I draw from a deck of 40 cards 9 cards (edit: without replacing them). 3 of the cards in the decks are a Joker. What is the chance I drew at least 1 Joker?
My initial thought was: "Easy!, the first draw chance for a Joker is 3/40, the chance for a Joker on the second draft is 3/39, etcetera." But if I just add up all this chances, something goes wrong. None of those chances is smaller than 3/40, so drawing 14 cards would add up to more than 42/40... above 100%. Experience says, that the chance to draw 15 or more cards out of 40 without drawing one of 3 Jokers is definitely non-zero. The chance will only be zero after drawing 38 cards, as all 3 Jokers could randomly end up on the bottom of the deck.
I believe (and find it hard to exactly explain why) that I have to multiply each chance for a Joker per draft with the chance, that I did NOT draw a Joker in a previous draft, before I add the chances up. So, for 3 drafts the chance for a Joker is
- (3/40) <-first draft
- +(1-(3/40)) *(3/39) <-the chance first draft failed, times second draft)
- +((1-(3/40)) +(1-(1-(3/40))*(3/39)) *(3/38) <-the combined chance the first two drafts failed, times third draft)
which makes for quite horribly long and complicated equations, if I want to calculate the chances for more drafts.
I also believe that I can simplify the calculation, by instead simply calculating the overall chance to NOT have drawn a Joker, which I then can simply deduct from 100% to get the chance of drawing a Joker.
So, if I am correct in my assumption, the chance to NOT draw a Joker at first draft is 37/40. On second draft, the chance to NOT draw a joker is 36/39, so the total chance would be (37/40) *(36/39) to NOT have drawn a joker, third chance would be (37/40) *(36/39) *(35/38) etcetera. Is that correct? Let's check with a deck size of 5, 3 Jokers and 3 drafts. (2/5) *(1/5) *(0/5)=0. The chance to not have drawn one of 3 Jokers from 5 cards after 3 drafts is 0. That seems correct.
Is there a way to generalize this equation with n as the final probability to NOT have drawn a Joker, s as the deck size, j for number of Jokers in deck and d for number of drafts? I assume it should be possible by using integral and sigma signs, but I am not very practiced at using them. Or maybe by using a minimum function?
And, I would be very thankful, if someone can point me to a free online calculator, which accepts that exact generalized equation, resp. lets me enter it with a finite number of clicks, even for a lot of drafts, so I can finally get to optimizing my deck strategies for collectible card games?
From your attempts, the draws are without replacement.
Your formula towards the end is also correct, but the same result can be obtained in a way that you may find easier, using the formula for combinations, for which combination calculators are easily available online.
Out of $40$ cards, $3$ are jokers, $37$ non-jokers, and the formula is
$Pr = \;^{37}C_5 /^{40}C_5$
which is the same as $\frac{37}{40}\frac{36}{39}\frac{35}{38}\frac{34}{37}\frac{33}{36}$
PS The above formulas are for not drawing any joker, that for drawing at least one joker would be the complement of the above, viz using the combinations formula,
$ 1 - [^{37}C_5 /^{40}C_5]$