Gambling game between A and B?

Unfortunately I do not know formulas with which to approach this problem, but we can model it pretty easily.

We have six possible states for the scores for each round the game is played.

For ($A,$B): (5,0), (4,1), (3,2), (2,3), (1,4), (0,5).

Each round the game is played, we can define how many of each of these states appear with a recursive formula. To illustrate, consider that we start at the state (2,3) and run two rounds.

Round 1:

(3,2): 1

(2,3): 1

(1,4): 1

Round 2:

(4,1): 1

(3,2): 2

(2,3): 3

(1,4): 2

(0,5): 1

If we study the pattern we see that the number of each entry is a sum of entries from the previous round.

For Round(n)

(5,0) = (4,1) from Round(n-1)

(4,1) = (4,1)+(3,2) from Round(n-1)

(3,2) = (4,1)+(3,2)+(2,3) from Round(n-1)

(2,3) = (3,2)+(2,3)+(1,4) from Round(n-1)

(1,4) = (2,3)+(1,4) from Round(n-1)

(0,5) = (1,4) from Round(n-1)

We can then use Excel to take the cumulative totals for ($A,$B) states of (5,0) and (0,5) over many successive rounds to find the probability of A winning.

After 10 rounds we have 2046 states where the game ends with A winning, out of 4180 total game ends so P(A Wins) = .489474

After 200 rounds P(A Wins) = .5

Intuitively this makes sense as we're playing until either A or B runs out of money. A tie is not considered an end-state. We have roughly a .063661 chance of the game ending on a particular round given sufficiently a sufficiently large number of trials, but in the end there are only two outcomes. A wins, or B wins, and with infinite trials the starting position becomes moot.

Let P(A wins) = 1/3, P(A lose) = P(B wins) = 1/3 and P( A doesn't ern anything) = 1/3.

Let Q2 be the probability that A will win the game with $2.

Therefore

Q2 = 1/3 * Q1 + 1/3 * Q2 + 1/3 * Q3

2/3 * Q2 = 1/3 * (Q1 + Q3)

Q2 = 1/2 * (Q1 + Q3).

(Since if you have two dollor then you will have three options if you bet. (1) you lose money, (2) you earn and (3) game may tie)

2/3 * Q2 = 1/3 * (Q1+Q3). This is your difference equation with Q5 = 1 and Q0 = 0.

Now solve this. The intermdiate steps Im giving.

Q1 = 1/2 * Q2.

Q2 = 2/3 * Q3

Q3 = 3/4 * Q4

Q4 = 4/5 (since Q5 =1)

Therefore Q2 = 2/5 (Ans)

However long you have been playing, it is clear that the probability of winning starting from $2 or whatever fortune between $1 and $4 is always the same.

So you can apply the rule of conditional probabilities. The probability to win starting with $2 is the

probability to win $1 at the next round times the probability to win starting with $3 plus

probability to win $0 at the next round times the probability to win starting with $2 plus

probability to lose $1 at the next round times the probability to win starting with $1

Q2 = 1/3(Q1+Q2+Q3) or 2Q2=Q1+Q3

For analogous reasons: Q1 = 1/3(Q1+Q2) or 2Q1 = Q2

On the other hand, the probability to win starting from $2 is equal to the probability to lose starting from $3 because of the equalities of the 'up' and 'down' transition probabilities

so Q2 = 1-Q3

Solving the system one gets: 2Q2= 1/2Q2+1-Q2 or Q2 = 2/5 , Q3 = 3/5, Q1 = 1/5 and clearly also

Q4 = 1-Q1 = 4/5