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:



Five in a row:


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!

  • 1
    $\begingroup$ 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?" $\endgroup$
    – Matti P.
    Jul 26, 2019 at 5:43
  • $\begingroup$ Fixed! Sorry about that. $\endgroup$
    – user690808
    Jul 26, 2019 at 5:44
  • $\begingroup$ 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. $\endgroup$
    – user690808
    Jul 26, 2019 at 5:57
  • 1
    $\begingroup$ 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? $\endgroup$
    – user690808
    Jul 26, 2019 at 6:22
  • 1
    $\begingroup$ @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 $\endgroup$
    – katatahito
    Jul 26, 2019 at 6:23

3 Answers 3


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

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




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 $$\dfrac{1^3\cdotp6^2+5\cdotp1^3\cdotp6+6\cdotp5\cdotp1^3}{6^5}=\dfrac{1}{81}$$


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