Two Dice Puzzle
A is #1 die, B is #2 die, C is "if A or B = 2 then cell = 1, else cell = 0", D is "if both A and B = 2 then cell = 1, else cell = 0"
A and B are both RANDBETWEEN(1,6)
Another cell is SUM(C2:C1000) -- this is the number of occurrences where at least one die is a deuce.
Some other cell is SUM(D2:D1000) -- this is the number of occurrences where both dice are deuces.
The results -- they are mighty shocking:
# at least one deuce / # of both deuces
325 / 33
306 / 33
300 / 31
309 / 20
317 / 29
274 / 27
292 / 27
293 / 26
320 / 32
302 / 34
315 / 27
No. It's a matter of math.Quote: AlanMendelson... It's a matter of English.
Let's try this.
You shake the dice in the cup. You ask someone to look and confirm that at least one die is showing a deuce. He looks, and says, "Yep. At least one of them is showing a deuce."
So, the person has confirmed that out of 36 possible random dice results, you got one of the eleven that have at least one deuce. Out of those eleven, only one is the double deuce.
----
On a note about the Monty Hall problem: Monty agreed with the math, but added that he cheated. He knew all the results and would only offer the option to switch on certain occasions based upon the contestant's choice and his opinion of the contestant.
Quote: ThatDonGuyThat's if a particular die has a 2 on it. We don't know which die/dice has a 2 - only that none of the 25 rolls that do not have any 2s were rolled...
It does not matter which is the "particular" die showing deuce. Out of the 2 dice that could be showing deuce, they're mirror image scenarios.
Quote: DJTeddyBearNo. It's a matter of math.
Let's try this.
You shake the dice in the cup. You ask someone to look and confirm that at least one die is showing a deuce. He looks, and says, "Yep. At least one of them is showing a deuce."
So, the person has confirmed that out of 36 possible random dice results, you got one of the eleven that have at least one deuce. Out of those eleven, only one is the double deuce.
Are you serious?
If one die is known to be a 2, your only concern is the second die of which there are six possibilities.
Folks: ONE DIE IS KNOWN. Hello???
If Die-A reveals a deuce, you are trapped in the blue circle, with 6 possible results.
If Die-B reveals a deuce, you are trapped in the red circle, with 6 possible results.
Each of these "reveals" are exactly equivalent, and mirror images of each other.
You do not get to have the UNION of both sets, once the culprit Die is cast.
You get to have 6 equal outcomes, but you do not get to have 11 outcomes. SCREAM!
Quote: AlanMendelsonAre you serious?
If one die is known to be a 2, your only concern is the second die of which there are six possibilities.
Folks: ONE DIE IS KNOWN. Hello???
No.
One die is not known. One or the other die is known.
If you have the time and if you have Excel, I suggest you [and other non-1/11-believers] do what I did in Excel and simulate the scenario.
A: =RANDBETWEEN(1,6)
B: =RANDBETWEEN(1,6)
C: =IF(OR(A1=2, B2=2), 1, 0)
D: = IF(AND(A1=2, B2=2), 1, 0)
*highlight, drag/pull them down to row #1000*
E1: = SUM(C2:C1000)
E2: =SUM(D2:D1000)
E3: =E2/E1
Look at cell E3. If you're a 1/6-believer, then it should be close to 16.66%. However, you'll notice that's not the case (maybe in an extreme circumstance it'll happen [ie: variance]). But most of the time it'll hover around 7-11% (9.0909% being the average). Of course if you do few trials that number will jump around quite a bit (1000 trials is a small sample, but, couldn't/didn't feel like doing more on my chromebook). If you do 10,000 trials (down to row #10,000), you'll see that number gets really really close to 9.09%.
So, if you do this little exercise and realize the number tends to be around 1/11.................why do you think it's not close to 1/6? Did the Excel gods bring out their magical anti-deuce monkey?
Quote: DJTeddyBear...
You shake the dice in the cup. You ask someone to look and confirm that at least one die is showing a deuce. He looks, and says, "Yep. At least one of them is showing a deuce."
So, the person has confirmed that out of 36 possible random dice results, you got one of the eleven that have at least one deuce. Out of those eleven, only one is the double deuce...
NO YOU DID NOT! What you got is either one set of six rolls (w/deuce on A) or a different set of six rolls (w/deuce on B). It doesn't flippin' matter which of the two dice had the infernal deuce - you're getting 6 possibilities after that.
Quote: indignant99NO YOU DID NOT! What you got is either one set of six rolls (w/deuce on A) or a different set of six rolls (w/deuce on B). It doesn't flippin' matter which of the two dice had the infernal deuce - you're getting 6 possibilities after that.
https://www.youtube.com/watch?v=MF283pORG2E
Quote: pewI'll pay seven to one. Any throw with no twos is a push the bet is on only when at least one two is showing. Any takers..... Allen?
I had just thought of that last night before falling asleep. I concluded they wouldn't take the bet because they wouldn't know how to figure out they have an advantage (33% edge on resolved bets).
Quote: indignant99
If Die-A reveals a deuce, you are trapped in the blue circle, with 6 possible results.
If Die-B reveals a deuce, you are trapped in the red circle, with 6 possible results.
Each of these "reveals" are exactly equivalent, and mirror images of each other.
You do not get to have the UNION of both sets, once the culprit Die is cast.
You get to have 6 equal outcomes, but you do not get to have 11 outcomes. SCREAM!
It has been my experience that the people that start screaming are usually the ones that are wrong.
How many combinations are there for two dice? I hope you say 36.
How many of those contain at least one deuce?
Quote: DJTeddyBearOk. Let's try this from yet another angle...
How many combinations are there for two dice? I hope you say 36.
How many of those contain at least one deuce?
*makes chart showing there's 11*
*uses poor logic to subtract away 5*
*come up with "6" as my answer*
Am I doing it right?
Quote: RS*uses poor logic to subtract away 5*
You mean, like this?
| 11 | 12 | 13 | 14 | 15 | 16 | 21 | 22 | 23 | 24 | 25 | 26 |
| 31 | 32 | 33 | 34 | 35 | 36 |
| 41 | 42 | 43 | 44 | 45 | 46 |
| 51 | 52 | 53 | 54 | 55 | 56 |
| 61 | 62 | 63 | 64 | 65 | 66 |
| 11 | 12 | 13 | 14 | 15 | 16 | 21 | 22 | 23 | 24 | 25 | 26 |
| 31 | 32 | 33 | 34 | 35 | 36 |
| 41 | 42 | 43 | 44 | 45 | 46 |
| 51 | 52 | 53 | 54 | 55 | 56 |
| 61 | 62 | 63 | 64 | 65 | 66 |
I can restate the current problem with fewer possibilites.
You have two normal coins.
You flip one coin. You get a head. What are the odds that when you flip the 2nd coin, it will be a head?
answer: H T : 1 in 2.
You flip one coin. you don't look at it. What are the odds that when you flip the 2nd coin, it will be a head?
answer: H T : 1 in 2.
you flip two coins. What are the odds that you have flipped two heads? TT HT TH HH: 1 in 4.
Here is the magic that is the same as the dice problem:
you flip two coins. your friend covers up one of the coins, so the only coin you can see is a head.
what are the chances that he is covering a head?
answer: HH TH HT 1 in 3
see here? your friend can be covering up either head from the first group, the Tail from the 2nd group, OR the tail from the 3rd group.
these are the only possibilities that will allow him to cover a coin, and you can still see a head. any trials that end up TT are discarded.
You might review the two coin problem I linked to in this post. In the 2nd one I argue for the wrong answer at first.
Quote: Dween
You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
I read the question and find this information: I am truthfully told that at least one of the two-sided dice is showing a 2.
Using that information that at least one of the dice is showing a two, I now have to determine the probability that both dice are showing a 2 -- which is what the original question is.
Since it is known that one die is a 2, and there is another six-sided die I am considering only that other six-sided die. if that six sided die is numbered on each side 1, 2, 3, 4, 5, or 6 then the probability that both dice are showing 2s is 1/6.
I can't understand why you would jump through hoops to come up with the 1/11 figure, using the possibilities for two dice, when in the original question we are already told that at least one of the dice is showing a 2?
Now if you want to think that way, that's fine. But if you were working for me I'd fire you. In fact, if I knew you even thought that way and ignored the basic information in the question, I wouldn't even hire you.
Carry on.
Nope. One you see one of the coins, that coin is no longer random. In this situation, the answer is 1/2.Quote: Dalex64you flip two coins. your friend covers up one of the coins, so the only coin you can see is a head.
what are the chances that he is covering a head?
answer: HH TH HT 1 in 3
Keep the coins hidden and it remains 1/3.
Because you don't know which die is showing the two, it can be either (or both).Quote: AlanMendelsonI can't understand why you would jump through hoops to come up with the 1/11 figure, using the possibilities for two dice, when in the original question we are already told that at least one of the dice is showing a 2?
Quote: AlanMendelson
I read the question and find this information: I am truthfully told that at least one of the two-sided dice is showing a 2.
Since it is known that one die is a 2, and there is another six-sided die I am considering only that other six-sided die. if that six sided die is numbered on each side 1, 2, 3, 4, 5, or 6 then the probability that both dice are showing 2s is 1/6.
Carry on.
You're looking at it like, "A specific die is a deuce". But that is different than "either die // at least one die is a deuce".
You seem very opionated on your 1/6 answer. As has been mentioned before, would you like to play a game? We keep rolling two dice. If it's a 2-2, I'll pay you 7:1. If either die is a 2 and the other is not a 2, I win 1. Every roll where neither die is a 2, is a push (no one wins nor loses). Surely, you would have a nice advantage, getting paid 7:1 on a wager that should (true odds) would pay 5:1.
Interested? I'll be back in LV in a few days. You live in LV, correct?
Quote: DJTeddyBearNope. One you see one of the coins, that coin is no longer random. In this situation, the answer is 1/2.
Keep the coins hidden and it remains 1/3.
Because you don't know which die is showing the two, it can be either (or both).
So in the dice question, instead of telling you at least one of the dice is a two, if he told you, and then pulled it out and showed it to you, it is no longer random and the odds that the other die is a two now change?
I'm afraid not.
Telling you that one coin is a head or one die is a two, and showing you one die is a two or one coin is a head, is the exact same thing.
Quote: AlanMendelsonI think the issue here is the question and how you interpret the question to develop the answers. Again, here is the question from the initial post:
I read the question and find this information: I am truthfully told that at least one of the two-sided dice is showing a 2.
Using that information that at least one of the dice is showing a two, I now have to determine the probability that both dice are showing a 2 -- which is what the original question is.
Since it is known that one die is a 2, and there is another six-sided die I am considering only that other six-sided die. if that six sided die is numbered on each side 1, 2, 3, 4, 5, or 6 then the probability that both dice are showing 2s is 1/6.
I can't understand why you would jump through hoops to come up with the 1/11 figure, using the possibilities for two dice, when in the original question we are already told that at least one of the dice is showing a 2?
Now if you want to think that way, that's fine. But if you were working for me I'd fire you. In fact, if I knew you even thought that way and ignored the basic information in the question, I wouldn't even hire you.
Carry on.
The simulations are not convincing?
My rehash:
Toss two fair coins simultaneously so that each one lands by itself in one of two seperate boxes, labelled L and R. You cannot see into the boxes, but your parter can. He will truthfully tell you information.
Before your partner says anything, the possible results are:
___L___|___R___
H | H
H | T
T | H
T | T
(or HH HT TH TT for short)
Probabilty of HH is 1/4 with no other information.
Partner says that one of the coins is H. Now the possible results are:
___L___|___R___
H | H
H | T
T | H
(HH HT TH)
Now the probability of HH is 1/3, since there cannot be two Ts.
New toss, but now your partner says the L coin is H, leaving these possible results:
___L___|___R___
H | H
H | T
(HH HT)
Now the probability of HH is 1/2.
So what if we expand this to the two dice problem. Same setup - two fair dice are thrown simultaneously so that each lands by itself in one of two separate boxes, labelled L and R. You cannot see into the boxes, but your partner can.
Before your partner says anything, the possible results are:
___L___|___R___||COUNT||
1 | 1 1
1 | 2 2
1 | 3 3
1 | 4 4
1 | 5 5
1 | 6 6
2 | 1 7
2 | 2 8
2 | 3 9
2 | 4 10
2 | 5 11
2 | 6 12
3 | 1 13
3 | 2 14
3 | 3 15
3 | 4 16
3 | 5 17
3 | 6 18
4 | 1 19
4 | 2 20
4 | 3 21
4 | 4 22
4 | 5 23
4 | 6 24
5 | 1 25
5 | 2 26
5 | 3 27
5 | 4 28
5 | 5 29
5 | 6 30
6 | 1 31
6 | 2 32
6 | 3 33
6 | 4 34
6 | 5 35
6 | 6 36
Then your partner announces that the L die is a 2.
Cool! Remove all combinations where the L die is not a 2:
___L___|___R___||COUNT||
2 | 1 1
2 | 2 2
2 | 3 3
2 | 4 4
2 | 5 5
2 | 6 6
Now there is one combination out of 6 with L2 R2, so 1/6.
But wait! Your partner screwed up! Due to the arbitrary rules of the game, you cannot be told which die contains a two, only that A two has been thrown. So the throws will be repeated, this time with an admonishment to your partner.
A few throws later, your partner announces at least one of the die shows a two. Now what is the probability of BOTH dice showing a two?
Based on your partner's information, you can eliminate all combinations without at least one 2:
___L___|___R___||COUNT||
1 | 2 1
2 | 1 2
2 | 2 3
2 | 3 4
2 | 4 5
2 | 5 6
2 | 6 7
3 | 2 8
4 | 2 9
5 | 2 10
6 | 2 11
Now there is one combination out of 11 with L2 R2 after being told that at least one of the die showed a 2, so 1/11.
Note- Results of 3.6MM (10 runs of 360,000) rolls in Excel:
Any two Both Two both/any ideal 1/11
Totals 1,099,165 100,140 0.091105521 0.09090909
Quote: LoquaciousMoFW
Then your partner announces that the L die is a 2.
Cool! Remove all combinations where the L die is not a 2:
___L___|___R___||COUNT||
2 | 1 1
2 | 2 2
2 | 3 3
2 | 4 4
2 | 5 5
2 | 6 6
Now there is one combination out of 6 with L2 R2, so 1/6.
Now, let's go back to what the original question asked and this is what the original question said, that at least one of the two dice is a 2:
Quote: Dween
You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
So the way the original question was worded, and as you showed, the correct answer is 1/6.
Had the question been worded differently the answer might have been 1/11.
It's all in the wording of the question.
Quote: AlanMendelsonThank you for posting this:
Quote: LoquaciousMoFW
Then your partner announces that the L die is a 2.
Cool! Remove all combinations where the L die is not a 2:
___L___|___R___||COUNT||
2 | 1 1
2 | 2 2
2 | 3 3
2 | 4 4
2 | 5 5
2 | 6 6
Now there is one combination out of 6 with L2 R2, so 1/6.
Now, let's go back to what the original question asked and this is what the original question said, that at least one of the two dice is a 2:
So the way the original question was worded, and as you showed, the correct answer is 1/6.
Had the question been worded differently the answer might have been 1/11.
It's all in the wording of the question.
The problem with your proof is you changed the question (Locquacios, and I quote, in his proof says "Then your partner announces that the L die is a 2.".
This is very heart of the discussion, the original problem clearly says "at least one of the dice are a 2". This subtle distinction makes a world of difference to the correct sollution.
If you can answer that with certainty, then it becomes a 1/6 problem.
Quote: LoquaciousMoFWHere's the key: Which die is a 2?
If you can answer that with certainty, then it becomes a 1/6 problem.
It doesn't matter which die is a 2. If you tell me at least one die is a 2 then the question is what is the other die? If you know there are only two dice it should be simple for you to figure this out:
2 dice show a number - 1 die which shows a 2 = 1 die to figure out what face it shows. That's 1/6.
Quote: RSAlan, why no response?
Call me the next time you have a question that needs an immediate response. I have a life outside of this forum.
Quote: Dalex64Quote: DJTeddyBearNope. Once you see one of the coins, that coin is no longer random. In this situation, the answer is 1/2.
Keep the coins hidden and it remains 1/3.
Because you don't know which die is showing the two, it can be either (or both).
So in the dice question, instead of telling you at least one of the dice is a two, if he told you, and then pulled it out and showed it to you, it is no longer random and the odds that the other die is a two now change?
I'm afraid not.
Telling you that one coin is a head or one die is a two, and showing you one die is a two or one coin is a head, is the exact same thing.
Sorry to disappoint you but pulling out one die doesn't help because you still don't know which die it was.
Quote: kenarmanI will be in Vegas next month Alan. If you want we can simulate the problem for real money. You roll the dice. I will check the dice. No 2 seen is a push. If I see at least one 2 you will pay me $1. If there are two 2's I will pay you $7. Larger stakes if you want but I will feel bad about cheating you.
I don't live in Vegas. I'm in LA. I don't know if I will be in Vegas when you are there, but it doesn't matter, I don't make "side bets" at the craps table.
Besides, your "bet" has nothing to do with this discussion. But perhaps you would like to answer the question:
If I roll two dice and one shows a "2" and the second die is a spinner, what are the chances that the "spinner" will settle on another "2"? I think 1/6. What do you think? Are there 11 sides on the spinner?
Quote: DJTeddyBearQuote: Dalex64Quote: DJTeddyBearNope. Once you see one of the coins, that coin is no longer random. In this situation, the answer is 1/2.
Keep the coins hidden and it remains 1/3.
Because you don't know which die is showing the two, it can be either (or both).
So in the dice question, instead of telling you at least one of the dice is a two, if he told you, and then pulled it out and showed it to you, it is no longer random and the odds that the other die is a two now change?
I'm afraid not.
Telling you that one coin is a head or one die is a two, and showing you one die is a two or one coin is a head, is the exact same thing.
Sorry to disappoint you but pulling out one die doesn't help because you still don't know which die it was.
Exactly what I am saying with the two coin problem.
You don't know which coin your partner is hiding, you can only see that one of the cons is a head.
So, HH HT TH. 1 in 3. It is the same if he covers one coin (you don't know which one) so you can see that at least one of them is a head, or if he covers both coins and tells you at least one of them is a head.
The odds in my scenario are not 50/50.
You're clearly not grasping the problem properly.
If you do make it to Vegas after Memorial Day I would be glad to meet you and play some craps, no math questions allowed :-)
Quote: RSIf the first one is a 2, then the spinner has 1/6 chance.
You're clearly not grasping the problem properly.
Correct.
In this situation, the dice are segregated and the first die is no longer random.
Quote: RSIf the first one is a 2, then the spinner has 1/6 chance.
You're clearly not grasping the problem properly.[/]
If the dice are thrown at the same time it doesn't matter if one stops before the other. Each die is always one of six but that's not the issue. The two dice thrown together is a single event.
Quote: pew"If I roll two dice and one shows a "2" and the second die is a spinner, what are the chances that the "spinner" will settle on another "2"? I think 1/6. What do you think? Are there 11 sides on the spinner?" One out of eleven times it will land on a two.
Really? If I have a "spinner" you are telling me that one out of 11 times the spinner will land on a two? Are you using a loaded die?
Quote: AlanMendelsonReally? If I have a "spinner" you are telling me that one out of 11 times the spinner will land on a two? Are you using a loaded die?
Pew is wrong. 1 in 6 times it will land on a 6.
This is different than the original question, though.
The original question is asking you, before the dice are known, what is the chance that you have double twos,
(1 in 36) with the wrinkle that you are discarding all of the trials that don't have a two (25 / 36) leaving only one way in 11 to have a double two.
It would be rather tiresome to play this out in real life -
Cup is slammed, friend peeks - is there a two? Nope, roll again.
25 times out of 36 you'll have to roll again.
Quote: DweenAlong the vein of the Two Coin Puzzle, I wanted to put this in another thread as to not clog up the other.
You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
Based on the discussion in the Two Coin thread, I'm interested to see how people solve this one.
Oh yeah, and the cat is dead.
Quote: Dalex64Pew is wrong. 1 in 6 times it will land on a 6.
This is different than the original question, though.
The original question is asking you, before the dice are known, what is the chance that you have double twos,
(1 in 36) with the wrinkle that you are discarding all of the trials that don't have a two (25 / 36) leaving only one way in 11 to have a double two.
It would be rather tiresome to play this out in real life -
Cup is slammed, friend peeks - is there a two? Nope, roll again.
25 times out of 36 you'll have to roll again.
I have to disagree with you. You didn't read the question carefully. The original question tells you one die is known to be a 2. And that is the same as throwing two dice on a craps table, one of them coming to rest as a 2, and the spinner being the unknown die. You have a 1/6 chance that the spinner will be another 2.
Quote: RSIf the first one is a 2, then the spinner has 1/6 chance.
You're clearly not grasping the problem properly.
But this is how the original question was given, in effect.
The original question tells you that there is one 2. And the question is what is the chance that the other die is also a 2.
It's the same thing as if you threw two dice on the table, and one comes to land as a 2, and the second is a spinner.
We know that one die came to rest as a two, the same way we know that one die under the cup is a 2.
The question is the second die.
The spinner on the craps table has a 1/6 chance. The second die under the cup also has a 1/6 chance.
Here is the original question again:
Quote: Dween
You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
Quote: Ayecarumba...there is a 16.67% probability the other die is also a two.
Oh yeah, and the cat is dead.
Unfortunately Ayecarumba, a vast majority of the forum members say you and I are wrong. They don't "see" in the original question that one of the dice is known to be a 2. They maintain it can't be 1/6 for various reasons such as we don't know which of the two dice is a 2 (as if that matters).
I just can't figure out why the simple answer and common sense doesnt apply here.
Did no one read the original question????
Quote: Dalex64Again, the simulations aren't convincing? They even tell you how to do it yourself. The wizard even clearly answered the same thing in the second post of this thread.
what simulations? ONE OF THE DICE IS KNOWN TO BE A 2. Therefore it's a 1/6 chance that the second die is also a two if the dice aren't loaded.
Frankly, I think the Wizard missed the point about one die being a 2 and truthfully said to be a 2.
- Wizard
- MathExtremist
- ThatDonGuy
- charliepatrick
- DJTeddyBear
- JB
- RS
- Joeman
- mustangsally
- teliot
- AlanMendelson
- waasnoday
- kenarman
- LoquaciousMoFW
- Dalex64
- pew
- Ayecarumba
So I am a convert to the 1/11.
My epiphany? Thought experiment about whether I would wager on this game in the casino. Here was my scenario, in the casino...
- 2 dice are shaken in an electronic cup.
- Electronic cup for honesty. No human deceit possible.
- Slam down the cup with the 2 dice, as yet unrevealed.
- Bottom screen display of the cup is now on top.
- Cup's display - an optical reader - now shows one and only one of the die's pip-count.
- Casino let's me bet whether there's a pair under there.
- No pair - my bet loses one unit. Pair - casino pays me 9-to-one.
- Duh. I think not.
Would anyone like to make a side wager that this thread was a joke?
Mango put his finger on the button. We don't have enough information to know the rules. In the absence of the rules, 'the probability that both dice are showing a 2 two cannot be known.'
Quote: MangoJ
..this would only hold if the player would say nothing when no 2 were spotted.
If (before peeking) your partner is determined to say "at least one of the dice is a X" where X is one of the numbers he may spot, the probability for a pair would be different.
If the friend sees any pair (1/6), probability of "X-X" where X is the named number, is one.
If the friend sees a non-pair (5/6), probability of "X-X" is zero whatever he choses for X.
Hence, like the Monty-Hall-Problem, if your friend must always say "At least one of the dice is a X" (while not lying), probability of a pair is 1/6.
We do not know if he always speaks: That is not given in the OP. It's central to the issue. Hell, it may be that he only makes the assertion 'At least one of the dice is a two' when a pair of twos are showing. Then 'the probability that both dice are showing a 2' is 100%. Maybe there's an unwritten rule that he only speaks when two different dice show.
* It might be that he always says something true, but arbitrary, about the dice and that something might be as simple as 'The two dice are blue'
