Dr.X has a house comprised of ten empty rooms. He flips a fair coin and if it comes up heads, he places a clock in one of the ten rooms at random. If it comes up tails he does nothing.
You do not know the outcome of the coin flip, but you begin to search the house anyway. You randomly choose room #6 and look inside; there is no clock in this room.
Question: At this point, what is the probability that there is no clock in the house?
1 in 2
do I get my prize now?
I get 10 in 19, or about 52.1%. Am I right?
1 in 2.
Tisme wins, but only because he found the thread before I did.
I’m with Biotop on this one.
There is a 1 in 2 chance that the clock is in the house, the other numbers you guys are getting refer to the odds of a clock being in a paticular room.
probability? 1 in 2
either he has placed the clock or not.
Now that I think about it more, I am not certain what the odds would be but I doubt they would be 1 in 2.
Think about all the possible combinations of coin flips, and place them on a tree diagram.
Knowing that at least one room doesn’t have a clock, you can eliminate the possibilty of getting 10 consecutive heads, thus scrapping that one OFF the tree diagram. This would slightly sway the odds? No?
MadHatter, as far as the OP goes, there’s only been one coin flip.
I’d have to go with 1/2 - the result of the coin flip was either placing a lamp, or not placing a lamp.
Biotop is correct. (.51)/(.51+.5*.9)
I would like to side with the 1 in 2 group. Not discovering the clock in one room doesn’t affect the outcome of the coin toss.
Now, I hope this makes me really popular with the 1 in 2 crowd. 1 in 2 Rocks!
There are, at most two random events. The first event is the coin flip that determines whether the lamp is placed in one of the rooms. The rules state that the flip is fair, indicating a 50-50 chance of heads (or tails).
If the result is heads, the lamp is placed (assumedly at random) in one of ten rooms. The chances of picking the correct room (at random) is 1 in 10 if the lamp WAS placed in a room and 0 if not.
Putting these facts together we find that the likelihood of a room containing the lamp is .05 (5 percent) and the likelihood that the lamp is outside the house is .50 (50 percent).
Going back to the OP, the probability that the clock was not placed in the house is 50-50 – the direct result of the coin flip.
However we’ve looked in room #6 and determined that the clock is not there.
9 rooms remain, each having a 5 percent chance of the clock having been placed in it. The 50-50 chance of the clock having not been placed in a room also exists.
So we determine that there is a 9 in 19 chance of the clock being in one of the rooms.
So the answer is there is a 50-50 chance that the clock was NOT placed in the house. However there is now a 10 in 19 chance that the clock will not now be found in the house.
CoyoteFish:
Not discovering the clock does impact the probability, because had there been a clock somewhere there would have been a possibility of it’s being in this room.
Imagine a case where the results of a coin toss determine wehether Dr X puts clocks in nine of ten rooms. Surely you will acknowledge that in this case when one does not find a clock in a given room, it is a good bet that there is no clock in any room. The reasoning in this case is similar, although the fact that there was only a one in 10 chance in any event makes the impact much smaller.
Howver, I do hope that you become popular with the 1 in 2 crowd, despite the fact that they do not rock.
I’ve heard something very similar to this. I will try to find it, and report back.
It seems to me that the opening of one door and not discovering the clock gives us no additional information. If the coin was heads, it stays heads. If it was tails, it stays tails. All that the empty room tells us is that its not in this particular room. The probability that the clock is in the house is dependant on the outcome of the coin toss, not on whether it is in this one room. Therefore, the probability that it’s in the house at all is still 50:50.
I’m just a punk lawyer, however. Not like I know jack about probabilities. This is just what seems right to me.
1 in 2 Rules!
I recently finished my statistics, combinations and permutations unit in math. We did tons of questions like this. Going by the same method I used on other coin tossing problems, the probability of 0 coins turning up tails on 10 flips should be 1/1024, or approximately .0098%. Since you already know that one room does not have a clock, you could eliminate that room and figure out the probability of Dr. X flipping tails in 9 consecutive rooms, which would be 1/512.
All of this is assuming that he has already been through each room of the house, and completed the coin flipping, and followed directions. This is simply the probability of producing all tails (or heads) on 9 or 10 consecutive flips.
Someone far more mathematical than I might have a different method, but this is how I learned to do these.
It gets kind of confusing when you start thinking of later actions. Our natural instinct is that a later action doesn’t effect a previous condition, so the instinctual answer is 50%. The problem is that it isn’t truly a subsequent choice, it is a forced initial condition(becuase it always happens in the universe of the problem).
Don’t think of it as a subsequent choice changing things. reword the problem into the initial conditions(which is hard to get a grasp on).
He is going to flip a coin to determine if he is going to put the clock in, then determine the room, but one of the combinations (clock-yes and room 6) can’t happen. so you are really left with only 19 possible combinations, 10 of which involve the clock not being in the house.
CoyoteFish:
You might wish to check under “posterior probability”, or “Bayes’ Formula”
SouthernStyle
This seems to mean that there is a .5 in 19 chance that the clock was placed in the house but will not be found.
Discovering there is no clock in room 6 most definitely does affect the probability. Suppose you don’t stop at checking just one room, you check 8 or 9 rooms and still don’t find the clock? Wouldn’t you now be inclined to think that maybe there isn’t a clock at all?
Anyway, the correct answer is 10/19, as has been said before.
Are you guys sure about this? Would I just make myself look like an ass if I continue to argue it?
Wolfman makes a pretty convincing case (and is most likely correct), but I still can’t help but see it as 50:50. The coin is tossed. The outcome determines if it is in the house at all. If the house had only two rooms, and if you know that one room doesn’t have the clock, it seems that it still is a fifty/fifty chance that there may be a clock in the other room. Similarly, if there were three rooms, and one room is opened and no clock is found, the chance that the clock is in one of the other two rooms is not affected by the discovery that it isn’t in the first room. If this is carried on up to 10 rooms, or even more, it still seems as if the chances of the clock being in the house at all are still fifty/fifty.
My theory relies on the assumption (which very well may be incorrect) that the only information gained by not finding a clock in room 6 is that there isn’t a clock in room 6, and is not indicative of the clock being in the house at all.
Please help me.
I think SouthernStyle tripped up on semantics. I’m sure what was meant was, “There was originally a 50-50 chance that the clock was NOT placed in the house.” The rest of the analysis, and the final conclusion: “However there is now a 10 in 19 chance that the clock will not now be found in the house.” look correct to me.
Seems to me that the “1 in 2” crowd is getting confused by one of the issues that dogs the Monty Hall problem, namely, that each additional bit of information affects the probability. For those of you who think the answer is 1 in 2, riddle me this: Suppose the OP was:
Now what is the answer? Is it one in two?
And another thing: Why isn’t this in General Questions?