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# Magoosh Brain Twister: Flipping Out – Explanation

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

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 โexactlyโ. This means we canโ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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### 22 Responses to Magoosh Brain Twister: Flipping Out – Explanation

1. Rasitha July 18, 2016 at 11:05 pm #

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

2. Inanje July 10, 2016 at 11:05 am #

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.

• Magoosh Test Prep Expert July 13, 2016 at 6:52 am #

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:

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 ๐

3. Pavan April 27, 2016 at 1:02 pm #

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

• Magoosh Test Prep Expert April 30, 2016 at 11:05 am #

Hi Pavan ๐

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 ๐ Happy studying!

4. Sid July 19, 2015 at 12:37 am #

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…

5. Sambuddha (SAM) July 3, 2015 at 5:47 am #

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!

6. Kshitiz February 17, 2015 at 11:48 pm #

Hello Chris,

Don’t we have four possibilities namely:
1)HHHTTT
2)TTTHHH
3)THHHTT
4)TTHHHT

So the answer should be 1/16

• Shashank February 21, 2015 at 2:00 am #

hi

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

Hope that helps.

• vivek March 17, 2015 at 10:45 pm #

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!

• Chris Lele March 18, 2015 at 2:43 pm #

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 ๐

7. filippo February 17, 2015 at 9:23 pm #

Hi!

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

• filippo February 17, 2015 at 9:24 pm #

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!

• Chris Lele February 18, 2015 at 10:56 am #

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!

8. Sanidhya Jain February 12, 2015 at 9:01 pm #

Hi Chris,

Its really a brain twister.
Can you please provide a simpler solution which might not involve this confusing listing of events.
Thanks

• Chris Lele February 18, 2015 at 11:01 am #

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).

• Anubhav May 24, 2015 at 10:59 am #

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

• Chris Lele May 26, 2015 at 5:23 pm #

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 ๐

• Anubhav May 27, 2015 at 1:07 am #

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?

• kutkut February 18, 2016 at 3:49 am #

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

• Akash April 28, 2016 at 5:30 pm #

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

• Magoosh Test Prep Expert April 30, 2016 at 10:46 am #

Hi Akash ๐ 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 ๐

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