Roll a die until a 6, P(first 6 on roll 2) Step by Step

Step by step

The question

Question: we roll a die until we get a 6: what is the probability that the first 6 comes on roll 2?

p = 16: first success on roll 2?

X counts the rolls up to the first success.

X = the roll of the first success

1 Name p and q

p, the chance of a success on one try, is 1/6.

p = 16

q, the chance of a fail, is 1 minus p: 5/6.

q = 1 − 16 = 56

Every try has the same chance, and one try does not change the next.

every try: the same p, and the tries do not affect each other

2 Fail, then win

First success on roll 2: the first 1 roll fail, then roll 2 wins.

rolls 1 to 2: F S

F = fail, S = success

Fail, then win: we multiply.

q for each of the 1 fails, then p for the win.

P(X = 2) = q × p = 56 × 16

Careful!

Only 1 fails come before the win, not 2.

Careful: not q² × p ✗

3 Multiply

Tops: 5 times 1 is 5.

Bottoms: 6 times 6 is 36.

5/36 is about 0.1389.

P(X = 2) = 536 ≈ 0.1389

4 Check

Let's check with 36 people trying.

On each roll, 1 in 6 of those left win, the rest go on.

Roll 1: 6 win, 30 go on.

36 people: roll 1: 6 win, 30 go on;

On roll 2, 5 win for the first time.

5 out of 36 is about 0.1389.

So it is right!

roll 2: 5 win: 5 ÷ 36 ≈ 0.1389

Let's look back.

First: name p and q.

Then: fail, then win.

Then: multiply.

Last: check.

Answer