game show hosts depend on the fact that most people's magical thinking make them stick with their original choice. They tend to think the offer to change is the trick, the "temptation", and they trust themselves for no valid reason. Without knowing the math an overwhelming majority will stick to their first choice. By the way, this applies to any number of "doors" or suitcases or any other type of reveal.
How does the player know the game show host would also ask them if they want to reconsider upon choosing a goat first time or only when they're in danger of choosing the car?
Or whether the game show host's behaviour is consistent.
Or whether the game show host themselves doesn't know what's behind the doors and is being coached through an earpiece?
Etc.
The player is quite likely for the first and only time on the show.
here is basically the math behind it, in the non swapping scenario you have a 33% chance of selecting a car and a 66% of selecting a goat since the revealing of the goat will not alter your decision in anyway and the chance will be dependent only on your original choice aka 33%. In the swapping scenario lets say you choose a goat, the most likely scenario since its 66%. since the game host reveals to you a goat that the chance of you getting a car in the swap is 100%. 66% * 100% = 66%
In terms of probabilities, you still have a 1/3 chance of "Not Winning", which is not the same as 1/2. The idea by swapping the initial selection is the shift from the reduced domain of loss (from 2/3 to 1/3) to a 2/3 chance of "Not Losing". (Bear in mind that the probabilities must add up to 1).
providing you ALWAYS swap. this means picking a goat first time means you'll win, and picking a car first time means you'll lose. And picking a goat first time has a 66% chance, thus you have a 66% chance of winning as long as you swap.
Both vssnfrankiej and Tpendragon want to say that it is a 1/3 vs 2/3 chance, however, you'll notice that my statistics are consistent with the experimental results. How do you explain this?
@Testytestacci It's called variance. Ever noticed how in a casino, a roulette wheel will sometimes have a string of red numbers or black numbers. It's not particularly unusual to get runs of 7-8 blacks or reds. Looking at those results in isolation, one could infer that the roulette wheel was biased. However, if you looked at the spins for the night you will find the probability to be close to 1/2. The same applies here. I won't bother going over it, as Sautoy explains it at 2:18 + others below.
@tornadoxp2 IIRC, the math instructor said that he had run the experiment with 40 cups, and he got about the same ratio. That's some kind of variance.
The "maths" instructor says that he expected to turn up twice as many cars as goats. Well, he's wrong.
You have two choices. The first choice is 2/3 likely to turn up a goat. Then one choice is taken away (information added). Now you choose between two options. Keeping your first choice will result in a car 1/2 x 1/3 = 1/6 of the time(For 20 events:1/6 x 20=3+ cars). With only two choices left, swapping gives 1-1/6 = 5/6 chance of a car.(For 20 events: 5/6 x 20=16 cars).
@strixin79 Thats not true, because if they see the problem from another perspective, it is quite a bit harder. These two dudes just explain it so well, that it looks alot easier.
game show hosts depend on the fact that most people's magical thinking make them stick with their original choice. They tend to think the offer to change is the trick, the "temptation", and they trust themselves for no valid reason. Without knowing the math an overwhelming majority will stick to their first choice. By the way, this applies to any number of "doors" or suitcases or any other type of reveal.
Grafight23 1 month ago
If this problem confounded some of the world's leading mathematicians, then i'm a genius right?
JitenChablani 1 month ago
HAHA! jokes on you! I'm gonna get 20 goats and start a goat farm and get rich! :]
saturnstar16 1 month ago 2
Obtuse, even stupid mathematician.
How does the player know the game show host would also ask them if they want to reconsider upon choosing a goat first time or only when they're in danger of choosing the car?
Or whether the game show host's behaviour is consistent.
Or whether the game show host themselves doesn't know what's behind the doors and is being coached through an earpiece?
Etc.
The player is quite likely for the first and only time on the show.
maersklandro 2 months ago
What a beautiful world we live in. Always something to challenge the head organ.
Atheeizm 2 months ago
here is basically the math behind it, in the non swapping scenario you have a 33% chance of selecting a car and a 66% of selecting a goat since the revealing of the goat will not alter your decision in anyway and the chance will be dependent only on your original choice aka 33%. In the swapping scenario lets say you choose a goat, the most likely scenario since its 66%. since the game host reveals to you a goat that the chance of you getting a car in the swap is 100%. 66% * 100% = 66%
S4disticJ0k3r 2 months ago
This has been flagged as spam show
@TopGoldVideos
In terms of probabilities, you still have a 1/3 chance of "Not Winning", which is not the same as 1/2. The idea by swapping the initial selection is the shift from the reduced domain of loss (from 2/3 to 1/3) to a 2/3 chance of "Not Losing". (Bear in mind that the probabilities must add up to 1).
swordpunisher 2 months ago
Comment removed
swordpunisher 2 months ago
This has been flagged as spam show
it actually is very simple, there are 4 possible events :
You first pick goat 1, they reveal goat 2, swapping=car, staying=goat
You first pick goat 2, they reveal goat 1, swapping=car, staying=goat
You first pick the car, they reveal goat 1, swapping=goat, staying=car
You first pick the car, they reveal goat 2, swapping=goat, staying=car
so 1/2 times when you swap you get the car and 1/2 times when you swap you get a goat.
ivanyeung177 2 months ago
after the math, yes, U can kind of double Ur winning ... but if you lose... you lose, no matter the math
ValmisFilm 3 months ago
if you choose goat first, it's goat-car
if you choose car first, it's goat-goat
change choice second time
Inigobalboa 3 months ago
Alan Davies FTW!
TheDavsto 3 months ago
isnt sheep worth more then a car now ?? even a cow! JK :D
KPGorKFC 5 months ago
What TV show is this?
ozipk 6 months ago
@ozipk I believe it's "Let's make a deal".
Romanstandrd 5 months ago
providing you ALWAYS swap. this means picking a goat first time means you'll win, and picking a car first time means you'll lose. And picking a goat first time has a 66% chance, thus you have a 66% chance of winning as long as you swap.
RichardTheThirdd 6 months ago
Both vssnfrankiej and Tpendragon want to say that it is a 1/3 vs 2/3 chance, however, you'll notice that my statistics are consistent with the experimental results. How do you explain this?
Testytestacci 6 months ago
@Testytestacci It's called variance. Ever noticed how in a casino, a roulette wheel will sometimes have a string of red numbers or black numbers. It's not particularly unusual to get runs of 7-8 blacks or reds. Looking at those results in isolation, one could infer that the roulette wheel was biased. However, if you looked at the spins for the night you will find the probability to be close to 1/2. The same applies here. I won't bother going over it, as Sautoy explains it at 2:18 + others below.
tornadoxp2 6 months ago
@tornadoxp2 IIRC, the math instructor said that he had run the experiment with 40 cups, and he got about the same ratio. That's some kind of variance.
Testytestacci 6 months ago
@Testytestacci I hope you're trolling mate. Just reread the post.
tornadoxp2 6 months ago
The reason this works is because you have a 2/3 chance of picking a goat to begin with, therefor switching gives you a 2/3 chance of picking the car.
TpendragonT 6 months ago
The "maths" instructor says that he expected to turn up twice as many cars as goats. Well, he's wrong.
You have two choices. The first choice is 2/3 likely to turn up a goat. Then one choice is taken away (information added). Now you choose between two options. Keeping your first choice will result in a car 1/2 x 1/3 = 1/6 of the time(For 20 events:1/6 x 20=3+ cars). With only two choices left, swapping gives 1-1/6 = 5/6 chance of a car.(For 20 events: 5/6 x 20=16 cars).
Testytestacci 6 months ago
@Testytestacci that's not right, it actually is very simple, there are 3 possible events :
You first pick goat 1, they reveal goat 2, swapping=car, staying=goat
You first pick goat 2, they reveal goat 1, swapping=car, staying=goat
You first pick the car, they reveal 1 of the 2 goats, swapping=goat, staying=car
so 2/3 times when you swap you get the car and 1/3 times when you swap you get a goat.
2/3*20=13.3333
vssnfrankiej 6 months ago
@Testytestacci you have a 1/3 chance of winning if you stay, 2/3 if you switch. Removing the other goat doesn't change the probability at all.
TpendragonT 6 months ago
mmmmm i didn't get any uv that
;(
fwalker1992 7 months ago
I prefer the goat.
D3m190d 8 months ago
omg I never got it but now I get it ! it's so easy ;d
Dierenfreak555 1 year ago
oh wow. i get it now
penguindiverz 1 year ago 22
ofc it doubles the chances of getting the car... people not getting this in 1sec are probably retarded... srsl wtf.
strixin79 1 year ago
@strixin79
you are an idiot
IronChefWannabe 1 year ago
@strixin79 Thats not true, because if they see the problem from another perspective, it is quite a bit harder. These two dudes just explain it so well, that it looks alot easier.
MyDefCount 10 months ago
perfect demonstration for monty hall theory*
piterXdT 1 year ago 31
(?) Alan is going by the probability of Success ... and Marcus is going by Probability of Failure.
jjpjimmy 1 year ago