Statistics 3302

Question

A weighted die has probability 0.102 of landing on a 6. If you roll the die until you get a 6 for the first time, what is the expected number of rolls it will take?

Round to the nearest integer.

Answer 1

Let's call P the probability of a six. Let's look at the contribution EV for as we roll: (Roll will be Prob * number of rolls.

Roll Probability of 6 EV contribution

first P 1*P = P one 6

second (1-P)*P 2*(1-P)*P one non-six, one 6

third (1-P)²*P 3*(1-P)²*P two non- sixes, one 6

nth (1-P)^n-1*P n*(1-P)^n-1*P

So our solution is EV = Σ Pi(1-P)^i-1 from i = 1,infinity

We need to figure out the infinite sum of the series. Let's solve a simpler series: Σir^i-1

Let S be the sum: S = 1 + 2r¹ + 3r² + 4r³ +...

rS = r¹ + 2r² + 3r³ +...

Subtract: ( 1-r)S = 1 + r + r² + r³ + ... = 1/(1-r) (geometric series)

Therefore S = 1(1-r)²

Almost there... Our sequence is PΣi(1-P)^i-1 where we multiply by P and r = (1-P)

Therefore, our sum is equal to P * 1(1-(1-P))² = 1/P Nice!

You could just brute force the solution by summing, but you'll have to do 20+ terms to feel like you're nearing the answer. This is exact. A normal die with P = 1/6 would have an EV of 6 rolls. I leave you the honor of plugging in the value.

Take care.