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Let the probability of Finn winning the game be 'F' and that of Sharky be 'S'. For Finn to win the game the possiblities are,

Finn winning the consecutive games in (1st & 2nd), or (2nd & 3rd) or (3rd & 4th game) or so on....

The probability of Finn winning the 1st& 2nd game is FF.

The probability of Finn winning the 2nd & 3rd game is SFF.

The probability of Finn winning the 3rd & 4th game is FSFF.

The probability of Finn winning the 4th & 5th game is SFSFF.

So, the total probability of Finn winning the game is FF+SFF+FSFF+SFSFF+FSFSFF+........ which is equal to,

FF( 1+S+FS+SFS+FSFS+......).

This can be split into 2 G.P's as, FF[(1+FS+FSFS+.....)+(S+SFS+SFSFS+....)] which on solving gives $\frac{FF(1+S)}{(1-FS)}$ .

Substituiting the values of F=4/7 and S=3/7, we get 160/259.

Therefore b-a=259-160=99.

I used a different method (neither using converging series, nor splitting the probability of winning into odd or even no. of games), so I thought I'll share my solution: using Equations.

By definition, during the first match, Finn has a $\frac{4}{7}$ chance to emerge winner - we call this Case 1 - and Sharky has $\frac{3}{7}$ - we call this Case 2 . Then after the first match, we let the probability of Finn winning in Case 1 be $Y$ and the probability of Finn winning in Case 2 be $X$ .

Case 1 Notice that in Case 1, if Finn were to win again, he is the overall winner. If not (Sharky wins instead), then it is back to Case 2 again. Hence, probability of Finn winning in Case 1 is $\frac{4}{7}$ + his probability of winning Case 2. $Y = \frac{4}{7} + \frac{3}{7}X.$ Case 2 Similarly notice that, Finn only has a chance of winning if Sharky doesn't win the next match. If Finn wins the next round, it is back to Case 1 again. Hence, $X = \frac{4}{7}Y.$ (can you see why?)

Solving these 2 equations, we get: $Y = \frac{28}{37}$ and $X = \frac{16}{37}$ . Thus, overall probability of Finn winning is $\frac{4}{7}Y + \frac{3}{7}X = \frac{160}{259}$ and $b-a = 259 - 160 = \boxed{99}$ . We're done.

Just quickly, (details to come tomorrow) .....

The probability that Finn wins the match in an even number of games is

$\displaystyle\sum_{k=1}^{\infty} (\dfrac{4}{7})^{k+1} * (\dfrac{3}{7})^{k-1} = \dfrac{16}{37}$ .

The probability that he wins in an odd number of games is

$\displaystyle\sum_{k=1}^{\infty} (\dfrac{4}{7})^{k+1} * (\dfrac{3}{7})^{k} = \dfrac{48}{259}$ .

Adding these together gives us a Finn-win probability of $\dfrac{160}{259}$ , and thus $b - a = 259 - 160 = \boxed{99}$ .

EDIT: For Finn to win in an even number of games, the possible sequences of game results will be

$FF, FSFF, FSFSFF, FSFSFSFF, ......$ ,

i.e., two more games won by Finn than by Sharky.

For Finn to win in an odd number of games, the possible sequences are

$SFF, SFSFF, SFSFSFF, SFSFSFF, ......$ ,

ie., one more game won by Finn than by Sharky.

Instead of solving an infinite series, we use the trick of getting an equation with the sum in it.

Set Y= probability of Finn winning the match if Finn won the previous game.

Set Z= probability of Finn winning the match if Sharky won the previous game. The following equations hold: $Y=4/7+3/7*Z$ $Z=4/7*Y$

The solution to these equations is

$Y = 28/37, Z = 16/37$

The probability of Finn winning the match, is equal to $4/7*Y+3/7*Z$ , or $160/259$ .

Therefore, the answer is $259-160=99$

Let's define some variables:

• Let $p_{0}$ be the probability that Finn wins the match before any games are played. We will solve for this variable.

• Let $p_{F}$ be the probability that Finn wins the match assuming the match has not yet been won, but at least one game has been played, and Finn won the previous game.

• Let $p_{S}$ be the probability that Finn wins the match assuming the match has not yet been won, but at least one game has been played, and Sharky won the previous game.

• Let $p = \frac{4}{7}$ be the probability that Finn wins any given game.

We can evaluate $p_{0}$ by conditioning on the outcome of the first game. Finn wins the first game with probability $p$ and loses with probability $1-p$ . Therefore:

• $p_0 = p\cdot p_{F} + (1-p)\cdot p_{S}$

We can similarly evaluate $p_{F}$ and $p_{S}$ by conditioning on the outcome of the second game:

• $p_{F} = p + (1-p)\cdot p_{S}$

• $p_{S} = p\cdot p_{F}$

We can use the expression for $p_{S}$ to express $p_{F}$ in terms of only $p$ :

• $p_{F} = \frac{p}{p^{2}-p+1}$

This also gives us an expression for $p_{S}$ in terms of only $p$ :

• $p_{S} = p\cdot p_{F} = \frac{p^{2}}{p^{2}-p+1}$

We can now express $p_{0}$ in terms of only $p$ :

• $p_0 = p\cdot p_{F} + (1-p)\cdot p_{S} = \frac{p^{2}\cdot(2-p)}{p^{2}-p+1}$

This makes some sense:

• If $p$ were equal to 0, then Finn would have no chance of winning the match.

• If $p$ were equal to $\frac{1}{2}$ , then Finn would have a 50% chance of winning the match.

• If $p$ were equal to 1, then Finn would have a 100% chance of winning the match.

We were asked to solve for the specific instance of $p = \frac{4}{7}$ ...

In that case, $p_{0} = \frac{160}{259}$ , and $a - b = 99$ .