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Death Probability! Really Hard One!


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29 replies to this topic

#1 roolstar

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Posted 26 December 2007 - 11:32 AM

Alfred, Bob and Charlie (A, B, C) are three prisoners.

One day, one of the guards came and told them: "Tomorrow, 2 among you will be executed, and 1 will be set free. I know who will die myself but I don't want to tell you now, I prefer to tease you during your last night here."

That night while the other prisoners were sleeping, Alfred was unable to: he was calculating the probaility of being executed the next day. According to his calculations, he had a 2/3 (66.67%) probability of being executed tomorrow.

But then "EUREKA"!

He called the guard and said: "I don't want you to tell me if I'm gonna die tomorrow, but can you tell me which among the other two will die?"

The guard answered him: "Mmmm, ok. Charlie will be one of the two prisoners who will get executed tomorrow. But I will not tell you if the second one will be you or Bob!"

Alfred with a smile: "Thank you so much man!'

Guard: "But I don't understand how this will make you feel any better."

Alfred replied with a huge smile: "Well, now I have a 1/2 (or 50%) chance of being executed instead of the 2/3 (66.67%) I had before your answer: either me or Bob will die tomorrow so 1 out of 2!"

Is Alfred right in his reasoning?
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#2 Martini

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Posted 26 December 2007 - 07:24 PM

One million people, including yourself, purchase lottery tickets of which there will be three winners. If you could get someone with inside info to rule out 999,994 losers, other than yourself if you should be one, is your probability of being a winner 1/2?
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#3 spoxjox

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Posted 26 December 2007 - 10:27 PM

Alfred replied with a huge smile: "Well, now I have a 1/2 (or 50%) chance of being executed instead of the 2/3 (66.67%) I had before your answer: either me or Bob will die tomorrow so 1 out of 2!"

Is Alfred right in his reasoning?


Spoiler for solution

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#4 bonanova

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Posted 27 December 2007 - 01:58 AM

Who's that standing over there in the corner, smiling at us?
Why, it's ... Monty Hall!

Let's see: Alfred is the door #1 that I chose,
Bob is the door #2 that Monty opened to show a losing choice,
and Charlie is the door #3 that I should trade my choice for.
If only I had that choice!
But I don't.
Ooops, poor Al still has only a 1/3 chance of surviving.

Charlie, on the other hand, now has a 2/3 chance of making it.

Interesting ... because if it had been Charlie who had asked the question,
and the guard had told Charlie that poor old Bob was gonna bite the big one,
then Al would have the 2/3 chance.

Right?
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#5 roolstar

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Posted 27 December 2007 - 09:59 AM

Let me just add that I started this new puzzle as a way to answer this old one:

viewtopic.php?f=7&t=325

Can anybody see the similarities (or differences) between the two?

Didn't we move from:

Before the answer:
Al - Bob ==> Dead
Al - Charlie ==> Dead
Bob - Charlie ==> Dead

After the answer:
Al - Charlie
Bob - Charlie

So from a probability of 1/3 of survival for Al to 1/2 now?

This is just a question for discussion and not necessarily the final answer to this riddle.

PS: And bonanova, nice call on the Monty Hall reference...
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#6 roolstar

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Posted 27 December 2007 - 01:29 PM

Before the answer:
Al - Bob ==> Dead
Al - Charlie ==> Dead
Bob - Charlie ==> Dead

After the answer:
Al - Charlie
Bob - Charlie



Just to be clearer on the above quote

Before the answer there were 3 possibilities:
Al - Bob will die
Al - Charlie will die
Bob - Charlie will die

This means Al has a 2/3 probability of dying (in 2 cases out of 3 above)

Now after the answer:
Al - Charlie will die
Bob - Charlie will die
(Bob & Al cannot both die or else the guard couldn't have said Charlie)

Hence, Al now has 1/2 probability of dying (in 1 case out of 2 above)

What's the error in this reasoning? (If there is one)
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#7 Scraff

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Posted 27 December 2007 - 01:37 PM

Hence, Al now has 1/2 probability of dying (in 1 case out of 2 above)

What's the error in this reasoning? (If there is one)


spoxjox already answered this and the replies of others seem to indicate that they agree. It's usually at this point that the OP either posts that he agrees or disagrees with the conclusions given.
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#8 roolstar

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Posted 27 December 2007 - 02:53 PM

I do agree with the reasoning presented in this post... This is in fact my own reasoning...

I'm just the kind of people who start with a puzzle and move to an even bigger application or project...

Example: From this puzzle we can make a reference like Monty Hall for example in the show where out of 3 doors, the guest picks one and then the host opens one of the remaining 2 (the one that has nothing behind it) and then the guest was presented with the choice to either keep his choice unchanged or open the remaining door instead. (Only one door had a prize behind it)

And we can compare that with the "Deal or No Deal" program today to see the differences between both!

Personal opinion: In the case of Monty Hall, the guest should always change his initial choice. This will in fact double his chances of winning (2/3 instead of 1/3).
While in Deal or No Deal, switching the box at the end will not affect the probabilities one bit. This also suggests that every time they open a box with a small or big amount of money in it, the guests should neither feel happy, nor nervous. He/She is not changing anything whatsoever in His/Her probabilities of success!!

Now where does "One Girl, One Boy" post fit into all this? And where in the calculations does the fact that the host knows which are the empty boxes (Or who will die the next day) affect the probabilities?

As far as spoxjox answer: I consider that the fact that Charlie will be among the executed IS new information (New Information: Charlie will die) and does not answer my dilemma of moving from a system of 3 possibilities to a system of two!
The remaining possibilities are:
Alfred and Charlie die
Bob and Charlie die

To be clear, I share the opinion of the rest of the people who replied on the post. But how can I clearly refute the other ones.

Thank you for your patience.
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#9 bonanova

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Posted 27 December 2007 - 04:22 PM

My comments in red.

in Deal or No Deal ... every time they open a box with a small or big amount of money in it, the guests should neither feel happy, nor nervous.
He/She is not changing anything whatsoever in His/Her probabilities of success!!
Since the offered amount is calculated from the undisclosed amounts, your payout is increased if large amounts remain undisclosed.

Now where does "One Girl, One Boy" post fit into all this?
It's not a direct fit.
The condition [one of the children is a girl] affects the outcome for a different reason from your puzzle, which is Monte Hall in disguise.
The outcome differs if the sex of a particular child is disclosed; and opening a door knowing [or not] that it's a loser affects the odds.
But that's the extent of the similarity. The odds change for different reasons.


And where in the calculations does the fact that the host knows which are the empty boxes
(Or who will die the next day) affect the probabilities?
In your puzzle [and in Monte Hall] if the guard [host] does not know, then there's no advantage to swapping:
1/3 of the time the opened door will have the car and 1/3 of the time you will have chosen the door with the car.
But if it is certain that a losing option is revealed, then your odds of winning double if you swap.


To be clear, I share the opinion of the rest of the people who replied on the post. But how can I clearly refute the other ones.
Give a clear explanation of the correct answer.

Thank you for your patience.


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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#10 spoxjox

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Posted 28 December 2007 - 01:01 AM

As far as spoxjox answer: I consider that the fact that Charlie will be among the executed IS new information (New Information: Charlie will die) and does not answer my dilemma of moving from a system of 3 possibilities to a system of two!


You may consider it new information, but it is not. To be more specific: It is not relevant new information. I mean, Alfred could have been told that Charlie's wife is pregnant, and that's new information, too -- but it doesn't help Alfred know whether he is going to be executed.

The remaining possibilities are:
Alfred and Charlie die
Bob and Charlie die


Here is the crux of the problem. You are assigning names to the possibilities, but in this case, the names don't matter.

Possibility One: Alfred and the first of the other two get executed.
Possibility Two: Alfred and the second of the other two get executed.
Possibility Three: Both of the other two get executed.

As we see, Alfred has 2 chances in 3 of being executed. Now, Alfred asks the guard to name which of the other two is being executed (or to pick one of the other two, if both are to be executed). But Alfred already knows that at least one of them is to be executed! Finding out the name of the victim doesn't give him any relevant new information. To put it another way: Alfred doesn't know whether Charlie is "the first" or "the second" of the "other two" in the possibilities listed above. The exact same three possibilities still exist, even after Alfred gets a name out of the guard.
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