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July 1st, 2018, 07:53 AM  #1 
Newbie Joined: Jul 2018 From: Long Beach Posts: 1 Thanks: 0  how not to get poisoned?
This morning I was watching an old edition of silent library, where six contestants choose between six cards, with one of them denoting the loser. To simplify the problem, I tried to think of it as six people picking from six vials of potion, with one vial being poison. Naturally, I started thinking, "is there any strategy that would help you given the way everyone picks?" (Everyone grabs at the same time at random). My question is: Does order of choice make any difference or alter the direct 1/6 odds of choosing the poisoned potion? For instance, I thought that by picking last, you are betting against the field that someone has chosen the poisoned potion out of the batch, because it would be like 5 people choosing linearly regardless of actual order. Then just generally I started thinking it would be best to choose first or last, because it would be essentially the same choice. You getting the most safe options, therefore the best probability. However trying to prove this mathematically, I keep finding that you can't escape the direct probability. Is this true? Statistics is not my strength. Last edited by skipjack; July 1st, 2018 at 10:00 AM. 
July 1st, 2018, 10:01 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,186 Thanks: 1648 
You're right (no escape).

July 1st, 2018, 01:06 PM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,244 Thanks: 887 
Suppose you were to choose first. Then the other people really do not matter. There are 6 vials, one containing poison. The probability of choosing the poison is 1/6. Suppose you choose second. Then one other person will choose before you. The probability this person chooses the poison is 1/6 in which case you cannot choose the poison. The probability this person does NOT choose the poison is 5/6. In that case there are 5 vials left so you probability of choosing the poison is 1/5. If you choose second the probability you will choose the poison is (1/6)(0)+ (5/6)(1/5)= 1/6. Suppose you choose third. There are two other people choosing before you. The probability the first person chooses the poison is 1/6 in which case neither the next person nor you can choose the poison. The probability the first person does not choose the poison is 5/6 in which case there are 5 vials left, one of which has the poison. In this case, the probability the second person chooses the poison is 1/5 and in that case you cannot choose the poison. But there is also a 4/5 chance the second person does not choose the poison. In this case there are 4 vials left, one of which contains the poison. In this situation, the probability you choose the poison is 1/4. Overall, if you choose third, the probability you will choose the poison is (1/6)(0)(0)+ (5/6)(1/5)(0)+ (5/6)(4/5)(1/4)= 1/6. Do I need to go on? 
July 1st, 2018, 02:08 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,765 Thanks: 860 
Label the vials 1 to 6. Shuffle and deal 6 cards numbered 1 to 6. Same thing, right? 