Your simulation is missing 1 variable, which is the host's choice when car is at door 1. In that case the host can choose to open door 2 or 3, which is also a random variable.
Hi lad, thanks for commenting. The contestant will always choose Door 1, which can also be known as the "original contestant door". Then Door 2 is always going to be defined as one of the other doors the game show host opens to display that a car is not in there. Door 3 is then can be considered the "only remaining door". According to the IF formula in the video, if the contestant STAYS with their original decision (i.e. Door 1), then this results in them winning about 33.3% of the time. However, if they switch to Door 3, or the "only remaining door", the car is in there 66.6% of the time. I hope that helps lad.
try starting from the fact that 1 door is always eliminated from the beginning as if you couldn't possibly choose the door with goat that gets revealed. it's just 50/50 dude.
The video was Well explained. I just have one question. How did you switch to Stay and Switch at the end of the video? where the % was changing by switching or staying. I would be glad if you explain.
Thanks for asking lad. In the video, I was just pressing backspace, or ‘delete’ on my keyboard, which is how I could stay in the same cell but refresh or ‘switch’ the simulations. The percentages are not constant, because it’s a simulation and we are gathering experimental data. We are counting how many times switching and staying using the COUNTIF function, and refreshing the data will change this result or percentage. Also, the shortcut SHIFT+F9 should recalculate all active cells in an Excel worksheet. Otherwise, every time you enter new data or delete a row or column, the %’S will change (because we are simulating the problem).
This is good stuff, except you've completely skipped the step where the "host" eliminates one of the negative outcomes before the contestant is allowed to switch!
Hi lad, thanks for watching 🙏 In this video I've carried it as an assumption. Consider a scenario where Door 1 is your initial door chosen. Either Door 2 or Door 3 will be empty, no matter the situation. As a result, when the host opens an empty door, the model instantly switches to the 'other' door. That is, if Door 2 is opened, then by switching, the model will select Door 3. Overall, the model is correct but my explanation in the vid is kinda poor haha 👍
You don't have to. If you choose to switch in every single game, 100% of the time, you will switch to the opposite of your original choice. That's sort of assumed in the video - but doesn't explain why. Example : If you choose a goat first, the host eliminates the only other goat - you will switch to the car If you choose the car first, the host eliminates a goat - you will switch to the other goat Since you will pick a goat first 2 out of 3 times, you will be switching to the car 2 out of 3 times.
This isn't an accurate sim. Love the effort, but to sim the monte hall problem you need to sim the decision tree. Step 1: choose 1. Step 2: remove one false answer. Step 3: choose to stay or switch between remaining 2. Step 4: compare results of staying vs switching
Let's assume 3 cards J,Q,K , who choose the card K is the winner, let us say I choose card 1 host open the card 3 which has Q. Okay now the probability between J and K, 50/50🎉 either in card 1 or 2
1/3 of the time you pick the correct one the first round. The only way to win when staying is if you were correct in the first round. That means you only win 1/3 of the time when staying. You pick the wrong one 2/3 of the time in the first round. You always win if you switch when your original pick was wrong. Which means you win 2/3 of the times when switching. You could think of it as choosing between the 1 you picked or both of the other. He just opens one of them before you've answered.
Yeah it's 2/3 if the host know where the winning door is and always chooses a non winning door to make you choose to switch or stay. It is 1/2 if the host does not know which door is the winning door. If you model this as such that the host randomly chooses a door and only get the instances where the host randomly picks a non winning door and the contestant choose stay or switch, the probability will be 1/2. Yeah you can try it for yourself if you know Excel. Kinda weird. lol.
Lol all you did here was make a test of how often 2 out of 3 numbers come up and act like it's surprising that the answer is 2 times out of 3. You're missing steps
Your simulation is missing 1 variable, which is the host's choice when car is at door 1. In that case the host can choose to open door 2 or 3, which is also a random variable.
Hi lad, thanks for commenting. The contestant will always choose Door 1, which can also be known as the "original contestant door". Then Door 2 is always going to be defined as one of the other doors the game show host opens to display that a car is not in there. Door 3 is then can be considered the "only remaining door". According to the IF formula in the video, if the contestant STAYS with their original decision (i.e. Door 1), then this results in them winning about 33.3% of the time. However, if they switch to Door 3, or the "only remaining door", the car is in there 66.6% of the time. I hope that helps lad.
try starting from the fact that 1 door is always eliminated from the beginning as if you couldn't possibly choose the door with goat that gets revealed. it's just 50/50 dude.
The video was Well explained. I just have one question.
How did you switch to Stay and Switch at the end of the video?
where the % was changing by switching or staying.
I would be glad if you explain.
Thanks for asking lad. In the video, I was just pressing backspace, or ‘delete’ on my keyboard, which is how I could stay in the same cell but refresh or ‘switch’ the simulations. The percentages are not constant, because it’s a simulation and we are gathering experimental data. We are counting how many times switching and staying using the COUNTIF function, and refreshing the data will change this result or percentage.
Also, the shortcut SHIFT+F9 should recalculate all active cells in an Excel worksheet.
Otherwise, every time you enter new data or delete a row or column, the %’S will change (because we are simulating the problem).
Hey dear friend awesome, but once door 3 is open, then your formula for random between should be reduced to 1-2 and not 1-3
No
This is good stuff, except you've completely skipped the step where the "host" eliminates one of the negative outcomes before the contestant is allowed to switch!
Hi lad, thanks for watching 🙏 In this video I've carried it as an assumption. Consider a scenario where Door 1 is your initial door chosen. Either Door 2 or Door 3 will be empty, no matter the situation. As a result, when the host opens an empty door, the model instantly switches to the 'other' door. That is, if Door 2 is opened, then by switching, the model will select Door 3.
Overall, the model is correct but my explanation in the vid is kinda poor haha 👍
You don't have to. If you choose to switch in every single game, 100% of the time, you will switch to the opposite of your original choice. That's sort of assumed in the video - but doesn't explain why.
Example :
If you choose a goat first, the host eliminates the only other goat - you will switch to the car
If you choose the car first, the host eliminates a goat - you will switch to the other goat
Since you will pick a goat first 2 out of 3 times, you will be switching to the car 2 out of 3 times.
Good video, very well explained, greetings from Spain :)
Thanks lad :)
thank you !
This isn't an accurate sim. Love the effort, but to sim the monte hall problem you need to sim the decision tree. Step 1: choose 1. Step 2: remove one false answer. Step 3: choose to stay or switch between remaining 2. Step 4: compare results of staying vs switching
What was the intrinsic value
Comment below if you have any questions!!!
Let's assume 3 cards J,Q,K , who choose the card K is the winner, let us say I choose card 1 host open the card 3 which has Q. Okay now the probability between J and K, 50/50🎉 either in card 1 or 2
1/3 of the time you pick the correct one the first round.
The only way to win when staying is if you were correct in the first round.
That means you only win 1/3 of the time when staying.
You pick the wrong one 2/3 of the time in the first round.
You always win if you switch when your original pick was wrong.
Which means you win 2/3 of the times when switching.
You could think of it as choosing between the 1 you picked or both of the other. He just opens one of them before you've answered.
LADZ
Hi! I still don't understand it. But nice attempt. I still think the probability is 50%.
Yeah it's 2/3 if the host know where the winning door is and always chooses a non winning door to make you choose to switch or stay. It is 1/2 if the host does not know which door is the winning door. If you model this as such that the host randomly chooses a door and only get the instances where the host randomly picks a non winning door and the contestant choose stay or switch, the probability will be 1/2. Yeah you can try it for yourself if you know Excel. Kinda weird. lol.
Lol all you did here was make a test of how often 2 out of 3 numbers come up and act like it's surprising that the answer is 2 times out of 3. You're missing steps