Math Question? *shrugs*

[mikem562]

hopefully someone here can help me out. I'm trying to figure out the odd's of something happening, but my math knowledge is falling quite short. If anyone could give me a hand that would be great or even a link to something that could potentially help. If you have a deck of cards I'm trying to figure out the odds of dealing out 13cards, 2sets of 6 & 1off having them end up like this : (b=black|r=red) :

set1 - b_king b_king r_jack b_5 b_a b_9

set2 - b_queen b_queen r_jack b_5 r_a b_9

1off - r_a

 

*shrugs* any thoughts would help

[Alex]

Does the order of the sets matter (e.g. set 2 before set 1). Does the order of the cards within a set matter?

[mikem562]

the order of the cards within the set does -NOT- matter as long as each set contains those 6 cards.

The order of the sets should be as being delt to two people.

 

1card to set one

1card to set two

continue until both sets are full

[Alex]

Does set 1 have to Go to person 1, or could who gets which set be switched? Also, is the "1 off" the next card after both people have 6 cards?

[mikem562]

which person gets which set dosn't matter. sorry for not including all this previously. Person1 could get set 1 or 2.

the one off will be delt last after each person has their set

[Alex]

It's late, and it's been a long day, so forgive me if I make a mistake.. my brain is half asleep. 

 

Let P(k) be the probably of the kth card drawn as we want. So then, we want to find , or the probability of the 13 successive events. We define the two sets as A and B. I use the notation: B4, for example to mean black 4, BK means black king, etc..

 

P(1) - This is the simple one, the first card just has to be any of the first 12 cards we're looking for. However, we have to be careful because there are two extra cards in the deck: the black ace and red ace. There are two of each of those, but we're only looking for one of each.

 

P(1) = P(any card in A or B) = P(BK ∪ BQ ∪ RJ ∪ B5 ∪ BA ∪ RA ∪ B9) = 14/52

 

 

P(2) - This is where it starts to get a little more complicated, because there are multiple cases to account for. The RJ, B5 and B9 are needed in both sets, so we need to keep this in mind. For example, if in the first pick we drew a BK, because the second draw wouldn't require a BK, that wouldn't affect the probability of that case; however, if we drew a RJ, the second draw also needs that, so we need to keep that in mind.

 

If RJ or B5 or B9 are not in the first draw, then the probability of getting the next card we want is 10/51. Otherwise, it would be 9/51:

 

P(2) = P(RJ or B5 or B9 not in first draw) * 10/51 + P(RJ or B5 or B9 in first draw) * 9/51

 

--

 

Those are the first two, it gets longer more tedious to compute the further you do down because there are more possible scenarios to account for. Hopefully that can get you started, but maybe I didn't explain well enough. I'll get some sleep and maybe continue with the problem later. Good luck.

[Alex]

Continuing where I left off..

 

For P(2) we have:

 

P(2) = P(RJ or B5 or B9 not in first draw) * 10/51 + P(RJ or B5 or B9 in first draw) * 9/51

 

Now we need to calculate P(RJ or B5 or B9 not in first draw) and P(RJ or B5 or B9 in first draw).

 

P(RJ or B5 or B9 not in first draw) = 8/14

 

P(RJ or B5 or B9 in first draw) = 6/14

 

So we get:

 

P(2) = 8/14 * 10/51 + 6/14 * 9/51 = 67/357

 

 

So, so far we have P(1) = 14/52; P(2) = 67/357. So the probability that the first two steps happen successively is: = P(1) * P(2) = 6/14 * 67/357 = 67/833.

 

I don't have time to continue now but hopefully that at least helps. I don't actually have much experience with probability, so it may look more complicated than it actually is, and it might be hard to follow my train of thought. You just need to know a few simple rules (that are more or less common sense anyway) and apply them (e.g. Probability of something happening = that outcome / total possible outcomes). It's a bit involved, but it all really stems from that simple idea.

[mikem562]

Thank you so much! I'm following your train of thought and me & my friend are going to sit on a patio today and try to work it all out. The information you gave was exactly what I was looking for, though I do have a question.

 

P(RJ or B5 or B9 not in first draw) = 8/14

P(RJ or B5 or B9 in first draw) = 6/14

How did you figure out the 8/14

 

[Alex]

If we made it to "step 2", then we know that one of the following cards was selected: BK, BQ, RJ, B5, BA, RA, or B9. In total that's 14 cards. That's where the /14 comes from. Now we need see how many of those aren't RJ or B5 or B9 to calculate P(RJ or B5 or B9 not in first draw). So that means it can be BK, BQ, BA, or RA. There are 8 of those in the deck, so 8/14.