Steply

Roll a die until a 6, P(first 6 within 3 rolls)

Step by step, with the reason for every step · The geometric distribution · University Statistics

Step by step

The question

Question: we roll a die until we get a 6: what is the probability that the first 6 comes within 3 rolls?

p = 16: first success within 3 rolls?

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 None in 3 rolls

Within 3 means on roll 1, or 2, up to 3.

The only other way: all 3 rolls fail.

within 3: on roll 1, or 2, ... or 3
the other way: all 3 fail

All 3 fail: q times itself 3 times.

That is 125/216, about 0.5787.

all fail = q3 = (56)3 = 125216 ≈ 0.5787

3 The rest

Everything else has a success within 3: 1 minus that is about 0.4213.

P(X ≤ 3) = 1 − 0.5787 ≈ 0.4213

Careful!

p to the power 3 would be 3 successes in a row.

We only need one.

Careful: not 0.16673 ✗

4 Check

Let's check by adding each roll: 0.1667 plus 0.1389 plus 0.1157.

P(1) + ... + P(3) = 0.1667 + 0.1389 + 0.1157

That is about 0.4213.

So it is right!

≈ 0.4213

Let's look back.

First: name p and q.

Then: none in 3 rolls.

Then: the rest.

Last: check.

Answer
Careful: not 0.16673 ✗

Your turn

Try this one yourself, then check:

p = 16: first success within 2 rolls?
Show the answer
P(X ≤ 2) ≈ 0.3056

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