Probability With Coins- Is There a Proof of This?

Oct 2013
690
90
New York, USA
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?
 
May 2013
94
10
i'm having a hard time parsing your question.
Let's say you want exactly X heads and get to choose how many flips you want.
Do you mean x consecutive heads, or total number of?
I'm going to assume total number of.
Let's start simple. 1 head. Then for two flips,
H H
H T
T H
T T

you have a 50% chance with 2 flips.
Now let's try 1 flip.
H
T
still 50%.
Now let's try 3 flips.
HHH
HHT
HTH
THH
HTT
THT
TTH
TTT
so 37.5% chance (3/8).
Now lets try 2 heads.
With 3 flips, 3/8
with 4 flips
HHHH
HHHT
HHTH
HTHH
HHTT
HTHT
HTTH
HTTT
THHH
THHT
THTH
TTHH
THTT
TTHT
TTTH
TTTT
6/16, or 3/8. so it seems you hypothesis is correct,
In reality, the probability of exactly X heads has a tie for the highest with 2X and 2X-1 flips.
I'm not well versed enough to comment on the Pascal's triangle relationship.
 
May 2013
94
10
To make it clear for others, if you take the 2*x row of Pascal's triangle, where x is the number of flips, the largest value on that row, divided by 2^(x+1), will be the probability of getting exactly that many heads.
for example, for two heads;
Code:
    1
   1 1
  1 2 1
 1 3 3 1  -> 3/8
1 4 6 4 1
 
Oct 2013
690
90
New York, USA
I meant total heads, so you wrote about what I meant.. The probability of getting exactly X consecutive heads from Y flips takes more work.