Note: In the video, I assumed that a and b are at most 35. If we allow a and b to be 36, we also obtain the pair (a, b) = (18, 36). This means that player B must win on their first roll, and player A has a 1/2 probability of winning on the first roll. Although this is also a fair scenario, where both players have a 1/2 probability of winning, I excluded this case so that the setting of taking consecutive rolls is meaningful.
If you flip a fair coin 99 times and it lands on tails, wouldn't there be a slightly higher chance for the next flip to be heads? or is it some gambling fallacy that my brain just made up
The odds are actually still 50/50! Each coin flip is independent so no matter what happened in previous flips, the odds of the next flip is not affected.
oh damn! well then what about the Monty Hall Problem? With the goats & cars, since odds are independent then shouldnt the second pick (between 2 doors) be a 50/50? But my mate keeps telling me its 2/3 if you switch doors? how does it work
Originally, you have a 1/3 probability of picking the car, and 2/3 probability of picking the goat. After a door with a goat is revealed to you, you know one unopened door has a car, and another has a goat. Therefore, if you switch doors, you have the following: P(originally chose a car) = P(choosing a goat after the switch) = 1/3 P(originally chose a goat) = P(choosing a car after the switch) = 2/3 Your probability of winning the car if you switch is P(choosing a car after the switch) = 2/3
Note: In the video, I assumed that a and b are at most 35. If we allow a and b to be 36, we also obtain the pair (a, b) = (18, 36). This means that player B must win on their first roll, and player A has a 1/2 probability of winning on the first roll. Although this is also a fair scenario, where both players have a 1/2 probability of winning, I excluded this case so that the setting of taking consecutive rolls is meaningful.
2 Problems?! It must be Christmas🎄!
If you flip a fair coin 99 times and it lands on tails, wouldn't there be a slightly higher chance for the next flip to be heads? or is it some gambling fallacy that my brain just made up
The odds are actually still 50/50! Each coin flip is independent so no matter what happened in previous flips, the odds of the next flip is not affected.
oh damn! well then what about the Monty Hall Problem?
With the goats & cars, since odds are independent then shouldnt the second pick (between 2 doors) be a 50/50? But my mate keeps telling me its 2/3 if you switch doors? how does it work
Originally, you have a 1/3 probability of picking the car, and 2/3 probability of picking the goat. After a door with a goat is revealed to you, you know one unopened door has a car, and another has a goat.
Therefore, if you switch doors, you have the following:
P(originally chose a car) = P(choosing a goat after the switch) = 1/3
P(originally chose a goat) = P(choosing a car after the switch) = 2/3
Your probability of winning the car if you switch is P(choosing a car after the switch) = 2/3
Dammit, I was wrong after all 😞
Well atleast I learned something!
Cheers mate! Subbed! 😄
Another mathematical classic
keep 'em comin' !