SIG QT Intern Interview Questions
What is the expected value to you of the following game: You and another person have 3 coins each. You both flip all of your coins and if your three coins show the same number of tails as the other person, you pay them $2, otherwise they pay you $1.
$(1/16) about 6 cents The above answer is correct, $1/16. 5/16 probability of the same number of tails, 11/16 that they're different. Therefore, 11/16 x $1 - 5/16 x $2 = $1/16
If we play a game where I start with 2 dollars and you start with one dollar, and I have a probability of 1/3 of winning a dollar from you and you have a 1/3 probability of winning a dollar from me, what is the probability that I win all of your money?
2/3
Four kids in a room of different height. What are the chances they will be in equal order?
2/4! = 1/12
We are going to bet on a football game for $10. Halfway through the game I can choose to raise the bet to $20. After I raise the bet to $20 you have a crystal ball that tells your chances of winning the game, what is the minimum probability of winning in order for you to take the bet. (note: if you decline the raise you forteit your original $10. IF you accept, $20 is on the line.
0.25
You have 17 coins and I have 16 coins, we flip all coins at the same time. If you have more heads then you win, if we have the same number of heads or if you have less then I win. What's your probability of winning?
0.5
if missile A has a 30% chance of hitting and B a 50% what is the chance of only one hitting.
0.5 0.3*0.5+0.5*0.7=0.5
Q: What's the probability of flipping 50 heads in a row on a fair coin?
0.5^50
You are playing a game and the probability of winning is 0.6. Would you rather play best of 3, or play until someone is ahead by 2 points?
0.6^2+0.6^2*0.4*2=0.648 vs 0.6^2/(0.6^2+0.4^2)=0.692 so Ahead by 2
If you have 4 people who need to cross the bridge at night. They take 1, 2, 5, 10 minutes respectively and 2 can cross at one time. The time it takes 2 people to cross it is the slower of the 2 people. Suppose there is only 1 flashlight which needs be used anytime someone crosses the bridge, how long will it take to cross the bridge.
1 min and 2 min cross. 1min comes back 10 min and 5 min cross. 2min comes back. 1min and 2 min cross. Total time is 17min.
A king wants to hold a party, for which he has 8 kegs of wine, one of which he knows is poisoned. If someone consumes any of the poisoned wine, he will drop dead in exactly one months time. The king has 3 servants who have volunteered their lives to test the kegs of wine. What is the shortest amount of time in which the king can identify which keg of wine is poisoned and dispose of it so he can hold the party.
1 month Label the kegs 1-8, each servant has a glass of wine which is a mixture of four of the kegs: Servant one has a glass consisting of wine from kegs 1, 2, 3 and 4. Servant two has a glass consisting of wine from kegs 2, 3, 5 and 6. Servant three has a glass consisting of wine from kegs 3, 4, 6 and 7. We then wait one month. If only servant one dies, keg 1 is poisoned. If servants one and two die, keg 2 is poisoned. If servants one, two and three die, keg 3 is poisoned. If servants one and three die, keg 4 is poisoned. If only servant two dies, keg 5 is poisoned. If servants two and three die, keg 6 is poisoned. If only servant three dies, keg 7 is poisoned. If no one dies, keg 8 is poisoned.
Find the heavier ball of the given 12 pool balls using a weight balance that can be used just 3 times.
1)4 vs 4. 2)if equal 2v2 the ignored 2)if one heavier, 2v2 the ignored 3)1v1 the heavier side
Bag with 4 marbles. Two red and two blue. Draw the balls one at a time, but before it comes out, try to predict the color. If correct, you get a dollar. If you play optimally, what is the expected value of the game?
1/2 + 2/3 + 2/3*1/2 + 1/3 * 1 + 1= 17/6
If you have a 3 sided dice, what is the probability that you will get a 6 and what is the probability that you will get a 6 three times
1/2, 1/216
We each flip three fair coins. I offer to pay you $1 if we do not get the same amount of heads, if you agree to pay me $2 if we do (get the same amount of heads). Will you agree to play this game?
1/2^6 + 9/2^6 + 9/2^6 + 1/2^6 = 20/64 the probability to get same number of heads. Expected value : 20/64 X (-2) + 44*/64 * (1) = 1/16 play the game
If you roll three fair dice, what is the probability that all three will land on the same any number.
1/36
Probability of rolling 3 die of all the same number?
1/36
Probability question: You roll three fair six sided dice, what is the probability of each die turning up with the same number.
1/36
Suppose you and I are playing a game, and it could be any game. We have bet 10 dollars which would go to the winner. At some point, I offer to double the bet to 20 dollars. If you accept, the game continues with the new bet. If you refuse. you lose the game, along with the original ten dollars. What is the minimum probability of winning the game that you would need to accept the increased bet?
1/4
We are racing, and can at any time signal to the other that we would like to double our bet. We've put down 100 to start and during the race I signal to you I want to double the bet, What is the minimum probability of winning for you to accept to continue?
1/4
The probability of three people have the birthday on the same day of a week?
1/49 Probability of three people having their brithday on the same day of the week is 1/49... This is because the first one has his bday on any given day - then there's a 1/7 chance the next two have the same day. Multiply 1/7 * 1/7 = 1/49
Suppose you have ten dies with six faces each. If you randomly throw these dies, what the probability that the sum of all the top faces is divisible by 6?
1/6
6 people sitting at a round table. What is a probability that they are in height order?
1/60
If you roll two dice whats the probability that they sum up to 5 ?
1/9
What is the expected number of rolls required for a fair die to have rolled all 6 values?
15 6/6+6/5+6/4+6/3+6/2+6/1=14.7
"You are a player in a basketball team.Your team loses with 2 points. You have the ball and the game ends in 3 seconds. You have 2 choices. Pass the ball for a 3 point shot or pass the ball for a 2 point shot. The 3 point shooter has 37% probability hitting the three (and winning the game) and the 2 point shooter has a probability of 77% probability hitting the 2 (and tie the game). In overtime there is 50-50% chance of winning/losing. What do you do and why?"
2 pointer. .77*.5 = .385 > .37
question about having 12 marble (1 weigh different) , and one balance. Ask to find the different weight one with least try.
3 tries This is really a complicate question. We should remember that we don't know the different marble is heavier or lighter. The correct answer should be: Separate 12 marbles into 3 groups, each one has 4 marbles. Mark the 12 marbles as ABCD, EFGH, IJKL. First weight ABCD and EFGH. Case one: If ABCD=EFGH, different one is in I,J,K,L. Second weight AIJK and EFGH If AIJK=EFGH, L is different, then the third time weight L and any other one, and know everything. If AIJK > EFGH, the different one is heavier and is in I,J,K. Third time weight I and J If I> J, I is the different; if I J, J is the different; if I EFGH, the different one is in A, B, C, D, E, F, G, H, and I=J=K=L Second time weight AEFI and GHJK If AEFI = GHJK, the different one is in B, C, D, and heavier. Third time weight B and C. If AEFI>GHJK, the different one is in A, E, F, G, H. Because ABCD>EFGH and AEFI>GHJK, Only A is heavier or one of G, H is lighter. Third time weight G and H. If G>H, H is different one and lighter; If GEFGH and AEFI
A submarine fires its both torpedoes. The probability of a torpedo hits the target is 3/7. if the target sunk what is the probability of both torpedoes has hit the target?
3/11
There are 2 boys and unknown number of girls in a nursery. A new baby is just born inside the room. We pick randomly a baby from the room, it turns out that the baby is a boy. What is the probability that the new baby just born is a boy?
3/5 Assume there are X girls. Let the event A be that the baby born is a boy. P(A) = 1/2 Let the event B be that the baby picked is a boy. P(A|B) = P(A&B)/P(B) by Bayes rule. P(B) = P(born is a boy)*P(picked is a boy) + P(born is a girl)*P(picked is a girl) = 0.5*(3/(3+X)+2/(3+X)) P(A&B) = P(born is a boy)*P(picked is a boy) = 0.5*(3/(3+X)) Therefore, P(A|B) = P(A&B)/P(B) = 3/5
If I have a fair, shuffled deck of cards without jokers, what are the odds that if I pull the top two cards from the deck, that they are a pair?
3/51 The first card does not matter, so it has a probability of 1 and the second card has a probability of 3/51 because there are now 3 cards of that number left and 51 cards left in the deck. Multiply 1*3/51 and you get 3/51
probability of being dealt a pair?
3/51 = 1/17
How many ways are there to place numbers on a six sided dice?
30
Bag of thirty balls, 10 black, 5 Orange, 10 green, 5 Blue, what's the probability of drawing a 2nd black in the 3rd draw, (without replacement)
30/203
Questions about dice rolling. Can roll die max 2 times and get whatever value is on the die on your last roll (be it first or second). What is expected value.
4.25 if roll 1,2,3--> roll again b/c below 3.5 expectation so expectation of first roll 50% of time that you keep is 5 b/c (4+5+6)/3 other 50% you reroll and have same 3.5 expectation overall .5x5 + .5x3.5
Hospital has three boys and an unknown number of girls. A mother has a baby. A nurse picks up a baby at random and it is a boy. What is the probability that the mother had a boy?
4/7 P(F) = .5, probability of giving birth to a female P(M) =.5, probability of giving birth to a male P(B) = ?, probability of a nurse choosing a boy x = number of babies in hospital after mother gave birth so the question wants you to find P(M|B) = conditional probability of mother giving birth to a male given that the nurse picked up a boy. so P(M|B) = P(M∩B)/P(B) If mother gave birth to a boy, then odds of picking a boy is 4/x If mother gave birth to a girl, odds 3/x Both these cases have a 50% chance of happening to total probability of picking a boy is: .5(4/x) + .5(3/x) = 3.5/x = P(B) P(M∩B) = probability that the woman gave birth to a boy AND that the nurse picked it which is .5 * 4/x = 2/x so P(M|B) = P(M∩B)/P(B) = (2/x)/(3.5/x) = 4/7
I have 10 cards face-down numbered 1 through 10. We play a game in which you choose a card and I give you the corresponding dollar amount. a) What is the fair price of this game? b) Now, after picking a card you can either take the dollar value on the card or $3.50. Also, cards worth less than 5 are now valued at $0. What is the maximum price you are now willing to pay for the game?
5.5 Total amount=3.5*4+(5+6+~+10) Expected value=total/10=5.9
Rolling two dice and adding their results, what is the probability of obtaining a 6?
5/36
Each has Two dices, probability of no match.
5/6
You have are offered a contract on a piece of land which is worth 1000000 70% of the time, 500000 20% percent of the time and 150000 10% of the time. The contract says you can pay x dollars for someone to determine the land's value from where you can decide whether or not to pay 300000 for the land. What is x? I.e. how much is this contract worth?
530,000 This is an options question. 70% of the time you will find out that it is worth 1,000,000 and you make 700,000 in profit. 20% of the time you find out that it is worth 500,000 and you make 200,000 in profit. 10% of the time you do not buy. Therefore the value of this contract is 70%*700,000+20%*200,000 = 530,000.
If a shooting star has an 80% chance of appearing in the next hour, what is the chance of it appearing in the next half hour?
55.3% 1-sqrt(0.2)
Say we have a pond with lily pads. The lily pads double every minute. After 60 minutes, the pond is completely covered. How long does it take for the pond to be 1/4 covered?
58 minutes
The chance it'll rain in the next hour is 84%. What is the chance it'll rain in the next half hour?
60% 1-√(.16)=0.60
The expectation of rolling two dices.
7
I remember being asked a question about 25 horses, and you want to find the fastest 2 or 3 or 4 (I forget) despite having no stopwatch, and only being able to race 5 horses at a time. (You wish to accomplish this in as few heats as possible.) I struggled with this problem but was able to solve it, in part because the interviewer gave me some helpful hints when I got stuck on the wrong track.
7 . chzwiz, your answer is incorrect. One of the first or second place horses could be 2nd fastest and/or 3rd fastest. I drew a grid to visualize the problem. First, run five races to establish 5 groups of internally ranked horses, and you can of course immediately eliminate 4 & 5 of each race. 1 2 3 x x 1 2 3 x x 1 2 3 x x 1 2 3 x x Then race 1st place horses, eliminate 4 & 5, and those they beat earlier. You can also eliminate the horses #3 beat, and the 3rd place horse from #2's first race. 1 ? ? X X 2 ? X X X 3 X X X X X X X X X X X X X X You know #1 is fastest. Race the remaining horses (2, 3 and ?'s), and the top two are 2nd and 3rd. After reading the above answers, this is the same as B's revised answer, but I found it easier to explain/visualize with the grid.
20% of the time a painting is real, 80% of the time it is fake. If it is fake you can sell for 10,000 if it's real it's worth 500,000 1. How much is the painting worth? 2. If you could buy the option to buy the painting for 100,000 after you found out if it was real or not, but not an obligation to buy, how much is this option
8.*10,000 + .2*500,000 = 108,000 400,000 profit if real, 400,000*0.2= 80,000
50% of population does not smoke, 20% are heavy smokers and 30% are light smokers. If heavy smokers are twice as likely to die as light smokers, and light smokers are twice as likely to die as non-smokers then what is the likelihood that if someone died they were a heavy smoker?
8/19 (4/7)*0.2/(4/7*.2 + 2/7*0.3 + 1/7*0.5) = 8/19
I have 20% chance to have cavity gene. If I do have the gene, there is 51% chance that I will have at least one cavity over 1 year. If I don't have the gene, there is 19% chance that I will have at least one cavity over 1 year. Given that I have a cavity in 6 months, what's the probability that I have at least a cavity over 1 year?
8/25 You can use the Bayes theorem to calculate the posterior probability of having the cavity gene. P(gene | cavity 1) = P( cavity | gene) * P( gene) / P(cavity) = 2/5 (rounding off all the probabilities) Now, using the posterior probability, you can calculate the revised probability of having a cavity P( cavity 2 | cavity 1) = P( cavity 2 | gene) * P( gene | cavity 1) + P( cavity 2 | no gene) * P( no gene | cavity 1) = 8/25
First was the 70% free throw shooter, if he shoots twice what are the chances he'll make at least one?
9%
You have a deck of 97 cards and I will pay you $10 if I draw 4 cards and they are in ascending order (not necessarily consecutive order) and you pay me $1 if they are not. Would you play?
EV = 10*(1/24) - 1*(23/24) = -13/24 Don't play the game ! It does not matter if it is 97 cards or 907 !
Suppose you were 80% sure of something, and someone was offering you 1:1 odds on it. What percentage of your wealth would you bet on it?
Bet 60%. It's a simple application of Kelly's formula. f = [p*(b+1)-1]/b = (0.8*2-1)/1 = 0.6.
If i toss 3 pieces on a tic-tac-toe board at random, and I pay you $9 if the pieces create tic-tac-toe, and you pay me $1 if they dont, do you want to play the game? (Expected value problem)
E(x) = 9x[8/84] - 1x[76/84] = -4/84 = -1/21. So no, you wouldn't want to play
I want to have a birthday party outside this weekend (i.e. need sunny weather). On Saturday, the chance of rain is 60 percent, sun 40 percent. On Sunday, rain is 80 percent, sun is 20 percent. From there, the interviewer can ask an array of questions: What are the chances I can have my party this weekend? If I have my party, what are the chances my party is on Saturday/Sunday? Etc.
First answer is quite correct. 1. P(party)=1-P(no parties)=1-(0.6)*(0.8)=0.52 2. P(Sa | party)= P(Sa & party)/P(party) = 0.4/0.52 = 10/13 (since, sun on Saturday guarantees party on that day, that's how we get the numerator) 3. P(Su | party) = P(Su & party)/P(party) = P(rain Sa and sun on Sun)/P(party) = (0.6)(0.2)/0.52 = .12/.52 = 3/13 OR: just use P(Su | party) + P(Sa | party) =1
Suppose you want to gamble in Vegas. In a game, you win $x if the number is prime and lose $x/2 if composite. The number is uniformly randomly generated by a machine between 1 and 10 inclusive. Will you play this game? Follow up: What if you can play n number of times and then stop. Will you play it?
I would not play. Still will not.
a family have 2 kids. if you have seen 1 girl , what is the proberbl the other kids is a boy
Outcome space is (b1,b2), (b1,g2), (g1,b2) and (g1,g2). Now - if you HAVE SEEN the girl (specific girl let's say girl 1) that removes outcomes (b1,b2)&(b1,g2) leaving us with p=1/2 of choice between outcomes (g1,b2) and (g1,g2). If you KNOW (was told) that one is a girl, but do not know which one, than only (b1,b2) is removed from outcome space and the chance is 2/3.
You and I throw a coin 3 times. What is the probability that we get the same number of heads
P = 1/64 + 9/64 + 9/64 + 1/64 = 5/16
There is a jar full of 10 coins. 9 are fair, and 1 has both faces showing heads. You draw a coin at random from the jar and flip it 5 times. If you flip heads 5 times in a row, what is the probability that you get heads on your next flip?
P(H | 5H) = P(H, 5H) / P(5H) = P(6H) / P(5H). Apply law of total probability, get 73/82 = 0.8902.
Q: Which is better, multiple lines each serving one register or one line funneling to all registers
Multiple registers, one line
I propose we play the following game. We each flip three fair coins. If the number of hears we get is equal, you win a dollar for each head. If the number of heads is different, you lose $1. Do you want to play?
No.
is 1599 a prime number?
No. /3
It costs $1 to play a game where you flip a fair coin four times and if you get four straight heads, you win $10. Do you play the game?
No. 1/16 chance of this happening so need to win $16
Two blind men are carrying 3 pairs of red rocks and 3 pairs of white socks. They hit each other in the mall and have to redistribute the socks so that they receive the same amount of each color they had before. They can't receive any outside help. How do they do it?
One man took all the socks and pulled the pairs apart. As he pulled them apart, he kept one sock for himself and gave the other to the other man so that each man ended up with the same colors of socks he started with: six red and six white. Hide
3 Horses: A,B, and C. For every dollar you wager, you get $2,4,6$ dollars respectively. Can you device a strategy that warranties money?
Optimal mix is 35:21:15 on A, B, C. This guarantees a positive return of 34, regardless of the outcome. The claim by the candidate is wrong - the return of that strategy depends on the outcome, but it is always positive, so it's a correct answer to the question asked.
E(Value of a dice roll) .. followed by what is the E(Value) if you can roll a second dice if you choose but you must keep the value on the second dice if you re-roll
P(zero hits) = 0.7 * 0.7 = 0.49 P(at least one hit) = 1 - P(zero hits) = 0.51
roll a dice, win 2* the value of the face if it is an even number, or win 1*the value of the face if it is an odd number. what is the amount of money you need to charge to play the game. What do you expect to get if you can reroll again?
Part 1: 1/6*(4+8+12+1+3+5) = 5.5 Part 2: Replay the game if we get paid less than $5.5, i.e. if we roll a 1, 2, 3, 5 (since we get paid $4 for rolling a 2). E(game) = E(game | play once)*P(play once) + E(game | play twice)*P(play twice) E(game | play once) = expectation of a game where we can only roll 4 or 6 with equal probability = 10 E(game | play twice) = expectation of game from part a => E(game) = 10 * 1/3 + 5.5 * 2/3 = 7
A game where two players each throw two dice each. If any one die matches the other, you win a dollar. If there is no match, you lose a dollar. Would you play this game or not and why?
Probability that none match is 6*5*4*3/6^4 = 5/18. Complement is 13/18
Would you rather, assuming you want to live, a) Re-spin, aim at your own head and pull the trigger. b) Do not spin, aim at your own head, and pull the trigger Second part, given I spun - what is the probability of me dying this time?
Spin 1/3 die, Not Spin 1/4, do not spin
Two people take turns counting to 50. Each person can add from 1 to 10 to the total sum. The first to count the number 50 wins. What is your strategy? Do you want to go first?
The answer above is right. Here is the logic: work it backwards, in order to pick 50 you would need him to pick a number between 40 to 49, in order to get him to pick a number between 40 and 49 you need to be able to say 39. to say 39 you need to be able to say 28, ... you need to start with 6
If you flip a coin until you decide to stop and you want to maximize the ratio of heads to total flips, what is that expected ratio?
The answer should be 3/4 because there's a 1/2 chance you will stop at 1:1, the maximum ratio. The other half of the time you will play until the ratio approaches 1:2. So (1/2)(1/1)+(1/2)(1/2)=3/4.
'If you throw a dice and can choose to take the initial score or roll again and will get $1 for 1, $2 for 2 etc, what is the fair price for that game'.
The expected value of the second roll is 3.5. Therefore, on the first roll you would only accept the values 4, 5, and 6. Expected value before the first roll is (3.5*3 + 4 + 5 + 6)/6 = (15+10.5)/6 = 25.5/6 = 4+ 1.5/6 = 4.25
How much would you pay to play the following game: I roll a dice and pay you the amount showing (so if I roll a 4, I give you $4).Same game, except this time you are given the choice to re-roll after the initial roll, making the initial result null you are paid based on the second roll. How much would you pay to play this game?
The first roll is worth $3.5. 2) Would pay 4.67 If you get to roll twice, you need some game theory. You wouldn't take roll 1 if it wasn't 4,5, or 6. So the average of what you would take is 5. You will get these numbers 1/2 of the time(4,5,6 make up half the numbers on the die) .5*5 + .5*3.5 = 2.5 + 1.75 = 4.25 The other .5 of the time you get the average roll of a die(.5*3.5) for the third roll(it will be asked sometimes), you know 4.25 is the value of 2 rolls. So for the first roll, you only keep 5 and 6, or 5.5 averaged. You get these 1/3 of the time. 1/3*5.5 + 2/3*4.25 = 1.8333 + 2.833 = 4.667
Every morning, Mike the security guard at CP high school opens all 1000 doors in the building. Let's assume the doors are numbered 1-1000. The next security guard closes all even numbered doors. The third security guard touches all doors that are multiples of 3. If a door is open, he'll close it and vice versa. The fourth guard changes the position of every fourth door (if it's open he'll close it etc.,) and the fifth guard changes the position of every fifth door and so on, until the 1000th guard changes only door 1000. HOW MANY DOORS ARE LEFT OPEN IN THE END?
There will be 31 doors left. This question is basically asking how many perfect squares there are between 1-1000.
You have five pirates, ranked from 5 to 1 in descending order. The top pirate has the right to propose how 100 gold coins should be divided among them. But the others get to vote on his plan, and if fewer than half agree with him, he gets killed. How will the coins end up being divided, assuming all the pirates are rational and want to end up alive?
This is just a classic game theory question and you have to work backwards through it starting with the base case to understand the pirates motivation. If you have two pirates, p1 and p2, and if p1 is the final voter then he will always vote negatively unless p2 gives him all $100 since he can kill him anyways and take the whole loot. This means p2 is in a compromised position and does not want the game to go down to 2 pirates and will take any value greater than 0 from any other pirate, or will vote yes if he receives at least $1. When p3 is introduced, he knows p2 will need at least $1 to vote for the plan therefore he keeps $99 and gives away the last dollar to p2. This means p3 is in a dominant position and will vote down any plan that grants him less than $99. When p4 is introduced then he needs two of the three voters to vote for his plan. Granted p1 will decline unless he receives all of it and p3 will decline unless he receives at least $99 of it then he will give p3 exactly that and p2 $1 otherwise he is killed. p4 is in a compromised position so he will accept any offer where he receives something greater than 0. When p5 is introduced he knows p4 and p2 are screwed and the maximum they can earn if it bypasses him is $1. Therefore granting them each that money will guarantee their vote leaving the remaining $98 for himself and half the votes are positive thus he is not killed and gets to keep $98. So the distribution for p5, p4, p3, p2, p1 should be 98, 1, 0, 1, 0
-If you have 8 balls with one is slightly heavier, how do you weight the least time to find that one?
Weigh 3v3. Do 1v1 from heaver side or if equal do 1v1 of ignored.
You and I are going to play a game where we alternate saying a number. The rules are that the starting person must say a number between 1 and 10, and then the next person can say any number that is between 1 and 10 higher than the previous number -- i.e., if I start and say 5, you can then say anything between 6 and 15. The goal of the game is for you to say the number 40. What is your strategy and why?
Work backwards from 40. In order to say 40, you must force the other player to say anything between 30 and 39. Therefore, you must say 29. In order to say 29, your opponent must say something between 19 and 28. To ensure they are within that range, you must say 18. In order to be able to say 18, your opponent can say anything between 8 and 17. Therefore, you want to start the game and say 7.
We are playing a game where the first person to say "40" wins, and each player must add 1-10 to the previous number. For example, I say 5; you can say any number 6-15 to which I have to react. For you to say "40" first, what is your strategy? Hint: You need to choose whether or not to go first, then your strategy going forward.
Work backwards. You want to say 29, so the opposition has to say 30-39, and you say 40. To say 29, you must say 18, and to say 18, you must say 7. So, you want to go first and say 7.
A bookie gives you certain odds for Team A (say 3:1) to win again Team B, and vice versa (say 4:1). The bookie also gives you odds for a draw (say 11:1). Would you bet? How would you bet
Yes
You have a total of 4 investments to make of 100 dollars. You can either invest in a stock which if it goes up in value, you gain an additional 100 dollars, if it goes down, you lose 50 dollars. Or you can bet on whether the stock will go up or down at the casino, where you either win 100 if you're right or lose everything if not. Can you devise a strategy where you will always make a profit?
You always want to make a profit p. You buy s stocks and you bet b on the stock going down, your payoff will be: 100s - 100b = p if the stock goes up 100b - 50s = p if the stock goes down Solve and you get s/b = 200/150 = 4/3. Buying 4 stocks and betting 3 on the stock going down will result in a profit of 100 no matter where the stock goes.
5 people all of different age are sitting around a round table, what is the probability they are sitting in age order.
there are in total 120 different ways people can sit, 10 of which they are sitting in age order(5 for clockwise and 5 counter clockwise) so 1/12
I flip a coin three times, and then another three times. If the number of heads is the same the first and second time, I win. If the number of heads is different, then I win. If I win, you pay me one dollar. If you win, I give you 3 dollars. Is this game worth playing?
this is definitely worth playing. Calculate E(v) E(V) = P(winning) * payout + p(losing) * loss = 3*20/64 + 44/64*-1 = +16/64
here were a few questions on probability- tossing a coin 5 times, what is the expected value, when the game is likely to end if the winner needs to get 'head' three times in a row.
filler
We're playing a game. Cards numbered from 1-10 are placed in a hat. Whatever card is drawn is the paid out amount. How much should I charge you to play this game (i.e. fair price)? If you can redraw by placing the first card back into the hat, what is the new price?
first game : 5.5 with redraw : 6.75
A piece of land has a 30% chance of being located over an oil reserve, in which case it is going to be worth 100M. If there is no oil, the land is worth 30M. You are offered an option to buy the land at 40M after inspecting it and ascertaining if there is oil. How much are you willing to pay for this option?
i think the answer should be 18 million. the payoff of the option is 60m or 0. 60*0.3=18
You start out with 1 dollar and your friend starts out with 2 dollars. You bet 1 dollar until one of you runs out of money. You have a 2/3 chance of winning each bet. What is your chance of winning?
p = (2/3)*(2/3) + (2/3)*(1/3)*p p = 4/7
if there are 2 coins in a bag, one biased, the other fair, and you pick a coin at random and flip it 6 times and it came up heads every time, what is the probability you picked the biased coin
p= 1*0.5/((1*0.5)+1/2^6*0.5) = 0.98
a group of people wants to determine their average salary on the condition that no individual would be able to find out anyone else's salary. How can they accomplish this?
the first person think of a random number, say K and add that number to his/her salary. Pass the result to the second person. The second person adds his/her salary in and pass to the third person. The third person do the same thing and pass the result back to the first person. The first person subtract the random number K and share the average with the others.
candy bar A produce 60%, of which 20%damage,B produce 40%, 10% damaged, see a damage bar, probability it come from b
using bayes theorem and let x be the number of energy bar produced, then, answer = (x*0.4*0.1)/(x*0.6*0.2 + x*0.4*0.1) = (0.04x)/(0.16x) = 0.25
