# Thread: Unfair coin

1. ## Unfair coin

I am trying to determine a probability. One side of my coin (heads) is "two and one-half times" heavier than the other side (which is tails).

Can anyone tell me what a probability would be for that? I am needing to do a simulation where I have to flip the coin 100 times and determine how many times it lands on heads .. which would be the heavier side.

Any suggestions?

2. ## Re: Unfair coin

Once I know the "numbers" for both sides, how will I determine how many times it would land on heads if I was to flip the coin 100 times? Is there any equation?

3. ## Re: Unfair coin

Hello, jthomp18!

I am trying to determine a probability.
One side of my coin (heads) is "two and one-half times" heavier than the other side (which is tails).

Can anyone tell me what a probability would be for that?
I am needing to do a simulation where I have to flip the coin 100 times
and determine how many times it lands on heads .. which would be the heavier side.

Any suggestions?

I had to baby-talk my way through this one.

Suppose that Heads is twice as heavy as Tails.
I would assume that Heads would be twice as likely to be on the bottom.
. . That is, the coin would show Tails.

So we have: .$\displaystyle P(T) \,=\,2\!\cdot\!P(H)$

We also know that: .$\displaystyle P(H) + P(T) \:=\:1$

Solve the system and we get: .$\displaystyle P(H) = \tfrac{1}{3},\;P(T) = \tfrac{2}{3}$

For your problem, Tails is $\displaystyle \tfrac{5}{2}$ times as likely as Heads.

Solve the system: .$\displaystyle \begin{array}{ccc} P(T) \:=\:\frac{5}{2}\!\cdot\!P(T) \\ P(H) + P(H) \:=\:1 \end{array}$

. . and we get: .$\displaystyle P(H) \,=\,\tfrac{2}{7},\;P(T) \,=\,\tfrac{5}{7}$

4. ## Re: Unfair coin

I don't believe Soroban answer.

If a coin with a heavier side it flip, it will rotate around an off center center of gravity, but will still rotate with a constant angular speed, and will have the same probability to touch the ground on either side.

Also, if this coin stay in the air long enough for the air friction stop its rotation, it will stop with the heavier side down. This is its lower potential state. And therefore will touch the ground with its heavier side first. This will happen with an almost 100% probability.

Now, what will give this coin more chance to stop the heavier side down is the fact that it will bounce. And in this bouncing process, the coin have more chance to finish in its lower potential position. (If the coin had touch the ground with the heavier side up, it will have tendency to rotate when bouncing. in the other case, it will have tendency not to rotate).

However, I am really far to believe that this process will be a linear function with the distribution of the weight inside the coin. (Specially due to the fact that a coin is thin)

If you ave this unfair coin, try to flip it many time and let us know the result.