i've been looking for help for some days for this,I have a few simple questions. Could you fill in the blancs where needed?
situation : i Pull 3 cards from a normal deck of 52 cards, simultaniously.
Question1 : How do i calculate the odds of drawing a street ? (3-4-5)
Question2 : How do i calculate the odds of drawing a flush? ( Diamond - diamond - diamond) This one i can figure out but i'd like to see the math behind it :)
Question 3 : How do the odds change (if atall) if more decks are added, lets say 3 total decks?
For Q1 : i can figure out theres 52c3 = 22100 possibilities to draw 3 cards differently.
These are the way you can make a street : "1-2-3" "2-3-4" "3-4-5" "4-5-6" "5-6-7" "6-7-8" "7-8-9" "8-9-10" "9-10-j" "10-j-v" "j-v-k" Wich comes to a total of 11 solutions.
This is where i am stuck, i assume you have to calculate in how many way you can make these 11 solutions.
Any help would be apreciated :)
For next post I would advise to post your try, not only your questions because it can be a useful guidline for anyone attempting to solve it, this is my try:
1) Since we want 3 consecutive cards, the first one could be any, but once you get that first one our possibilities are reduced. This is a conditioned probability where $$P(\text{3 consecutives cards})=P(\text{Any card})\times P(\text{Any card+1 (any suit)})\times P(\text{Any card+2 (any suit)})$$
Therefore, the probability of drawing any card would be $\frac{52}{52}=1$, the probability of the next card being 4 possible choices (if my first card was 1, the second one must be 2, and there are 4 suits) out of 51, and the probability of the next card would be also 4, but out of 50, leaving:
$$P(\text{3 consecutive cards})=\frac{52}{52}\times \frac{4}{51}\times\frac{4}{50}\approx0.63\%$$
2)This is done with the same line of thought:
$$P(\text{flush})=\frac{52}{52}\times\frac{11}{51}\times\frac{10}{50}\approx4.31\%$$
3) I would say the odds would change but you should try it and see what happened
P.S. On Q1 I assumed you could draw any card on your first draw, but you must substrac two cases which would be Q and K