# What is wrong with my logic in this dice and coin problem?

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

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

• It's asking for the probability of $4$ successive heads, not $4$ heads total. Commented May 6 at 5:50

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.
• 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? Commented May 6 at 6:11
• There are $8$ ways to get (at least) $4$ successive heads in $6$ coin tosses, so the probability that I get is $3/64$. Commented May 6 at 6:20
• @Madly_Maths I get $HHHHHH$, $HHHHHT$, $HHHHTH$, $HHHHTT$, $HTHHHH$, $THHHHH$, $THHHHT$, $TTHHHH$. Commented May 6 at 6:55