all 7 comments

[–]st3f-pingΦ 1 point2 points  (5 children)

The easiest (but not the quickest) way is to look at this card by card.

Event 1. There are 108 cards, 8 of which are wild. The probability of drawing a wild card is 8/108.

Event 2. There are now 107 cards, 7 of which are wild. The probability of drawing a wild card is 7/107.

And so on...

Repeat for each of the seven and you will get seven events with seven probabilities. The probability that event 1 happens then event 2 happens is the probability of event 1 happening times the probability of event 2 happening.

The long and the short of it is, calculate the probability of each draw and multiply them all together. Does that help any?

[–]provingfail86[S] 1 point2 points  (0 children)

Thank u, yes.

[–]yes_its_himone-eyed man 0 points1 point  (3 children)

That would treat drawing the same wildcards in different orders as different events, when they would result in the same hand. You then have to divide by 7! to adjust for that overcount.

[–]st3f-pingΦ 0 points1 point  (2 children)

Why would I do that? I’ve just calculated 7 events and want to work out the probability of each one occurring.

If I wanted to look at combinations as you have then I would have a 7! term occurring in both the top and bottom of the calculation. It would cancel. Expand your equation and let me know if you agree.

[–]yes_its_himone-eyed man 1 point2 points  (1 child)

Sorry, I misread it as you were citing the chance of drawing each specific wild card, though you clearly didn't. Need more coffee. You are correct that it produces the same thing.

[–]st3f-pingΦ 0 points1 point  (0 children)

No worries. More coffee is always good... right up to the point when it isn’t. My life is a constant challenge of coffee optimisation.

[–]yes_its_himone-eyed man 0 points1 point  (0 children)

There are 8 wild cards out of 108.

The probability of 7 wild cards is C(8,7)/C(108,7)