I am tutoring two people: one's in high school, the other in a polytechnic school. Your style promotes the kind of discovery-based learning approach I really want to instill in my students. I'll be sure to link your videos when applicable!
I am tutoring two people: one's in high school, the other in a polytechnic school. My student in polytechnic school is much better. On average, they score twice as high as my other student, so, to make it fair, I have decided that the polytechnic school student has to win by two points, while the high school student only has to win by one point. That's it, those are the rules, and there's no limit on how long my students' education goes. In theory they could just keep alternating until one of them collapses from the stressful academic environment, but that's probably pretty unlikely. So, who gets a college scholorship? Who has the advantage?
I did give this a try but found problem the rules didn't clarify: the fact that alice has to win *twice*, I originally thought that the points didn't reset after a tie, I came with an answer to that but I'm not good with probabilities, here it is: bob has 2 ways of winning, either by scoring the first point which is 1/3 OR after alice scores her first point (the score is 1-0 alice to bob with alice needing another point), the probability of this would be the regular 1/3 but multiplied by the 2/3 which is the alice probability of scoring a point (which gives this current situation) so in the and the two probabilities end up being 1/3 + 1/3 * 2/3 which is 5/9 to bob, and 4/9 to alice. Either way, great video
He never said that Alice has to score two points he said she needed to win BY two points. i.e. number of As - number of Bs has to equal 2 in order for Alice to win.
@@abdulrahimss english is not my native language so I guess this more adequate interpretation flew over me, but still, the thumbnail still says "needs 2 points to win" and probably I followed the thumbnail blindly, still, thanks for the correction
I think the more clean way to solve this is the following way: The only difference between the two is that Alice always ends with AA and Bob with B. Now, the chances of these two events are always the same likelihood multiplied by P(AA) or P(B) respectively. Now: P("Alice wins")+P("Bob wins")=1. Let us shorten: y=1+x+x^2+x^3+x^4+... with x=P(AB). Since 0
I'll write my answer and then edit after watching video: it's actually 0.55(5) (~56%) that Bob wins. Ok, let's go. Edit 1: I actually listened to the problem, and it's different to what I thought by thumbnail. I now think Bob has ~43% chances. BRB xD Edit 2: Hey, I was right, it's 4/7 vs 3/7 :D
It would be more interesting to calculate how we should modify the winning rule to make the game fair instead of looking at what scoring rates would make your fixed 2-1 rule to be fair. For example Alice wins if she scores A times and Bob wins if he scores B times. What values must A and B have for the game to be fair?
That was my first thought for generalization after I solved the initial problem (w/ statespace and whatnot), too, but after following the process of the OP, it occur to me, that it might also tied to the golden ratio, ie requireing F_n and F_{n-1} points from each player will bring the chances closer and closer to equal, but it semms to a lot of calculation now and I'm too lazy. Also, bojler eladó ;)
If X and Y are the number of points that Alice and Bob must win by (eg. 8-5 would be a win by 3). Then there is no such pair (X,Y) which results in a completely fair game. In fact, as soon as Y=1 (Bob must be up by 1 point for him to win), there exists no X such that Alice does not have a slight advantage. If we let X go to infinity, then the probability of Bob winning does approach 0.5 so the game is fair, but it would take an infinite amount of time for Alice to win. Also note that as long as X is finite, while the game can run on forever, it has a probability 0 of doing so.
@@Triple_Trouble739 interesting. Can you show me why? I'm stuck in my mindset and incapable to figure out a scalable solution to derive a general formula needed to the limit.
Aye ima get the first comment and first like. Love this channel bro this shit is fire 🔥 Your content is already on par with 3b1b and number file my boi.
What I think is an easier method than infinite series and is more applicable to larger versions: let P(i) be the probability of alice winning given they are at "state" i, where i is the imbalance of points (-1 in favor of bob to 2 in favor of alice) We start tied, so our goal is P(0) we can do this because of its Markov property (fancy words for "it doesnt matter how you got to a state, just that youre there now") Obviously: P(-1) = 0 since alice cannot win after bob has already P(2) = 1 since alice has won so what is in between is i = 0, 1 (other values dont really make sense, like alice having 7 points) P(i) = (what happens if alice wins) * (probability alice wins) + (what happens if bob wins) * (probability if bob wins) [total law of probability] if alice wins, we move up a state, and equally if bob wins, we move down a state P(i) = P(i+1) * 2/3 + P(i-1) * 1/3 plug in i = 0,1: P(0) = P(1) * 2/3 + P(-1) * 1/3 = P(1) * 2/3 + 0 = P(1) * 2/3 P(1) = P(2) * 2/3 + P(0) * 1/3 = 2/3 + P(0) * 1/3 This is just a system of equations where: P(0) = 4/7 (and P(1) = 6/7 if you were interested) I'll skip doing a generic solve with a probability (in the video, x) to win an individual round. This is actually a famous math problem: "the gambler's ruin problem"
# Easy peezy lemon squeezy import random a=0 b=0 count=0 while 1: i,j=0,0 while i-j!=2 and j-i!=1: if random.randint(0,2): i+=1 else: j+=1 if i-j==2: a+=1 else: b+=1 count+=1 print(str(100*(a-b)/count)[:5]) # answer is alice would win ~14.31% more.
4/9 Alice wins
3/9 Bob wins
2/9 a reset happens
Overall I think alice has the Advantage.
I am tutoring two people: one's in high school, the other in a polytechnic school. Your style promotes the kind of discovery-based learning approach I really want to instill in my students. I'll be sure to link your videos when applicable!
I am tutoring two people: one's in high school, the other in a polytechnic school. My student in polytechnic school is much better. On average, they score twice as high as my other student, so, to make it fair, I have decided that the polytechnic school student has to win by two points, while the high school student only has to win by one point. That's it, those are the rules, and there's no limit on how long my students' education goes. In theory they could just keep alternating until one of them collapses from the stressful academic environment, but that's probably pretty unlikely. So, who gets a college scholorship? Who has the advantage?
As an applied math graduate, I just ran a Markov process on this thing and found a stable distribution using matrix multiplication 😅
As a pure maths graduate, I just drew a probability tree and don't mess about with matrices unless I have no other choice.
With no majors in hand, I just counted one by one.
I did give this a try but found problem the rules didn't clarify: the fact that alice has to win *twice*, I originally thought that the points didn't reset after a tie, I came with an answer to that but I'm not good with probabilities, here it is: bob has 2 ways of winning, either by scoring the first point which is 1/3 OR after alice scores her first point (the score is 1-0 alice to bob with alice needing another point), the probability of this would be the regular 1/3 but multiplied by the 2/3 which is the alice probability of scoring a point (which gives this current situation) so in the and the two probabilities end up being 1/3 + 1/3 * 2/3 which is 5/9 to bob, and 4/9 to alice.
Either way, great video
He never said that Alice has to score two points he said she needed to win BY two points. i.e. number of As - number of Bs has to equal 2 in order for Alice to win.
@@abdulrahimss english is not my native language so I guess this more adequate interpretation flew over me, but still, the thumbnail still says "needs 2 points to win" and probably I followed the thumbnail blindly, still, thanks for the correction
@@EtevoldoEnglish is my native language, and I thought the same thing.
@@abdulrahimss The problem is that the image before opening the video says it the other way around
I think the more clean way to solve this is the following way: The only difference between the two is that Alice always ends with AA and Bob with B. Now, the chances of these two events are always the same likelihood multiplied by P(AA) or P(B) respectively. Now: P("Alice wins")+P("Bob wins")=1. Let us shorten: y=1+x+x^2+x^3+x^4+... with x=P(AB). Since 0
P(B wins game) = P(B wins 1st pt) + P(A wins 1st pt) * P(B wins 2nd pt) * P(B wins game)
P(B wins game) = 1/3 + 2/3 * 1/3 * P(B wins game)
7/9 * P(B wins game) = 1/3
P(B wins game) = 3/7
as soon as I saw sqrt5, I knew it was the golden ratio.
wtf, thumbnail didn't say that 1-1 is a draw
I thought it meant "if Bob scores at any moment, he wins"
maybe change it to "needs to be X points ahead"?
"needs X points advantage"
Golden ratio, called it!
I'll write my answer and then edit after watching video: it's actually 0.55(5) (~56%) that Bob wins. Ok, let's go.
Edit 1: I actually listened to the problem, and it's different to what I thought by thumbnail. I now think Bob has ~43% chances. BRB xD
Edit 2: Hey, I was right, it's 4/7 vs 3/7 :D
alice, bob, and the third party
Great video! Love the content and your video style, keep it up dude!
It would be more interesting to calculate how we should modify the winning rule to make the game fair instead of looking at what scoring rates would make your fixed 2-1 rule to be fair. For example Alice wins if she scores A times and Bob wins if he scores B times. What values must A and B have for the game to be fair?
That was my first thought for generalization after I solved the initial problem (w/ statespace and whatnot), too, but after following the process of the OP, it occur to me, that it might also tied to the golden ratio, ie requireing F_n and F_{n-1} points from each player will bring the chances closer and closer to equal, but it semms to a lot of calculation now and I'm too lazy.
Also, bojler eladó ;)
If X and Y are the number of points that Alice and Bob must win by (eg. 8-5 would be a win by 3). Then there is no such pair (X,Y) which results in a completely fair game. In fact, as soon as Y=1 (Bob must be up by 1 point for him to win), there exists no X such that Alice does not have a slight advantage. If we let X go to infinity, then the probability of Bob winning does approach 0.5 so the game is fair, but it would take an infinite amount of time for Alice to win. Also note that as long as X is finite, while the game can run on forever, it has a probability 0 of doing so.
@@Triple_Trouble739 interesting. Can you show me why? I'm stuck in my mindset and incapable to figure out a scalable solution to derive a general formula needed to the limit.
Nice videos. You will be popular soon
14/27 vs 13/27 it looks like to me.
Nvm, he changed the rules without realizing it. He originally said bob to 1 is a win or alice to 2 is a win.
Aye ima get the first comment and first like. Love this channel bro this shit is fire 🔥 Your content is already on par with 3b1b and number file my boi.
The Thumbnail and the video describe different situations.
What I think is an easier method than infinite series and is more applicable to larger versions:
let P(i) be the probability of alice winning given they are at "state" i, where i is the imbalance of points (-1 in favor of bob to 2 in favor of alice)
We start tied, so our goal is P(0)
we can do this because of its Markov property (fancy words for "it doesnt matter how you got to a state, just that youre there now")
Obviously:
P(-1) = 0 since alice cannot win after bob has already
P(2) = 1 since alice has won
so what is in between is i = 0, 1 (other values dont really make sense, like alice having 7 points)
P(i) = (what happens if alice wins) * (probability alice wins) + (what happens if bob wins) * (probability if bob wins) [total law of probability]
if alice wins, we move up a state, and equally if bob wins, we move down a state
P(i) = P(i+1) * 2/3 + P(i-1) * 1/3
plug in i = 0,1:
P(0) = P(1) * 2/3 + P(-1) * 1/3 = P(1) * 2/3 + 0 = P(1) * 2/3
P(1) = P(2) * 2/3 + P(0) * 1/3 = 2/3 + P(0) * 1/3
This is just a system of equations where:
P(0) = 4/7 (and P(1) = 6/7 if you were interested)
I'll skip doing a generic solve with a probability (in the video, x) to win an individual round.
This is actually a famous math problem: "the gambler's ruin problem"
Does P(0) mean nobody wins?
@@vikraal6974 am guessing P(0) means equal, aka their scores are ‘tied’
Alice and Bob? clickbait, this should be a crypto video lol
6:55
So what is the name of this thing
I want to learn more about this...
It's called a Geometric series
@@YOM2_UB Thanks
you should really change the thumbnail, but nice problem
# Easy peezy lemon squeezy
import random
a=0
b=0
count=0
while 1:
i,j=0,0
while i-j!=2 and j-i!=1:
if random.randint(0,2):
i+=1
else:
j+=1
if i-j==2:
a+=1
else:
b+=1
count+=1
print(str(100*(a-b)/count)[:5])
# answer is alice would win ~14.31% more.