bonanova

Bayes, revisited. Easy: A magician shows four fair coins. One is two-headed; the others have both heads and tails. He places the coins in a black bag and blindly removes one. Repeated flips of the selected coin show "heads". After how many flips is it safe to bet with even odds that the selected coin has two heads? . Interesting: A magician shows four fair coins. Among their eight faces are five heads and three tails. He places the coins in a black bag and blindly removes one. Repeated flips of the selected coin show "heads". After how many flips is it safe to bet with even odds that the selected coin has two heads? . Enjoy!

As Melanie Griffith said to her boyfriend in Working Girl, when he didn't like her "Maybe" after he proposed, "Want a different answer? Ask a different girl." Here's the puzzle I posed: Three doors; one car, two goats. You point to a door but do not open it. Monty discloses the contents of a different door. You choose a door and win its contents. Assume Monty acts as your adversary when he can. Assume nothing else. Find a single strategy that always gives p[car] = 2/3. Answer: Pick the car if Monty shows it, otherwise open the third door. Want a different result? Pose a different puzzle.

It doesn't, and we don't disagree. We agree that p[7|"one of your dice is 6"] = 2/11. It's rather the 100th contestant that makes a statement we disagree with. by specious argument, that p[7|No Statement] = 2/11. That's clearly wrong: p[7] unconstrained is 1/6. So, what's wrong with his argument? Where's the hole? The hole is that C100 makes an assumption. One that is not warranted. So let's find it. Without loss of generality, say the dice are 3 5.

Don't see why you say that. The decimal point is not there of course. Otherwise,

Sort of. Well, yes. He's directing Monty's choice. Same as if Monty chooses randomly. The contestant never opens two doors, tho. If Monty certainly shows a goat, swap [third door] gives p[win] = 2/3. But if Monty randomly opens a door, then if a car appears, swap [car] gives p[win] = 1. if a goat appears, swap [third door] gives p[win] = 1/2. Given that a goat appears, the contestant can always swap with the third door. If Monty knows where the car is, he doubles his chances from 1/3 to 2/3. If Monty doesn't know, his original choice and the third door both go to 1/2 and it doesn't hurt. So there's an interesting point, and I don't remember ever hearing it. The contestant's expectation of winning is 2/3 regardless of Monty's algorithm of choice. . . . . . If Monty knows the car's location and shows a goat, contestant is certain of 2/3. . . . . . If he doesn't know, contestant gets a weighted average of 1 x 1/3 + 1/2 x 2/3 = 1/3 + 1/3 = 2/3. The only thing that changes with Monty's method is p[win] for "swap with the third door." If the strategy includes "grab the car if it appears", then p[win] is always 2/3.

I agree with those statements. Let's talk about how the algorithm of the informant must be considered. Premiss: When we are given data, we have to consider the probabilities that lie behind them. [1] The classic Monty Hall problem There is a 2 to 1 favorability to switch doors when we know that Monty knows the extra door he opens will show a goat. Otherwise, we have to consider the probability that he would open a door that had a goat. If he DIDN'T know, and showed a goat, that fact increases the probability that our first choice has the car. Because if we chose the car door, it would be CERTAIN that his door would show a goat. In fact, it increases the probability that we initially chose the car to EXACTLY 1/2 [exercise left to the student], making it pointless to switch - both of the other doors have 1/2 probability. Now let's consider what happens when the informant is known to have NO CHOICE when he gives us information. This happens when he simply responds to a question. A question that WE choose. In Monty Hall, I choose a door. Before I open it, I point to a second door and ask Monty to open it. If it shows a car, I'm going to win. If it shows a goat, then I'm back to even odds again. The two outcomes [certainty and even odds] are weighted by the probability that I pointed to a door that has the car. That probability is 1/3. So the total probability is 1/3 x certainty + 2/3 x even odds = 1/3 + 1/3 = 2/3. That is, I will win the car 2/3 of the time if I get a peek at the second door. And that's the same result as when Monty knows that he will show us a goat. [2] The child gender problem Without knowing how the informant chose to tell us about his girls, if any, the probability is unknown. We know only that it's 1/3, 1/2, or something in between. Let's remove the uncertainty. Informant: I have two children. Me: Is one of them a girl? Informant: Yes. The probability of Informant having two girls is 1/3. There is no ambiguity here, because the informant completely describes the situation I asked about. I asked him, in effect, whether I could eliminate the BB case from the general population. He told me that I could. There is no other relevant information that he could have given. Since he had no choice, we don't have to guess his algorithm. His algorithm, as a trusted observer, is simply to give a truthful answer. His answer permits the simple calculation that p[GG] = GG/[GG + BG + GB] = 1/3. If he had simply ventured the information that one is a girl, he might have been thinking only of the older, or the taller, or the one whose name came first in the alphabet. Then it would be the absurdly trivial case of p[GG] = p[other child = girl] = 1/2. If I don't know why he gave me the information, I know only that 1/3 <= p[GG] <= 1/2. [3] The informed two dice totaling 7 problem The unconstrained probability is 1/6 that fair dice total 7. If we learn from the informant that one of the dice is 6, then p[7] changes. How it changes depends on why the informant chose to tell us about sixes, rather than about fives, or threes. More to the point, since those cases are symmetric, why he chose to tell us about a number that appeared, rather than a number that was absent. Since we don't know, the probability is unknown. Let's remove the ambiguity. Informant: I just rolled two fair dice Me: Is at least one of the dice a 6? Informant: Yes. The probability now that the dice total 7 is 2/11. Again no ambiguity. There is no other relevant information that could be given in response to the question that I chose to ask. For example, he could not tell me about fives, or threes. So we simply count the cases where fair dice show 6 [there are 11 cases] and of them, the number [two] that total 7: 1-6 and 6-1. Does this make sense? Yes. Consider that of the 36 possible cases, 11 show a 6 and 25 do not. So, we know the probabilities that lie behind that data we got from the Informant: p[Yes] = 11/36; p[No] = 25/36. Of those 25, there are four that total 7: 2-5, 5-2, 3-4, and 4-3. So if Informant had answered No, then the 7-total probability would have been 4/25. Note that 2/11 is slightly greater, and 4/25 slightly less, than 1/6. So now we have the whole picture, and the weighted sum is seen to be what unconstrained result: p[7] = 1/6 p[7] = p[7|Yes] x p[Yes] + p[7|No] x p[No] p[7] = p[7|Yes] x 11/36 + p[7|No] x 25/36 p[7] = 2/11 x 11/36 + 4/25 x 25/36 p[7] = 2/36 + 4/36 = 6/36. What's the takeaway here? The takeaway is that algorithms matter. Unconstrained, p[7] = 1/6. When Informant tells us one die is a 6, we don't know p[7] any more. Why did he choose to talk about sixes? We don't know. That leaves the answer unknown. But when WE make the choice to talk about sixes, we remove the ambiguity. We know that it will change, and we know how it will change. If he says Yes, it will become 2/11. If he says No, it will become 4/25. And the sum of these results, weighted by the known probabilities of the informant's possible responses, brings us back to the 1/6 case. We need to know the informant's algorithm when he ventures information. What is the probability he chose [to talk about sons rather than to talk about daughters] [to talk about sixes rather than to talk about fives] and then what is the probability of the data itself [believing what he says it true.] When we chose to talk about sixes, we can calculate them - they're 11/36 for a Yes answer and 25/36 for a No answer. With all that information, the ambiguity is gone. Without it, the answer in unknown, although we can compute bounds, so long as the information is relevant. The phase of the moon, for example, wouldn't be helpful in any of these problems.

Dej Mar, when you speak of one of the children and the other child, you are making the mistake of identifying one of them, specifying its gender, then finding the gender probability of the other. That probability is always 1/2. I have two children, named Leslie and Pat. Leslie is a boy. What's the probability that both my children are boys? That reduces to: I have a child named Pat. What is the probability that Pat is a boy? Read the

Write out the alphabet starting with J: JKLMNOPQRSTUVWXYZABCDEFGHI Erase all letters that have left-right symmetry (such as A) and count the letters in each of the five groups that remain. Enjoy.

You're asking about Ramsey numbers. You can Google the subject to read more about it. The question you ask is the only nontrivial, multicolor case for which a solution is known. The solution is discussed here and here. - bn

Similar to my second suggestion, Does this give all numbers an equal chance? I haven't though through it.

It's been forever, it seems, since we downed a cold one at Morty's, but a quick visit last night did not disappoint, in more ways than one. As usual, Alex was holding court, over in the corner table. Gather round me, boys, he began, and try to learn something. And if it costs ya a quid or two, you'll remember it all the more. With that, he took the four Aces from a deck of cards, shuffled, and dealt them face down on the table. Now one of you lads grab a couple bottle caps from the floor and place them, one each, on two of these here cards. Jamie obliged. Now here's the deal, Alex continued, there's a certain probability that the caps are sitting on cards of the same color [red or black]. I'll bet even money with anyone who can say what that probability is. Jamie was first to respond. It's quite simple, ya see? There are three cases. They're sitting on black cards, red cards, or mixed cards. The probability of the same color is clearly 2/3. You've been at the tap too long, laughed Ian, you forgot that mixed cards can be red-black or black-red. There are actually four cases, and the same-color probability is 1/2. They either match, or they don't. Davey sat for a moment, stroked his bushy red beard, and then walked over to Alex and whispered his answer so none of the others could hear. Given that only one of the boys had it right, who went home that night richer than he came?

Welcome back, peanut!

The storied probability question of the two-child family with at least one son was a while ago in original form, and again lately with a few twists. In the latter post, I suggested /page__view__findpost__p__247003 about the information given in the puzzle, and in some cases the probability asked for might be 1/2. In the first topic I had argued most vehemently for 1/3, although I never stated the assumptions that are needed to support that answer. Assumptions were also debated in where I asked the probability of rolling a 7 with a pair of fair dice, making a specious argument that it is 2/11, not 1/6. I invited a refutation. It comprised more than one hundred posts that demonstrated assumptions affect the answer. If assumptions do affect answers, then either (a) puzzles of this type are hopelessly indeterminate, or (b) there is a best, unwritten perhaps, set of assumptions that we should invoke, when none are stated, that will give these puzzles well defined answers. We don't like (a), so let's assume (b). This topic asks: what is that best set of assumptions? Consider: An trusted observer looks at a situation [children, dice, whatever] and makes a statement: e.g. "At least one child is a girl," or "at least one die is a six." What might we now assume? Here are some possibilities. [1] It is the only statement the observer could make; that is, it's a complete description. [2] There are other relevant statements that are true, and the observer chose from them at random when he spoke. [3] There are other relevant statements that are true, and the observer chose from them with some type of bias. Alternatively, there is a position that says "We should make no assumption at all." But does that position serve to remove ambiguity? A starting point could be /page__view__findpost__p__247110. Enjoy.

Monkey wrench, anyone?

Ten prisoners are selected and told by the warden that next morning a hat with a single digit 0-9 will be placed on each of their heads. The digits can be repeated; not all digits need be used. Each prisoner can see the other nine numbered hats. After the hats are placed, each prisoner will be brought to the warden, one by one and out of sight and hearing of the others, to guess his own number. If at least one prisoner gets it right, all ten will be released. Otherwise they all will be executed. A strategy may be agreed upon tonight. The prisoners then cannot communicate with anyone else until they have all spoken to the warden. What strategy gives them the best chance of being released?

Can you escape an even faster ogre if you paddle in a spiral?

What do players learn from an accepted or rejected handshake? Is there a strategy agreed upon before this process starts? Or is that all part of the puzzle?

is the solution for quickest escape [already given in this thread by darmelpitts, Tuckleton and Hugo], and also the solution for the greatest distance escape. The second solution permits escape from a faster opponent - one that is 4.6 times faster than you.

Good job everyone. I posted a a while back, and a newbie named polynym got it on his first and only post in in this forum! Probably took longer to create than to solve. Maybe I'll try another one.