6
$\begingroup$

A man speaks the truth on only two days a week. The days are the same for each week and they are in the TFT form (TFT = True False True. This means that the days on which he speaks the truth are Monday and Wednesday or Friday and Sunday or Saturday and Monday, etc). On all the other 5 days, he lies.

He gives the following statements on 3 consecutive days.

  • Day 1: I lie on Monday and Tuesday.
  • Day 2: It's Thursday, Saturday or Sunday today.
  • Day 3: I lie on Wednesday and/or Friday.

What is the probability that he speaks the truth on Thursdays?

This is a very similar puzzle to the one above. The difference is is that the other puzzle is very ambiguous. Also, it does not have the rule that the man lies on all the other 5 days.

Clarifications: all combination of truth days (Monday and Wednesday, Tuesday and Thursday, ..., Saturday and Monday) are equally probable. Also, it is equally probable that the statement on day 1 was spoken on any one of the 7 days of the week. The probability he makes that statement today is the same regardless of which days he speaks the truth.

$\endgroup$
12
  • 3
    $\begingroup$ What do you mean by "TFT form"? The days on which he speaks the truth are exactly two days apart? $\endgroup$ Apr 13 at 11:31
  • 7
    $\begingroup$ What exactly does "probability" mean here? Presumably one has to start with a certain given distribution of probabilities of the various possible truth/lies combinations, and filter out the ones incompatible with the given data. But without that pre-existing distribution, asking about the residual probability is meaningless. $\endgroup$ Apr 13 at 22:08
  • 3
    $\begingroup$ @GregMartin I would guess we have to assume a uniform prior for whether he lies on Mondays and Wednesdays, Tuesdays and Thursdays, etc. (7 possibilities, each with probability 1/7), and also a uniform prior for what day it is when he makes the first statement. Then I think you probably have enough information to give a unique answer. Rand al'Thor's answer is implicitly reasoning this way I think. $\endgroup$
    – N. Virgo
    Apr 14 at 5:39
  • 3
    $\begingroup$ @N.Virgo you also need a prior under which for each statement that the man might make, letting A be the subset of the 49 possible combinations of which day it is today and which days he lies on which are logically consistent with the statement, the probability he makes that statement today is the same regardless of which of the elements of A is the true one. $\endgroup$
    – fblundun
    Apr 14 at 10:39
  • 2
    $\begingroup$ Please add the needed assumptions regarding probabilities. Otherwise the question cannot be answered. (1) Is there an assumption of an underlying prior that the 7 possibilities for the truth-days all had probability 1/7, as asked by @GregMartin; (2) Please take into account fblundun's answer and in particular the suggestion in its last paragraph. $\endgroup$
    – Stef
    Apr 14 at 11:03

7 Answers 7

8
$\begingroup$

Look at the Days 1 and 3 statements:

If the Day 3 statement is false, then he tells the truth on Wednesday and Friday and lies on the other days, so the Day 1 statement is true. So the Days 1 and 3 statements can't both be lies, which means at least 1 of them is true.

Now, the Day 2 statement

must be a lie, so Day 2 must be Monday, Tuesday, Wednesday, or Friday.

Going back to Days 1 and 3:

If the Day 1 statement is true and the Day 3 statement is false, then the truth days are Wednesday and Friday, so Day 1 is Wednesday or Friday, so Day 2 is Thursday or Saturday, contradiction.

Now we know that the Day 3 statement

is true, so the truth days are not Wednesday and Friday (although they might still be Monday and Wednesday, or Friday and Sunday, etc.)

  • If the Day 1 statement is true, then

    the truth days are Days 1 and 3, which must be {Sunday,Tuesday} or {Monday,Wednesday} or {Tuesday,Thursday} or {Thursday,Saturday} according to what we know about Day 2, but the first three of these are not possible since he lies on Monday and Tuesday. So, in this case,
    Days 1,2,3 are Thursday, Friday, Saturday and the truth days are Thursday and Saturday.

  • If the Day 1 statement is false, then

    one of the truth days must be Monday or Tuesday. We already know Day 3 is a truth day and that it is Tuesday, Wednesday, Thursday, or Saturday, and now we know that the other truth day is Day 5. So there are only two options:
    Days 1,2,3 are Sunday, Monday, Tuesday and the truth days are Tuesday and Thursday, or
    Days 1,2,3 are Thursday, Friday, Saturday and the truth days are Saturday and Monday.

Now we have found the complete list of possibilities. I don't know exactly what is meant by "probability", but assuming the intent of the question is just to figure out how many of the possible scenarios have Thursday as a truth day, the answer is

$2/3$.

$\endgroup$
5
  • 3
    $\begingroup$ Why must Day 2 be a lie? EDIT: Ooh, got it. Because can't tell truth two days simultaneously, and either Day 1 is true, or Day 3 is true $\endgroup$
    – justhalf
    Apr 13 at 15:39
  • 1
    $\begingroup$ There appears to be a bit of a jump in the probability conclusion. How do you got from the list of possibilities that you've identified, to a probability? There is nothing in your answer that suggests these possibilities should be equiprobable. $\endgroup$
    – Stef
    Apr 14 at 10:56
  • $\begingroup$ @Stef There's no way to know what else is affecting the likelihoods of events, so I just assumed the intent of the question was to find out what proportion of the possible situations involve the truth being spoken on Thursdays. $\endgroup$ Apr 14 at 11:22
  • $\begingroup$ Making extra assumptions can be good, especially if the question sounds like it's expecting a precise numerical answer but lacks assumptions to arrive at such a number. But it would be good to state those assumptions explicitly as part of your answer. In particular, there is a difference between assuming "the intent of the question was to find out what proportion of the possible situations involve the truth being spoken on Thursdays" (then the answer is 2/3) and assuming "before the man spoke, the 7 possibilities were uniformly probable" (then the answer might be something else). $\endgroup$
    – Stef
    Apr 14 at 12:45
  • $\begingroup$ @Stef fair point, edited. $\endgroup$ Apr 14 at 12:49
5
$\begingroup$

All possibilities of the game before looking at the statements are

Sun M Tu W Th F Sat
T F T F F F F
F T F T F F F
F F T F T F F
F F F T F T F
F F F F T F T
T F F F F T F
F T F F F F T

The TF possibilities for any three consecutive statements are FFF, TFF, FTF, FFT, TFT

Now given the statements,

FFF, TFF, and FTF are impossible.
You can try any FFF (2 per row), TFF (1 per row), or FTF (2 per row) sequence in the table and check they are invalid.

So we are left with

FFT or TFT
FFT is only possible with truth days on {Tu, Th} or {Sat, M}.
TFT is only possible with truth days on {Th, Sat}.

So the probability of a Thursday truth day

After assuming the probability is based on equal chances of the statements being FFT or TFT,
(P(FFT) * P(Th | FFT)) + (P(TFT) * P(Th | TFT)) =
(1/2) * (1/2) + (1/2) * 1 =
3/4

Or if assuming each possibility is equally probable,
2/3 possibilities have Thursday as a truth day.

$\endgroup$
3
$\begingroup$

We cannot answer the question without insight into how the man decides what to say.

A simpler example: suppose there is a woman who lies on Mondays, tells the truth on Tuesdays, and is silent on other days. You hear her say: "I just won the lottery four times in a row." This is logically compatible with today being either Monday or Tuesday. But it's very rare for a person to win the lottery four times in a row, so her statement was probably false: it was something that she was more likely to say on a Monday than on a Tuesday. So today is probably a Monday.

Similarly, it's possible that the man from the original question enjoys coming up with convoluted sets of statements which trick the listener into believing that false things are likely and true things are unlikely, and any attempt to calculate the probability that he lies on Thursdays would have to take this possibility into account.

This issue could be fixed by having the man answer yes/no questions rather than make statements. So the listener would ask him on day 1 "Is it the case that you lie on both Monday and Tuesday?", and so on.

$\endgroup$
2
$\begingroup$

My first stab at the solution:

If the third statement is a lie, then the two days of the week that he tells the truth are Wednesday and Friday. This makes the first statement true, which means the first statement is made on a Wednesday or a Friday. But if it was made on Wednesday, then the third statement would be made on Friday, which is a contradiction with the assumption that the third statement is a lie; if the first statement was made on Friday, then the second statement would be made on a Saturday, but this is also a contradiction since that statement must be a lie. Therefore the third statement cannot be a lie. Since it is true, then the second statement must be a lie (the first statement could still be truth or lie).

If the first statement is true, then the two “truth” days could be Wed-Fri, Thu-Sat, or Fri-Sun. The first of those contradicts the third statement; The third of those is a contradiction with the second statement. So in this case, the two truth days must be Thu-Sat.

On the other hand, if the first statement is false, then he tells the truth on either Monday or Tuesday. Then the second statement would be made on Monday, Friday, Saturday or Sunday. The last two of those are a contradiction with the second statement, so the second statement is made on Monday or Friday. In one of those cases, Thursday is a truth day, in the other it is not.

So assuming all else is equal, there's a 50% chance the first statement is false. If it is, there's a 50% chance that he lies on Thursdays, otherwise he tells the truth on Thursdays. So the chance that he speaks the truth on Thursdays is 75%.

$\endgroup$
1
  • 4
    $\begingroup$ I agree with your analysis of the possible scenarios, but considering the Boy or Girl paradox leads me to think your stated probability might not be correct - it may not be safe to be "assuming all else is equal". $\endgroup$
    – Rob Watts
    Apr 13 at 20:26
0
$\begingroup$

A math/programming approach

Disclaimer: I am not a logician and may not use the standard syntax expected by those in the field. I'm happy to accept edits to adjust this where it improves readability.

Start by defining

Monday = 0, Tuesday = 1, etc.
0 ≤ D ≤ 6 is the first day a statement is given (day 1)
0 ≤ T ≤ 6 is the first day the truth is told. The truth is also told on T+2.
Days are always Mod7 to ensure they fit within the ranges given.

Now we encode each of the three statements as logic:

Each statement has two components: the statement itself and the day that it was said on. We must translate the statement into something that we know about T or D. These represent when the statements are true:
A (Day 1): T ∈ [2, 3, 4] && (D = T || D = T+2)
To lie on both Monday and Tuesday, the first truth day must be between Wednesday and Friday.
B (Day 2): D ∈ [2, 4, 5] && (D+1 = T || D+1 = T+2)
Yesterday must have been Wednesday, Friday, or Saturday.
C (Day 3): T ≠ 2 && (D+2 = T || D+2 = T+2)
The only way this statement can be false is if the first truth day is Wednesday.
The false versions of the statements are very similar, but are inverted as so (not a true inversion, since the two components must still both be true):
~A (Day 1): T ∈ [0, 1, 5, 6] && (D ≠ T && D ≠ T+2)
~B (Day 2): D ∈ [0, 1, 3, 6] && (D+1 ≠ T && D+1 ≠ T+2)
~C (Day 3): T = 2 && (D+2 ≠ T && D+2 ≠ T+2)

We consider the following possibilities:

FFF (~A, ~B, ~C)
FFT (~A, ~B, C)
FTF (~A, B, ~C)
TFF (A, ~B, ~C)
TFT (A, ~B, C)

Finally, we can put these together:

By searching {0 ≤ D ≤ 6, 0 ≤ T ≤ 6} across the five possibilities described above, we find only three logically consistent solutions:
TFT, D=3, T=3 → Day 1 is Thursday, truth is told on Thursday and Saturday
FFT, D=3, T=5 → Day 1 is Thursday, truth is told on Saturday and Monday
FFT, D=6, T=1 → Day 1 is Saturday, truth is told on Tuesday and Thursday

To find the probability of truth on Thursday:

Truth on Thursday means (T = 1 || T = 3) (telling the truth on Tuesday/Thursday or Thursday/Saturday), so our final probability is 2/3.

This may be a bit of overkill...

But having model let me answer other questions, such as: what happens if the Day 2 statement becomes "Day 2: It's Friday, Saturday or Sunday today."?

Answer:

There's a 100% chance of telling the truth on Thursday now!

$\endgroup$
0
$\begingroup$

There are two independent questions to consider here:

  • Which day of the week was “Day 1”?
  • Which day of the week is the first day of the TFT sequence?

Each one has 7 possible answers, for a total of 49 possible combinations. I'll evaluate them brute-force with a computer program.

# Define constants for the days of the week.
SUN, MON, TUE, WED, THU, FRI, SAT = range(7)

def is_consistent(day1, truth1):
    """
    Check if a combination of 

    day1   -- Weekday (0-7) of the first day the man speaks.
    truth1 -- First weekday (0-7) of the TFT pattern.
    """
    # Which day of the week are Day 2 and Day 3?
    day2 = (day1 + 1) % 7
    day3 = (day1 + 2) % 7
    # truth_days[i] = whether man speaks truth on weekday i
    truth_days = [False] * 7
    truth_days[truth1] = True
    truth_days[(truth1 + 2) % 7] = True
    # Evaluate the truth of the man's statement for each day
    day1_truth = (not truth_days[MON]) and (not truth_days[TUE])
    day2_truth = day2 in (THU, SAT, SUN)
    day3_truth = (not truth_days[WED]) or (not truth_days[FRI])
    # Check if the statement truths match truth_days
    return (day1_truth == truth_days[day1]) and (day2_truth == truth_days[day2]) and (day3_truth == truth_days[day3])

for day1 in range(7):
    for truth1 in range(7):
        if is_consistent(day1, truth1):
            print(day1, truth1)

This prints the valid combinations:

* 0 2 = Day 1 is Sunday, and the man tells the truth on Tuesday and Thursday. Truth of the three statements is FFT. * 4 4 = Day 1 is Thursday, and the man tells the truth on Thursday and Saturday. Truth of the three statements is TFT. * 4 6 = Day 1 is Thursday, and the man tells the truth on Saturday and Monday. Truth of the three statements is FFT.

Assuming that all of these possibilities are equally likely:

Two of them have the man telling the truth on Thursdays, and one has him lying on Thursdays. Therefore, the probability that he tells the truth on Thursday is 2/3.

$\endgroup$
-4
$\begingroup$

The probability is either 1 or 0.

It's the same as the probability of any observable current/past event. E.g. the probability that I am wearing pants right now: I either am or am not, there is no uncertainty around it.

$\endgroup$
12
  • 5
    $\begingroup$ From my perspective whether you are wearing pants or not is not an observable event. I would say the probability that you are wearing pants is about 0.2. $\endgroup$
    – N. Virgo
    Apr 14 at 5:38
  • 3
    $\begingroup$ I think you may be confused about the subjective nature of probability. $\endgroup$
    – Sneftel
    Apr 14 at 15:49
  • 1
    $\begingroup$ @fectin 1 of course. Yet if you had not told me the ball you drew was a red one, then my answer would have been 1/2. $\endgroup$
    – N. Virgo
    Apr 15 at 4:03
  • 1
    $\begingroup$ @fectin 1/2 Is referring to the probability that you picked a red ball, not the full outcome. In the case of this riddle, the point is you don't know whether the man is lying on Thursday. This means you need a probability value that tells you approximately if you were to run the situation repeatedly what the chance he's lying on Thursday is. The problem with your population portion comparison is that you do not just calculate probability by purely the amount of options available $\endgroup$
    – TMK
    Apr 15 at 16:04
  • 2
    $\begingroup$ @fectin I think you are much less likely to offer me that bet if you are shorter than average than if you are not. If you offer the bet then I have a posterior probability $p(\text{fectin is short}\mid\text{fectin offers the bet}) = p(\text{fectin offers bet}\mid\text{fectin is short})p(\text{fectin is shorter than average})/p(\text{fectin offers bet})$. I think the ratio $p(\text{fectin offers bet}\mid\text{fectin is short})/p(\text{fectin offers bet})$ is very small, let's say 0.01, and so my posterior probability for you being short is 0.5*0.01=0.005, and I would not like to take that bet. $\endgroup$
    – N. Virgo
    Apr 16 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.