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Stumped joined the community
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My guess is, “Will you NOT marry me?” This yes/no question works because either way you answer it, the answer is “no”, ensuring you do not marry the middle daughter: “no, I will not marry you” and “yes, I will not marry you”
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Jan Japan joined the community
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Not sure if contractions count but I got: A I Am I’m An In
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Vaibhavi joined the community
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loved the an elephant in a refrigerator puzzle... Have seen another variant here- https://youtu.be/kXKeLBwSzPI
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impulse joined the community
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there is a way however to permute through all possibilities by swapping two letters. ABC -> ACB -> BCA -> BAC -> CAB -> CBA ABCD -> ABDC -> ACDB -.> ACBD -> ADBC -> ADCB -> BDCA -> BDAC -> BADC -> BACD -> BCAD -> BCDA -> CBDA -> CBAD -> CABD -> CADB -> CDAB -> CDBA-> DCBA -> DCAB -> DBAC -> DBCA -> DACB -> DABC
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madhanraj joined the community
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Hats on a death row!! One of my favorites puzzles!
Hannah replied to roolstar's question in New Logic/Math Puzzles
This will only work if there were exactly 10 of each colour. -
Playing Chess with a Coin
rocdocmac replied to TimeSpaceLightForce's question in New Logic/Math Puzzles
Ah-ha! The first toss is irrelevant. Thus the second player has the best chance ... -
There's a way to understand that the odds are equal for both bowls without doing any math (for both the one poison apple = death and the two poison apples = death variants). Label the apples in each bowl as either common (C) or rare (R), so each bowl has CCCRR. For bowl 1, the common apples are poisonous and the rare apples are not. For bowl 2 the opposite. When you choose the apples that you are going to eat (from whichever bowl) you are going to divide the apples into two groups - those you will eat, and those you won't. In the one apple = death variant, the only way to survive is to divide the apples into two specific groups: CCC in one group and RR in the other. This is true for bowl 1 and bowl 2 - which means that the probability is the same for both. For bowl 1, you will then eat the two apples (both rare/nonpoisonous). For bowl 2, you will then eat the three apples (all common/nonpoisonous apples). In the two apples = death variant, you will only die in one way: if you divide the apples such that the group of two apples contains two common apples. That is, CRR in one group and CC in the other. Again, this is true for bowl 1 and bowl 2, so the probabilities are the same. (The order of apples doesn't matter within the groups because we will either eat all or none of the apples in a given group.) If we choose bowl 1, we eat the smaller group, and only die if that smaller group contains two common/poisonous apples. If we choose bowl 2, we eat the larger group and will only die if the smaller group (the one we didn't eat) contains two common/nonpoisonous apples (as this means that both of the poisonous apples were in the group we ate). So, without doing any math, you can see that the probabilities are equal, even if you can't tell what the probabilities are. This understanding of the problem also makes calculating the probability of death easier, if you want to calculate it. For one apple = death, you only survive if the small group has two rare apples. 2/5 * 1/4 = 1/10 (10% survival rate/90% survival rate) For two apples = death, you only die if the small group has two common apples. 3/5 * 2/4 = 3/10 (30% death rate/70% survival rate)
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You know the riddle: https://sites.math.washington.edu/~morrow/336_11/papers/yisong.pdf We have the exact same scenario & assumptions but with a different question to be answered: "Are you the last called prisoner out of the bunch?" In other words: "Are you the 100th prisoner?" Obviously, all prisoners called in the days 1-99 know they arent -
Event Horizon, you have shed a lot of light on this puzzle! Your observations about the 4-triangle and 1+ 3^n are quite surprising and pleasing. It is I who brute forced it; you actually came to a deeper understanding. I’ll send you kudos as well.
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Playing Chess with a Coin
TimeSpaceLightForce replied to TimeSpaceLightForce's question in New Logic/Math Puzzles
Welcome here @ Silver. Note that the fool's mate requires 4 tosses -
Playing Chess with a Coin
TimeSpaceLightForce replied to TimeSpaceLightForce's question in New Logic/Math Puzzles
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Beat me to it, kudos CaptainEd. I guessed the solution and came up with a proof while trying to get to sleep last night. The first step was inelegant, so I was going to work on it a bit before posting. This problem reminded me a lot of
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KrystalGamer246 changed their profile photo
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I once created this paradox with a notecard, then I threw it at people that I knew as I screamed "PARADOX!!"
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Sorry, typographical error in line (3) Should end: ...EG-F;FH-G;GI-H;HJ-I
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Answer and proof:
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Form a "Triangle" With 10 blocks in its top row, 9 blocks in the next row, etc., until the bottom row has one block. Each row is centered below the row above it. Color the blocks in the top row red, white, or green (or any three colors of your choosing) in any way. Use these two rules to color the remaining rows of the triangle: - If two consecutive blocks in a row have the same color, the block between them in the row below has the same color - If two consecutive blocks in a row have different colors, the block between them in the row below has the third color Tell how you can always predict the color of the bottom block after seeing only the top row (and not constructing the intermediate rows) PROVE your answer.
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the next letters are E N T E T T F F S...... should i go on?
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Playing Chess with a Coin
Silver replied to TimeSpaceLightForce's question in New Logic/Math Puzzles
I’d say you have at least a 68.75% (11/16) chance of winning against yourself. Because you can make “stupid” moves when helping yourself win, right? So black can win against white in three moves if they go like this: 1. W: Pawn to F4 (or F3) 2. B: Pawn to E5 (or E6) 3. B: Queen to H4 White needs to win the first coin toss (probability 1/2) and black needs at least two of the remaining four (probability (1/2)^5 * 11 = 11/32 ). The same goes for black winning against white, so the first toss decides which side will try to win. The probability then would be 2 * 11/32 = 11/16 = 68.75%. At least that much, because I haven’t explored all possibilities. -
Ur nan
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Hey doods merry Xmas!!!
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Hi. I have a question of nomenclature. Is there a name given for a particular set of 4 cells from an otherwise empty 4x4 grid, having the property that they include ONE cell in each row, ONE cell in each column, and ONE cell in each of the two main diagonals? An example: ╔════╤════╤════╤════╗ ║ ▓▓ │ │ │ ║ ╟────┼────┼────┼────╢ ║ │ │ ▓▓ │ ║ ╟────┼────┼────┼────╢ ║ │ │ │ ▓▓ ║ ╟────┼────┼────┼────╢ ║ │ ▓▓ │ │ ║ ╚════╧════╧════╧════╝ I hope the above figure shows up well in your display. You may wish to copy all this and paste it into a simple text editor and using the UTF-8 text encoding, as well as a fixed-pitch font. Anyway, I've been referring to such sets of 4 as challenger configurations, which I named after a daily puzzle I used to do, carried in some newspapers, called the Challenger. They would show a mostly empty 4x4 grid with totals displayed for each row, column, and both diagonals, as well as 4 given numbers. The challenge was to fill in the remaining 4x4 grid cells each with a choice of number from 1 to 9 so as to make the totals correct. There was also a challenge time in which to complete the task. Now I figure that about 99% of the time the POSITIONS of the grid for the given numbers would, like the configuration shown above, include one of the given numbers by itself in a corner, one of the given numbers by itself on an outer edge, and the other two given numbers diagonally adjacent to each other, both of their positions on the grid being a chess knight's jump from the lone edge, and one of them also being a knight's jump from the lone corner. For there to be ONE given number in each row, and in each column, and in both of the two main diagonals, the POSITIONS would HAVE to be as I've described. Anyway, shall I continue to refer to it as a challenger configuration, or have mathematicians already chosen a name for one of these configurations? -
Playing Chess with a Coin
rocdocmac replied to TimeSpaceLightForce's question in New Logic/Math Puzzles
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the armor/bullet thing is about sticks n stones. words only scar if they weigh within truth. but since our society is greedy and violent... truck bed liner actually was proven to be practically blast proof if applied to a building. at least within their scale modeling. mjolnir was the hammer which could only be wielded if the sentient alloy it was comprised of deemed them worthy to have its accompaniment. this hammer. Excalibur was a similar type weapon from mythos forged by dwarves at the request of Loki, whom also had an enchantment stating his neck was ne'er to be harmed in any way as part of the price of the deal to build the commissioned pieces. that technically isn't a rock. but the weird green meteor in Kansas they found sure did lots of weird stuff...
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Will it be more interesting to play none turn-base chess with balanced skill and luck? With all regular rules applied, starting with initial position, the first move depends on head or tail result of the coin toss. Next move likewise and so on until white/black wins or draw. While in case a player's King is in check, it is his turn to make a move unless it is a checkmate. Suppose I want to help win against myself in five or less coin spins, what is the best probability that I can make a checkmate?
