# Unfair coin

• Sep 20th 2012, 05:25 PM
jthomp18
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?
• Sep 20th 2012, 05:27 PM
jthomp18
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?
• Sep 20th 2012, 10:00 PM
Soroban
Re: Unfair coin
Hello, jthomp18!

Quote:

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: . $P(T) \,=\,2\!\cdot\!P(H)$

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

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

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

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

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

• Sep 24th 2012, 11:45 AM
LChenier
Re: Unfair coin