what is the probability of getting a 1) 3 card sequence 2) triplet from a pack of 52 cards?

Question

I want a real mathematecian to answer it

Answer 1

Your question doesn't seem to say whether order matters or not.

Below, nCr = n!/{r!(n-r)!}.

If you are drawing 3 cards...

1) for any particular 3 card sequence, the probability is

1/(52*51*50) = 1/132,600 (if order matters)
1/(52C3) = 1/22,100 (if order doesn't matter)

2) for any three cards...

52*(4*3*2)/(52*51*50) (if order matters)
52*(4C3)/(52C3) (if order doesn't matter)

Answer 2

1 in 3

Answer 3

1) 3 card sequence.

Total number of ways in which any 3 cards can be drawn from a deck of 52 cards = 52C3 (order irrelevant)
There are 4x4x4x11 distinct 3-card sequences.
Therefore probability of getting a 3-card sequence = (4x4x4x11)/52C3 = 0.03186
Assumes that Ace is either treated as "1" or follows King, but not both.

2) Triplet (assumes triplet refers to 3 cards of same rank).

Total possible ways of drawing 3 cards = 52C3
Total number of ways of drawing triplet = 13x4 = 52
Therefore probability of triplet = 52/(52C3) = 0.00235