[Misc] Probability problem

[Simster]

in a fair game, if the other 999998 box had not revealed the prize, the final two are still 50/50. the only "some reason" one has been overlooked is a rigged game, so yeah, switch.

I don't think it is to do with it being a fair game. Monty has to choose boxes that don't reveal the prize, so you absolutely know that there is a 50% chance that the one he left alone is the winning door.

 

[Berty23]

In the 100 example, the original box you selected has a 1% chance of being correct, so there is a 99% chance it is in one of the other 99 boxes.

So if you remove 98 empty boxes from the 99, there is a 99% chance it is in the remaining box, isn't it?

Edit: assuming empty boxes were removed knowing they were empty, or regardless

Yes if you have all of the information. Without the info it is a 50/50 chance. Sorry I didn’t explain that at all well.

 

[Arthritic Toe]

In actual fact it doesn't matter whether Monty knows or not. We are where we are, with one door having been revealed not to have the prize. Monty may have opened the door with the prize and the game would have been over, but he didn't, and here we are with two closed doors, one of which hides the prize.

Oh, so Monty wasn't deliberately picking an empty door to prolong the game after the ad break... he was making an arbitrary choice , so might as well have done it randomly...?

Having dredged the mathematical archives I'm no longer certain if its relevant that Monty knew where the prize is. Assume he did. My head hurts.

Edit again - It definitely IS relevant that Monty knows the location. Apologies for earlier incorrect statement.

 

[beorhthelm]

Compare the 'stick' and 'swap' strategies...
Stick
To win you MUST be right in the first place... 1in3 chance.
If you pick wrong first time and stick then you lose... 2in3 chance.


Swap
To win you must be WRONG in the first place... 2in3 chance.
If you pick right first time and swap then you lose... 1in3 chance.


Clearly the 'swap' strategy is best. It doesn't guarantee a win because you might have picked right in the first place. But chances are higher you picked wrong first time and with the removal of all the other wrong options, you can only be left with the right option... So swap.

that explaination makes some sence.

 

[wealdgull]

Switching will always give you a 50% chance. Not switching gives you a 1 in however boxes were there chance.

Not correct with the 50% bit. If there are 3 doors then you have a 1/3 chance of winning if you stick to your original choice and a 2/3 chance of winning if you switch your choice.

 

[father_and_son]

that explaination makes some sence.

The "solution" is to stop thinking about the second half of the problem as some sort of "random" event. There is no probability, only a strategy. There is NEVER a 50/50 chance anywhere in the puzzle (even though I said there was in my first reply - its been a while since I actually had to write anything vaguely academic about Game Theory).

The entire problem is 1.What is your initial choice? and 2.What is your strategy?

The first is a probability (1 in 3 that we pick the right door first time).
The second is NOT a probability,it is game theory,... it just gives the chance to change the payoff to the problem - but we talk about ALWAYS stick or ALWAYS swap - there is no chance element.

In the problem there is no second decision, no 50/50, no two-doors, etc. The only decision is the original problem (pick the right door).

When deciding on your strategy, the ONLY thing that has happened that was random is the original guess, so everything boils down to whether you can change the "payoff". What we do in choosing an "always swap" strategy is reverse the payoff ... so instead of winning if you guess right, you win when you guess wrong (but the probabilities of these two options have not changed at all - because we made that decision right at the beginning of the show).

Clearer?

 

[yxee]

The easiest way I can explain this is this

If you picked the right one, you'll swap and lose your prize. That'll happen 1/3 of the time.

If you picked the wrong one, you'll swap to the right one, since the other wrong one was removed. That'll happen 2/3 of the time.

So this strategy gives you a 2/3 chance of winning. If you don't switch, you're stuck with a 1/3 chance of winning.

 

[wealdgull]

The way I've always thought about it is:

- you pick one door. The chance of having picked the door with the prize behind it is 1/3.
- you now have two choices: either open the door you picked, or open both of the doors you didn't. It should be obvious that it makes more sense to open both of the doors you didn't choose, as that gives you 2/3 chance of finding the prize.

The fact that the order goes "you pick a door, Monty opens one of the other doors (without the prize), you open the remaining other door" rather than "you pick a door, you open two other doors" is irrelevant to the probabilities; either gives you a 2/3 chance of winning.

 

[DFL JCL]

I've not read the entire thread but the answer is B

Sent from my SM-G960F using Tapatalk

 

[beorhthelm]

Clearer?

honestly, not really. the previous explaination seemed ok how the calculation is made, as a forecast with the player strategy preset. but there is a second decision as described.

so if player A choses a door, host reveals a empty door (assuming not knowingly, they could have revealed the prize), a second player makes the call to stick to change. does this change the probability and outcome?

 

[neilbard]

Stick 1/3 chance
Swap 2/3 chance

b) is the answer

 

[father_and_son]

honestly, not really. the previous explaination seemed ok how the calculation is made, as a forecast with the player strategy preset. but there is a second decision as described.

so if player A choses a door, host reveals a empty door (assuming not knowingly, they could have revealed the prize), a second player makes the call to stick to change. does this change the probability and outcome?

No. It doesn't change any probabilities. You still have a 1in3 chance of picking the correct door at the beginning.

Think of it like this... You have seen the show before and know the format, so you are starting the finale knowing you are going to stick or knowing you are going to swap. This isn't a decision made after doors are revealed, it was a decision made before you even went on the show. The only question is which wins you the prize more often, swapping or sticking.... but the decision is made and fixed - ALWAYS stick or ALWAYS swap.

 

[yxee]

honestly, not really. the previous explaination seemed ok how the calculation is made, as a forecast with the player strategy preset. but there is a second decision as described.

so if player A choses a door, host reveals a empty door (assuming not knowingly, they could have revealed the prize), a second player makes the call to stick to change. does this change the probability and outcome?

If the host doesn't know, and just opens a door, then it creates a 3rd outcome where he picks the right one (depriving you of the prize). Whereas before, when you picked wrong and switched you were guaranteed the prize (2/3 of the time) now in this case the host takes the prize off you half the time (i.e. half of the 2/3 so 1/3 of the time) that means you're back to 1/3 probability winning with the switching strategy.

 

[Baldseagull]

honestly, not really. the previous explaination seemed ok how the calculation is made, as a forecast with the player strategy preset. but there is a second decision as described.

so if player A choses a door, host reveals a empty door (assuming not knowingly, they could have revealed the prize), a second player makes the call to stick to change. does this change the probability and outcome?

Imagine no one opens any door, they just say, do you want to stick with what is behind your one door, or do you want to take what is behind the other two doors? You know without anyone opening anything that at least one of the other 2 doors is going to have nothing behind it, but you have 2 chances v 1 chance of getting the winning door if you switch.

 

[Notters]

This always used to involve a lion. Why did it have to be americanised?!

 

[beorhthelm]

Imagine no one opens any door, they just say, do you want to stick with what is behind your one door, or do you want to take what is behind the other two doors? You know without anyone opening anything that at least one of the other 2 doors is going to have nothing behind it, but you have 2 chances v 1 chance of getting the winning door if you switch.

i'd ask my cat to decide.

 

[nickjhs]

This is a classic problem that had Maths professors wrongly having a shot at Vos Savant when she said you should switch doors. She was correct. The chance of you getting it right the first time is 1/3, monty knows that the door he chooses does not have the car which means there is a 2/3 chance the car is in the last door vs your 1/3 chane. You switch.
If you like probability and stats read The drunkards Walk by Leonard Mlodinow https://www.amazon.com.au/Drunkards-Walk-Randomness-Rules-Lives/dp/0307275175

 

[Dick Head]

 

[Dick Head]

Imagine no one opens any door, they just say, do you want to stick with what is behind your one door, or do you want to take what is behind the other two doors? You know without anyone opening anything that at least one of the other 2 doors is going to have nothing behind it, but you have 2 chances v 1 chance of getting the winning door if you switch.

I've puzzled over this periodically over the years, thinking I've understood it and often doubting that I actually have, but this is the clearest explanation I've seen so far.

 

[Seagull kimchi]

Or look at it this way.

When the empty door is gone - saying stik or swap is akin to saying pik any of the 2 doors you like - a 50/50 bet. The 1/3 bet is washed away by new info that now presents a 1/2 bet.