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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.

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  • $\begingroup$ explain the rules a bit more who is in or out? $\endgroup$ – Willemien Sep 8 '13 at 6:45
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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".

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  • $\begingroup$ This just shows the reality that I am back in school, all this math haha. This sounds pathetic that I can't argue with you in terms of actual math, but I still think that you would rather go first. If you went second the odds are that the first person did not pick the 'It' number now you have to only pick between 2 numbers and the bad one is probably still in there but now because you go second you have a worse chance picking between only two. like if you picked between 100, then the next person picked between 2 (it was reduced down for this explaining purposes) you would want to pick first. $\endgroup$ – Brandon Pickert Sep 8 '13 at 19:50
  • $\begingroup$ @BrandonPickert If the first person picks between $100$ items, and the second between $2$ items, what happens to the $97$ items which are not picked by the first, and aren't available to the second? You do have to be very careful with these questions. You may have an intuition about the "right answer" at the moment, but clear thinking and accurate calculation will help to improve your intuitive grasp of these situations (if you let it). $\endgroup$ – Mark Bennet Sep 8 '13 at 19:56
  • $\begingroup$ I guess from what your saying directly or indirectly, I'm forgetting to add the probability of the first person picking the bad one, to the favor of the second person getting to choose between two of the last ones, which adds up to again a favorable 2/3 chance of picking a good one. And the third person gets the chance of 2/3 that the other two picked the bad one too. Did I put your math into words? And thank you for being awesome and smart =). So basically I am wrong and my friends are right, winning them the bet and forcing me to go swing dancing? $\endgroup$ – Brandon Pickert Sep 8 '13 at 23:54
  • $\begingroup$ I also did a tree diagram (I looked back at my Algebra 2 notes from last year). I now see this is what you did and they all have an equal chance of being 'it'. Thank you Conditional Probability for your existence, unless I am actually right and hundreds of mathmaticians are wrong, which is a possibility (but not a probability =) ) I guess I will have to go swing dancing which i find no enjoyment in. Now the new problem is that they think I will begin to like swing, but I have already done it twice and still am bored swinging girls around me. Worst part is they know my dislike :/ Thanks though! $\endgroup$ – Brandon Pickert Sep 9 '13 at 1:55
  • $\begingroup$ From this answer one would conclude that all this "A picks a number first" stuff irrelevant -- one might as well have them draw cards numbered 1, 2 and 3 up front, flip them at the same time, and the one with the "#1" card will be "it". $\endgroup$ – CompuChip May 27 '16 at 17:32
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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.

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  • $\begingroup$ Yeah but this whole question is about how whether the probability changes if someone from one or two might pick it. "assuming person two did not pick it." $\endgroup$ – Brandon Pickert Sep 8 '13 at 19:39
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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 $\frac{1}{3}$.

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  • $\begingroup$ Thanks for making it easier to understand. But wouldn't you think that because person A probably didn't choose it, B and C would have a harder time. Giving the most advantage to A. $\endgroup$ – Brandon Pickert Sep 8 '13 at 19:44
  • $\begingroup$ As you pointed out yourself, the probability that A doesn't get the number is about $66.66\%$, so the probability she gets the number is about $33.34\%$, or more exactly, $\frac{1}{3}$. That is exactly what my answer says. So you in fact agree with your friends. $\endgroup$ – André Nicolas Sep 8 '13 at 19:53

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