# Looking for terms/formuals to use.

#### irons

##### New Member
I am seeking a way to express a problem (and know very little).

I have a 60 card deck (MTG) that can contain up to 4 of the same cards. What is the probability that I will select 4 of the same cards within 17 successive draws. Assume, cards are removed and not replaced. Nothing else happens to the deck outside of the draw.

At first I started with (4/60)*(3/59).... This seemed to be a best case scenario.

I moved to n=1,17. (4/60-n)*(3/59-n)... However, this didn't seem to fit but I can't put my finger on it. Outside that doesn't seem to account for number of draws when a failure state happens.

Next I considered I am only working with 17 cards. Of which I have 4/60+1/17 of even 1 of them being the card I need. .066*(4/17)*(3/17)... Still not accounting for the success/fail at any point.

I am sure I am over simplifying the issue. What terms/formulas should I be looking for to express this problem? Thank you for any assistance.

Irons.

#### asterisk

##### New Member
We would need to know the exact makeup of the deck to answer this. Are there exactly 15 sets of four in the deck, or are there say 10 sets of four and then 20 other cards that don't have any four of a kinds among them?

Here is a link that shows how to calculate poker hands.

If you had a standard deck of cards, and wanted to know the probability of getting four of a kind in 17 cards, it would be

$$\frac{13*{{48}\choose{13}}}{{{52}\choose{17}}}$$

= $$\frac{13*\frac{48!}{13! * 35!}}{\frac{52!}{17! * 35!}}$$

= $$\frac{13 *192928249296}{21945588357420}$$

= 0.114285714

btw this will include hands that have more than 1 four of a kind

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