Question

2. A deck of 16 cards, from the bottom up, consists of 4 ♣, 4 ♠, 4 ♦, and 4 ♥. The cards are cut at a random place. Let Tk be the event that the all the cards of every suit are together after cut k.

(a) Assuming that all possible cuts are equally likely, find P(T2).

(b) Find P(Tk+1) in terms of P(Tk).

(c) If the deck is repeatedly cut many, many times, what is the approximate probability the cards will be together by suit?

Answer 1

A deck of 16 cards, from the bottom up, consists of 4 ♣, 4 ♠, 4 ♦, and 4 ♥. The cards are cut at a random place.

Number of permutations = 16!

All the cards are together = 4!*4!

A)

Assuming that all possible cuts are equally likely

P(1) =P(0) * 4/15

If the first cut does not results in all the suits together, the second cut can result in all the cards together with probability = 1/16* (1-P(T1))

=> P(2) = P(1) * 4/16 + 1/16(1 - P(1))

substitute P(1) values and P(0)

P(2) = 1/16 + 1/774918144000

B)

Recursive relation

= P(Tk+1) = P(Tk) * 3/16 + 1/16; k>=1, P(T1=) = 1/ 4*36324288000

C)

R code

PT_k <- function(k)
{
if(k==1)
{
    return (1/36324288000/4)
}
else
{
    return(3*PT_k(k-1)/16+1/16)
  
}
}

PT_k (89)

lin (k->.infinity) P(Tk) = 0.07692308