bubbled
-
Posts
93 -
Joined
-
Last visited
-
Days Won
2
Content Type
Profiles
Forums
Events
Gallery
Blogs
Posts posted by bubbled
-
-
There is a well known gambling hustler from a time gone by name Titanic Thompson. He was a great golfer, but lived during a time when you could make more money hustling amateurs than playing on tour.
One of his classic hustles was he'd go down to Florida during the winter and play for weeks with a group of players and play right-handed (he was a very good right-handed golfer). After beating them out of a fair amount of money, he'd offer to play them double or nothing left-handed. Of course, he was a natural lefty.
Here's the hustle that will form my puzzle. He once bet a group of businessmen that he could hit a golfball a mile. He went to the Grand Canyon and simply hit the golf ball off the edge. He told them that he felt bad and then offered to bet double of nothing that he could hit a ball over 500 yards horizontally. They took the bet. This is a time when given the hickory clubs of the time, it was hard to hit a ball over 200 yards.
How did he win?
-
I think this strategy is better
Hidden Content
Jasen, you are on the right track, but you can't move the boxes. Imagine that the boxes are wooden with a hinged lid. The boxes are nailed to the table in position. You can open the boxes and look in, but you can't touch the names inside and when you've finished looking in your 50 boxes, you have to close all the boxes. Also, for Rainman, the problem is definitely achievable. The best strategy is not on the order of 1 / 2*100
-
what you mean "they can open at most 50 boxes" ?
total boxes they can open or each prisoner can open 50 boxesIf each prisoner can open 50 boxes, I think, surviving probability is more than 50%
The percentage bonanova is looking to get above 30% is the probability that the ENTIRE group of 100 prisoners will all find their names. Without a powerful group strategy, each prisoner will be exactly 1/2 to find their name, thus the chances the entire group will succeed would be 1 / (2^100).
-
Here's one where you only flip the coin once:
The girl secretly writes one of the friend's names and H or T on a piece of paper. As long as she keeps this secret, the game is fair.
Then the friends discuss who flips. As long as they don't know what is on the paper, the game is fair either way. After deciding, one of the friends flips the coin. If their name matches, the coin has to match. If their name doesn't match, the coin needs to not match for them to win.
I admit that there are other factors besides the coin that decides their fate, but the coin is the deciding factor.
-
Based on this exacting 1000 time simulation of the best night of your high school life, you got very lucky!! BTW, this code works fine.
from random import randint
def he_gets_the_bottle():
if randint(0,9) == 1:
return Truedef it_points_to_uma():
if randint(0, 9) == 1:
return Truedef bonanova_gets_a_kiss():
if he_gets_the_bottle() and it_points_to_uma():
return Truekisses = 0
for i in range(1000):
if bonanova_gets_a_kiss():
kisses += 1
print kisses-
1
-
-
Just for giggles I ran a simulation for 1,000,000 games. Here's my output and Python code:
As expected, the results are very close to 1/8 for each spot.
{1: 125123, 2: 124812, 3: 125169, 4: 125239, 5: 125046, 6: 125019, 7: 125028, 8: 124564}
from random import randint
wins = {1 : 0, 2 : 0, 3 : 0, 4 : 0, 5 : 0, 6 : 0, 7 : 0, 8 : 0}
def play_game():
locations = [8, 7, 6, 5, 4, 3, 2, 1, 0, 8, 7, 6, 5, 4, 3, 2, 1]
where = 8
visited = [0]
direction = [-1, 1]
while len(visited) < 8:
move = direction[randint(0, 1)]
where = where + move
guest = locations[where]
if guest not in visited:
visited.append(guest)
for i in range(9):
if i not in visited:
return i
for i in range(1000000):
winner = play_game()
wins[winner] += 1
print wins -
I believe I posted a very similar problem years ago involving a roulette wheel and a ball with paint on it. It can be solved with just logic. No math is needed:
The seating position of the guests does not matter. All guests are equally likely to win.
For a guest to win, exactly two things must happen:
1. At some point the bottle must move to the location 1 to their right or 1 to their left.
2. Then the bottle must travel completely around the circle, in the opposite direction, until it has arrived at that guest's location. There is a chance the bottle could do 1, move all the way around to the other side of the guest and then reverse directions again, but by then the game is decided, as all other guests have been visited and we are just waiting for that guest to eventually get the bottle.
Here is an example of how 3 might win. 1, 2, 1, Host, 8, Host, 8, 7, 6, 7, 6, 5, 6, 5, 4, 3. All wins will look something like this, or its opposite.
It's clear that for any guest, if 1 happens, there is a fixed percentage chance that 2 will happen, and that percentage is the same no matter where that guest sits. And if you think about how the game works, the probability that 1 will happen for any given guest is 100%. At some point, all guests will have the bottle one to their right or left before they get the bottle. Therefore, all guest are equally likely to win the game.
-
A general solution:
Hidden Content
Clarification:
To be clear, H is the number of heads in the your side pile BEFORE you flip those coins.
-
A general solution:
Let S = the starting number of heads face up, where S = any number from 1 to 9, inclusive. Move S coins into a separate pile and flip them all. Let H = the actual number of the coins in the side pile you created that happened to be heads. After the flip, you will have two piles with exactly h = S - H of the coins showing heads. No matter what S is, h could be as little as 0. h can never be more than 5, and is limited by 10 - S when S > 5.
-
Should be:
By a lock-box that can hold two locks. John puts the ring in the box, locks it and sends off to Mary. When Mary gets it, she adds her lock to the box and send back to John. When John gets the box back, he takes off his lock and sends back to Mary.
-
My guess:
You up 2-love in sets, and you are in a tiebreaker and up 6-love. The set score doesn't really to matter, as once your powers are gone, you have about a 0% chance of winning a set from the start. This gives you 6 points to win the match with just one bad stroke from Fed. I'd rather this than up 5-love and 40-love, in a set. In that scenario, you only have 3 points to win directly. Once you get to deuce and your powers are gone, your win percentage will go down to about 0%.
Against a decent amateur, even with this advantage, I'd love to bet on Fed.
-
I think:
It's completely fair.
-
OK. Here's my attempt at a formal inductive proof:
I will show that for all N = r + b, where N = total number of cards, r = number of red cards and b = number of black cards, that no matter whether you stop immediately, or look at another card, expected chances of winning is P(win) = r / (r + b).
Base case: Clearly, if N = 1, P(win) = r / (r + b). If the one card left is red, you are P(win) = 1 / (1 + 0) = 1 and if it's black, P(win) = 0 / (0 + 1) = 0. I'm a little uncomfortable limiting the base case to just 1 card, as you don't really have the option to stop or continue. So, let's include 2 cards into the base case.
If there are 2 red cards, P(win_stop) = 2 / (2 + 0) = 1. If you let a card turn over, P(win) = 1 / (1 + 0) = 1. If there are 2 black cards, P(win_stop) = 0 / (0 + 2) = 0. If you let a card turn over, P(win) = 0 / (0 + 1) = 0. If there is 1 red and 1 black, then P(win_stop) = 1 / (1 + 1) = .5. If you let a card turn over, P(win) = 0 * P(a red card shows) + 1 * P(a black card shows) = (0 / (0 + 1) )* (1 / (1 + 1)) + (1 /(1 + 0)) * (1 / (1 + 1)) = .5.
Assumption: We assume that if r + b = N, then whether you stop or continue, P(win) = r / (r + b).
Inductive Step: I will try to show that if r + b = N +1, whether you stop or continue, P(win) = r / (r + b). Clearly, if you stop, P(win_stop) = r / (r + b). If you let a card turn, P(win) = P(win if black card shows) * P(a black card shows) + P(win if red card shows) * P(a red card shows) = r / (r + b - 1) * b / (r + b) + ((r - 1) / (r + b -1)) * (r / (r + b )) = (r * (r + b -1)) / (r + b) * (r + b -1) = r / (r + b).
It's been a while since I've done an inductive proof, but I think this should hold up.
-
Here's my answer:
I think there is no strategy that can do better than 50%. There may be a way to do a formal inductive proof. But, in lieu of that here's my thinking.
Let's say for some reason we have continued with the game until there are 2 cards. If there are two black cards or two red cards left, it doesn't matter what you do. If there 1 of each, you can say stop and be 50-50. Or, if you let the 51st card turn, you will either be 0% or 100% with total odds of 50%. No optimal strategy.
Now, let's see what happens if there are 3 card left. Again, if there are 3 red or 3 black, it doesn't matter what you do. If there are 2 red and 1 black, you can stop and be 66%, or you can wait another card. 66% of the time you will be 50%, and 33% of the time, you will be 100%, which totals 66%. If you are down to 1 red and 2 black, you can stop and be 33%, or wait another card and 33% of the time you will be 0%, and 66% of the time you will be 50%. Again, that all adds up to total EV of 33%. Again, no optimal strategy.
I'm sure I could build up from there to 4, 5, ..., 52 cards.
But, let's look at your decision with 52 cards left. You could stop immediately, and be exactly 50%. But, if you let a card turn over, it is 50% to be red and 50% to be black. So, you'd be 50% to be 26/51 or 50% to be 25/51. Again, that adds up to 50% total EV.
I'm totally open to the idea that there could be a strategy, but it appears to me, you never know what card will turn over next. Your odds of winning will change from card to card, but that change is random, and in the long run will result in a 50% win rate, no matter what you do.
-
It appears that there's a general solution for how many barrels/bottles (B) of wine you can test given a number of soldiers (N) and a number of testing time periods (P).
B = (P + 1)^N.
If the poison killed in a day and you had 7 days and 3 soldiers, you could test 512 barrels. Cool.
-
Ok I try again
Hidden Content
Beautifully done. I had never considered going with a base-3 approach. That now answers a question I couldn't figure out. Given N soldiers, why is the maximum number of barrels one can investigate in two time periods equal to exactly 3^N?
-
from bubbled question, I need only 4 soldier !!
Hidden Content
If you read the question closely, you will see you can't use smaller slices of time than 24 hours. A person is guaranteed to die within 24 hours, but when they die, is unknown. I assure you, you will need all 5 soldiers.
-
I think I posted this one before. But it's certainly similar to the OQ.
The king has 240 barrels of wine, one of which has been poisoned. After drinking the poisoned wine, one dies within 24 hours. The king has 5 soldiers whom he is willing to sacrifice in order to determine which barrel contains the poisoned wine. How does he achieve this in 48 hours?
-
Here's another approach:
For a given number of bottles B, minimize M such that:
B <= (M choose 0) + (M choose 1) + (M choose 2) + ... + (M choose M)
You are assigning a unique combination of soldiers (both identity and number) to each bottle. Note it and see who dies. Then have a great party!
Gruesome and easy.
-
My attempt...
Hidden Content
Well done k-man. The explanation I heard from the lecturer was very similar:
Let's say you have a room with 30 pupils in it. You then figure out all the ways you can choose 10 students. For kicks, you also see how many ways you can choose 9 students. Now imagine the professor walks into the room, you now have 31 people in the room. If you were to add the professor to all the groupings of 9 students, combined with the earlier groupings of 30 choose 10, you have all the groupings of 31 choose 10.
-
Express (M choose N) + (M choose N-1) as a single (X choose Y) in terms of M and N.
The problem is solvable using algebra, but the result can be explained using logic. I'm looking for that approach.
-
There is a conjecture for the probability p0(n) where n=2m for all n bullets to be annihilated. I do not have a proof. Now that there are two programs that are giving the answers for n=4 and n=10 that agree with p0(4) and with p0(10) obtained from the conjectured formula, I think I will give the formula. It can be tested against simulations for other values of n, and its simple form may suggest a line of attack to derive it analytically.
Hidden Content
bonanova, it certainly looks like your formula is right. I wrote some new quicker code using suggestions from CaptainEd. I then simulated 100,000 trials for each even number from 4-100 bullets. I then compare the simulation to the result predicted by your formula. The results track the formula very well:
Results for 4 bullets
Total annihilation in simulation 0.3727
Predicted annihilation rate 0.375
Results for 6 bullets
Total annihilation in simulation 0.31259
Predicted annihilation rate 0.3125
Results for 8 bullets
Total annihilation in simulation 0.27502
Predicted annihilation rate 0.2734375
Results for 10 bullets
Total annihilation in simulation 0.24536
Predicted annihilation rate 0.24609375
Results for 12 bullets
Total annihilation in simulation 0.22711
Predicted annihilation rate 0.2255859375
Results for 14 bullets
Total annihilation in simulation 0.20922
Predicted annihilation rate 0.20947265625
Results for 16 bullets
Total annihilation in simulation 0.19769
Predicted annihilation rate 0.196380615234
Results for 18 bullets
Total annihilation in simulation 0.18393
Predicted annihilation rate 0.185470581055
Results for 20 bullets
Total annihilation in simulation 0.17622
Predicted annihilation rate 0.176197052002
Results for 22 bullets
Total annihilation in simulation 0.16714
Predicted annihilation rate 0.168188095093
Results for 24 bullets
Total annihilation in simulation 0.16207
Predicted annihilation rate 0.161180257797
Results for 26 bullets
Total annihilation in simulation 0.15443
Predicted annihilation rate 0.154981017113
Results for 28 bullets
Total annihilation in simulation 0.15071
Predicted annihilation rate 0.149445980787
Results for 30 bullets
Total annihilation in simulation 0.1459
Predicted annihilation rate 0.144464448094
Results for 32 bullets
Total annihilation in simulation 0.13952
Predicted annihilation rate 0.139949934091
Results for 34 bullets
Total annihilation in simulation 0.13503
Predicted annihilation rate 0.135833759559
Results for 36 bullets
Total annihilation in simulation 0.13288
Predicted annihilation rate 0.132060599572
Results for 38 bullets
Total annihilation in simulation 0.12941
Predicted annihilation rate 0.128585320635
Results for 40 bullets
Total annihilation in simulation 0.12712
Predicted annihilation rate 0.12537068762
Results for 42 bullets
Total annihilation in simulation 0.12247
Predicted annihilation rate 0.122385671248
Results for 44 bullets
Total annihilation in simulation 0.12085
Predicted annihilation rate 0.119604178719
Results for 46 bullets
Total annihilation in simulation 0.11745
Predicted annihilation rate 0.117004087878
Results for 48 bullets
Total annihilation in simulation 0.11474
Predicted annihilation rate 0.114566502713
Results for 50 bullets
Total annihilation in simulation 0.11162
Predicted annihilation rate 0.112275172659
Results for 52 bullets
Total annihilation in simulation 0.10913
Predicted annihilation rate 0.110116034723
Results for 54 bullets
Total annihilation in simulation 0.10976
Predicted annihilation rate 0.108076848895
Results for 56 bullets
Total annihilation in simulation 0.10705
Predicted annihilation rate 0.106146905165
Results for 58 bullets
Total annihilation in simulation 0.10369
Predicted annihilation rate 0.10431678611
Results for 60 bullets
Total annihilation in simulation 0.10212
Predicted annihilation rate 0.102578173009
Results for 62 bullets
Total annihilation in simulation 0.10062
Predicted annihilation rate 0.100923686347
Results for 64 bullets
Total annihilation in simulation 0.09913
Predicted annihilation rate 0.099346753748
Results for 66 bullets
Total annihilation in simulation 0.098
Predicted annihilation rate 0.0978414999033
Results for 68 bullets
Total annihilation in simulation 0.09683
Predicted annihilation rate 0.0964026543165
Results for 70 bullets
Total annihilation in simulation 0.09548
Predicted annihilation rate 0.0950254735405
Results for 72 bullets
Total annihilation in simulation 0.0936
Predicted annihilation rate 0.0937056752969
Results for 74 bullets
Total annihilation in simulation 0.09156
Predicted annihilation rate 0.0924393823875
Results for 76 bullets
Total annihilation in simulation 0.0905
Predicted annihilation rate 0.0912230747245
Results for 78 bullets
Total annihilation in simulation 0.09127
Predicted annihilation rate 0.0900535481255
Results for 80 bullets
Total annihilation in simulation 0.08921
Predicted annihilation rate 0.0889278787739
Results for 82 bullets
Total annihilation in simulation 0.08814
Predicted annihilation rate 0.0878433924474
Results for 84 bullets
Total annihilation in simulation 0.08683
Predicted annihilation rate 0.0867976377754
Results for 86 bullets
Total annihilation in simulation 0.08456
Predicted annihilation rate 0.0857883629175
Results for 88 bullets
Total annihilation in simulation 0.08574
Predicted annihilation rate 0.0848134951571
Results for 90 bullets
Total annihilation in simulation 0.084
Predicted annihilation rate 0.0838711229887
Results for 92 bullets
Total annihilation in simulation 0.0821
Predicted annihilation rate 0.0829594803475
Results for 94 bullets
Total annihilation in simulation 0.0834
Predicted annihilation rate 0.0820769326843
Results for 96 bullets
Total annihilation in simulation 0.08145
Predicted annihilation rate 0.0812219646355
Results for 98 bullets
Total annihilation in simulation 0.08119
Predicted annihilation rate 0.080393169078
Results for 100 bullets
Total annihilation in simulation 0.07821
Predicted annihilation rate 0.0795892373872 -
I agree with bubbled about Jasen's statement:
Hidden Content
Bubbled, I agree that you have to recognize that bullets disappear, but I disagree that you have to simulate each second; merely calculate collision timestamps
Hidden Content
I agree with bubbled about Jasen's statement:
Hidden Content
Bubbled, I agree that you have to recognize that bullets disappear, but I disagree that you have to simulate each second; merely calculate collision timestamps
Hidden Content
OK, I'm trying to wrap my head around you statement of calculating collision times, without simulating the the actual firing of the bullets. How would you calculate the result of this 4-bullet sequence. b_0 = v 0.9; b_1 = v 0.1; b_2 = v 0.2; b_3 = v 0.99. Unless you account for the firing of the bullets and the timing thereof, it might appear that b_3 annihilates b_2 and b_0 and b_1 survive. But taking into account the firing phase, b_2 annihilates b_1, and later b_3 annihilates b_0.
I agree with bubbled about Jasen's statement:
Hidden Content
Bubbled, I agree that you have to recognize that bullets disappear, but I disagree that you have to simulate each second; merely calculate collision timestamps
Hidden Content
OK, I think I figured out what you meant.
1. I create n bullets all at once.
2. I store them in a list of the form [(n-1, V_n-1), (n-2, V_n-2),...,(0, V_0)]
3. For each pair in order starting from the n-1 bullet, I calculate time of collision. Example for n-1 bullet and n-2 bullet: t = (((n -1) - (n -2)) * V_n-2) / (V_n-1 - V_n-2). The two velocities can't be equal (division by zero) which is very unlikely. In the example, the ((n-1) - (n-2)) term is obviously 1, but in later iterations it might be higher.
4. If t < 0 ignore.
5. If t > 0, then actual collision time is t + n-1.
6. Find min of all actual collision times and remove those two bullets. Then repeat the process with remaining bullets until all bullets gone or there are no collisions left.
I'd be afraid to try to optimize further. CaptainEd, do you have a way of removing all bullets will eventually collide in one pass?
-
I'm going with:
1 shot
If p = probability of success with eyes open -- in the range 0 < p < 9; (1 - p) < (1 - (p/3))^3. p larger than 1 doesn't make sense.
IQ test question
in New Logic/Math Puzzles
Posted
Here is my best reasoning: