M15-31

Question

If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?

A.  \frac{1}{16}  
B.  \frac{1}{8}  
C.  \frac{3}{16}  
D.  \frac{1}{4}  
E.  \frac{3}{8}

Bunuel · Accepted Answer

Official Solution:

If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?

A.  \frac{1}{16}  
B.  \frac{1}{8}  
C.  \frac{3}{16}  
D.  \frac{1}{4}  
E.  \frac{3}{8}

Calculate the probability that either of the two sequences,  THHH  or  HTTT , will occur. The probability of  THHH  occurring =  (\frac{1}{2})^4 = \frac{1}{16} . The probability of  HTTT  occurring =  (\frac{1}{2})^4 = \frac{1}{16} . The probability of either  THHH  or  HTTT  occurring =  \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8} .

Answer: B

mejia401 · Answer

Since there are two options, Beroulli's formula can also be applied:

\frac{2!}{1!1!}*\frac{1}{2}^{4-1}*\frac{1}{2}^{4-3}

2*\frac{1}{8}*\frac{1}{2}

\frac{2}{16}

B.

vivekgautam1 · Answer

Hi Bunuel,

Here's how i am approaching this question.

The question is asking - out of 4 throws, what's the probability of getting 3 consecutive heads or 3 consecutive tails.

Total # of arrangements = 2 * 2 * 2 * 2 = 16

Now, 3 consecutive heads can be like - THHH or HHHT
Similarly, 3 consecutive tails can be like - HTTT or TTTH.

So, total # of favourable outcomes = 2 + 2 = 4.

So, the probability =  \frac{4}{16}

Can you please point to where i am wrong ?

Bunuel · Answer

HHHT and TTTH does not work because in these cases the game ends on third throw not on the fourth.

Hope it helps.

vivekgautam1 · Answer

awesome !
i missed that.

Thanks Bunuel.

tejyr · Answer

Hi Experts,

I agree that the favorable outcomes are htttt or thhhh and total is 16.
I have done favorable calculation as (4!/3!)*2 and d/r is 16 which comes as 8/16.
I was wondering where I went wrong.

Bunuel · Answer

The game ends when three consecutive heads or three consecutive tails appears. So, the game to end on the fourth throw, we should have first T and then three H's or first H and then three T's. Thus, there are only two favorable cases THHH and HTTT (no need for factorial correction which accounts for different arrangements).

tejyr · Answer

Got it Thank you Bunuel for clarifying

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

commodimaiores · Answer

I don’t quite agree with the solution. Total possible outcomes are 4: THHH, HHHT, HTTT, TTTH.

So probability will be 4/16 = 14

chetan2u · Answer

In HHHT, the game will get over in 3rd throw itself and it will be just HHH. Similarly it will be TTT

M15-31

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