Same or Higher
An interesting quant puzzle asked in WorldQuant Interview
Anton rolls a fair die and records the value he rolls. Afterwards, he continues rolling the die until he obtains a value at least as large as the first roll. Let N be the number of rolls after the first he performs. Find E[N]
Try to solve this problem yourself before moving on to the solution below
.
.
.
.
Solution
This problem may seem daunting but its fairly easy as we partition the cases.
If Anton rolls 1, he gets a number greater than 1 with a probability of 6/6 (i.e. 1). Now this is a geometric distribution with the desired event coming with the probability p.
The mean of the geometric distribution is also the expected value of the geometric distribution. The expected value of a random variable, X, can be defined as the weighted average of all values of X. The formula for the mean of a geometric distribution is given as follows: E[X] = 1 / p
Let Y be the first roll outcome
So E[X | Y = 1] = 6/6 = 1
Similarly
E[X | Y = 2] = 6/5
and so on.
Finally using the law of total expectation:
E[X] = 1/6 * (6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1)
E[X] = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6)
E[X] = 147/60
.
.
😃 Over to you: Were you able to solve this?
👉 If you liked this post, don’t forget to leave a like ❤️. It helps more people discover this newsletter on Substack and tells us that you appreciate solving these weekly questions. The button is located towards the bottom of this email.
👉 If you love reading this newsletter, feel free to share it with friends!
Correction in solution : In the second line, it should be greater than or equal to 1 with probability 6/6
My approach: (I don't remember results. So had to derive this 1/p result.)
The problem can be reduced to heads and tails as follows. Say on the first roll, you get x. Essentially from the next roll, you are solving this problem: Expected number of tosses before an head comes where Pr(head) = (7-x/6). Because this is nothing but success to you. So each situation boils down to a heads and tails toss situation. So say you get a 3 on the first roll. Now getting {3, 4, 5, 6} is success to you. Pr(getting 3,4,5 or 6) = 4/6. So E[tosses before a head comes] where the Pr(head) = 4/6 is 6/4.
Since we know that expected number of tosses before an head comes = 1/p where Pr(head) = p, we just have to calculate the summation.
E[N] = Summation ( 6/(7-x) * 1/6 ) where (1/6) is nothing but the probability of x coming on the first roll. x goes from 1 to 6.
E[N] = Summation( (1/7-x) ) where x ranges from 1 to 6
E[N] = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6