Probability in picking Me and my 2 friends were trying to pick who would be it for a game. A number was picked and recorded on a phone, it was 1, 2 or 3. If the first person who picks a number takes away that option from the next two who pick, does the first picker have a higher probability of picking the number that doesn't make him it? What my thinking is is that the first person has a 66.6% chance of picking a good one to not be it. Therefore probably leaving the wrong number left for the other two to pick. So now they only have two to chose from and one of them is the probable number left over from the first pick. My friends say no matter what they all have a 1/3 chance of being picked as the 'it' number, no matter if one was taken out by the first person at the start.
I did a test the day after with 3 cards. I drew for each person 100 times. It just so happened that person A had the best % of not being 'it'.  The example only ended up as a tie twice in the whole game, and person A never was in the lead for getting the 'it' card. I told my friends that i did not think it would be a significant difference but a difference there none the less. At 45 trials A~28%, B~33%, C~37%. At 50 trials A~26%, B~40%, C~34%. At 70 trials A~28.6%, B~35.7%, C~35.7%. At 90 trials all 33.3%. At 100 trials 34%, 35%, 31%.... The lengths people will go to for 40 bucks...
And yes, I did get this idea from the Monty hall problem. I heard that a while before we came to this problem.
 A: The first person has $\frac 23$ chance of picking a good number, and $\frac 13$ chance of picking the bad one.
In the first of these cases the second person has $\frac 12$ chance of picking a good number and $\frac 12$ chance of picking the bad one. In the second case, there is no chance of picking a bad one, as it has already been picked.
So the second person has $\frac 23 \cdot \frac 12=\frac 13$ chance of picking the bad number, and $\frac 23 \cdot \frac 12+\frac 13\cdot 1 = \frac 23$ chance of picking a good one.
The third person has no choice. Either the bad one has already been picked - chance $\frac 13+\frac 13=\frac 23$, or it hasn't - chance $\frac 23 \cdot \frac 12=\frac 13$ - reality check, these probabilities add up to $1$.
So each person has a $\frac 13$ chance of being "it".
A: The first person chooses from three numbers, he has a 1/3 probability of picking the 'in' number. If the number he picked is removed from the rat race, the other players are left with a 1/2 for picking the correct number. After the next number is taken, it is a 1/1 probability of picking it, assuming person two did not pick it.
A: The process has not been described in full detail. So we give one interpretation. One of the numbers $1,2,3$ is chosen at random by a referee, with all choices equally likely. 
We have contestants A, B, and C, who make a guess as to what the hidden number is, in the order A then B then C. The rule is that B must not repeat A's guess, and $C$ must not repeat A's guess or B's guess. What is the probability A is the one to guess the hidden number?  That is $\frac{1}{3}$. And the probability $B$ is the one who guesses the hidden number is also $\frac{1}{3}$. 
