If an unbiased coin is flipped 5 times, the probability that

Question

If an unbiased coin is flipped 5 times, the probability that the same face does not show up in any three consecutive flips is

A. 1/2
B. 5/8 
C. 3/8
D. 7/8
E. 9/8

Bunuel · Accepted Answer

Probably the easiest way would be just to write down the cases with at least 3 consecutive faces:

HHHHH
HHHHT
HHHTH
HHHTT

TTHHH
THHHH
HTHHH

THHHT

So we have 8 cases for at least 3 consecutive heads, which means that there will be also 8 cases for at least 3 consecutive tails. As there are total of 2^5 outcomes then the probability would be  P=\frac{8+8}{2^5}=\frac{1}{2} .

Answer: A.

ankitranjan · Answer

Bunuel that was the right answer.Thanks for the explanation.

144144 · Answer

Bunuel, i know it will be the same answer, but shouldnt it be 1-p here?

Bunuel · Answer

Yes (as we counted the probability of opposite event).

KarishmaB · Answer

I agree with Bunuel. The easiest solution I could think of was enumerating in a sequence. I try and find how I can have the same face three or more times.

Start with 3 Hs and 2 Ts, go on to 4Hs and 1T and finally 5Hs
The 3 Hs have to be together so: HHHTT, TTHHH, THHHT
Then, 4 Hs and 1T. Place the 4 Hs first: H H H H. Now look for the places where you can place the T: THHHH, HTHHH, HHHTH and HHHHT
Finally 5 Hs: HHHHH
Total 8 ways to get 3 or more Heads in a row. Ways to get 3 or more Tails in a row is also 8.
Total number of ways outcomes with 5 coins = 2^5 = 32
Probability of same face occurring 3 or more times = 16/32 = 1/2
Probability of same face occurring less than 3 times = 1 - 1/2 = 1/2

Kconfused · Answer

Is there a straighter way to solve this problem? Without going the opposite way
My attempt: Of the five slots, three straight slots are reserved for opposite results. (HTH, THT). The remaining two slots can yield a head or a tail. 
So the possible number of outcomes is 2*2 = 4. Now these two slots are always clubbed together and can be treated as one slot that can be adjusted anywhere among the 4 possible slots in 4c1 ways. So the total possible outcomes under the given conditions will be 2*2* 4c1 = 16

Probability is 16/32 = 1/2

Is the thinking behind reaching this solution correct? Or did I just get lucky with the answer?

TIA
KC

Feb2024 · Answer

gmatophobia   Bunuel   KarishmaB   MartyMurray

Hello Experts!

Can someone please tell me why the following approach is incorrect:

The probability that the same face does not show up in any three consecutive flips is
= 1 -P (same face shows up in any 3 consecutive flips +
same face shows up in any 4 consecutive flips +
same face shows up in all 5 consecutive flips)------------------------------------( Eq. 1)

Now, number of times 3 consecutive flips can be chosen is:
i) Flip number 1,2 & 3
ii) Flip number 2,3 & 4
iii) Flip number 3,4 & 5

Similarly, number of times 4 consecutive flips can be chosen is:
i) Flip number 1,2,3 & 4
ii) Flip number 2,3,4 & 5

And, number of times 5 consecutive flips can be chosen is:
i) Flip number 1,2,3,4 & 5

NOW, each of the consecutive flips may have Head or Tail as an outcome.

So, summarising the above information into Eq. 1,

The probability that the same face does not show up in any three consecutive flips is
= 1 -P (same face shows up in any 3 consecutive flips +
same face shows up in any 4 consecutive flips +
same face shows up in all 5 consecutive flips)

= 1- (3*2*(1/2)^5 + 2*2*(1/2)^5 + 1*2*(1/2)^5)

= 1- (  )

=

Hence, Ichoose option B could be the answer.

Please tell me where is the mistake in this approach.

Thanks in Advance!

MartyMurray · Answer

You're close, but you left out some cases.

For three consecutive flips, there are more than what you said.

As you said, there are the following:
 
At the same time, in using just those cases, you are assuming that the other flips are all the opposite sides of the coin. However, they are not necessarily.

For example for heads 1, 2, & 3, we could have TT for 4 and 5, as you implied, or TH, which  you didn't consider. So, there's one case you missed.

Then for heads 3, 4, & 5, we could have TT for 1 and 2, as you implied, or HT, which you didn't consider. So, there's another case you missed.

Similarly, you missed two cases for TTT as well.

Thus, there are four more cases of exactly three consecutive flips, meaning that 1/8 more cases have exactly three consecutive flips.

So, the answer is not 1 - ((6+4+2)/32). It's 1 - ((10+4+2)/32) = 1 - 1/2 = 1/2.

Tricky question for sure!

Kinshook · Answer

ankitranjan

Asked: If an unbiased coin is flipped 5 times, the probability that the same face does not show up in any three consecutive flips is
Total ways = 2ˆ5 = 32 ways

Unfavorable ways: 
HHHHH, TTTTT, HHHHT, THHHH, TTTTH, HTTTT, HHHTT, HHHTH, THHHT, HHHTT, HHHTH, TTTHH, TTTHT, HTTTH, TTTHH, TTTHT
= 16 ways

Favorable ways = 32 - 16 = 16 ways

The probability that the same face does not show up in any three consecutive flips = 16/32 = 1/2

IMO A

Feb2024 · Answer

It is clear now. Thank you very much MartyMurray for your time.

jassonadder · Answer

When you  flip a coin  5 times, and you want to make sure that the same result (either heads or tails) doesn’t appear three times in a row, things get interesting. You have to carefully think about which outcomes are allowed. For example, a sequence like HHH or TTT would break the rule, so those need to be avoided.

If you try different combinations using a simulator—like imagining or using a tool that lets you  flip a coin online —you’ll quickly notice that not all possible sequences work. You have to skip the ones that contain three heads or three tails in a row.
After working through the possibilities (or simulating them), you’ll find that the chance of avoiding three of the same face in a row over 5 flips is  5 out of 8 , or 62.5%. It's a fun example of how probability works with real rules and constraints.

So, the correct answer is  B. 5/8 !

If an unbiased coin is flipped 5 times, the probability that

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