Timeline for What's Wrong With My Math - Odds of 3 Cards of the Same Suit When Drawing 10 Cards

8 events

when toggle format	what		by	license	comment
Jun 29 at 19:13	comment	added	Acccumulation		@IanClayman The most generalized for is [Inclusion-Exclusion[(en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle), but often there are ways to simplify to calculations. For instance, in this case, as lulu pointed out in a comment to your question, you can instead calculate the probability on not getting three of the same suit.
Jun 29 at 3:25	comment	added	David K		The calculation in the question also counts each hand with suites DDDDHHCCSS four times: once for each way you can choose one of the diamonds to make the remaining $7$ cards along with the HHCCSS cards. Any hand of the form DDDDDHHCCS is counted ten times.
Jun 28 at 23:57	comment	added	MrPuffer		@N.F.Taussig Thanks for mentioning, completely missed it, I fixed it!
S Jun 28 at 23:55	review	First answers
Jun 28 at 23:56
S Jun 28 at 23:55	history	edited	MrPuffer	CC BY-SA 4.0	added 22 characters in body
Jun 28 at 23:17	comment	added	Ian Clayman		Is there a way to generalize this, or would the conditions change for every number of cards drawn? For example, obvious there’s no risk of drawing a second set of 3 when drawing less than 6 cards. Do I need to split my calculations into less than 6 and 6+?
S Jun 28 at 22:14	review	First answers
Jun 28 at 22:21
S Jun 28 at 22:14	history	answered	MrPuffer	CC BY-SA 4.0