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I was solving this problem:

There is a die and a coin. The dice is rolled and the coin is flipped according to the number the die is rolled. If the die is rolled only once, what is the probability of 4 successive heads?

My methodology:

Case 1: Die rolls $4$

P(4Heads) = $\frac{1}{6}.\frac{1} {2}\frac{1}{2}\frac{1} {2}\frac{1} {2}=\frac{1}{6}.\frac{1} {2^4}$

Case 2: Die rolls $5$

P(4Heads) = $5.\frac{1}{6}.\frac{1} {2}\frac{1}{2}\frac{1} {2}\frac{1} {2}\frac{1} {2}=5.\frac{1}{6}.\frac{1} {2^5}$

{Multiplied by $5$ because there are $\frac{5!}{4!}$ different arrangements possible for heads in $5$ coin throws}

Case 3: Die rolls $6$

P(4Heads) = $5.\frac{1}{6}.\frac{1} {2}\frac{1}{2}\frac{1} {2}\frac{1} {2}\frac{1} {2}\frac{1} {2}=15.\frac{1}{6}.\frac{1} {2^6}$

{Multiplied by $15$ because there are $\frac{6!}{4!.2!}$ different arrangements possible for heads in $6$ coin throws}

Net probability = $\frac{1}{6.2^4}.\frac{29}{4}$

However the answer in the book is different from my answer. So, please help me understand if there is some lapse in my logic.

Edit:

The answer in my book is given as $\frac{1}{16}$ which I'm unable to get.

I also found this Quora post which arrived at the same answer but I couldn't understand its methodology

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    $\begingroup$ It's asking for the probability of $4$ successive heads, not $4$ heads total. $\endgroup$
    – K. Jiang
    Commented May 6 at 5:50

1 Answer 1

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Outcomes such as $HHHTH$ and $HTHHTH$ are not permissible because they do not have four successive heads, yet they are included in your computation.

To get $4$ successive heads, there are only $3$ ways to do it in $5$ coin tosses: $HHHHT$, $THHHH$, $HHHHH$. There can't be any tails in between the heads, so any tail must be at the end.

How many arrangements are possible for $6$ coin tosses? Try enumerating them like I showed for $5$ coin tosses. Then recompute your probability.

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  • $\begingroup$ Thank you. I did so and the answer I'm getting is $\frac{1}{6.2^4}.\frac{16}{4}$ = $\frac{1}{24}$ when I multiplied the Case 3 with 6 instead of 15 and added the probabilities. But the answer in the book is given as $\frac{1}{16}$. Is there another mistake that I have made or is the answer in the book wrong? $\endgroup$ Commented May 6 at 6:11
  • $\begingroup$ There are $8$ ways to get (at least) $4$ successive heads in $6$ coin tosses, so the probability that I get is $3/64$. $\endgroup$
    – heropup
    Commented May 6 at 6:20
  • $\begingroup$ Can you please tell which cases I'm missing as for me I had 3 for 4 heads, 2 for 5 heads and 1 for 6 heads ? $\endgroup$ Commented May 6 at 6:30
  • $\begingroup$ I have also added link to a quora post which arrived at the same answer as the one in my problem book but the methodology of which I couldn't understand. $\endgroup$ Commented May 6 at 6:37
  • $\begingroup$ @Madly_Maths I get $HHHHHH$, $HHHHHT$, $HHHHTH$, $HHHHTT$, $HTHHHH$, $THHHHH$, $THHHHT$, $TTHHHH$. $\endgroup$
    – heropup
    Commented May 6 at 6:55

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