I found that the probability of drawing three cards and have them be of the same suit i.e. HHH, DDD, CCC,SSS from a 53 card deck (one joker included) is $(52/53)*(12/52)*(11/51)=0.048835$
I want to find the probability of drawing any two suit sets in a row. I believe I would find the probability of drawing the same suit set as before and add the probability of drawing any of the other three possible suit sets.
Is the probability of drawing another same set as the previous suit $10/50*9/49*8/48 = .006122$ ?
If so then do I add that to the probability of it instead being any of the other suit sets $39/50*12/49*11/48=0.043776$
$.006122+.043776=0.049898$ This doesn't make sense to me since the probability would be higher than drawing just the first suit set.
I can calculate the independent probability of drawing two suit sets in a row as $((52/53)*(12/52)*(11/51))^2=0.002385$ so I feel common sense says the dependent probability should be higher than this value but lower than drawing a single suit set.