Rahul flips 850 fair coins. Rohit flips 851.What is the probability that Rohit gets strictly more heads than Rahul?

This is a complex problem that involves combinatorics and probability theory. Here's a step-by-step solution:

Step 1: Understand the problem Rahul flips 850 coins and Rohit flips 851 coins. We need to find the probability that Rohit gets more heads than Rahul. Since the coins are fair, the probability of getting a head (H) or a tail (T) is the same, 1/2.

Step 2: Define the events Let's denote by H_R the event that Rahul gets a head and by H_Rohit the event that Rohit gets a head. We are interested in the probability P(H_Rohit > H_R).

Step 3: Use symmetry Because the coins are fair, the number of heads and tails are symmetric. This means that the probability of getting a head is the same as the probability of getting a tail. Therefore, P(H_Rohit > H_R) = P(T_Rohit < T_R) = P(H_Rohit + T_Rohit > H_R + T_R) = P(H_Rohit > H_R + 1), because Rohit flips one more coin than Rahul.

Step 4: Use combinatorics The number of ways to get k heads out of n flips is given by the binomial coefficient "n choose k". Therefore, the probability P(H_Rohit > H_R) is the sum of the probabilities P(H_Rohit = k) for k = 426 to 851. Each of these probabilities can be calculated as P(H_Rohit = k) = "851 choose k" * (1/2)^851.

Step 5: Calculate the probability The final probability is the sum of the probabilities calculated in step 4. This is a large sum that can be calculated using a computer.

This problem is quite complex and requires a good understanding of probability theory and combinatorics. It might be easier to solve it using a computer program.