Call911 498 Posted April 15, 2021 Report Share Posted April 15, 2021 Maybe 30? :D Quote "The Art of war is simple enough. Find out where your enemy is. Get at him as soon as you can. Strike him as hard as you can, and keep moving on." Ulysses S. Grant Link to post Share on other sites
JillyJill 5336 Posted April 15, 2021 Author Report Share Posted April 15, 2021 Nope Quote Link to post Share on other sites
Teddy 330 Posted April 15, 2021 Report Share Posted April 15, 2021 11 minutes ago, JillyJill said: You need to x before + my calculator does this automatically :P Quote Link to post Share on other sites
Substanz 4964 Posted April 15, 2021 Report Share Posted April 15, 2021 46 ^^ Quote Link to post Share on other sites
Brasil_66 368 Posted April 16, 2021 Report Share Posted April 16, 2021 43 my final answer JillyJill likes this Quote Link to post Share on other sites
JillyJill 5336 Posted April 16, 2021 Author Report Share Posted April 16, 2021 @Brasil_66 Yes! You can make your riddle. Solution: Pair of shoes - 10 Bunny - 5 Cones - 4 But attention! In the last line there is only 1 shoe and 1 cone while bunny is wearing shoes and has 2 cones in his hands. So it will be: 5 + (5+10+4) x 2 = 43 Teddy likes this Quote Link to post Share on other sites
Brasil_66 368 Posted April 16, 2021 Report Share Posted April 16, 2021 I leave it to first next person that shows up here.. JillyJill likes this Quote Link to post Share on other sites
stelthii 606 Posted April 17, 2021 Report Share Posted April 17, 2021 Answer is 100 (if we assume that the images of green apples in equation 2 & 3 = 2 apples, instead of 1.5). Spoiler Equation 1: 3x = 90 => x = 30 (Red Apple = 30) Equation 2: x + 2y + 2y = 230 => 4y = 200 => y = 50 (Green Apple = 50) Equation 3: z + 2y + x = 210 => z + 100 + 30 = 210 => z = 80 (Yellow Apple = 80) Equation 4: y - x + z = A => 50 - 30 + 80 => A = 100 Substanz and JillyJill like this Quote Link to post Share on other sites
stelthii 606 Posted April 18, 2021 Report Share Posted April 18, 2021 A woman needs a new dress and so she enters a store. Upon trying many dresses, she likes 3 of them the most. She can't decide between the 3 and ends up leaving it to luck, so she reaches for her purse and takes out a single coin. Using that coin only, how can she pick a dress randomly while being fair (equal probability for each of the 3 choices)? Good luck! SOLVED Click on the spoiler below to reveal the correct answer. Spoiler Started with the common purpose that we need to choose a dress using toss (yes/no) for each dress : Positive way (we find a direct tree to the goal) : 3 (from) > 3 (yes) > 2 (yes) > 1(yes) or gives : 3 > 1 (means 2 no at first shot) or gives : 3 > 2 > 1 So 9 moves for a first direct approach to the goal, that we can x3 to let a chance to each entity to get its tree ^^ = 27 moves for something fair (let's close the eyes and take one dress randomly, it's easier ^^) Alternative way: Assuming that she knows binary code i would think she converts the head and number to 0 and 1. now she toss the coin 2 times. she can get 00, 01, 10, and 11. 00 - first dress 01 - second dress 10 - third dress 11 - try again thats equal chance for each - 25% JillyJill and Substanz like this Quote Link to post Share on other sites
JillyJill 5336 Posted April 19, 2021 Author Report Share Posted April 19, 2021 I think we could use a little hint. Does she actually throw this coin or does something else with it? Quote Link to post Share on other sites
stelthii 606 Posted April 19, 2021 Report Share Posted April 19, 2021 7 hours ago, JillyJill said: I think we could use a little hint. Does she actually throw this coin or does something else with it? When there's "luck", "coin" and "probability" in the same context, then yes, it's almost always assumed that there's a coin toss. So you're completely right. :) The above is not a hint though, so I'll provide one below: Spoiler As we know, a coin has only 2 sides. A single coin toss can only represent 2 outcomes (example: Heads = Dress1, Tails = Dress2). Pay attention to the word in bold. :) Quote Link to post Share on other sites
JillyJill 5336 Posted April 19, 2021 Author Report Share Posted April 19, 2021 @Substanz maybe your guess? Quote Link to post Share on other sites
Substanz 4964 Posted April 19, 2021 Report Share Posted April 19, 2021 Started with the common purpose that we need to choose a dress using toss (yes/no) for each dress : Positive way (we find a direct tree to the goal) : 3 (from) > 3 (yes) > 2 (yes) > 1(yes) or gives : 3 > 1 (means 2 no at first shot) or gives : 3 > 2 > 1 So 9 moves for a first direct approach to the goal, that we can x3 to let a chance to each entity to get its tree ^^ = 27 moves for something fair (let's close the eyes and take one dress randomly, it's easier ^^) First intention, be clement pls :D ps: *see > as a toss* stelthii and JillyJill like this Quote Link to post Share on other sites
Teddy 330 Posted April 19, 2021 Report Share Posted April 19, 2021 assuming that she knows binary code i would think she converts the head and number to 0 and 1. now she toss the coin 2 times. she can get 00, 01, 10, and 11. 00 - first dress 01 - second dress 10 - third dress 11 - try again thats equal chance for each - 25% stelthii, JillyJill and Substanz like this Quote Link to post Share on other sites
stelthii 606 Posted April 20, 2021 Report Share Posted April 20, 2021 @Substanz You're correct! I didn't expect that solution also. @Teddy Your reasoning(binary code) while not right, happens to make your execution both completely correct and optimal(lowest number of tosses while 100% fair)! You're both technically correct with both answers having their pros and cons. Apologies, but I can't choose a winner, I leave it up to you. JillyJill likes this Quote Link to post Share on other sites
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