### Brain Teasers

# Itchy and Scratchy

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

Itchy and Scratchy are playing Backgammon, which requires two identical dice to be rolled on each turn. Scratchy rolls the dice but just as they come to a halt on the board he covers them up with his hands.

"You can't look," screams Scratchy.

"Too late," yells Itchy. "I saw one of them and it was six! Show me the other one." Scratchy at first declines the offer but Itchy applies a gratuitous amount of violence with a chainsaw resulting in the amputation of both of Scratchy's arms, revealing the dice. If the first die revealed is a six, what is the probability that the second die is also a six?

"You can't look," screams Scratchy.

"Too late," yells Itchy. "I saw one of them and it was six! Show me the other one." Scratchy at first declines the offer but Itchy applies a gratuitous amount of violence with a chainsaw resulting in the amputation of both of Scratchy's arms, revealing the dice. If the first die revealed is a six, what is the probability that the second die is also a six?

### Hint

The answer is not 1/6.### Answer

Since Itchy saw one of the dice, the other one is either a six (1/6 chance) or a non-six (5/6 chance.)This means that if the scenario is repeated several times over, you would expect that on 5 out of every six occasions there would be one six and one non-six on the board in which case the hidden die is not a six. On the sixth occasion there are 2 sixes on the board so that either one may be revealed and still leave one six hidden. That means that out of every 7 occasions where a six can be revealed, 2 will result in a six remaining hidden.

Answer: Probability = 2/7

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## Comments

uh........wha........?

Apr 21, 2003

I completely bombed on this one, and I'm not sure I follow the answer. I got a probability of 1/11. I counted 11 possible outcomes that have at least one 6 and only one of those that has two 6s. Where did I go wrong?

Since we know that one is a six for sure because it was seen, then the possibilities are 6-1,6-2,6-3,6-4,6-5,6-6. Imagine for the first five occasions the first six is revealed. that would leave a non six behind. For the 6-6 result, either one can be revealed. That makes seven ways in which a six can be revealed. Only 2 of these (the 6-6 case) permit another six to be still hidden. Hence 2 in 7.

Apr 22, 2003

OK, now I understand. Thanks for explaining. It's a good thing my job doesn't involve too much math. Tough teaser!

May 01, 2003

These problems drive me nuts! To me they assume that the outcome of one dice affects the outcome of another. One dice does not affect the other, it is a seperate entity unto itself. Say, for example, the first dice turned over was a 5...why does this change the probability of the second? It doesn't...at least not in any way that I can see.

While the outcome of one dice cannot change the other, knowledge of one dice effects your knowledge of the entire outcome. If we know nothing, then there are 36 possibilties for the dice. If we know that one of them is a six, there are now only 6 possibilties. Conditional probability is tricky because we know something, but not enough to eliminate all possibilities. The trick is to figure out what remaining outcomes are possible.

I love the simpsons, great teaser. just so ya know I'm a fan of yours jimbo.

I don't agree with the given answer. Two dice are thrown. One is seen, and that one has no effect on the outcome of the other. The probability of the second die being a six is one in six.

That is correct Asimov but now the two dice are hidden again. Is it the six you first saw or is it the other one that is now revealed as a six. There is a second part to it.

Why is the answer not 7/12? Let's call the first dice we see Dice A. Dice A is 6, so the probability that the other one is a 6 is equal to 1/6. However, we have the additional information that the second dice we see is also a 6. There is a 1/2 probability that the second dice we see is the same as the first, and a 1/2 probability that it is different. If it is the same, then the probability that the other one is a 6 is equal to 1/6. If it is different, then the probability that the other one is a 6 is equal to 1. So, the total probability that both dice are 6 is:

(1/2)*(1/6) + (1/2)*(1) = 7/12.

I would be very interested to know why this solution is inconsistent with the posted solution (2/7). Thanks!

(1/2)*(1/6) + (1/2)*(1) = 7/12.

I would be very interested to know why this solution is inconsistent with the posted solution (2/7). Thanks!

There is not a fifty fifty chance that the second is the same as the first. The first die is a six. We saw it. The probability that the second one is the same (ie a six) is one sixth (not a half). Here is the bit that confuses some people at the end. The probability that we have two sixes is 1/6 but now one is revealed. Which one? If we have 2 sixes there are 2 ways that a six cane be shown. If we have a six and any other of the 5 numbers there is only one way in which a six can be revealed (the first die). Hence the seven cases.

Here's how I see it:

Let's call the dice die A and die B. There are 36 ways for them to land (listing A first): 1-1, 1-2, 1-3, 1-4, 1-5, 1-6, 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 3-1, 3-2, 3-3, 3-4, 3-5, 3-6, 4-1, 4-2, 4-3, 4-4, 4-5, 4-6, 5-1, 5-2, 5-3, 5-4, 5-5, 5-6, 6-1, 6-2, 6-3, 6-4, 6-5, and 6-6. Since we do not know whether die A or die B was revealed, but it was a six, that leaves eleven ways (again listing A first): 1-6, 2-6, 3-6, 4-6, 5-6, 6-1, 6-2, 6-3, 6-4, 6-5, and 6-6. Of those eleven, only one can have the hidden die be a 6.

Where is my mistake?

Let's call the dice die A and die B. There are 36 ways for them to land (listing A first): 1-1, 1-2, 1-3, 1-4, 1-5, 1-6, 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 3-1, 3-2, 3-3, 3-4, 3-5, 3-6, 4-1, 4-2, 4-3, 4-4, 4-5, 4-6, 5-1, 5-2, 5-3, 5-4, 5-5, 5-6, 6-1, 6-2, 6-3, 6-4, 6-5, and 6-6. Since we do not know whether die A or die B was revealed, but it was a six, that leaves eleven ways (again listing A first): 1-6, 2-6, 3-6, 4-6, 5-6, 6-1, 6-2, 6-3, 6-4, 6-5, and 6-6. Of those eleven, only one can have the hidden die be a 6.

Where is my mistake?

But we do know which die is a six. I doesn't matter whether you call it A or B but THAT die IS a six. So the possibilities are 6-1, 6-2, 6-3, 6-4, 6-5, and 6-6.

Now for the second part. If you see a six and the other one is 1 , 2 , 3 , 4 or 5 then the hidden one can't be a six (5 cases). If the hidden one is a six, then you can see either A or B revealed and the hidden one will still be a six. (2 cases). Thus in 2 out of the possible 7 cases, the remaiming one will be a six.

Now for the second part. If you see a six and the other one is 1 , 2 , 3 , 4 or 5 then the hidden one can't be a six (5 cases). If the hidden one is a six, then you can see either A or B revealed and the hidden one will still be a six. (2 cases). Thus in 2 out of the possible 7 cases, the remaiming one will be a six.

Oh, I see now. Sorry.

May 07, 2004

Do NOT EEVVEERR say something when you don't know what you are talking about. It will make you feel like this guy:

I have never cared much for playing backgammon. It is much too violent.

For those looking for a Bayes' Theorem approach... (okay, so nobody was, but this is how I solved this):

I am saying the 1st die is the 6 you saw, the second unknown.

Pr(1st|6) = [(1/2)*1]/[(1/2)*1 + (1/2)*(1/6)] = 6/7

Therefore, Pr(2nd|6) = 1/7

If you saw the first die, the probability you have two sixes would be 1/6 (for the probability for the second die) * 1 (the probability the first die was a 6) * 6/7 (the probability it is the first die) = 1/7

If you saw the second die, the probability you have two sixes would be 1 (the probability the second die was a six) * 1 (the probability the first die was) * 1/7 (the probability it was the second die that you saw) = 1/7

The sum of these two is your net probability. 1/7 + 1/7 = 2/7.

I am saying the 1st die is the 6 you saw, the second unknown.

Pr(1st|6) = [(1/2)*1]/[(1/2)*1 + (1/2)*(1/6)] = 6/7

Therefore, Pr(2nd|6) = 1/7

If you saw the first die, the probability you have two sixes would be 1/6 (for the probability for the second die) * 1 (the probability the first die was a 6) * 6/7 (the probability it is the first die) = 1/7

If you saw the second die, the probability you have two sixes would be 1 (the probability the second die was a six) * 1 (the probability the first die was) * 1/7 (the probability it was the second die that you saw) = 1/7

The sum of these two is your net probability. 1/7 + 1/7 = 2/7.

I'm disregarding sockmonkey's advice, because I *want* to understand!!

Jimbo states that if we know one of them is a six, there are only six possibilities. But you somehow turn one of those six possibilities into two different possibilities?!?!?

Does this have something to do with the glimpse we got of a six at first? Are we supposed to take that into account or not?

How is this different from "what's the chance of rolling a six now that I just rolled a six?"

I'm going to go look up Bayes' Theorem!

Jimbo states that if we know one of them is a six, there are only six possibilities. But you somehow turn one of those six possibilities into two different possibilities?!?!?

Does this have something to do with the glimpse we got of a six at first? Are we supposed to take that into account or not?

How is this different from "what's the chance of rolling a six now that I just rolled a six?"

I'm going to go look up Bayes' Theorem!

OK, I've figured this a few different ways, and keep coming up with 7/12. Here's my logic:

When the dice were rolled, you saw that one of them was a '6'. The other die still has a 1/6 chance of being a '6'.

After employing the chain saw, one die was revealed (randomly). It was a '6'. It was either the one you saw before, or not.

If it was the same die you saw earlier, then the still-hidden die still has a 1/6 chance of being a '6'.

If it (the chainsaw-revealed die) is not the same one you saw earlier, then we know the hidden die is a '6' (because you saw it) -> probability 6/6 that it is a '6'.

With me?

Since the chainsaw randomly revealed one of the two die, it's a 50/50 shot (1/2 chance) as to which of the above two statements is true (are we seeing the same die we saw before, or the other one?). So:

(1/2 * 1/6) + (1/2 * 6/6) = 7/12

Where is my flaw?

When the dice were rolled, you saw that one of them was a '6'. The other die still has a 1/6 chance of being a '6'.

After employing the chain saw, one die was revealed (randomly). It was a '6'. It was either the one you saw before, or not.

If it was the same die you saw earlier, then the still-hidden die still has a 1/6 chance of being a '6'.

If it (the chainsaw-revealed die) is not the same one you saw earlier, then we know the hidden die is a '6' (because you saw it) -> probability 6/6 that it is a '6'.

With me?

Since the chainsaw randomly revealed one of the two die, it's a 50/50 shot (1/2 chance) as to which of the above two statements is true (are we seeing the same die we saw before, or the other one?). So:

(1/2 * 1/6) + (1/2 * 6/6) = 7/12

Where is my flaw?

RT -- the flaw is that you are assuming it is equally likely to have been either die that you saw when you applied the chainsaw. Given that it was a 6 o nthe die, it is more likely to be the die that you already saw had a 6 on it. PM me if you have more questions about this.

Good one. Had me stumped there when you gave the hint.

Let me get this right, none of you does really understand the concept of distinct events (except for profwynde).

Take the following case, I rolled two dice, one of them is kept hidden and the other is kept hidden too. And I asked: "for the die on the left, what is the probability that it is 6?" You answer 1/6. After I hear your answer I uncover the other die revealing a 6, and I repeat the question.

According to you the probabilty of the left die being a 6 is 2/7. I affected the probability without even affecting the problem(it is still about the 2nd die). So now I flip the second die to make it 5, the probability is 1/6 again, then I throw the second die in the trash can, and I bring 256 other dice all are 6s.

What is happening? the answer is nothing. You're probability problem about the hidden die hasn't changed, it is still 1/6.

now think about this, what if I flip a coin and it is TAILS, how would that affect the hidden die.

Take the following case, I rolled two dice, one of them is kept hidden and the other is kept hidden too. And I asked: "for the die on the left, what is the probability that it is 6?" You answer 1/6. After I hear your answer I uncover the other die revealing a 6, and I repeat the question.

According to you the probabilty of the left die being a 6 is 2/7. I affected the probability without even affecting the problem(it is still about the 2nd die). So now I flip the second die to make it 5, the probability is 1/6 again, then I throw the second die in the trash can, and I bring 256 other dice all are 6s.

What is happening? the answer is nothing. You're probability problem about the hidden die hasn't changed, it is still 1/6.

now think about this, what if I flip a coin and it is TAILS, how would that affect the hidden die.

Now if the problem was like this, I saw that one of them is 6, and then I uncovered one of the two dice revealing a 6(without actually being sure whether this was the die I saw in the beginning) then the probability of them being both 6's is:

P(6,6) = 1/2(I uncovered the other die) * 1(I saw this one) + 1/2(I uncovered what I saw)*1/6(The die I know nothing about) = 7/12

P(6,6) = 1/2(I uncovered the other die) * 1(I saw this one) + 1/2(I uncovered what I saw)*1/6(The die I know nothing about) = 7/12

Actually, here's the way it works:

We know that one die is a six. I roll the dice, then I show you a six.

The six possibilities are as follows:

1-6

2-6

3-6

4-6

5-6

6-6

As you can see, there are seven 6's, and I could have showed you any one of these with equal likelihood. Two of those have a six opposite. Hence, 2/7.

Your problem is that you assume that the six mentioned first is the same as the six mentioned second. While that is a possibility, it might not be. It could be that the first six is not the same six as the second six.

We know that one die is a six. I roll the dice, then I show you a six.

The six possibilities are as follows:

1-6

2-6

3-6

4-6

5-6

6-6

As you can see, there are seven 6's, and I could have showed you any one of these with equal likelihood. Two of those have a six opposite. Hence, 2/7.

Your problem is that you assume that the six mentioned first is the same as the six mentioned second. While that is a possibility, it might not be. It could be that the first six is not the same six as the second six.

Answer is incorrect, If we are to believe Sratchy (or Itchy or whatever) then at least 1 dice is a 6. We are looking at a 6. Boy are we lucky! Either we are looking at the first dice with a 6 on it, or the second die is also a 6. In any case, it does not matter. We have to worry only about 1 dice, the hidden one. It's as if it had not been cast yet, since we don't know what it is. So the question is: if I cast this die, what is the probability that it will be a 6? it's 1/6 of course. If I look at the 2nd dice, probability that it is a 6 is 1/6. Another way to look at it: there are 6 possibilities: the cat has one of the following in his hands: 6&1, 6&2, 6&3, 6&4, 6&5, or 6&6. They all have a 6 in them. I am looking at a 6. Am I looking at the first dice of 6&1 or am I looking at the 2nd dice of 6&6? The probability that one dice has a 6 on it is 7/12, but that is not the question asked here. The question is: I am looking at a 6, what is the chance that the other die is a 6: it's one is 6.

YVAU -- you are missing the very important fact that you are more likely to be looking at the 6&6 combination (it is twice as likely as the 5 other combinations that have a 6 in them, since there are two that you could have seen). So, with 5 equally-likely events, and one event that is twice as likely, the probabilities are 1/7 (5 times) and 2/7.

Imagine that one die is red and the other is white. There are 36 possible outcomes denoted by R6-1, R6-2, R6-3, R6-4,R6-5, R6-6, R5-1, R5-2,..... R1-4, R1-5, R1-6.

I now show you one of the dice is a six. If the die you see is red then the only possiblities are R6-1, R6-2, R6-3, R6-4, R6-5, R6-6.

If the die you see is white then the only possiblities are 6-R1, 6-R2, 6-R3, 6-R4, 6-R5, 6-R6.

In either case, there are only 6 of the 36 original combinations that are possible. If the dice were coloured and I showed you a red, then the answer would be 1 in six BUT say I now paint them both yellow.You do not know whether the one you are looking at is the one that used to be red or the one that used to be white. In either case there are only six combinations but one of those combinations gives me 2 chances to show you another six, while the other 5 combinations allow me to show you a six and keep a non-six hidden. Thus total chances = 5 + 2 = 7 while only 2 of these have a hidden 6. Probability = 2/7.

I now show you one of the dice is a six. If the die you see is red then the only possiblities are R6-1, R6-2, R6-3, R6-4, R6-5, R6-6.

If the die you see is white then the only possiblities are 6-R1, 6-R2, 6-R3, 6-R4, 6-R5, 6-R6.

In either case, there are only 6 of the 36 original combinations that are possible. If the dice were coloured and I showed you a red, then the answer would be 1 in six BUT say I now paint them both yellow.You do not know whether the one you are looking at is the one that used to be red or the one that used to be white. In either case there are only six combinations but one of those combinations gives me 2 chances to show you another six, while the other 5 combinations allow me to show you a six and keep a non-six hidden. Thus total chances = 5 + 2 = 7 while only 2 of these have a hidden 6. Probability = 2/7.

May 05, 2006

Hi all,

We have: two dice one showing, one hidden. the question refers to the hidden one. We know that one of the two dice is a six, it could be either with equal probability (without further information).Since the open dice is a six it could have been the dice that was seen and this option will be considered.

If we saw the hidden die with probability 1/2 then it will definitely be a six. If we saw the uncovered die then the probability (of the 2nd being a six) is reduced to 1/6.

1/2 x 1 = 1/2

1/2 x 1/6 = 1/12

Probability of hidden die being a six = 7/12

We have: two dice one showing, one hidden. the question refers to the hidden one. We know that one of the two dice is a six, it could be either with equal probability (without further information).Since the open dice is a six it could have been the dice that was seen and this option will be considered.

If we saw the hidden die with probability 1/2 then it will definitely be a six. If we saw the uncovered die then the probability (of the 2nd being a six) is reduced to 1/6.

1/2 x 1 = 1/2

1/2 x 1/6 = 1/12

Probability of hidden die being a six = 7/12

Pretty Good. I originally came up with the 7 out of 12 as well. Until I wrote a quick program to do a sanity check on my answer. And lo and behold the sample kept coming up about 28% or roughly 2 in 7 times. I finally got it. There are only 6 possible die combinations, but you have to count one of them twice. When first uncovered by the severed hand the die combination 1st die = 6, 2nd die = 6 would reveal a six both showing die 1 first or die 2 first. Therefore, out of the seven posibilities of the 6 being shown first, only two would reveal a 6 when the second hand is removed. Or in other words, 2 in 7 times. Hats off.

two dice are rolled one is definatly six (we saw it) so the options are:

1 6

2 6

3 6

4 6

5 6

6 6

now they are both hidden again, now here is the twist, we uncover a dice under one arm and its a six now it splits into 7 options

*6* 1

*6* 2

*6* 3

*6* 4

*6* 5

*6* 6

6 *6*

* * being the dice that we uncovered after they have both been hidden again

the confusion arises when you assume that there is one die where you can always uncover a 6 at 1/2 each and the other has only 1/6 chance being a 6 and has a /2chance to be uncovered. but as it is there are 12 possible outcomes 5 of which are irelevant since we have already uncovered a 6

1 *6*

2 *6*

3 *6*

4 *6*

5 *6*

6 *6*

*1* 6

1 6

2 6

3 6

4 6

5 6

6 6

now they are both hidden again, now here is the twist, we uncover a dice under one arm and its a six now it splits into 7 options

*6* 1

*6* 2

*6* 3

*6* 4

*6* 5

*6* 6

6 *6*

* * being the dice that we uncovered after they have both been hidden again

the confusion arises when you assume that there is one die where you can always uncover a 6 at 1/2 each and the other has only 1/6 chance being a 6 and has a /2chance to be uncovered. but as it is there are 12 possible outcomes 5 of which are irelevant since we have already uncovered a 6

1 *6*

2 *6*

3 *6*

4 *6*

5 *6*

6 *6*

*1* 6

Here is probably the easiest approach:

There are seven sixes you could have seen (6-1, 6-2, ... , 6-6). 2 of them have a 6 on the other die. Hence, the answer is 2/7.

There are seven sixes you could have seen (6-1, 6-2, ... , 6-6). 2 of them have a 6 on the other die. Hence, the answer is 2/7.

The answer is not 2/7 as each of the 7 cases are not equally likely, for instance getting 1&6 is twice more likely than 6&6 as u may get (1,6) or (6,1). Thus there are 11 possible cases but only 1 favorable case, thus 1/11

Oh hell! You need to read the rest of the comments.When the dice are tossed, yes you may expect to get 6-1 or 1 - 6. But after you see one of the dice it is a six. Now only 6-1 is possible. The one you have seen can no longer be a "1"! BTW the answer is correct. Get 2 dice and throw them a number of times and observe what happens.

Not sure, but still think the correct answer is 1/6.

Of the 36 possibilities, once one dice is revealed to be 6, 12 possibilities remain.

1) 6 1 (first dice is revealed)

2) 6 2 (first dice is revealed)

3) 6 3 (first dice is revealed)

4) 6 4 (first dice is revealed)

5) 6 5 (first dice is revealed)

6) 6 6 (first dice is revealed)

7) 1 6 (second dice is revealed)

2 6 (second dice is revealed)

9) 3 6 (second dice is revealed)

10) 4 6 (second dice is revealed)

11) 5 6 (second dice is revealed)

12) 6 6 (second dice is revealed)

Of the 12 outcomes, in only 2 (6 and 12) will the other dice be 6. So the probability is 2/12 = 1/6.

The 2/7 answer assumes that (in the 7th possibility), if second dice is revealed as 6, the first dice MUST BE 6 - which is not the case. The first dice could still be any from 1 to 6.

Of the 36 possibilities, once one dice is revealed to be 6, 12 possibilities remain.

1) 6 1 (first dice is revealed)

2) 6 2 (first dice is revealed)

3) 6 3 (first dice is revealed)

4) 6 4 (first dice is revealed)

5) 6 5 (first dice is revealed)

6) 6 6 (first dice is revealed)

7) 1 6 (second dice is revealed)

2 6 (second dice is revealed)

9) 3 6 (second dice is revealed)

10) 4 6 (second dice is revealed)

11) 5 6 (second dice is revealed)

12) 6 6 (second dice is revealed)

Of the 12 outcomes, in only 2 (6 and 12) will the other dice be 6. So the probability is 2/12 = 1/6.

The 2/7 answer assumes that (in the 7th possibility), if second dice is revealed as 6, the first dice MUST BE 6 - which is not the case. The first dice could still be any from 1 to 6.

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