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A puzzle

#1 User is offline   lamford 

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Posted 2016-October-13, 04:29

Two fair dice are rolled together once covertly, and you are only told that "at least one of the dice is a 6". A 6 is removed, and you are then shown the other die. What is the probability of the other die showing a 6? [ We assume that the person rolling is obliged to make a true statement of the from "at least on of the dice is an X".]

The above generated heated discussion among some top bridge and backgammon players on Facebook!
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#2 User is online   StevenG 

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Posted 2016-October-13, 05:01

Off the top of my head - it depends ...

If the person providing the statement will always make it when there is a 6 showing, the probability is 1/11 as there are 11 combinations showing a 6, and only one is the (6,6).

On the other hand, if the person picks one of the numbers randomly, then the answer is (I think) 1/6.
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#3 User is offline   jandrew 

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Posted 2016-October-13, 05:20

View Postlamford, on 2016-October-13, 04:29, said:

Two fair dice are rolled together once covertly, and you are only told that "at least one of the dice is a 6". A 6 is removed, and you are then shown the other die. What is the probability of the other die showing a 6? [ We assume that the person rolling is obliged to make a true statement of the from "at least on of the dice is an X".]

The above generated heated discussion among some top bridge and backgammon players on Facebook!



Two dice have been rolled and one removed. The removed die was a 6 proving the truth of the statement "at least one of the dice is a 6". There are no further inferences to be drawn from the truthful statement. Thus the remaining die behaves as if it was the only one thrown with a probability of 1:6 for each number. Therefore, the probability of the other die showing a 6 is 1:6.
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#4 User is offline   lamford 

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Posted 2016-October-13, 06:04

View PostStevenG, on 2016-October-13, 05:01, said:

Off the top of my head - it depends ...

If the person providing the statement will always make it when there is a 6 showing, the probability is 1/11 as there are 11 combinations showing a 6, and only one is the (6,6).

On the other hand, if the person picks one of the numbers randomly, then the answer is (I think) 1/6.

We don't know whether the roller chose a 6 from 6-5 or was forced to choose a 6 from 6-6, however, so it is still 1/6 - a slight variation on the principle of restricted choice with QJ doubleton.
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#5 User is offline   Fluffy 

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Posted 2016-October-13, 06:12

View Postlamford, on 2016-October-13, 06:04, said:

We don't know whether the roller chose a 6 from 6-5 or was forced to choose a 6 from 6-6, however, so it is still 1/6 - a slight variation on the principle of restricted choice with QJ doubleton.


You are assuming the pick from 6-5 is totally free, which is nowhere stated in the initial problem. It is done by a human after all.
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#6 User is offline   shyams 

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Posted 2016-October-13, 06:13

View Postlamford, on 2016-October-13, 04:29, said:

Two fair dice are rolled together once covertly, and you are only told that "at least one of the dice is a 6". A 6 is removed, and you are then shown the other die. What is the probability of the other die showing a 6? [ We assume that the person rolling is obliged to make a true statement of the from "at least on of the dice is an X".]

The above generated heated discussion among some top bridge and backgammon players on Facebook!


I think, the odds of the second dice also being a 6 is 1:11.

We start off with a 6x6 grid where every combination is equally probable. However, once the event occurs ("at least one of the dice is a 6"), 25 of the 36 combinations are no longer possible. That leaves 11 live combinations of which only one is a 6+6. So, the odds are 1:11
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#7 User is offline   billw55 

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Posted 2016-October-13, 06:45

First reply got it.

If the puzzlemaster is only interested in sixes, and thus will do nothing if there is no six, then 1/11.

If the puzzlemaster will always pick one of the numbers rolled at random, and then make such a statement, then 1/6. In this case her action carries no relevant information because she can always do it. Therefore the original probability of rolling doubles is unconstrained.

This is similar to the Monty Hall problem or Bertrand's box.
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#8 User is offline   kenberg 

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Posted 2016-October-13, 07:25

i agree with StephenG, especially the part where he says it depends on exactly what the rules are. But taking a reasonable view of what the rules are, 1 in 11 seems right.

The fact that "one 6 is removed" is, as he says, a big issue.

Suppose it goes like this.

A pair of dice are rolled. If neither is a 6 we go have a beer. If at least one of them is a 6, a genie announces that this is so, he removes a 6, and we are to guess whether the other is a 6. The removal of the 6 is irrelevant. He could just as well have said, "At least one of them is a 6, what are the chances they both are 6?" When they both are 6, then the remaining one after the removal of one 6 will be a 6. If only one is a 6, the remaining one after the 6 is removed will not be a 6. So the question comes down to: Given that at least one is a 6, what are the chances both are 6?

But that assumes something about what the rules are. Which is usually the case in these problems.
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#9 User is offline   alok c 

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Posted 2016-October-13, 07:38

If atleast one dice is a six then it is a constant not a variable as if it is not rolled at all.The chance of other dice to hit a six is 1/6.
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#10 User is offline   helene_t 

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Posted 2016-October-13, 07:40

View PostFluffy, on 2016-October-13, 06:12, said:

You are assuming the pick from 6-5 is totally free, which is nowhere stated in the initial problem. It is done by a human after all.

This.

"At least one is a six" covers 11 possibilities but since he could instead have said "at least one is a 5" with 6-5, we will give 10 of them half weight.

Now, a six is removed. If he was bound to remove a die of the kind that he had already named, then this gives us no extra information, and hence we have 1:5 odds for double six. OTOH, if he removed a die at random, the 10 mixed combos should again be given half weight, so we now have 1:2.5 odds.
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#11 User is offline   kenberg 

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Posted 2016-October-13, 08:07

View Posthelene_t, on 2016-October-13, 07:40, said:

This.

"At least one is a six" covers 11 possibilities but since he could instead have said "at least one is a 5" with 6-5, we will give 10 of them half weight.

Now, a six is removed. If he was bound to remove a die of the kind that he had already removed, then this gives us no extra information, and hence we have 1:5 odds for double six. OTOH, if he removed a die at random, the 10 mixed combos should again be given half weight, so we now have 1:2.5 odds.


Well, yes. At first I thought this way. But what was he allowed to say and do if a 6-5 had been rolled?

Sooner or later these things always roll back to the Monty Hall problem. Three doors, a car behind one, goats behind the others. A contestant chooses a door, the host opens another door to display a goat, the host offers the contestant a chance to change her choice. Question: Why did he open a door? Why did he offer her a chance to change. Maybe she is his mistress, she made the wrong choice, and he wants her to get the car. Maybe she chose correctly but she is someone he wants off the show. Maybe, maybe, maybe. In some formulations the rules are set out unambiguously and the answer is clear. But not always, and I think not in the current problem.

We are told that one is a 6. Why are we told that? If the roll is 6,5 will we always be told that at least one is a 6? Rather than at least one is a 5? Do the rules require that we be told that? "A 6 is removed". Randomly? By choice? Required? I lean toward the interpretation that if there is a 6 that could be removed, than a 6 is always removed. But I would not want my life to depend on this interpretation.

As I said, the part of StephenG's answer that I support the most strongly is t"it depends".
Ken
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#12 User is offline   billw55 

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Posted 2016-October-13, 08:16

Saying that "at least one of the dice is an X" and then showing a die that is an X, is equivalent to just showing one die at random. It should then be obvious that the other die is unconstrained, and will therefore match X 1/6 of the time.

To answer such puzzles correctly requires a rigorous explanation of the puzzlemaster's behavior in multiple trials. For example, consider a variant where the PM will always name the higher number die, or where she prefers odd numbers. The answers are quite different.
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#13 User is offline   nige1 

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Posted 2016-October-13, 10:39

View Postlamford, on 2016-October-13, 04:29, said:

Two fair dice are rolled together once covertly, and you are only told that "at least one of the dice is a 6". A 6 is removed, and you are then shown the other die. What is the probability of the other die showing a 6? [ We assume that the person rolling is obliged to make a true statement of the from "at least one of the dice is an X".]

The above generated heated discussion among some top bridge and backgammon players on Facebook!

Agree with everybody.
It depends on roller-psychology. e.g. If you know that the roller nominates as X:
  • The lower-numbered die, then the probability that he threw a double-six is one :).
  • The rightmost die, then the probability that the other die is a pair is one sixth.



Perhaps Cherdano can put us out of our misery :)
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#14 User is offline   jjbrr 

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Posted 2016-October-13, 11:03

I have two children; at least one of them is a boy.
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#15 User is offline   billw55 

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Posted 2016-October-13, 12:33

View Postlamford, on 2016-October-13, 04:29, said:

The above generated heated discussion among some top bridge and backgammon players on Facebook!

I'd be interested in reading that discussion. Can you post or link?
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#16 User is offline   shyams 

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Posted 2016-October-13, 12:46

View Postjjbrr, on 2016-October-13, 11:03, said:

I have two children; at least one of them is a boy.

The probability that the other one is also a boy is 33.33%
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#17 User is offline   1eyedjack 

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Posted 2016-October-13, 13:11

View Postshyams, on 2016-October-13, 12:46, said:

The probability that the other one is also a boy is 33.33%

If you are a gambler, and I offered you evens, presumably you would bet that the other is a girl.
In that case I might only 'fess up that at least one of them is a boy, if I have two boys.
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#18 User is offline   Phil 

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Posted 2016-October-13, 14:01

View Postjjbrr, on 2016-October-13, 11:03, said:

I have two children; at least one of them is a boy.


I have three children. One of each.
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#19 User is offline   Fluffy 

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Posted 2016-October-14, 00:53

View PostPhil, on 2016-October-13, 14:01, said:

I have three children. One of each.


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