Probability of Surviving Pill Selection

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SUMMARY

The discussion centers on a probability problem involving Mr. Hope and two indistinguishable pills, one of which is poisoned. The odds of a victim selecting a pill are 0.3, with a 0.5 chance of survival if a pill is chosen, and a 0.65 chance of survival if no pill is selected. The correct calculation for the probability of survival in one encounter is 0.3 * 0.5 + 0.7 * 0.65, while the probability for two independent encounters is the square of the survival probability from one encounter. Participants express confusion over the scenario's logic and the implications of allowing victims to choose.

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transmini
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I'll start off by saying this is technically a coursework question, just not mine, nor of the present. I work in a tutoring center and it's been a while since I have taken stats and don't quite understand this person's professor's reasoning.

The general idea of the question was that there was a guy named Mr. Hope who had 2 pills which he could not distinguish. One of which was laced with poison, and the other was not. For one of Hope's victims, he offered them the choice of pills. The odds were 0.3 that the victim selected a pill. If the victim selected the pill, the odds were 0.5 that Hope would survive. If the victim did not select the pill, the odds were 0.65 that Hope would survive.

What are the odds of Hope surviving this endeavor with 2 victims, where the scenario is the exact same for each?What we did was say the odds of Hope surviving one encounter were ##0.3*0.5+0.7*0.65##, because there was a 30% chance of getting to the scenario of 50% chance of survival OR 70% chance of a 65% chance of survival, and then square this result to get it for surviving both of 2 encounters. But apparently the first step is wrong, so could anyone explain? He had ##0.3*0.65+0.7*0.5## if I recall correctly. I know he transposed 2 of the numbers in that sum.
 
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Your answer to the first case is correct, but it is unclear how the scenario with 2 victims is supposed to look.
The problem is weird in general. If Hope wants to survive, and has a better chance to do so if he chooses the pill himself, why would he let someone else make a random choice?
 
mfb said:
Your answer to the first case is correct, but it is unclear how the scenario with 2 victims is supposed to look.
The problem is weird in general. If Hope wants to survive, and has a better chance to do so if he chooses the pill himself, why would he let someone else make a random choice?

I totally forgot I asked this, whoops. But that's what I thought. I asked someone else I work with as well and they thought the same thing. I'm not sure what the person's professor was thinking then. But as far as 2 victims, the trial conducted with the first is the exact same as the second. But since the victims are independent of each I just said to multiply the probability with itself. Now as far as the back story, sure it's flawed but maybe he wanted it to seem more authentic and less "I'm trying to kill you"?
 
What does “the trial conducted is the same” mean? Mr. Hope has a set of pills for each one? Mr. Hope has just one set and picks one only if both prisoners don’t choose one? What happens if both prisoners choose a pill - but not the same one? Way too many open questions.
 
transmini said:
The odds were 0.3 that the victim selected a pill.
Selected a pill (so the possibility of not selecting either) or selected a particular pill?
If the latter and they are indistinguishable, how come the odds are not 50-50?
transmini said:
If the victim selected the pill, the odds were 0.5 that Hope would survive.
Who is taking the selected pill, the "victim" or Hope?
transmini said:
He had 0.3∗0.65+0.7∗0.5
Sounds like he just got confused and swapped two numbers somewhere.
 

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