Monty Hall, we have a PROBLEM

[owslachief]

Re: Monty Hall, we have a PROBLEM

Hah! my first thought too!


You do point out a lack of clarity in the initial problem: how many bottles of wine are you trying to save? Your solution works, with two or three bottles tossed. Unofun's works with only one bottle tossed. Mine may or may not work depending upon how immediately effective the poison is (I like his solution better than mine as it doesn't depend upon the time limit; his would work even if the wedding was 26 hours away).

Excellent point - (1) The bride's father would like to save 12 bottles if at all possible. (2) I can narrow this to "the reception is in 25 hours" and the solution is still the same .

[owslachief]

Re: Monty Hall, we have a PROBLEM

Assuming that there is no problem associated with opening a bottle and then re-corking it right away, then

Answer in white:

Open four bottles and serve each of the rats enough wine from one bottle to be lethal in 24 hours. 5-1/2 hours later or so, open four more bottles and serve each of the rats wine from one of those bottles. another 5-1/2 hours later or so, open four more bottles and serve each of the rats wine from one of those bottles.

If one of the rats is dead 24 hours later, the poison was in the first batch of four. If one of the rats is dead 29-1/2 hours later, the poison was in the 2nd batch of four. If one of rats is dead 35 hours later, the poison was in the 3rd batch. If they are all alive at the start of the reception, the poison is in the unopened bottle.

I apologize because it wasn't stated clearly that the guy wants to keep the 12 good bottles'o'wine. He doesn't want to toss 4 of them - just one.

[owslachief]

Re: Monty Hall, we have a PROBLEM

Is the answer Jesus?

*****http://media.avclub.com/images/425/425916/16x9/627.jpg?1836******

[joecct]

Re: Monty Hall, we have a PROBLEM

*****http://media.avclub.com/images/425/425916/16x9/627.jpg?1836******

OK, you stumped me. What is that?

[LynahFan]

Re: Monty Hall, we have a PROBLEM

AT&T Commercial.

[owslachief]

Re: Monty Hall, we have a PROBLEM

AT&T Commercial.

Well what that guy's really thinking is "ssssso I pick another door and maybe I won't get the goat, but ..."

[FreshFish]

Re: Monty Hall, we have a PROBLEM

I apologize because it wasn't stated clearly that the guy wants to keep the 12 good bottles'o'wine. He doesn't want to toss 4 of them - just one.

The solution I proposed would allow him to keep 12. I like Unofun's solution better though, it is more reliable and more elegant and far more clever.

Maybe I didn't phrase mine properly before: reprised in white

open bottles 1 - 4, so rat A gets sample from bottle 1, rat B gets sample from bottle 2, etc.

5-1/2 hours later, repeat with bottles 5 - 8, so rate A gets sample from bottle 5, rate B from bottle 6 etc.

5-1/2 hours later, repeat with botles 9 - 12, and so on. If Rat A dies, it is from 1 or 5 or 9; and the timing of Rat A's death determines which of the three bottles is responsible.

[sagard]

Re: Monty Hall, we have a PROBLEM

I apologize because it wasn't stated clearly that the guy wants to keep the 12 good bottles'o'wine. He doesn't want to toss 4 of them - just one.

Is mixing samples from the bottles allowed?

[owslachief]

Re: Monty Hall, we have a PROBLEM

Is mixing samples from the bottles allowed?

Knock yourself out, though I'm not sure what can be gained.

[owslachief]

Re: Monty Hall, we have a PROBLEM

FreshFish your answer in #82 is a correct one, as correct as the 4-digit binary digit-rat solution from unofan ... given I mentioned the reception was "about 36 hours away". I should have said at the outset however that "even if the reception is 25 hours away, the solution remains the same". Since I didn't state that until several posts later, however, Bravo for your answer based on initial story.

But the concepts are along the same lines:

Unique combinations. There are more unique combinations among 4 rats than there are bottles to "test" (2 more in fact). And there are 12 combinations of rat-times (Three, 5 1/2-hour intervals for 4 rats), meaning either one rat-time combination results in death, or no rats die. = 13.

I wasn't looking for the latter answer, but I left myself open to it .

[sagard]

Knock yourself out, though I'm not sure what can be gained.

Mix two samples, each from six bottles. Give to two rats and observe.

If one dies, split that group of six bottles into two samples of three bottles each. Now give each sample to a rat.

One rat will die narrowing it to three bottles. Two rats left, give them each a sample from a single bottle of the three.

If either rat dies, you have the poisoned bottle. If neither die, give the rat you like least a taste from the last poisoned bottle just to be sure!

[owslachief]

Re: Monty Hall, we have a PROBLEM

Mix two samples, each from six bottles. Give to two rats and observe.

If one dies, split that group of six bottles into two samples of three bottles each. Now give each sample to a rat.

One rat will die narrowing it to three bottles. Two rats left, give them each a sample from a single bottle of the three.

If either rat dies, you have the poisoned bottle. If neither die, give the rat you like least a taste from the last poisoned bottle just to be sure!

The only issue is we need time to do this, and I think we ran way over 36 hours.

[Twitch Boy]

Re: Monty Hall, we have a PROBLEM

In front of you are three boxes. Each box has two drawers on it, each drawer containing a coin. One box has two gold coins. Another has two silver coins. The third has one gold and one silver. You randomly pick a box and randomly open a drawer. You find a gold coin. At this point, what are the odds you have the box with two gold coins?

2/3. We can obviously rule out the box with two silvers. Now there are two boxes in play, but there are THREE gold coins available to select from them. It must be either gold coin A from the two gold box (win,) gold coin B from the two gold box (win,) or the gold coin from the gold/silver box (lose.)

[owslachief]

Re: Monty Hall, we have a PROBLEM

In front of you are three boxes. Each box has two drawers on it, each drawer containing a coin. One box has two gold coins. Another has two silver coins. The third has one gold and one silver. You randomly pick a box and randomly open a drawer. You find a gold coin. At this point, what are the odds you have the box with two gold coins?

I had to avert my eyes oh Lord when I unwittingly Replied With Quote

My answer:
I have a 2/3 probability of having picked the box with two gold coins. This is because I either (a) picked drawer 1 from the Gold-Gold box, (b) drawer 2 from the Gold-Gold box, or (c) the Gold-coin drawer from the Gold-Silver box. That last one had a 1/3 probability, so the remaining 2/3 chance is that I picked one of the Gold-Gold box drawers.

[The Rube]

Re: Monty Hall, we have a PROBLEM

owls: it's a variation on the Monty Hall prob, I think:

You have a 50/50 shot at the two golds. Why? You eliminated the silver/silver at this point (much like removing one door that contains the goat).

Crap. You are right. I just looked it up. Stupid math.