probabilityCan a man legally marry his widow's sister in the state of California?
NO since she is a 'widow', the guy's dead!
Similar riddles
See also best riddles or new riddles.
logicprobabilityHussey has been caught stealing goats, and is brought into court for justice. The judge is his ex-wife Amy Hussey, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Hussey from close range.
To make things a little better for Hussey, Amy Hussey tells him she will place two bullets into a six-chambered revolver in successive order. She will spin the chamber, close it, and take one shot.
If Hussey is still alive, she will then either take another shot, or spin the chamber again before shooting. Hussey is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower.
He steels himself as Amy Hussey loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Amy Hussey asks, 'Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?'
What should Hussey choose?
Hussey should have Amy Hussey pull the trigger again without spinning.
We know that the first chamber Amy Hussey fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Hussey has Amy Hussey pull the trigger again, the probability that a bullet will be fired is 1/4.
If Amy Hussey spins the chamber again, the probability that she shoots Hussey would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.
mysteryprobabilityImagine you're in a room that is filling with water. There are no windows or doors. How do you get out?
Stop imagining!
mathprobabilityA girl was ten on her last birthday, and will be twelve on her next birthday. How is this possible?
Today is her 11th Birthday.
logicprobabilityYour friend shows you two jars, one with 100 red marbles in it, the other with 100 blue marbles in it.
He proposes a game. He'll put the two jars behind his back and tell you to pick one of them at random. You'll then close your eyes, he'll hand you the jar you picked, and you'll pick a random marble from that jar.
You win if the marble you pick is blue, and you lose otherwise.
To give you the best shot at winning, your friend gives you the two jars before the game starts and says you can move the marbles around however you'd like, as long as all 200 marbles are in the 2 jars (that is, you can't throw any marbles away).
How should you move the marbles around to give yourself the best chance of picking a blue marble?
Put one blue marble in one jar, and put the rest of the marbles in the other jar. This will give you just about a 75% chance of picking a blue marble.
logicmathprobabilityWhat is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday?
Only 23 people need to be in the room.
Our first observation in solving this problem is the following:
(the probability that at least 2 people have the same birthday + the probability that nobody has the same birthday) = 1.0
What this means is that there is a 100% chance that EITHER everybody in the room has a different birthday, OR at least two people in the room have the same birthday (and these probabilities don't add up to more than 1.0 because they cover mutually exclusive situations).
With some simple re-arranging of the formula, we get:
the probability that at least 2 people have the same birthday = (1.0 - the probability that nobody has the same birthday)
So now if we can find the probability that nobody in the room has the same birthday, we just subtract this value from 1.0 and we'll have our answer.
The probability that nobody in the room has the same birthday is fairly straightforward to calculate. We can think of this as a "selection without replacement" problem, where each person "selects" a birthday at random, and we then have to figure out the probability that no two people select the same birthday. The first selection has a 365/365 chance of being different than the other birthdays (since none have been selected yet). The next selection has a 364/365 chance of being different than the 1 birthday that has been selected so far. The next selection has a 363/365 chance of being different than the 2 birthdays that have been selected so far.
These probabilities are multiplied together since each is conditional on the previous. So for example, the probability that nobody in a room of 3 people have the same birthday is (365/365 * 364/365 * 363/365) =~ 0.9918
More generally, if there are n people in a room, then the probability that nobody has the same birthday is (365/365 * 364/365 * ... * (365-n+2)/365 * (365-n+1)/365)
We can plug in values for n. For n=22, we get that the probability that nobody has the same birthday is 0.524, and thus the probabilty that at least two people have the same birthday is (1.0 - 0.524) = 0.476 = 47.6%.
Then for n=23, we get that the probability that nobody has the same birthday is 0.493, and thus the probabilty that at least two people have the same birthday is 1.0 - 0.493) = 0.507 = 50.7%. Thus, once we get to 23 people we have reached the 50% threshold.
logicprobabilityshortWhy are ghosts bad liars?
Because you can see right through them.
logicmysteryprobabilityA man puts on a clean shirt every night before bed. On the first nigh he puts on a blue shirt. He than sleeps for 5 hours. Every one hour more he sleeps than the night before he put on a different color shirt the next night according to this scale: blue, black, red, green, white, pink, orange, brown, purple, yellow, grey, neon green, tan, and teal. Every one hour less he sleeps than the last night he put on a different color shirt the next night going backwards on his scale. If he were to wear a blue shirt because he slept more hours than the last night he does. If it was because he slept less hours than the night before he skips it and wears a teal shirt instead. If he goes backwards on the scale and goes to blue but would not wear a blue shirt he still counts blue in his going backwards on his scale. The second night the man wears a blue shirt because he did not sleep any more or less hours than the last night. The man sleeps for six hours that night. The next night he sleeps for five hours. Night number four he sleeps for eight hours. The next night he sleeps for seven hours. The next night he sleeps so well he sleeps for 11 hours. Night number seven he stays up so late he only sleeps for four hours. The next night he is so tired he sleeps for eight hours. The next night he sleeps for eight hours again. Night number ten he sleeps for 14 hours because he is sick. Since he slept so long the last night he only sleeps for seven hours. The next night he is a little bit tired so he sleeps for eight hours. The night after that he had to do so much work he only slept five hours. The next night at work they let him out early and he slept for nine hours. The next night he slept for eight hours. And the last night the man did he slept for ten hours. The next night he put on a different color shirt according to his scale, but the next night he randomly picked a shirt. At what night will the man wear a blue shirt again?
12th Night.
logicprobabilityYour enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.
Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?
You are better off shooting again without spinning the barrel.
Given that the gun didn't fire the first time, it was pointing to one of the four empty slots. Because your enemy spun the cylinder randomly, it would have been pointing to any of these empty slots with equal probability. Three of these slots would not fire again after an additional trigger-pull, and one of them would. Thus, by not spinning the barrel, there is a 1/4 chance that pulling the trigger again would fire the gun.
Alternatively, if you spin the barrel, it will point to each of the 6 slots with equal probability. Because 2 of these 6 slots have bullets in them, there would be a 2/6 = 1/3 chance that the gun would fire after spinning the barrel.
Thus, you are better off not spinning the barrel.
logicprobabilityYou and a friend are standing in front of two houses. In each house lives a family with two children.
"The family on the left has a boy who loves history, but their other child prefers math," your friend tells you.
"The family on the right has a 7-year old boy, and they just had a new baby," he explains.
"Does either family have a girl?" you ask.
"I'm not sure," your friend says. "But pick the family that you think is more likely to have a girl. If they do have a girl, I'll give you $100."
Which family should you pick, or does it not matter?
You should pick the house on the left. Specifically, there is a 2/3 chance that the family on the left has a girl, whereas there's only a 1/2 chance that the house on the right has a girl.
This is a very counterintuitive riddle. It seems like there should always be a 1/2 chance that a given child is a girl. And in fact there is. The key word there is "given". Because we are not asking about a "given" child for the house on the left. We are asking about what could be either child. Whereas for the house on the right, we are asking about a "given" child...specifically, we're asking about the younger child.
There are 3 possibilities for the children in the first house:
Younger Older
Girl Boy
Boy Girl
Boy Boy
There is no "Girl, Girl" option because we know the house on the left has at least one boy. Since each of these 3 options is equally likely, and 2 of them have one girl, there is a 2/3 chance of there being a girl in the house on the left.
For the house on the right, because we already know the older child is a boy, there are only two possibilities:
Younger Older
Girl Boy
Boy Boy
And as we can see, there is a 1/2 chance for the house on the right having a girl.
cleanfunnyprobabilityThere was once a college that offered a class on probability applied to the real world.
The class was relatively easy, but there was a catch.
There were no homework assignments or tests, but there was a final exam that would have only one question on it.
When everyone received the test it was a blank sheet of paper with a solitary question on it: "What is risk?"
Most students were able to pass, but only one student received 100% for the class!
Even stranger was that he only wrote down one word!
What did he write?
He, brilliant student, wrote down: "This".
