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#101




If you roll two dice for one die to show a six won't that be odds of 1 in 3?

#102




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http://www.math.hawaii.edu/~ramsey/P...y/TwoDice.html As borschevsky is alluding to in the question he asked you, you will get a 1 and a 6 twice as often as a 6 and a 6. 
#103




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Lay out 36 pairs of dice in a grid showing all 36 possible results. Of the 6 pairs in which the die on the right side of the pair is six, the left die will also be 6 once (one chance in 6). Of the 6 pairs in which the die on the left side of the pair is six, the right die will also be 6 once (one chance in 6). Of the 11 pairs in which at least one of the dice in the pair is 6, the other die will be 6 in one case (1 in 11). Last edited by Andy L; 10062016 at 06:46 PM. 
#104




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Ok, I just rolled two dice for a good chunk of time. I discarded all rolls that did not include a 6 and I stopped after 20 double sixes. I had 132 rolls that included at least one 6. Here are the ratios for the rolls. 4 22 5.5 5 26 5.2 6 29 4.833333333 7 30 4.285714286 8 35 4.375 9 41 4.555555556 10 85 8.5 11 86 7.818181818 12 96 8 13 98 7.538461538 14 99 7.071428571 15 111 7.4 16 112 7 17 117 6.882352941 18 118 6.555555556 19 129 6.789473684 20 132 6.6 


#105




I'm afraid you'll have to explain to me what those numbers represent.

#106




octopus, then you think with a roll of two dice, you have an equal chance of rolling two sixes as you do rolling a 1 and a 6?
Do you agree that the probability of rolling a 1 and a 6 is 1/18? 
#107




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The sum of 1 and 6 is 7 and that is a 1/6 chance of occurring. So what are we looking for sums or particular probabilities of particular events. In that example he is summing events. If 1,6 is considered exactly equivalent to 6,1 than yes those 2 have twice the probability of 6,6. Let's make the math easier and use the coin example. Throw out all double tails. Use a penny and a nickel. What are the outcomes Penny Nickel H T T H H H How many possible ways can I show one head? I can't show one head with double tails so that isn't even in the denominator. It's not counted. Correct? So how many ways can I show you a head? A) I can show you Penny's head if Penny comes up head and Nickel comes up tails. Correct? That is one count. B) I can show you Penny's head if Penny comes up head Nickel comes up head. That is two. C) I can show you Nickel's head if Nickel comes up head and Penny comes up tails. That is three. D) I can show you Nickel's head if Nickel comes up head and Penny comes up head. That is four. Ah  ha ha ha. Four ways to show you a head in which at least one head was flipped. These are four permutations of a state in which one coin is revealed and one is concealed with the constraint that there will only be a revelation if at least one is a head. Correct? Now how many of those permutations satisfy the second case where the concealed coin is also a head? Two events B and D. 2/4 is 1/2 which is what you'd expect with a fair coin. 
#108




Cool. Now please Google probability of rolling snake eyes.

#109




Yes it is, it's just more detailed. If the red die is a 6, the odds of a green 6 are 1/6. If the green die is a 6, the odds of a red 6 is 1/6. That covers all possibilities, and in every case, it is 1/6. It doesn't mater if the dice are colored  it's 1/6 in every possible case.



#110




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#111




What I'm enjoying now are the folks writing simulations and doing manual dice throwing trials.
The ambiguity is in the problem statement. Once you decide on your preferred interpretation of the problem statement then your result (1/11 or 1/6) is foreordained. And can be proven with a few short lines of small words. So any effort spent to simulate is foolish. It's merely proving the trivial conclusion you already (perhaps unwittingly) assumed. Last edited by LSLGuy; 10062016 at 07:23 PM. 
#112




Snake eyes is 1/36.
1,1 is one ordered pair out of 36. Simple. I'm not the one confusing what the actual constraints of the question are. Shoot let's say we only count the probability of the die roll after we rolled a 6. Instead of revealing a 6 and asking we are only counting 6s after we rolled a 6. What's the probability of that? Why would the probability of the die change? It only has 6 possible states regardless of what was rolled before. 
#113




I'm confused. You agree the probabilty of rolling a 1 and a 6 is 1/18, and that rolling a 6 and a 6 is 1/36, yet you count 6,6 twice here:
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#114




Do you agree with the following?:
http://www.math.hawaii.edu/~ramsey/P...y/TwoDice.html There are 36 equal rolls of two dice. 


#115




That's not the OP's problem. It didn't say anything about BOTH dice, but what would happen IF one condition was already met. That condition is no longer a factor, as we are not considering multiple conditions, nor conditional conditions.
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And just in case, running my program for 10,000,000 iterations confirms a 1:6 ratio, within 3 decimal places. I doubt if running it longer will change much, so it's the premise that counts here. I stand by my post #89, more than ever. Anyone want 10 million data points to look at? 
#116




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#117




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#118




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FWIW, I prefer your interpretation of the problem too. But it's just that: one interpretation of the subtly (and deliberately) ambiguous sentences in the original problem statement. The fact the multiple choice answers have choices which are correct for each interpretation and for the common confusion of [is R6G6 different from G6R6?] pretty well proves the problem was carefully deliberately designed to produce these incompatible lines of thinking. Last edited by LSLGuy; 10062016 at 07:46 PM. 
#119




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#120




Because what is being revealed and what is being concealed are different states. Look at it closely. There are 11 different pairs.
1,6 2,6 3,6 4,6 5,6 6,6 6,1 6,2 6,3 6,4 6,5 That's 11 things right? 6,6 is 1/11 correct? That is out of this set of ordered pairs 6/6 occurs 1 time out of 11. Makes sense right? But we have 11 ordered pairs that could exist. The other 25 are discarded. Each of these 11 have at least one 6. If you reveal the 6 you see what the other number could be. But look! There are two different 6s in the 6,6 example that could be revealed. The left or the right, the first or second, the red or green, etc. So there are actually 12 different sixes that could be revealed. And out of that 12 only 2 of the revealed have another 6 concealed and that is reveal red 6 conceal green 6 and reveal green 6 conceal red 6. That's why the denominator is not 11. Because we aren't dealing with a set of 11 ordered pairs. We are dealing with revealing a particular 6 from a set of 11 ordered pairs. But the question isn't what is the probability of one of these to occur. The question is what is the probability if I am shown a 6 that the other number is a 6. Well with colored dice there are 12 combinations that can occur because what matters is what is the state of the dice. Are they revealed, concealed? What number is on them. So if I can reveal a green 6 with 6 equally likely permutations on a red die that's 6 events. If I can reveal a red 6 with 6 equally likely permutations on a green die that's 6 different events. Yes or no? Is a revealing a green 6 the same as revealing a red 6? No. isn't. If I show you a green 6 what are states the red die can be in? 1,2,3,4,5,6. Correct? Why would one number be favored over another. If I show you a red 6 what are the states the green die can be in? 1,2,3,4,5,6. Correct? Why should 6 be a lessor probability than the 1,2,3,4 or 5? There's no logical reason to treat 6 differently. 
#121




Instead of rolling the dice simultaneously, do it sequentially. Keep throwing the red one until it is a 6. Then rolll the green one. What are the odds that the green one will be six? Just superimpose the two events in an overelap time frame..
In what way would the results be skewed by the dice not being thrown simultaneously? Or by the dice being thrown in separate rooms by collaborating experimenters who then communicate the events to each other with a walkie talkie? Last edited by jtur88; 10062016 at 07:50 PM. 
#122




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#123




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#124




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#125




Different answers for each boy/girl problem. Also different answers depending on how this problem is phrased (see post 68).

#126




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(B)"Mr. Smith says: 'I have two children and it is not the case that they are both girls.' Given this information, what is the probability that both children are boys?" So what's the paradox? It's a paradox because people don't read what is written? Child 1 is C1. Child 2 is C2. Here's a table of what the children can be. C1 C2 B B B G G B G G Now we have to exclude GG right? So we are left with C1 C2 B B B G G B You have 6 possible known, unknown pairs Child 1 being a boy, child 2 boy Child 1 being a boy, child 2 girl Child 1 being a girl, child 2 boy. Child 2 being a boy, child 1 boy Child 2 being a girl, child 1 boy Child 2 being a boy, child 1 girl. 2 of these 6 are the other child being a boy if the known child is a boy. So the answer is 1/3. Example 2 is the same as example 1 just written differently. Still get 1/3. I honestly don't see what's hard about that. Last edited by octopus; 10062016 at 09:09 PM. 
#127




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octopus, the two statements you give of the boygirl problem are equivalent, but those aren't the same two statements that xray vision just gave. 
#128




That's why I gave up.

#129




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Look, the question is so poorly worded it is unanswerable. 


#130




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Two dice are rolled. The implication is that ANY of the 36 possible combinations may have turned up. Not, you keep rolling until you have a six. Why even bother rolling two dice if you only count the results when one of them lands on a specific face? That's just like rolling one die in the first place and is a trivial question. So, you have one of 36 possible combinations. Then the omniscient viewer says, "Look, you rolled at least one six." At that point, what is the chance your other die is a six? It's 1/11. 
#131




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With the examples xray vision gave the answers would be. A) Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls? Possibility table. Older child is OC. Younger child is YC. OC YC G B G G 2 permutations to satisfy first condition of OC being a girl. 1 of the permutation satisfies the condition of 2 girls. So: 1/2 B) Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys? 1/3 Same table as my previous post. Where is the paradox? Different questions have different answers. Last edited by octopus; 10062016 at 10:36 PM. 
#132




Then maybe you should start a new thread?

#133




What's hard about calculating number of events that meet criteria A divided by number of events that meet criteria B?
Plus I answered your questions. Which since they are different than the OP's shouldn't they be in their own thread as well? 
#134




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#135




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Older child being a girl and figuring out the probability that the 2nd is as well isn't ambiguous at all. Being shown that one die is a 6 after a pair of dice have been rolled is ambiguous in what way precisely? If I flip a coin and roll a die and I show you heads on the coin how does that change what the die rolled? It's still a 1/6 chance on the die if you get shown a head, an ace of spades, a Draw 4 card, or another 6. Do that with a coin. Flip one and roll a die. What's the probability if heads comes up you roll a 6? How's it different with 2 different dice? This is a repeatable experiment. Last edited by octopus; 10062016 at 11:00 PM. 
#136




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Did the guy remove a die at random and it happened to be a six? (Then the answer is 1/6.) Did he remove any duplicated die? (The the answer is 1.) Did he remove a single six, whenever there was at least one six? (The the answer is 1/11.) If he always removes a single six when he can, but chooses the red six when both red and green are present, did he show you the color of the removed six? (Now the answers are 1/6 or 0, for red, green, resp.) As usual, "running a simulation" makes sense only when the problem is welldefined. But if it's welldefined, why run a simulation when you can just treat 11 cases? (or even just 3 or 4 cases.) 
#137




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The only way to get a 1/11 result is by NOT counting some of the results. 
#138




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If you do the process as described. That is have one person roll two dice and show the other person a 6 whenever a 6 is present in the pair of dice you can figure out an unambiguous answer. Obviously, you cannot show a 6 from a pair of rolled, fair dice when a 6 is not present. Where is the ambiguity? Just do this with someone else as explained and it's not hard. 
#139




Problem statement #1:
"Two fair dice are rolled together, and you are 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 six?"Problem statement #2: "Two fair dice are rolled together. A 6 is removed, and you are then shown the other die. What is the probability of the other die showing a six?"How do these problems differ from each other? Do they have the same answer? Quote:
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As for "you cannot show a 6 from a pair of rolled, fair dice when a 6 is not present" — that was my point. Read the two problems at the beginning of this post and tell us if they have the same answer. 


#140




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#141




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And those people who keep saying 1/11 can't count. The person rolling, showing, and removing a 6 has 2 6's to show in the double 6 result. I rolled dice til I got 20 sets of double 6s and I took the result of how many rolls it took that included at least one 6. My ratio is 20/132. Which is 1/6.6. I'm going to keep rolling til I get 20 more. Brb. 
#142




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Are you saying that you rolled a pair of dice many times and had 20 instances of doublesix and 112 instances (13220) of a single six? Your result is not impossible, but is surprising. On average singlesixes will outnumber doublesixes ten to one! 
#143




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Once a die of any color is a 6, the experiment proceeds with an examination of the other die, which will always be independent of the first die. Say the two dice are thrown in separate rooms, and throws continue until one of the operators announces "I have a 6". The operator in the other room examines his single die roll. What are the chances that it will be a 6? But one of the operators falls asleep, and is awakened by the first one saying "I have a 6", and suddenly rolls his die for the first time. Do the odds change? 
#144




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It takes awhile to roll that many dice. This time my results look like this. Double 6 At least one 6 Odds Inverse 1 9 0.111111111 9 2 13 0.153846154 6.5 3 41 0.073170732 13.66666667 4 48 0.083333333 12 5 66 0.075757576 13.2 6 80 0.075 13.33333333 7 85 0.082352941 12.14285714 8 86 0.093023256 10.75 9 95 0.094736842 10.55555556 10 97 0.103092784 9.7 11 106 0.103773585 9.636363636 12 135 0.088888889 11.25 13 149 0.087248322 11.46153846 14 150 0.093333333 10.71428571 15 154 0.097402597 10.26666667 16 181 0.08839779 11.3125 17 182 0.093406593 10.70588235 18 201 0.089552239 11.16666667 19 207 0.09178744 10.89473684 20 227 0.088105727 11.35 This looks a lot closer to 1/11. Hmm, maybe you don't count the green and red die separately and do treat that as one event. Maybe I'm the one who can't count! Sorry for the ugly formatting but padding spaces seem to be removed. Last edited by octopus; 10072016 at 12:45 AM. 


#145




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Skammer: " Not, you keep rolling until you have a six." xray: "We don't count any of the 36 possible rolls that don't include at least one 6." You can't have it both ways.The way I interpret Skammer in post 130 is roll two dice, and use whatever comes up. No throwing anything out. 
#146




I use Monte Carlo simulations a lot — in problems where the number of possibilities is too numerous for exhaustive counting.
Here there are only 11 (or, depending on how you count, 36, or 4, or just 3) possibilities. Why not just list them: 6 & Not6 ... 5 cases(Similarly, when I tell you a pair of red Kings will make a better hand than a pair of black Aces 18.553597959240882460% of the time in allin Texas Holdem, that's the result of enumeration, not simulation.) 
#147




I can't believe this thread is still going. People are looking for ambiguity where there is none.
Roll two dice. Is at least one of them a six? P (25/36) No > end game P (11/36) Yes > There are now 11 possibilities: 6 1 6 2 6 3 6 4 6 5 6 6 1 6 2 6 3 6 4 6 5 6 Is the other die a six? P (10/11) No P (1/11) Yes. Answer is 1/11. Where is the ambiguity please? 
#148




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#149




"A 6 is removed." This might mean "Monty picks one of the two dice at random and shows it to you. It happens to be a six." Now the answer is 1/6.
It's similar to the ambiguity in the Monty Hall problem. "Monty opens a curtain. This exposes a goat." Did Monty deliberately avoid the curtain with the brandnew Maserati? 


#150




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This is the second most common mistake made in the Monty Hall problem. (The first is misstating the problem.) You see two remaining doors and think they are equally likely to have the big prize. Look at any of the tables of eleven possible situations in this thread. Note that "the other is a 6" happens only one out of eleven times. For each of the others, they happen 2 out of eleven times. Given that the setup leaves 11 possibilities, how on Earth do you assign 1/6 to the (6,6) case but only 1/12 to each of (3,6) and (6,3). How can that possibly be???? The chances of rolling an (a,b) and a (c,d) are the same. You don't need to write any stupid code either. The listing of the possibilities, all equally likely, proves the answer without any further debate. 
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