Probability Distribution for the Sum of Coin Flips?

Question

Suppose I have a coin that has:

• A 0.3 probability of Heads and a 0.7 probability of Tails
• If Heads, score = score + 1.3 and if Tails, score = score - 0.9
• Each flip is i.i.d

I am interested in understanding how to analyze the following problems:

• Problem 1: Suppose the initial score = 0. What is the probability of the score = $n$ after $t$ turns?
• Problem 2: Given that the score at time $t$ is $n$ (i.e. conditional probability, what is the probability of the score being $m$ at time $t+s$?

Originally, I thought this problem would have had to be analyzed using Random Walks since we are trying to find out the Probability Distribution for the sum of a Binomial Distribution - but now I think that there might be an easier answer.

Here is my attempt:

Problem 1:

Define $ X_i$ as the result of each coin flip .Define the score after $ t$ steps by $ S_t$. Since, $ S_0 = 0$, the score after $ t$ turns can be expressed as: $$ S_t = X_1 + X_2 + \cdots + X_t $$

To find the probability that $S_t = n$, we need to determine the number of Heads ($ k$) and Tails ($ t - k$) that produces a total score of $n$. I tried to write out the algebra as:

$$ 1.3k - 0.9(t - k) = n $$ $$ 1.3k + 0.9k = n + 0.9t $$ $$ 2.2k = n + 0.9t $$ $$ k = \frac{n + 0.9t}{2.2} $$

There is a constraint in this problem that the number of Heads $ k$ must be an integer. If $ k$ is not an integer, the probability is zero. If $ k$ is an integer, the probability should just be a regular Binomial Distribution: $$ P(S_t = n) = P\left(H_t = k\right) = \binom{t}{k} (0.3)^k (0.7)^{t-k} $$

I think we can also drop the constraint and write this as an inequality:

$$ P(S_t \leq n) = P\left(H_t = k\right) = \binom{t}{k} (0.3)^k (0.7)^{t-k} $$

Problem 2: I think that this problem is identical to Problem 1, just with a different constraint equation.

The score difference between time $ t$ and time $ t+s$: $$ S_{t+s} - S_t = m - n $$

Define $ X_s$ as the score change over the next $ s$ flips: $$ X_s = 1.3H_s - 0.9(T_s) $$ where $ H_s$ is the number of Heads and $ T_s = s - H_s$ is the number of Tails in $ s$ flips.

Here is where the equation from Part 1 changes:

$$ 1.3H_s - 0.9(s - H_s) = m - n $$ $$ 1.3H_s + 0.9H_s = m - n + 0.9s $$ $$ 2.2H_s = m - n + 0.9s $$ $$ H_s = \frac{m - n + 0.9s}{2.2} $$

The probability of Heads is given by the Binomial distribution: $$ P(S_{t+s} = m \mid S_t = n) = P\left(H_s = \frac{m - n + 0.9s}{2.2}\right) = \binom{s}{k} (0.3)^k (0.7)^{s-k} $$ where $ k = \frac{m - n + 0.9s}{2.2}$.

We can also write this as an inequality without loss of generality.

Is my reasoning correct? It seems like Random Walks were not necessary in this problem?

• Note: I think somehow the answers from Question 1 and Question 2 can also be analyzed using a Normal Distribution?

Answer 1

Each heads observed increases the score by $1.3$, and each tails observed decreases the score by $0.9$. Thus, if there are total of $t$ turns, the random score for an initial starting value of $0$ will be $$S(t) = 1.3 X - 0.9 (t-X) = 2.2X - 0.9t,$$ where $$X \sim \operatorname{Binomial}(t, p = 0.3)$$ characterizes the random number of heads observed in $t$ turns. Since $$\Pr[X = x] = \binom{t}{x} (0.3)^x (0.7)^{t-x}, \quad x \in \{0, 1, \ldots, t\},$$ the set of permissible scores is $$S(t) \in \Omega(t) = \{2.2x - 0.9t\}_{x=0}^t;$$ that is, $\Omega(t)$ is the support of $S(t)$. Hence the probability that we have a score $n$ after $t$ turns is $$\begin{align} \Pr[S(t) = n] &= \Pr[2.2X - 0.9t = n] \\ &= \Pr\left[X = \frac{10n + 9t}{22}\right] \\ &= \binom{t}{\frac{10n + 9t}{22}} (0.3)^{(10n + 9t)/22} (0.7)^{(13t - 10n)/22}, \quad n \in \Omega(t), \end{align}$$ and $0$ otherwise.

So for instance, if $t = 7$, we have $\Omega(7) = \{-6.3, -4.1, -1.9, 0.3, 2.5, 4.7, 6.9, 9.1\}$, and $$\Pr[S(7) = 0.3] = \Pr[X = 3] = \binom{7}{3}(0.3)^3 (0.7)^4 = \frac{453789}{2000000} \approx 0.226895.$$ In other words, it is easier to conceptualize the desired probability in terms of the corresponding $X$ for which $S(t) = n$.

Regarding the conditional probability of the second question, it is $$\Pr[S(t+s) = m \mid S(t) = n] = \frac{\Pr[S(t+s) = m \cap S(t) = n]}{\Pr[S(t) = n]} = \frac{\Pr[S(s) = m-n]}{\Pr[S(t) = n]}.$$ Then we apply the formula above.