Probability Puzzle

Miscellaneous and Personal Stuff I Must Share
21

Yeah, I’m sure. I think you’re confusing point-of-view. Consider a simpler problem: Suppose I hide a gumdrop in one of my hands (mmmmm… gumdrop). Now, you get to choose which hand it’s in. You choose left. What’s your chances of being correct?

From your point of view, it’s 50/50, because you imagine there is an equal chance of the gumdrop being in either hand. From my point of view, it’s zero, because I know the gumdrop is in my right hand.

Who’s right? We both are. The probablity depends on what information we possess. In the current clock problem, you are conflating your point of view now (where additional information will help narrow down the possibilities) with that of Dr. X previously (just before he flips the coin, knowing it’s a fair 50-50 toss).

22

Sorry. I was in a hurry to get to lunch. I didn’t finish my explanation.

The clock was placed on the outcome of a fair coin toss. Assuming a uniform distribution, the probability of the clock being place in any one room is 5%. The probability that the clock is not placed in ANY room is 50%.

Examining room 6 doesn’t impact the placement. Nor does failing to find a clock in room 6.

When we examined room 6 there was a 5% chance of finding the clock. Since the clock was not found we must continue searching to determine it’s location. But the clock is not reassigned a position based on a model with one less room.

When we examined room 6 exactly one of two things could occur. The clock would be found or it wouldn’t. Since the clock was NOT found we must continue looking if we wish to determine its location. The probability of the clock being in any one room is still exact the same as the probability of it being in any other room. However, the probability of it not being in any room (10 in 20) is now greater than the probability of it being in any room (9 in 20) due to the fact that we’ve eliminated a possible outcome (1 in 20).

23

Zut: why wouldn’t it still be 1 in 2 if you had opened 999,999 rooms? It’s either in the room, or it isn’t.

I believe that I am being abandoned by my fellow 1-in-2’ers. I see y’all’s point of view, and I see the logic behind it. And, what you say seems right. Nonetheless, I will persist if for no other reason than to keep my 1-in-2 cause alive.

24

Well, sure, it either is or it isn’t, but that doesn’t mean there’s a 50-50 chance either way. For example: is SouthernStyle left-handed? I haven’t the faintest idea, but either he is, or he isn’t. Care to argue that the chances are 50-50?

Or better yet, when’s my birthday? Is it April 4th? Clearly it either is, or it isn’t, but the chances aren’t 50-50.

Back to the clock problem. Suppose you knew there was, originally, a 50-50 chance that a clock was placed in one of 1,000,000 rooms. So then you search 999,999 of them and don’t find a thing. Before you open the last room, I offer a sporting proposition. I’ll bet you 10:1 that the room is empty. You slap down a dollar, I slap down ten. If the chances were 50-50 (like a coin toss), it seems like a good gamble. Do you take that bet?

Or, another analogy: We should agree that (from your point of view) there is a 50-50 chance that my birthday is in January, February, March, April, May, or June. Agreed? OK, suppose you randomly ask me each date, i.e., “Zut, is your birthday February 11th? Zut, is your birthday June 22nd?” and so forth.

I answer no to every single date, until you get to the last one (say it’s May 12th). Do you still think there’s a 50-50 chance my birthday is on May 12th? If you do, then we can set up an experiment with other willing Dopers. I’m eager to take bets…

25

Damn, my last post didn’t…well, post.
Don’t become frustrated with me, zut. I have finally seen the light, have had an epiphany, found my ass with both hands and a flashlight, whatever. I (man, this is hard) was wrong. Thank you for your explanation.

26

Take a simpler approach to the same problem where there are only two rooms:


  Toss  Location  Odds
  Head  Room 1    1 in 4
  Head  Room 2    1 in 4
  Tail  Outside   2 in 4

We start our test by examining room 1 and find that the clock is NOT there. Searching room 2 will reveal whether the clock was placed in room 2. And since it’s absence from room 2 will indicate that the clock was not placed in ANY room, the puzzle will be solved.

But the probability that the clock is in room 2 is NOT 50-50. There was a 1 in 4 chance that the clock was placed in room 2. There was also a 2 in 4 chance that the clock was not placed in any room. (We’ve examined room 1 and eliminated the 1 in 4 chance that the clock was placed there.) The probability of the clock being in room 2 is clearly 1 in 3.

Expand the puzzle to 3 rooms, then 4, etc and the same principals alway apply.

27

Hey; no problem. The first criterion for learning new things is the willingness to, uh, learn. Or something like that. I’ve certainly been wrong a time or two in my short tenure here.

28

I’m still with my original post. Let me explain. Your favourable odds of Dr. X flipping tails 10 consecutive times (hence, no clocks in any room) is what we’re looking for. Probabilities are determined by putting favourable outcomes over all possible outcomes. There is one favourable outcome. All tails. All possible outcomes are determined by two possible outcomes (heads or tails - 2) to the power of the number of coin tosses, 10. Two the the power of 10 is 1024. 1/1024. Divide that out to get your percentage.

C’mon. Anyone behind me on this?

29

Lolagranola, you don’t seem to have interpreted the problem in the same way the rest of us have. I don’t believe he is flipping a coin for each room to determine if the coin is in the room. There is only one coin flip, heads he’ll put the clock in “a” room, tails he won’t put a clock in any room. then he puts the coin in his pocket and doesn’t use it anymore. If the coin flip was “put the clock in a room”, then he uses a random method to see which room.

30

Ok lola,

Your math is fine, it’s the application that’s faulty. Reread the OP. The setup is that there is an initial event with two outcomes having equal likelihood. (A coin toss has the same chance of landing on heads as on tails.) If the coin comes up heads, a second event with ten outcomes of equal likelihood occurs.

1/(2^N) doesn’t apply to the puzzle.
(1/2)*(1/10) does apply.

31

Just realized that your name is Iola, sorry 'bout the misread.

32

:: note to self – type faster! ::

33

:: MUCH faster ::

Darn you, wolfman!

34

I’m a bit late, but I just want to jump in and say that later information does affect probability. For instance: I have three boxes of cereal in my cabinet (Rasin Bran, Kix and Life), and let’s assume that on any given morning, the odds are 1/3 each of me picking a given cereal for breakfast. What’s the probability that I had Raisin Bran this morning? If you answer now, before reading any further, you’ll say (correctly) that it’s 1/3. However, I now tell you that I did, in fact, have raisin bran. What’s the probability now?

By the way, I think the reason that this was here, rather than GQ, is that the OP already knew the answer, but wanted to let us have the fun of figuring it out, and to watch our reactions to it. Had he not known, and was asking in order to find out the answer, it would be appropriate for GQ. Am I correct, APB9999?

35

Actual statistical answer aside, some comments about the relationship between the question and the information provided:

The information about a clock not being in a certain room of the house is relevant (ie, mathematically affects) the answer to the question, because the question specifically asks about the contents of the rooms of the house.

If the question had been “what are the chances that he flipped heads”, the information about the house would not have explicitly pertained to the question, so the answer would have to be 50%. But the question is not about the coin, it’s about the house.

The flipping of the coin clearly affects the probability of finding a clock in the house.

Let’s say we take the coin flip out of it, assume there’s a clock, and just deal with the house. We have a one in ten chance of there being a clock. That is not the same probability as there would be if a coin were flipped first, because the coin flip allows for the possibility of no clock at all.

The searching of rooms affects the probability of the coin flip having turned out a certain way.

To bring Chronos’ analogy in, if there were only one room in the house, and you went into that room, and found no clock, that would tell you about how the coin flip turned out. Adding rooms dilutes the amount of information received when you inspect a room but does not eliminate it.

Based on the above, I don’t see how the answer can be 50%.

36

A guy flips a coin. If it comes up heads, he randomly selects one room out of ten to put a clock in. You open the door to room #6. The clock is in the room. What is the probability that the clock is not in the house? Remember, you looked in room #6 after the coin was flipped, so the result of looking can’t affect the coin toss.

37

BTW, I’m planning on going outside soon to see if the clock is out there. I don’t think it is. Anyone want to bet that it is? I’ll offer 3:1 odds. Remember, you looked in room #6 after the coin was flipped, so the result of looking can’t affect the coin toss.

38

To answer this we need Bayes’ Theorem, P(A|B)*P(B) = P(B|A)*P(A).

Let A = there are no clocks in the house.
B= there is no clock in the first room examined.

So, the probability that there is no clock in the house to start with depends on the flip of a coin, P(A) = 0.50.

There being no clock in the first room examined could have happened two ways: (1) the coin came up tails (P=0.50) , in which case the probability of finding no clock in the room is 100%, and (2) the coin came up heads (P=0.50), in which case there is still a nine out of ten chance the room will be empty. So P(B) = 0.501.00 + 0.500.90 = 0.95.

The probability that there is no clock in the first room examined given that there is no clock in the house is, of course, 100%. P(B|A) = 1.00

So the probability that there is no clock in the house, given that it is absent from the first room examined, is
P(A|B) = P(B|A)P(A)/P(B)
or, substituting the values above, 10.50/0.95 = 0.526.

So the probability that the house is empty goes up to 53%. So Biotop was close. I’m curious why your third digit is different, though? What was you reasoning?

It seems the absence of evidence can be evidence of absence, at least a little …

39

APB9999:

Biotop was right, your answers are the same, 10/19 is 52.6…%. Biotop had 52.1%, but I think that was just a simple mistake.

40

The puzzle states that the house has ten empty rooms but doesn’t state how many non-empty rooms. The clock must have started out in one of the non-empty rooms and possibly moved to an empty room after the coin toss. Either way it’s in the house.

So, the way the puzzle is stated, my answer is that there is a 0% chance that there is no clock in the house.