Prime

(9/16)*(1/3) + (7/16)*(3/7) = 3/8 averages the same as my estimated probability of 3/8. But it does not follow that "Both statements agree 100%"... In fact...

This is a very interesting problem. After checking the posts, I see that Writersblock had solved it a year ago. The solution that I found is the same in essence. Though, I found a different kind of relation between the three men (other than LEFT RIGHT).

I did say, one of us is right and the other is wrong.

Assume, they shoot at each other with paintballs, and they compete to win.

This probability problem was born of this forum. Upon meeting at a gambling establishment, CR and I got into a dispute over the probability of a certain event. CR believed that a particular slot machine paid off 1 time out of 3, when it flickered a blue light, and 3 times out of 7 when it was illuminated pink. I argued that it paid on average 3 times out of 8, no matter what light was on. We both agreed, however, that the blue light is on 9/16 of the time, while pink glows the remaining 7/16 of the time. The dispute raged on, and neither of us could be swayed by the other's argument. Exasperated, CR offered to beat me with the whiteboard bearing his calculations. But I suggested a wager instead: We'd set the odds as CR calculated; and I would choose which side to bet on. Thus we'd set our betting odds on "machine payoff" 1 ruble against 2 for a blue light and 3 rubles against 4 for a pink light. And I would, naturally, bet on a "payoff" in the case of "blue", and against a "payoff" in the case of "pink." This being a popular slot machine and many casino guests waiting in line to get to it, we’d have no shortage of statistical samples. "This way, each of us can stick to their opinion, but at the end of the day one of us will walk away richer," says I. Is that bet a fair resolution for the dispute (statistically)? If not, how is it uneven? Assume for the purposes of this problem that one of us is right and another is wrong (not both wrong). Make no assumptions regarding our respective faculties to estimate probabilities.

Are you offering to beat me with your whiteboard? I decline. Don't take me wrong, I love violence, but it's not the most effective way to solve probability problems. Bearing the whiteboard threat in mind, I have to concede my point. After you switch your first choice and AFTER Monty opens second door, the probability of your door to hide the prize becomes 1/3 if Monty opened your originally chosen door second time around and 3/7 if he opened another one. See my post of another probability problem, which I derived from our exchange.

I thought of that. But the tribunal decided that you have to use each number exactly once. And you didn't use "7".

Here is somewhat heuristic proof. But it is less than a page long.

And let me preempt possibility for further confusion. Let's say Monty opened door C second time around. That excludes case 3 described above and case 2 inherets its probability, becoming 3/16 + 3/16 = 3/8. The choice has been made and no longer its probability is 1/2. Similarly, after door C is opened, case 4 gets excluded and case 1 inherits its probability becoming 1/4 + 3/8 = 5/8. Since those two cases were mutually exclusive.

Darts example is not a suitable analogy. Each dart player can either hit or miss, whereas in case of Monty -- he must always open non-winning door. Until he opens the very last one and shows your high school sweetheart hiding behind it. But at that point it's too late for the player to make calculations. With probability problems, when in doubt, go through all possible permutations. And there are but 4 in your scenario. We agreed that after you make the initial switch from door A to B, the probabilities for the doors to hide the prize are A: 1/4, B: 3/8, and C: 3/8. Then there are 4 mutually exclusive and collectively exhaustive cases: 1. A has the prize/Monty opens C (forced move). --------------------> P1=(1/4)*1 = 1/4. 2. B has the prize/Monty opens C (random choice from A and C).--> P2=(3/8)*1/2=3/16 3. B has the prize/Monty opens A (random choice from A and C).--> P3=(3/8)*1/2=3/16 4. C has the prize/Monty opens A (forced). ---------------------------> P4=(3/8)*1=3/8 Now case 2 and case 3 are the only two where door B is the winner. The probability P2 + P3 = 3/16 + 3/16 = 3/8. And that's the total probability for B to be the winner after Monty opens his second door. Cases 1 and 2 are the only cases when door C is opened. P1 + P2 = 1/4 + 3/16 = 7/16. Cases 3 and 4 are the only cases when door A is opened. P3 + P4 = 3/16 + 3/8 = 9/16. (Note how the four cases above add up to 1). I believe you can figure out where your formula is wrong/inapplicable without my help. When you multiply probabilities of two events, those must be dependent. When you add two probabilities -- their events must be mutually exclusive. E.g., you cannot multiply probability of door A being open second time around by probability of door B being a winner. Such operation has no meaning in this problem.

I am under the impression that you suggest that if you switch the door at your first opportunity, then probability of your newly chosen door (B) to have the prize would vary depending on which door Monty opened second time around -- your original choice, or the other one (A or C). I disagree with that assessment and state as following: In your scenario, SWITCH/?, the probability of your second choice (B) will remain 3/8 regardless of Monty's consequent manipulations. Whereas, the probability of Monty opening door A second time around is 9/16, versus door C -- 7/16. If you disagree with that, or have calculated different values, I suggest check your line of reasoning/calculations for error. (Note, that I gave exact numbers for probabilities as I calculate them.)

I already posted a derivative problem, where Monty teases player who swithched the door. (See my previous post.) Whereas, OP did not mention any criteria for Monty to send one or the other bill collector home, we must assume his choice is random in that situation. And for that case the formula that you showed is still incorrect. As well as it is incorrect for any other case with Monty's non-random decision between two bill collectors. (I have already explained how several times.) If I explained how Monty's partiality with respect to bill collectors would affect probabilities, I'd be giving away clues on solving my derivative problem.

Do you mean, you asked to find decimal 21, not octal!? What are the chances of that?

I posted a proof that you get (N-1)/N chance to win in this game by sticking to your first choice and switching when Monty has only one door left AND that you cannot do better than (N-1)/N with any other strategy. Here is a derivative from this problem: Say there are N doors, N>4. Suppose, you knew that if you switched your original choice, Monty would keep that door closed to tease you, except when he has to open it (he is down to two doors and has to open your originally chosen door, because the other one hides the prize.) What would be your chances and best strategy in that situation?

I understand the notation. In your scenario SWITCH/? the probability of your door to hide the prize is 3/8 and the probability of Monty's last remaining door is 5/8 regardless which door he opened 2nd time (A or C). If any equations/and or reasoning show otherwise, they are erroneus. (Like the equation you showed in the previous post.) And I am convinced at this point that you are quite capable of finding errors in your equation. They say you can't add apples and oranges. Well, sometimes you can't multiply them either.

Whoever started first will have spent 40 seconds for his posts, the other -- remaining 20 seconds. As far as the last word goes... If there is a minimal time in which you can still make a post, then probability of the last word is 1/2 for either person. If length of each post can be divided indefinitely (timewise) -- then neither can have the last word. Because after infinite number of replies, the length for each reply for either person becomes zero. And when you have zero time for your reply, you cannot make one. P.S. Since I took an entire minute for this missive -- not half a minute, no one can argue with that.

Oh, I see. Each leg can support 200 lb, not each bed. Then taking into account that each leg already supports 50 lb. of its own bed weight to begin with, the answer is

It seems, the bottom layer is already overloaded. 25 beds having to support the weight of 30 at 200 lb each. Am I missing the trick?

What do you mean:

Sorry, CR. But the formula that you show appear to be invalid and the numerical values for probabilities incorrect. The formula uses P(B winner) both on the left and the right side of the equation. And they seem to disagree. Both P(B winner|A opened) and P(B winner|C opened) are equal to 3/16 each -- not 1/3 and 3/7 as you stated. Or if you consider the probabilities of A or C open inside the case when B is the winner -- then it is 1/2 each. Your P(A open) and P(C open) on the right side use overall probability including the case when B is not a winner. Therefore, multiplication of that by the probability of the cases when B can only be a winner makes no sense. Your formula should be: P(Bw) = P(Bw)*P(Ao) + P(Bw)*P(Co), or substituting 3/8 = (3/8)*(1/2) + (3/8)*(1/2). And of course, it is not anything useful, because all it is saying: 1 = P(Ao) + P(Co) and we know that Monty will surely open one of his doors without any calculations. After checking the other posts, I found that "general solution" (N-1)/N, although without proof, has been posted already as far back as post #4.

Agree on that. Still, I think, my solution (post 9) is valid too.

CR, I commend your attempts to confuse the issue. They are clever and of some complexity. But they will not work. Once again your line of reasoning mixes up probabilities of Monty having some particular door unopened in the end and probability of HS hiding behind that door. When you switch your choice on the first opportunity, probability that Monty leaves your originally picked door unopened is 7/16, probability for the other door to be the last unopened is 9/16; whereas probability of the HS the behind last unopened door is 5/8 in either case. Do you disagree? Note, how the first two probabilities add up to 1 with each other, whereas 5/8 adds up to 1 with the probability of your second pick (3/8). When you break down your variation into two sub-variations, where in one case you have HS and in another – Monty does, no longer can you use the previous probabilities (1/4 and 3/8) of having HS for any of Monty’s doors. Since in one case you have already presumed that probability to be zero; and in the other case Monty’s choice is not random, but forced. Your “extreme example” changes the statement of the problem, making it different from what OP meant. I have already done that humorously, drawing some skepticism from Bonanova. (See posts 19, 20, and 21). Think of it this way: When you chose a door of probability P of hiding HS, all the remaining doors in Monty’s possession have total probability of 1-P. Monty’s manipulations change probabilities of his individual doors, but their total remains the same. Until you make another switch, that is. See my general solution in post #18.

Me think it is null. The problema say the pole is in a lake. It say nothihg about no pole in feet. If you swim up to it and hold with your feet, mayhap you can tell.

Here is one possible way: