# Thread: Game of "Odd man out", a binomial distribution problem?

1. ## Game of "Odd man out", a binomial distribution problem?

I find this question challenging, hope to find an answer here,

Situation:
3 people play the game" odd men out". Each flips a coin simutaneously, if one face is different from the other two, its owner is the odd man, and he loses.

Question:
1) What is the probability that there is an odd men on an given turn, assuming all three coins are fair?

2) If there is no odd man on the 1st turn, all coins are flipped again, until an odd man is determined. What is the probability that even number of turns are required to determined the odd man?

1)Let X be the number of odd man
X~Bin(3,0.5)
P(X=1) = (3C1)(0.5*2)(0.5)
= 0.375

2) this is where I get lost, should I be using geometric series here?

2. Originally Posted by csy0961743
3 people play the game" odd men out". Each flips a coin simutaneously, if one face is different from the other two, its owner is the odd man, and he loses.
1) What is the probability that there is an odd men on an given turn, assuming all three coins are fair?

2) If there is no odd man on the 1st turn, all coins are flipped again, until an odd man is determined. What is the probability that even number of turns are required to determined the odd man?
It seems to me as if you have misunderstood this game.
On any one tern there are only eight outcomes.
(H,H,H),(H,H,T),(H,T,H),(H,T,T),(T,H,H),(T,H,T),(T ,T,H),(T,T,T)

In all but two of those there will be an odd man out
(H,H,H),(H,H,T),(H,T,H),(H,T,T),(T,H,H),(T,H,T),(T,T,H),(T,T,T)
Thus the probability of odd man out on any turn is $\frac{3}{4}$.

In order that the first odd man out happens on the second turn is $\left(\frac{1}{4}\right) \left(\frac{3}{4}\right)$.

In order that the first odd man out happens on the fourth turn is $\left(\frac{1}{4}\right)^3 \left(\frac{3}{4}\right)$.

Can you finish?

3. this is a geometric rv, not a binomial