Question

# a fair die was rolled repeatedly. a) Let X denote the number of rolls until you...

a fair die was rolled repeatedly.

a) Let X denote the number of rolls until you get at least 3 different results. Find E(X) without calculating the distribution of X.

b) Let S denote the number of rolls until you get a repeated result. Find E(S).

Expected value of an event = Sum of all possible values each multiplied by the probability of its occurrence

a) Now X= number of rolls until you get at least 3 different results

clearly X>=3 since you need atleast 3 rolls to get 3 different resuls

X < infinity since technically we can have infinite number of rolls with same numbers

now, P(X=3) = 5/6 * 4/6

Probability of getting 3 different rolls in 3 attempts will be : in first attempt we will get  a number X (0<X<7) , in second attempt we will get Y (0<Y<7) which is any number but X so its probability will be 5/6 , similarly in third attempt we will get Z (0<Z<7) which is any number but X.Y so its probability will be 4/6 . Since all attempts are independent events we will multiply 5/6 and 4/6.

now P(X=4) = 1/6 * 5/6* 4/6

here in one of the attempts we get a number similar to one of the number in previous attempt , so for example if we got '6' in first attempt we might get '6' again in 3rd attempt whose probability will be 1/6. which is multiplied in calculating the total probability

similarly P(X=5) = 1/6* 1/6 * 5/6* 4/6

with 2 repetations of previous attempt.

This series will go on till infinity . hence the expected value of event (X) will be

E(X) =X * P(X) where X=3,4,5,.......infinity

now E(X) = 3* 5/6 * 4/6 + 4 * 1/6 * 5/6* 4/6 +  5* 1/6* 1/6 * 5/6* 4/6 + .....infinity - (1)

1/6 *  E(X) = 3 * 1.6* 5/6 * 4/6 +  4 * 1/6 * 1.6* 5/6* 4/6 + + .....infinity - (2)

multiplied eq (1) with 1/6 and shifting all terms to right for creating a set , Now eq (1) - eq (2)

E(X) - 1/6 *  E(X) = 3* 5/6 * 4/6 + (4-3) * 1/6 * 5/6* 4/6 + (5-4)  * 1/6* 1/6 * 5/6* 4/6 +....

-> 5/6 * E(X) = 3*5/6 * 4/6 + 1* 1/6 * 5/6* 4/6 + 1*1/6* 1/6 * 5/6* 4/6+....infinity

-> 5/6 * E(X) = 3 * 5/6 *4/6 + 5/6* 4/6 *( 1/6 + 1/62 + 1.63+----infinity)

-> 5/6 * E(X) = 60/36 +  5/6* 4/6 * (1/6 / [1-1/6]) ( formulla for sum of infinite series is a/(1-r) where 'a' is first term and 'r' is the multiplicative term here a=r=1/6)

-> 5/6 * E(X) = 60/36 + 4/36 =64/36

-> E (X) = 6/5* 64/36 = 32/15 = 2.13

b)  S= number of rolls until you get a repeated result

S >=2 you need atleast 2 rolls to repeat result

S <=7 longest you can go is 7 rolls, on 7th roll there has to be repeated result. after all 1-6 digits have appeared in the previous 6 rolls

P(S=2) = 1/6

P(S=3) = 5/6 *1/6

P(S=4) = 5/6*4/6*1/6

P(S=5) = 5/6*4/6*3/6*1/6

P(S=6) = 5/6*4/6*3/6*2/6*1/6 [Example: suppose you get 6 in first attempt followed by 1,2,3,4 in any order on the 6th draw the probability of drawing a 6 will be 1/6

P(S=7) = 5/6*4/6*3/6*2/6*1/6 [Example suppose you get 6 in first attempt followed by 1,2,3,4,5 in any order now on the seventh draw it doesnt matter whatever number will come you will achieve the event]

E(S) =[1/6]*2+[5/6 *1/6]*3 +[5/6*4/6*1/6]*4+[5/6*4/6*3/6*1/6]*5 + [5/6*4/6*3/6*2/6*1/6]*6+ [5/6*4/6*3/6*2/6*1/6]*7

E(S) =2/6 + 15/36+ 80/216 +300/1296 +720/7776 + 840/7776 =  1.55