4. A die is loaded in such a way that the probability of each face turning up is

Question

4. A die is loaded in such a way that the probability of each face turning up is proportional to the
number of dots on that face. (For example, a six is three times as probable as a two.) You get to roll
the die until you roll a 3, then you must stop.
a. What is the expected number of rolls before you are stopped?
b. Which is more likely: that you only get one roll (because your first roll is a 3) or that you
get exactly 7 rolls?

(answers should be a. 7
b. P(one roll) = .1429, P(21 rolls) = .0567 but I don't know how to reach these answers)

4. A die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on that face. (For example, a six is three times as probable as a two.) You get to roll the die until you roll a 3, then you must stop. a. What is the expected number of rolls before you are stopped? b. Which is more likely: that you only get one roll (because your first roll is a 3) or that you get exactly 7 rolls?

Explanation / Answer

Let the probability of getting 1 in a roll is p. Then probability of getting 2, 3, 4, 5, 6 are 2p, 3p, 4p, 5p, 6p respectively.
Now, p+2p+3p+4p+5p+6p = 1 ; or, 21p = 1 ; p = 1/21
Hence probability of getting 1, 2, 3, 4, 5, 6 in a roll are 1/21, 2/21, 3/21(=1/7), 4/21, 5/21, 6/21 respectively.

a) The expected value of any random experiment is given by: E(X) = x P(x) ; here x is the no of rolls before getting a 3
Probability of getting 3 in the 1st roll = 1/7 (hence, probability of not getting 3 in any roll = 1-1/7 = 6/7)
Probability of getting 3 in the 2nd roll = 6/7 * 1/7
Probability of getting 3 in the 3rd roll = (6/7)^2 * 1/7
.
.
Probability of getting 3 in the n'th roll = (6/7)^n-1 * 1/7
Now, E(X) = x P(x) = 1* 1/7 + 2* 6/7 * 1/7 + 3* (6/7)^2 * 1/7 + 4* (6/7)^3 * 1/7 +.............upto inf
= 1/7 * n* (6/7)^(n-1) (lower limit n is 1, upper limit n is inf)
=1/7 * 49 = 7
So, expected no of rolls before getting 3 is 7.

b) Probability of getting 3 in the 1st roll = 1/7 = 0.1429
Probability of getting 3 in the 7th roll = 1/7 * (6/7)^(7-1) = 0.0567
Obviously getting 3 in one roll is more likely compared to getting 3 exactly on 7th rolls.

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