1. ## Rigged Game of Heads or Tails

Hi guys. I've been pretty inactive in the past few weeks due to finals and other things coming up. Anyways, here's my question:

The problem I'm about to describe might seem a little obscure. I'll give an example after so that it's more clear.

If there are two events that can occur, event A and event B, and the likelyhood of event A happening is n (where n is some rational number such that 1 < n < 0) and the odds of B happening is 1 - n. What are the odds of event B happening "r" times out of "m" trials, where the order of wins and losses does not matter?

Example: In a rigged game of "Heads or Tails," the odds of "winning" (you choose heads and it comes heads or you choose tails and it comes tails) is 75% or 3/4 (so that 3 out of 4 times that you play you will win no matter which side of the coin you chose). What are the odds of loosing 10 times (total, not necessarily in a row) if you play the game 12 times?

Hope that made sense, and thanks for the help.

2. Originally Posted by ecMathGeek
Hi guys. I've been pretty inactive in the past few weeks due to finals and other things coming up. Anyways, here's my question:

The problem I'm about to describe might seem a little obscure. I'll give an example after so that it's more clear.

If there are two events that can occur, event A and event B, and the likelyhood of event A happening is n (where n is some rational number such that 1 < n < 0) and the odds of B happening is 1 - n. What are the odds of event B happening "r" times out of "m" trials, where the order of wins and losses does not matter?

Example: In a rigged game of "Heads or Tails," the odds of "winning" (you choose heads and it comes heads or you choose tails and it comes tails) is 75% or 3/4 (so that 3 out of 4 times that you play you will win no matter which side of the coin you chose). What are the odds of loosing 10 times (total, not necessarily in a row) if you play the game 12 times?

Hope that made sense, and thanks for the help.
I will presume you mean the probability of B happening is 1-n.

Then the number of wins for B in m trials has a binimial distribution $\displaystyle B(m,1-n)$ so:

$\displaystyle p(r\mbox{ wins for B}) = b(r,m,n) = \frac{m!}{(m-r)! r!} (1-n)^r n^{m-r}$

RonL

3. Originally Posted by CaptainBlack
I will presume you mean the probability of B happening is 1-n.
That is what I meant. Thank you very much, CaptainBlack.