Probability - good argument can help

$\begin{aligned} &Pr( \text{Ann's heads = Boris heads} ) \\ &= Pr( \text{Ann's tails = Boris' heads} ) \hspace{14mm} \text{ since it is a fair coin, both Probability equals } \\ &= Pr( \text{4 - Ann's heads = Boris' heads} ) \\ &= Pr( \text{Ann's heads + Boris' heads} = 4 ) \hspace{5mm} \text{ that is total 4 heads in 7 flips } \\ &= { {7}\choose{4} } \cdot \left( \cfrac{1}{2} \right)^7 \\ &= \cfrac{35}{2^7} \\ \therefore A \times B \times C &= 35 \times 2 \times 7 = \boxed{490} \end{aligned}$

If A = the number of heads Ann tosses and B = the number of heads Boris tosses, then A and B follow independent binomial distributions. Therefore,

P(A = k and B = k) = P(A=k) * P(B=k) = ((4 choose k) (1/2)^4) * ((3 choose k) (1/2)^3)

Substituting values of k = 0, 1, 2, and 3 and summing shows the required probability to be 35/(2^7). Therefore, ABC = 35*(2) * 7 = 490

Out of $2^4$ moves of Ann, $1$ contain $4$ heads, $4$ heads are found in $0$ of Boris's $2^3$ moves.

Out of $2^4$ moves of Ann, $4$ contain $3$ heads, $3$ heads are found in $1$ of Boris's $2^3$ moves.

Out of $2^4$ moves of Ann, $6$ contain $2$ heads, $2$ heads are found in $3$ of Boris's $2^3$ moves.

Out of $2^4$ moves of Ann, $4$ contain $1$ head, $1$ head is found in $3$ of Boris's $2^3$ moves.

Out of $2^4$ moves of Ann, $1$ contain $0$ heads $0$ heads are found in $1$ of Boris's $2^3$ moves.

Total Probability = $\frac{1}{16} * \frac{0}{8} + \frac{4}{16} * \frac{1}{8} + \frac{6}{16} * \frac{3}{8} + \frac{4}{16} * \frac{3}{8} + \frac{1}{16} * \frac{1}{8} \rightarrow \frac{35}{2^7}$

$35*2*7= \boxed{490}$

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Note that the number of heads is Ann's four dice vary as ${1, 4, 6, 4, 1}$ and Boris's vary ${1, 3, 3, 1}$ , which are steps of Pascal's triangle.

Probability of all possible outcomes for Ann’s three flips.

1 Possible 0 heads outcome is 3 tails flipped $P(0h)=\frac { 1 }{ { 2 }^{ 3 } } =\frac { 1 }{ 8 }$

3 Possible 1 heads outcomes are tails occurring at any 1 of the 3 flips
$P(1h)=\frac { 1 }{ 8 } +\frac { 1 }{ 8 } +\frac { 1 }{ 8 } =\frac { 3 }{ 8 }$

2 heads in three flips means 1 tail

3 Possible 2 heads outcomes
$P(2h)=P(1t)=P(1h)=\frac { 3 }{ 8 }$ since it is a far coin, both probabilities are equal

3 heads in three flips means 0 tails
1 Possible 0 heads outcomes
$P(3h)=P(0t)=P(0h)=\frac { 1 }{ 8 }$ since it is a far coin, both probabilities are equal

Probability of all possible outcomes for Boris’s four flips.

1 Possible 0 heads outcome is 4 tails flipped
$P(0h)=\frac { 1 }{ { 2 }^{ 4 } } =\frac { 1 }{ 16 }$

4 Possible 1 heads outcomes are tails occurring at any 1 of the 4 flips
$P(1h)=4\frac { 1 }{ { 2 }^{ 4 } } =\frac { 4 }{ 16 }$

6 Possible 2 heads outcomes =3! = 6 (2 heads occurring at any combination of 2 of the four flips)
$P(2h)=6\frac { 1 }{ { 2 }^{ 4 } } =\frac { 6 }{ 16 }$

4 Possible 3 heads outcomes
$P(3h)=P(2t)=P(2h)=\frac { 4 }{ 16 }$ since it is a far coin, both probabilities are equal

There are four possible outcomes where Ann and Boris could possibly match. 4 heads is not possible for Ann.

Probability that they both have 0 heads $P(0h)=\frac { 1 }{ 8 } *\frac { 1 }{ 16 } =\frac {1}{ 128 }$

Probability that they both have 1 heads $P(1h)=\frac { 3 }{ 8 } *\frac { 4 }{ 16 } =\frac { 12 }{ 128 }$

Probability that they both have 2 heads $P(2h)=\frac { 3 }{ 8 } *\frac { 6 }{ 16 } =\frac { 18 }{ 128 }$

Probability that they both have 3 heads $P(3h)=\frac { 1 }{ 8 } *\frac { 4 }{ 16 } =\frac { 4 }{ 128 }$

The sum of all possible matching outcomes $\frac { 35 }{ 128 } =\frac { 35 }{ { 2 }^{ 7 } }$

$35\times 2\times 7=490$