What is the probability of making a poker straight?

Question

Probability theory

I had a task to calculate the probability of making a straight, I solved the problem, but there are doubts whether I calculated correctly.
I have 5 and 6 in my hands (any suit)
. 3 cards should be laid out on the table from a deck in which there are already 50 cards left.
To make a straight these 3 cards must be: 7, 8 and 9 of any suit.
In my understanding, the final formula is:
(64*47!*3!)/50! \u003d 4 / 1.225 \u003d 0.0032 or 0.32%
64 is 4 cubed, since each of the cards 7,8 and 9 can be drawn in 4 different ways.
Are my calculations correct or am I missing something?

Answer 1

In fact, poker has many varieties and the chances vary.
www.poker-wiki.ru/poker/%D0%A1%D1%82%D1%80%D0%B8%D1%82

Answer 2

If the hands are 5 and 6, then the straights will be (2,3,4,5,6), (3,4,5,6,7), (4,5,6,7,8) and (5,6 ,7,8,9). For each particular street, the probability of drawing the correct card for the first card is 12/50, for the second - 8/49, for the third - 4/48. Total 0.0032653
The probability of getting one of the straights = 1-(1-0.032653) ⁴ = 0.013 or 1.3%
But in real poker some of the cards are not in the deck, but on the hands of opponents and on the table, depending on the specific option, the probability will change.

Answer 3

Remember that you are not playing poker with yourself. You need to know the number of opponents, which in the case of hold'em, for example, can be up to 7, and this is already 14 cards in hand.