Roll a die until a 6, P(first 6 on roll 4) 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 4?

p = 16: first success on roll 4?

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 4: the first 3 rolls fail, then roll 4 wins.

rolls 1 to 4: F F F S

F = fail, S = success

Fail, fail, and so on, then win: we multiply.

q for each of the 3 fails, then p for the win.

P(X = 4) = q × q × q × p = 56 × 56 × 56 × 16

Careful!

Only 3 fails come before the win, not 4.

Careful: not q⁴ × p ✗

3 Multiply

Tops: 5 times 5 times 5 times 1 is 125.

Bottoms: 6 times 6 times 6 times 6 is 1296.

125/1296 is about 0.0965.

P(X = 4) = 1251296 ≈ 0.0965

4 Check

Let's check with 1296 people trying.

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

Roll 1: 216 win, 1080 go on.

Roll 2: 180 win, 900 go on.

Roll 3: 150 win, 750 go on.

1296 people: roll 1: 216 win, 1080 go on; roll 2: 180 win, 900 go on; roll 3: 150 win, 750 go on;

On roll 4, 125 win for the first time.

125 out of 1296 is about 0.0965.

So it is right!

roll 4: 125 win: 125 ÷ 1296 ≈ 0.0965

Let's look back.

First: name p and q.

Then: fail, then win.

Then: multiply.

Last: check.

Answer