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So let's say I want to know the probability of not getting three heads in a row while flipping 5 coins. Would I use combinations or permutations?

I was thinking permutations, since the heads need to be next to each other, but also maybe combination, since the order of the heads next to each other does not matter.

Joe Larkian
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2 Answers2

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I would just do the math.
There are $2^5$ possible outcomes. a) all heads - one possibility, yes. b) four heads - five possibilities, one no four yes. c) three heads - ten possibilities, seven no three yes. Net result $P(3\ in\ row)=\frac{8}{2^5}=\frac{1}{4}$. You asked for $1-P=\frac{3}{4}$.

I use both permutations and combinations. To get this answer - combinations to get number of heads and permutations to get number of three in a row.

herb steinberg
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    Thanks for the help, but I really need to know the difference for a problem I am working on, as I don't know whether to use permutations or combinations to find the probability of not getting three heads in a row for 21 coin flips. It's a much larger situation so I want to start small and work my way up. – Joe Larkian Apr 18 '21 at 03:34
  • I revised my answer to address your question directly. As I said you need both. Combination to get number of heads and permutation to get count of three in a row. – herb steinberg Apr 18 '21 at 03:43
  • I will look at it, thanks! – Joe Larkian Apr 18 '21 at 03:45
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I find it easiest to take a counting approach to problems like these. Especially when the numbers are small, sometimes it's best to just count.

The total number of possible outcomes for $5$ coin flips is $2^5=32$, since each of the $5$ flips can have one of $2$ outcomes (heads or tails).

The total number of possible outcomes in which $3$ heads appear, and they are all in a row, is $3$. Since the numbers are small, I found by simply counting fingers: with 5 fingers on one hand, how many groups of 3 adjacent fingers can you count? The answer is $3$. (In general, it will be $n-k+1$; where here $n=5$ and $k=3$).

The total number of possible outcomes in which $4$ heads appear, and $3$ are in a row, is $4$. It may be easier to think of this case as $1$ tails appearing. Out of the $5$ spots that that tail can be in, all of them except $1$ leave $3$ heads in a row. (The one exception is when the tails is in the middle, $HHTHH$).

The total number of possible outcomes in which $5$ heads appear, and $3$ are in a row, is $1$. That's because there's only one outcome in which 5 heads appear.

So all in all, you have $3+4+1=8$ outcomes in which $3$ heads are in a row.

That means there are $32-8=24$ outcomes in which there are not $3$ heads in a row.

So your answer is $\frac{24}{32} = \frac{3}{4}$.

Note: I see that you care about generalizing this for higher numbers. It turns out the answer is pretty ugly. You should check out this post for more on that.

roddur
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