If you have $100 and bet $1 on an even-money game, what’s the probability that you reach $200 before going broke?

The Gambler’s Ruin is a classic problem in probability theory. In this case, we are interested in the probability of reaching a target amount of $200 before going broke, where we start with an initial amount of $100 and place a $1 bet on an even-money game each time.

Let p be the probability of winning each bet, and let q be the probability of losing each bet, with p + q = 1. In an even-money game, p = q = 1/2. Let P_n denote the probability of reaching $200 before going broke when starting with $n. So, we want to calculate P₁₀₀.

The gambler’s ruin problem can be solved using a recursive equation. Since the gambler either wins or loses each bet, the recursive equation can be written as:

P_n = p * P_n + 1 + q * P_n − 1

Plugging in the values for p and q for an even-money game, we get:

P_n = (1/2) * P_n + 1 + (1/2) * P_n − 1

Rearranging the terms, we get:

P_n + 1 − 2P_n + P_n − 1 = 0

This is a linear homogeneous difference equation with constant coefficients. The general solution of this equation can be expressed as:

P_n = A + Bn

Here, A and B are constants. The boundary conditions for our problem are:

1. P₀ = 0 (The probability of reaching $200 when starting with $0 is 0.)

2. P₂₀₀ = 1 (The probability of reaching $200 when starting with $200 is 1.)

Applying these boundary conditions, we have:

P₀ = A = 0,  =  > A = 0.

P₂₀₀ = B * 200 = 1,  =  > B = 1/200.

Thus, the solution to the difference equation is:

P_n = (1/200) * n

Finally, we can find P₁₀₀, the probability of reaching $200 before going broke when starting with $100:

P₁₀₀ = (1/200) * 100 = 1/2

So, the probability of reaching $200 before going broke in this even-money game is 1/2 or 50%.