Pilf

We can either have

Case 1

3 Heads and 3 Tails

${6 \choose 3} \times 3 = 60$ where ${6 \choose 3}$ is the number of ways to get 3 Heads and 3 is number of coins to flip over

Case 2

4 of the same face and 2 of the opposite

${6 \choose 2} \times 2 \times 2 = 60$ . We multiply by two because we have to account for either 4 Heads 2 Tails or 4 Tails 2 Heads

Case 3

5 of the same face and 1 of the opposite

${6 \choose 1} \times 2 \times 1 = 12$ .

Case 4

6 of the same face

${6 \choose 0} \times 2 \times 0 = 0$ .

Summing all this up and dividing by $2^6$ for the probability of each coin toss we get $\frac{60+60+12+0}{2^6} = \frac{33}{16}$ which gives an answer of $33+16 =\boxed{49}$

The set of possible minimum number of times that we will have to turn coins over is $\{0,1,2,3\}$ . It doesn't include $\{4,5,6\}$ because, for example, if turned the coin $4$ times, we wouldn't be doing it the minimum number of times. Instead we could just turn the remaining $2$ coins. For example, if the coins flips are

HTTHHH

we have to turn the $2$ T's and not the $4$ H's.

We would have to turn the coin $0$ number of times if the sequences are

HHHHHH
TTTTTT

This has a probability of $\dfrac{2}{64}$ or we can write this as $\dfrac{\dbinom{6}{0} \times 2}{64}$

For turning the coin $1$ number of times, we have a probability of $\dfrac{\dbinom{6}{1} \times 2}{64}$ . The binomial is multiplied by $2$ because we have to consider both cases, for example,

HTTTTT and THHHHH

For turning the coin $2$ number of times, we have a probability of $\dfrac{\dbinom{6}{2} \times 2}{64}$ . The binomial is multiplied by $2$ because we have to consider both cases, for example,

HTTTTH and THHHHT

Finally, for turning the coins $3$ times, we have a probability of $\dfrac{\dbinom{6}{3}}{64}$ . We don't multiply by $2$ because

HHHTTT is the same as TTTHHH

The answer, therefore is

$\left( 0 \times \dfrac{2}{64} \right) + \left( 1 \times \dfrac{12}{64} \right) + \left( 2 \times \dfrac{30}{64} \right) + \left( 3 \times \dfrac{20}{64} \right)$

Solving this gives us $\dfrac{33}{16}$ and thus $33 + 16 = \boxed{49}$

When there are either $6$ heads or tails, then there will be no flips required. The probability of getting $6$ heads or tails is: $\frac{{6 \choose 0}}{64} = \frac1{64}$

When there are either $5$ heads and $1$ tail or $1$ head and $5$ tails, then there will be $1$ flip required. The probability of this scenario is: $\frac{{6 \choose 1}}{64} = \frac6{64}$

When there are either $4$ heads and $2$ tails or $2$ heads and $4$ tails, then there will be $2$ flips required. The probability of this scenario is: $\frac{{6 \choose 2}}{64} = \frac{15}{64}$

Finally, if $3$ heads and tails are chosen, then $3$ flips will be required. The probability of this scenario is: $\frac{{6 \choose 3}}{64} = \frac{20}{64}$

Therefore, the expected value of the number of flips required can be calculated by: $2 \cdot \frac{1}{64} \cdot 0 + 2 \cdot \frac{6}{64} \cdot 1 + 2 \cdot \frac{15}{64} \cdot 2 +\frac{20}{64} \cdot 3 = \frac{33}{16}$

Hence, the final answer is: $33 + 16 = \fbox{49}$

The expected value is given by,

$E(x) = p_1 x_1 + p_2 x_2 + p_3 x_3$ ,

where $p_n$ is the probability of $x_n$ random variable.

The values that random variable can take are $3, 2, 1$ .

The probabilities of these are $\frac{6!}{3! 3! \times 2^6}$ , $2\times\frac{6!}{4! 2! \times 2^6}$ , $2\times\frac{6!}{5!1! \times 2^6}$ , respectively.

Putting these value in the first equation, we get the right answer.

A single toss can be represented by the polynomial $1 + x$ so that the results of six successive tosses can be represented by $\left( 1 + x \right) ^ {6} = x^{6} + 6 x^{5} + 15 x^{4} + 20 x^{3} + 15 x^{2} + 6 x + 1$ . Individual terms in this polynomial give rise to the following cases:

$x^{6}$ : 1 instance of all heads (say), no turnovers

$6 x^{5}$ : 6 instances of one tails, thus 1 turnover

$15 x^{4}$ : 15 instances of two tails, thus 2 turnovers

$20 x^{3}$ : 20 instances of three tails, thus 3 turnovers

$15 x^{2}$ : 15 instances of two heads, thus 2 turnovers

$6 x$ : 6 instances of one heads, thus 1 turnover

$1$ : 1 instance of no heads, no turnovers

Result follows.

It's not hard to see that the minimum number of moves required is one of: 0,1,2,3. How can we require 0 flips? For this to happen, all coins must be facing the same way, i.e, they are either all heads or all tails. This happens in exactly $$ 2\binom{6}{0}$$ ways. For only 1 flip, we must either have 5 heads and 1 tails, or 5 tails and 1 heads. This happens in exactly $$ 2\binom{6}{1} $$ ways.

Repeating this logic, we find there are $$2\binom{6}{2}$$ ways for two flips and $$\binom{6}{3}$$ ways for 3 flips. Note there is only one way for this to happen: we have exactly 3 heads and exactly 3 tails.

To find the expected value, we sum up the number of flips times the likelyhood it happens. Since there are 2^6= 64 ways to flip 6 coins, our expected value is $$ 0 \frac{\binom{6}{0} }{64} + 1 \frac{2\binom{6}{1} }{64} + 2 \frac{2\binom{6}{0} }{64} + 3 \frac{\binom{6}{3} }{64}$$.

This simplifies down to $$\frac{33}{16}$$ giving an answer of $$\boxed{49}$$

There are $2^{6} =64$ results of flipping and landing six coins. H=Head, T=Tail. We' ll have 1 case with 6 of H, so 0 turned over ; 6 cases with 1 H and 5 T, 1 turned over ; 15 cases with 2 H and 4 T, 2 turned over ; 20 cases with 3 H and 3 T, 3 turned over ; 15 cases with 4 H and 2 T, 2 turned over ; 6 cases with 5 H and 1 T, 1 turned over ; 1 case with 6 T, 0 turned over The expected value will be $(1 \times 0 + 6 \times 1 + 15 \times 2 + 20 \times 3 + 15 \times 2 + 6 \times 1 + 1 \times 0) / 64$ = 132/ 64 = 33/16. So, * a+b = 49 *