Question

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).

Answer #1

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

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