Four salesmen play "odd man out" to see who pays for lunch. They each flip a
coin, and if there is a salesman whose coin doesn't match the others he pays
for lunch. (For instance, he might get heads while the other three get tails.)
What is the probability that there is an "odd man" the first time they flip?
One way to do it is by listing all 16 possible cases:
1st 2nd 3rd 4th odd
man man man man man
---------------------------------------
case 1 H H H H NONE
case 2 H H H T 4th man
case 3 H H T H 3rd man
case 4 H H T T NONE
case 5 H T H H 2nd man
case 6 H T H T NONE
case 7 H T T H NONE
case 8 H T T T 1st man
case 9 T H H H 1st man
case 10 T H H T NONE
case 11 T H T H NONE
case 12 T H T T 2nd man
case 13 T T H H NONE
case 14 T T H T 3rd man
case 15 T T T H 4th man
case 16 T T T T NONE
Then we count and find there are 8 out of the 16
cases in which there is an odd man. So the
probability is or .
Method 2. There are 4 coins.
We can choose 3 to assign
heads to and choose 1 to assign tails to
OR
We can choose 3 to assign
tails to and choose 1 to assign heads to
OR
We can choose 3 to assign
heads to and choose 1 to assign tails to
OR indicates addition:
So there are (4C3)(4C1)+(4C3)(4C1) = (4)(1)+(4)(1) = 4+4 = 8
That's the numerator of the probability.
Then for the denominator, which is all the ways
the coins can be tossed.
We can choose heads or tails for the first one 2 ways
AND
We can choose heads or tails for the second one 2 ways
AND
We can choose heads or tails for the third one 2 ways
AND
We can choose heads or tails for the fourth one 2 ways
AND indicated multiplication.
So that's (2)(2)(2)(2) = 16 ways
That's the denominator of the probability.
So that also gives us 8 out of 16 or or
You can also do this problem by a tree diagram.
Edwin