Let's say you want exactly X heads and get to choose how many flips you want. It sounds logical that you would 2X flips. In reality, the probability of exactly X heads has a tie for the highest with 2X and 2X-1 flips. This can be shown with Pascal's triangle. As the simplest example, the probability of exactly 1 head is 0.5 with 1 coin (2X-1) or 2 coins (2X) and less than 0.5 for all other amounts. The amount of numbers in the row of Pascal's triangle is one more than the amount of flips because X flips has X+1 possibilities with all the numbers from 0 to X including both endpoints. The top row, which is just 1, would be for 0 flips if that mattered. When you go from a row with an even amount of terms and a tie for the highest number to the row below it, the row below it has one highest number that's double the highest in the row above it. An example is 1-3-3-1 to 1-4-6-4-1 where 6 = 2*3. When you go from a row with an odd amount of terms and one highest value to the row below it, the row below it has a tie for the highest value with two numbers than are less than double the highest number in the row above it. An example is 1-4-6-4-1 to 1-5-10-10-5-1 where 10 < 2*6. For the 1-3-3-1 to 1-4-6-4-1 case, 1-3-3-1 is for 3 flips, and 1-4-6-4-1 is for 4 flips. If you wanted exactly 2 heads, 1-3-3-1 would be 2X-1 flips, 1-4-6-4-1 would be 2X flips, and in both cases the probability is 3/8. The sum of the values doubles each time, so if the highest value doubles, the probability remains the same, such as 3/8 = 6/16, and if the highest value less than doubles, the probability decreases, such is 6/16 > 10/32. Is there a short proof or explanation of this?