# If a die is thrown $5$ times, what is the probability of throwing at least $3$ ones continuously?

The Question is:

Throw a die $$5$$ times, what is the probability of throwing at least $$3$$ ones continuously.

I wanted to verify my solution in which I let X denote the positions which the ones could be to count the number of total cases in which there was at least $$3$$ consecutive items.

Three in a row:

XXX_ _

_ _ XXX

_ XXX _

Four in a row:

_ XXXX

XXXX _

Five in a row:

XXXXX

With this, I assumed that the total number of possible outcomes would be $$6$$ and the total sample space would be $$6^5$$ giving us a probability of $$1/6^4$$.

Additionally, if the question just asked for at least three of any number would that be $$36/6^5$$ as there are $$6$$ numbers and $$6$$ ways each can satisfy the criteria?

Thank you!

• If you're throwing dice, why is the title "Coin flipping (...)" ? Anyway, I would approach this question by reformulating it as: "A die is thrown five times. What is the probability that there are at least three 1's, and that the three are consecutive throws?" Jul 26, 2019 at 5:43
• Fixed! Sorry about that. Jul 26, 2019 at 5:44
• Finding the probability that there are three ones can be solved through binomial distributions as 5choose3* (1/6)^5 but I'm not sure how to go about the second constraint. Jul 26, 2019 at 5:57
• Oh! Thank you so much. So correct me if I'm wrong but the total would be 90+10+1/6^5 for the first question? Jul 26, 2019 at 6:22
• @user690808 except for the third "pattern" with 3-in-a-row neither box can be a 1 so the total number for that is 25 Jul 26, 2019 at 6:23

Let $$Y$$ denote not getting a 1. Let $$X$$ denote getting a 1. Being careful to avoid double counting.

$$N(XXXY\_ )=N(\_YXXX)=5*6$$

$$N(YXXXY)=5*5$$

$$N(YXXXX)=N(XXXXY)=5$$

$$N(XXXXX)=1$$

Which gives a total of 96 agreeing with drhab's answer. And so the probability is $$\frac{96}{6^5}$$. If you want the probability of getting at least three consecutive numbers. Then it would be this probability multiplied by 6. Because getting a number say x, at least 3 times, precludes getting any other number y at least 3 times. And the events are disjoint.

If I understand well then the dice are thrown one by one.

Give the dice chronological numbers $$1,2,3,4,5$$.

• Let $$E$$ denote the event that the dice with numbers $$1,2,3$$ show face 1.
• Let $$F$$ denote the event that the dice with numbers $$2,3,4$$ show face 1.
• Let $$G$$ denote the event that the dice with numbers $$3,4,5$$ show face 1.

Then to be found is $$P(E\cup F\cup G)$$ and applying the principle of inclusion/exclusion we find:$$P\left(E\cup F\cup G\right)=P\left(E\right)+P\left(F\right)+P\left(G\right)-P\left(E\cap F\right)-P\left(F\cap G\right)-P\left(E\cap G\right)+P\left(E\cap F\cap G\right)=$$$$3\times6^{-3}-2\times6^{-4}-6^{-5}+6^{-5}=16\times6^{-4}$$

This tells us also that there are $$16\times6=96$$ possibilities.

Let $$\sf X$$ be a result of 1, $$\sf Y$$ be a result of something else, and $$\sf Z$$ be a result of anything. The disjoint outcome patterns for "at least three consecutive 1 among five rolls" are: $$\sf XXXZZ~, YXXXZ~,ZYXXX$$ So the probability is $$\dfrac{1^3\cdotp6^2+5\cdotp1^3\cdotp6+6\cdotp5\cdotp1^3}{6^5}=\dfrac{1}{81}$$