Nonstandard Card-Based Probablility Question

This isn't a homework problem that I need help with or anything of that nature. This is more of a personal math-related problem.

I play a collectable card game, and I am trying to determine which of my decks would benefit the most from a certain rare card that I possess. As part of this analysis I am trying to figure out the probability of drawing certain cards needed for turn-one or turn-two combos that can be played only if I draw the rare card in question. In all likelihood I cannot remotely rely upon these cards being in my opening hand with any reliability given the fact that the chances of drawing a five or six card combo will be ludicrously low, so this is really something I am looking at more for fun than anything else.

I already know the probabilities of drawing *one* of the given cards of a combo in question in a seven card opening hand, but unfortunately the course I took that dealt with probability did not deal with anything as complex as figuring out the odds of having multiple events occur at once. Even if it did I no longer own the textbook and it was years ago.

Let's say that one of the combos requires five specific cards in an opening hand of seven cards, and the probabilities of drawing each card on the first card drawn are as follows:

- Card 1 (the rare card in question): 1/60 chance of being drawn
- Card 2: 21/60 chance of being drawn. Two of this card in an opening hand are required for the combo, so the probability of a second one being drawn ranges from 20/59 to 20/54 depending on the number of cards drawn so far
- Card 4: 4/60 chance of being drawn
- Card 5: 3/60 chance of being drawn

Obviously the probability of drawing one of the desired cards increases with each card drawn. Additionally, only five out of seven cards in an opening hand need to be the desired cards. The other two cards drawn can be anything without affecting the combo's success. Finally, the five cards can be drawn in any order without affecting the combo's success.

Most of the combos I am looking at are five cards, but there is one that requires six cards (which has eight drawn cards instead of seven to draw from since it is a second-turn combo) and a couple that require four cards. I am hoping that whatever solution I am looking for can somehow be applied to any number of needed cards.

I'm afraid that I am wholly unsure of where to begin, and I'm hoping that the solution I am looking for is not horribly complicated.

Thank you!

Re: Nonstandard Card-Based Probablility Question

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Re: Nonstandard Card-Based Probablility Question

Hi Moxen,

This is a fairly complicated problem if I understand it correctly. To recap my understanding, you have a non-standard 60-card deck with

1 One

21 Twos

4 Fours

3 Fives

31 Others

and you would like to know the probability of drawing a 7-card hand with at least 1 One, 2 Twos, 1 Four, and 1 Five.

There are $\displaystyle \binom{60}{7} = 382,606,920$ possible 7-card hands, all of which are equally likely.

We need to count the number of acceptable hands. As a first step, let's list all the possibilities in terms of the numbers of Ones, Twos, etc. Let's say there are a Ones, b Twos, c Fours, d Fives, and e Others. Then the possibilities for an acceptable 7-card hand are

Code:

`a b c d e`

1 2 1 1 2

1 4 1 1 0

1 2 3 1 0

1 2 1 3 0

1 3 1 1 1

1 2 2 1 1

1 2 1 2 1

1 3 2 1 0

1 3 1 2 0

1 2 2 2 0

As you can see, there are 10 combinations of (a,b,c,d,e) possible. For each of these possibilities, the number of hands is

$\displaystyle \binom{1}{a} \binom{21}{b} \binom{4}{c} \binom{3}{d} \binom{31}{e} $

Sum up these products for the 10 (a,b,c,d,e) possibilities. (I used a spreadsheet.) If you do it correctly, I think you will find the sum is 1,980,720. So the probability of drawing an acceptable hand is

$\displaystyle \frac{1,980,720}{382,606,920}$,

which is approximately 0.00512.

If you are not familiar with the notation above,

$\displaystyle \binom{n}{m} = \frac{n!}{m! (n-m)!}$

is the number of combinations of n objects taken m at a time.