middle card and it gets messy again.