Make sure you remember this week’s question before moving on to the explanation.

## Question

If a fair coin is tossed six times, what is the probability of getting exactly three heads in a row?

(A) 1/16

(B) 3/16

(C) 1/8

(D) 3/8

(E) 1/2

## Answer and Explanation

A good way to think about this problem is to map out a valid instance of what the question is asking for:

H H H T T T

T H H H T T

T T H H H T

We could go on like this but in doing so we might miss something. See, the thing we want to pay attention to is the word “exactly”. This means we can’t have four heads in a row.

We also have to be aware of the fact that having three heads in a row can happen in more than one way, even in the first position above (HHHTTT).

For instance, successful trials with heads in the first position consist of the following:

H H H T H T

H H H T T H

H H H T H H

It is important to remember that three heads can also come at the end, even when the first coin is heads:

H T H H H T

H T T H H H

H H T H H H

With a tails in the first position we get the following:

T H H H T T

T H H H T H

T H T H H H

T T H H H T

T T T H H H

This gives us a total of 12 possible instances out of a total of 2^6 possibilities:

12/64 = 3/16

See you again soon!

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Where is HHHHHH case, please?

Hi Karina,

The key to this question is the word “exactly.” We can ONLY have 3 heads in a row–we can’t have four, five or (as in your example) six heads in a row. This scenario is not considered in this question because it isn’t relevant to finding the correct answer. If we didn’t have this limitation in the question, there would be far too many possibilities for this coin toss and it would make this question much more difficult and time-consuming. Since it’s not possible in the answer choice, we don’t consider this scenario 🙂

My approach was

getting a probability of head is 1/2 . If we consider getting 3H in a row as a single event and check the no. of ways to arrange these (_)(_)(_)(3H) distinct item is 4! ways.So numerator would be 1/2 * 24 = 12. For denominator no. of outcomes would be 2^6. So ans=12/64 or 3/16.

Is this correct ?

Hi Ani,

Great job identifying an alternative approach! 😀

This reasoning doesn’t seem correct to me. An important thing to consider here is that the sequence of 3H must be bordered on all sides by tails. In other words, we have these possibilities:

(3H) T _ _

T (3H) T _

_ T (3H) T

_ _ T (3H)

It happens here that we have 4! / 2 = 12 ways to fill these blanks, but this seems to be coincidence. (There are 4 + 2 + 2 + 4 = 12 ways to fill the blanks by the method outlined above.)

If you extend this problem to 4H in 8 tosses, you can see this method (and similar three-heads-as-one-event methods from posters in previous methods) break down. Then you’d have these possible outcomes:

(4H) T _ _ _

T (4H) T _ _

_ T (4H) T _

_ _ T (4H) T

_ _ _ T (4H)

Now there are 8 + 4 + 4 + 4 + 8 = 28 ways to fill these blanks. A factorial-based solution no longer works nicely.

In general, it seems if we wanted to consider these sequences as single events, we’d need to split up the cases where the sequence of heads is at the beginning or end of the string (where we’re looking for HHHT or THHH, respectively) vs. in the middle (THHHT).

Then we could consider the more general question of how to get exactly n heads in a row in 2n tosses. There will always be 2 cases corresponding to the beginning and end. For each of these, there are 2n – (n+1) = n-1 other coin tosses that can be either H or T, yielding 2^(n-1) outcomes.

Then for the middle cases, you need to fit the sequence T (nH) T (a sequence of length n+2) somewhere in a sequence of 2n tosses. Again, if you consider the T (nH) T sequence as one event that needs to occur exactly once, there will be 2n – (n+2) = n-2 other coin tosses in the sequence for a total of n-2 + 1 = n-1 spaces, one of which must be filled with the T (nH) T sequence. If you have n-1 spaces, and you’re looking to fill one of them, there are n-1 distinct ways to do it. In other words, we have n-1 middle cases. (When n=3 as in the example problem, this yields 2 middle cases. When n=4 as in the modified problem above, this yields 3 middle cases.) Finally, for each middle case, there were n-2 other coin tosses, which could be either H or T, yielding 2^(n-2) outcomes.

Putting this all together, you have 2 end cases with 2^(n-1) outcomes and n-1 middle cases with 2^(n-2) outcomes. So you have in total 2*2^(n-1) + (n-1)*2^(n-2) = 2^n + (n-1)*2^(n-2) outcomes. The sample space consists of 2^(2n) possibilities, so after dividing those two numbers, you get your probability.

To verify, consider n=3 (the original case, 3H in 6 tosses). The sample space consists of 2^(2n) = 2^6 = 64 total events, and there are 2^n + (n-1)*2^(n-2) = 2^3 + 2 * 2^1 = 8 + 2*2 = 8 + 4 = 12 outcomes that yield exactly n=3 heads. In the n=4 case (4H in 8 tosses), the sample space consists of 2^(2n) = 2^8 = 256 events, and there are 2^n + (n-1)*2^(n-2) = 2^4 + 3 * 2^2 = 16 + 3*4 = 16 + 12 = 28 cases that yield exactly n=4 heads.

All this to say: don’t get too fancy with the math. In this case, it’s by far the easiest to count the possible outcomes explicitly and move on.

hey

could you tell me why you multiplied 24 by 1/2 ?

I can clarify for you, Ani! 🙂 Basically, in the sequence (_)(_)(_)(3H), there is a 50% chance that the (_) slot right before (3H) will come up heads, in which case there are actually four heads in a row rather than exactly 3. So that means that half of the possible results for (_)(_)(_)(3H) will not fit the criteria for *exactly* three heads in a row. So to get the possible outcomes that fit the limits of the problem, you want half of the total possibilities for (_)(_)(_)(3H). The total possibilities for (_)(_)(_)(3H) are 4!, or 4*3*2*1, or 24. And half of those total possibilities would be– of course– 24 multiplied by 1/2.

This is my solution to the problem.

The possible ways of getting exactly three heads are as follows.

H H H T _ _

T H H H T _

_ T H H H T

_ _ T H H H

Where the _ represents any Head or Tail outcome being possible.

Then the total number of outcomes that can have exactly three heads in a row becomes.

H H H T _ _ x 4

T H H H T _ x 2

_ T H H H T x 2

_ _ T H H H x 4

Therefore 4 + 2 + 2 + 4 = 12

Total outcomes = 64

Therefore the probability = 12/64 = 3/16

excellent explanation.thanks

Hi. I understand that the 2^6=64 indicates the total number of possibilities. However, I’m having a difficult time understanding where the 12 came from.

Hi Inanje,

Happy to clarify! In the explanation, Chris divides the12 different ways it is possible to have three heads in a row into three categories:

1. Three heads first

2. Heads first, three heads at the end

3. Tails first

Together, we have the following 12 ways to have exactly three heads in a row:

1. H H H T T T

2. H H H T H T

3. H H H T T H

4. H H H T H H

5. H T H H H T

6. H T T H H H

7. H H T H H H

8. T H H H T T

9. T H H H T H

10. T H T H H H

11. T T H H H T

12. T T T H H H

Hope this helps 😀

my approach 10c3 (1/2)^3*(1/2)^3=5/16..Where iam getting wrong?

Hi Pavan 🙂

I’m happy to help. First it looks like you have a small typo and meant to write “6C3.” That said, your denominator, 64 is correct, as there is a total of 64 possible outcomes. We are flipping the coin 6 times and each time, there is the possibility that it lands on head or tails:

2^6 = 64

This part is correct in your solution, as (1/2)^3*(1/2)^3 = 1/64.

From there, though, we need to figure out how many of those 64 possibilities follow what the question is asking for, mainly that exactly 3 H are flipped in a row. While at first 6C3 seems like a good way to figure out this quantity, 6C3 ends up providing a number of possibilities that is greater than the actual answer. Why? This is because 6C3 includes combinations in which 4 heads are rolled in a row. These outcomes are not desired, since, as Chris mentions, the question specifically refers to “exactly 3 heads in a row,” which means we cannot have 4 heads in a row. For that reason, it is best to solve this question following the explanation Chris provides above. By doing so, we avoid counting situations in which we have more that 3 heads in a row.

I hope this clears up your doubts 🙂 Happy studying!

Hi Chris…. I think this can also be done in another way….

Let us consider the arrangement as: HHHT _ _

for exactly 3 in a row, HHH has to be paired with a tail… so let us take HHH as a single occurrence K. This paired with T and the rest of the 2 blank possibilities results in the chance of occurrence as 3! / 2!.

again as the 2 blanks can be filled by either H or T, its count becomes 2^2. hence the probability is then:

( (3! / 2!) X 2^2 ) / 2^6 = 3 / 16…

hope the solution is correct…thanks…

Hey Chris,

I think I may have solved this problem by using FCP, can you verify the validity of the solution?

Total outcomes = 2^6 = 64;

Now for the desired outcomes, there are 6 outcome spaces from the six tosses. We need 3 heads in a row, therefore we can club these 3 outcomes together into 1 outcome, so now we have a total of 4 outcome spaces( 6 – 3 +1 ).

Order matters so we use the FCP which gives us 4! ways to arrange the outcomes.

T and H are identical in terms of probability therefore we the actual number of ways is 4!/2!

hence, the desired outcomes are = 4!/2! = 24/2 = 12

hence P(3 heads in a row) = 12/64 = 3/16

Hope this solution is correct conceptually!

Cheers!

Hello Chris,

Don’t we have four possibilities namely:

1)HHHTTT

2)TTTHHH

3)THHHTT

4)TTHHHT

So the answer should be 1/16

hi

HHHTTT is the same as TTTHHH mathematically speaking. But THHHTT and TTHHHT are to be considered separate events.

Hope that helps.

Question asks for possibility of that event happening in a row. So in this sense, HHHTTT and TTTHHH are different. Even if their outcome is same, looking at the sequence of events happening, they are different cases!

Yes, those are definitely different cases. The twist here is that THHHTH is a different case than THHHTT. And HHHTTT is a different case than HHHTHT.

Hope that helps clear things up 🙂

Hi!

Is there an easy way (through combinations?) to calculate the “12” ways of having three heads in a row?

I mean, really I can’t think I will be counting all the possible way while sitting for the GRE!

Thanks a lot, you guys are great!

Not really :(. The key is noticing that you can have four–and even five–heads, as long as there is a tail in between the heads. A GRE question would probably have a slightly simpler set up, but with a similar twist.

Hope that helps!

Hi Chris,

Its really a brain twister.

Can you please provide a simpler solution which might not involve this confusing listing of events.

Thanks

So there is a logical way to go about it, eliminating the answers that are too obvious or too extreme. (A), 4 out of 64, is a trap for those who just count HHHTTT, THHHTT, etc. There has to be more to the problem than that, so eliminate (A). (D) and (E) are too high; having around 30 possibilities clearly isn’t possible. So as long as you can count more than eight possibilities, the answer has to be (B).

Hi,

Many patterns provided in you solutions are same such as

TTHHHT andTHHHTT

HHHTHT and THTHHH

HHHTTH and HTTHHH

HHHTHH and HHHTHH

and few you have missed

HHHHTT

etc

I have created these cases total making upto 10 not 12

with 3h’s two cases+with 4h’s 5 cases +and with 5h’s 3 cases=10 cases

PLease let me know I am going wrong?

Thanks

Hi Anubhav,

For the problem, HHHTTT and TTTHHH are very different in terms of the actual flip of the coin. In each case, a different outcome results. I think you were confusing mathematical probability (the likelihood of each is the same) with the actual presence of a given outcome (heads or tails) for a specific trails.

As for HHHHTT, it has four heads in a row; the question asks for exactly three heads in a row.

Hope that helps clear up any confusion 🙂

Hi Chris,

So do you want to say that

there are 12 cases in the following way:

HHHTTT

THHHTT

HHHTHT

HHHTTH

HTHHHT

HHHTHH

Total 6 cases in two as their mirror image is different.Therefore 6*2=12.

Please correct if I am wrong?

my approach (1/2)^6 * 6C3 = 5/16. wher i am geting wrong??

Answer has to be 1/16.

Event(E)-3 heads to occur in a row together alongwith 3 tails>>there are just 4 such instances(HHHTTT/THHHTT/TTHHHT/TTTHHH).

Sample space(S)>>2^6

P(E)=E/S

=4/2^6

=1/16

Hi Akash 🙂 In this question, while we’re concerned with getting three heads in a roll, that doesn’t mean that we can’t flip more than three heads in a couple of situations. For example, HHHTTH meets the condition of the problem (3 H in a row) while overall, 4 H are flipped. You can see other possible combinations in the explanation Chris provides. With that idea in mind, there are actually more possibilities than the once you’ve listed, and the answer is 12/64 = 3/16.

Hope that helps 🙂