Probability / odds question

[foxbody-87]

I was pondering this earlier - let’s assume you have an arbitrary game where you can pick numbers out of a hat from a finite range of numbers, and then a winning number is selected at random.

Let’s say there are 100 to choose from and you decide to pick 50 of them. Your chance of winning is 50%.

If you had your numbers in a block (like 1-50) vs spread out (for example odd numbers) then statistically it must be the same chance but it feels like the latter should be more reliable. Or is this just bullst?

[Jonboy_t]

Nope, still 50/50. There’s exactly the same chance of numbers 1-50 being picked out as there is numbers 51-100.

[underwhelmist]

Yep, 1-50 just as likely as any other combination.

I read somewhere that the most popular lottery pick is 1,2,3,4,5,6,7...people pick it because they think they’re being contrary, but it’s just as likely as any other combination.

[KungFuPanda]

Riding on the coat tails of this first post:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

[Russ35]

KungFuPanda said:

Riding on the coat tails of this first post:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

You switch.

[underwhelmist]

Russ35 said:

You switch.

Agree, but can’t remember the rationale.

[SlowcoachIII]

There are three doors to choose from. Two with goats behind and the other with the car.

If you choose one door, there is a 1 in 3 chance you have correctly picked the door with the car behind. There is a 2 in 3 chance it is behind the other doors you haven’t chosen.

If the host opened one of the other doors and there is a goat behind it, the car is either behind your original door or the remaining other door. However, the probabilities haven’t changed so there is still a 1 in 3 chance it is behind your original door or a 2 in 3 chance of being behind the other doors (one of which we know contains a goat). Hence, you switch to take the higher probably of it being behind the other remaining door.

[DanielSan]

underwhelmist said:

Russ35 said:

You switch.

Agree, but can’t remember the rationale.

I can't remember the exact explanation but it boils down to the odds of picking the winner with all the numbers there vs the odds being better with their being less options to pick the winner at the end. Or something like that, but statistically you're always better to switch. It doesn't always work but it works more often than not.

[LosingGrip]

The lottery is 50/50.

You either win or you don't.

[alorotom]

KungFuPanda said:

Riding on the coat tails of this first post:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

There’s a short sequence not long into the film ‘21’ that explains this very specific one perfectly

[davek_964]

SlowcoachIII said:

There are three doors to choose from. Two with goats behind and the other with the car.

If you choose one door, there is a 1 in 3 chance you have correctly picked the door with the car behind. There is a 2 in 3 chance it is behind the other doors you haven’t chosen.

If the host opened one of the other doors and there is a goat behind it, the car is either behind your original door or the remaining other door. However, the probabilities haven’t changed so there is still a 1 in 3 chance it is behind your original door or a 2 in 3 chance of being behind the other doors (one of which we know contains a goat). Hence, you switch to take the higher probably of it being behind the other remaining door.

I can see the argument, but I'm not 100% convinced. If we scale it up - and say you chose one door out of 100 - then it's far far more likely that it's in the 99 you didn't choose.
But at the point where you're down to just two doors remaining, I'm unconvinced that the previous probability is still relevant.

It is probable that the prize was behind one of the previous 98 doors you opened - but it wasn't. Now you only have two doors left, it's surely equal odds - no different than if you had started with just two doors.

[blinkythefish]

davek_964 said:

SlowcoachIII said:

There are three doors to choose from. Two with goats behind and the other with the car.

If you choose one door, there is a 1 in 3 chance you have correctly picked the door with the car behind. There is a 2 in 3 chance it is behind the other doors you haven’t chosen.

If the host opened one of the other doors and there is a goat behind it, the car is either behind your original door or the remaining other door. However, the probabilities haven’t changed so there is still a 1 in 3 chance it is behind your original door or a 2 in 3 chance of being behind the other doors (one of which we know contains a goat). Hence, you switch to take the higher probably of it being behind the other remaining door.

I can see the argument, but I'm not 100% convinced. If we scale it up - and say you chose one door out of 100 - then it's far far more likely that it's in the 99 you didn't choose.
But at the point where you're down to just two doors remaining, I'm unconvinced that the previous probability is still relevant.

It is probable that the prize was behind one of the previous 98 doors you opened - but it wasn't. Now you only have two doors left, it's surely equal odds - no different than if you had started with just two doors.

Id say it becomes more obvious with more doors. At the start you picked a random door. The host then opens 98 he knows are empty, leaving one goat and one car. It is unlikely(1%) you picked correctly the first time, meaning the prize is almost certainly(99%) behind the other door.

The key is the fact the host knows which doors to open. It makes the 99 door probability collapse onto the last door.

If the host didn't know and was randomly guessing, then at the two door position the choice would be 50:50, however the chance of getting to the two door position in this scenario would be small: 2×98!/(99!)=2%.

Edited by blinkythefish on Tuesday 9th April 07:23

[Cold]

But I've got a car already. I want a goat. Open the damn door and give me a goat.

[SCEtoAUX]

The Monty Hall problem, it's a classic.

[Antony Moxey]

underwhelmist said:

Russ35 said:

You switch.

Agree, but can’t remember the rationale.

Isn’t it because at first there’s a 1:3 chance of finding a car but once one door is opened the chance of it being behind the remaining door has increased to 1:2?

I imagine if you repeated the experiment, say, 100 times, it would give a reasonable answer as to whether it actually worked or not.

[Gompo]

SCEtoAUX said:

The Monty Hall problem, it's a classic.

I was trying to remember the name of it, the Wikipedia page is quite interesting..

[bristolracer]

underwhelmist said:

I read somewhere that the most popular lottery pick is 1,2,3,4,5,6,7...people pick it because they think they’re being contrary, but it’s just as likely as any other combination.

And they are all going to be gutted when they do win and have to share the jackpot with thousands of others.

[davek_964]

blinkythefish said:

Id say it becomes more obvious with more doors. At the start you picked a random door. The host then opens 98 he knows are empty, leaving one goat and one car. It is unlikely(1%) you picked correctly the first time, meaning the prize is almost certainly(99%) behind the other door.

The key is the fact the host knows which doors to open. It makes the 99 door probability collapse onto the last door.

If the host didn't know and was randomly guessing, then at the two door position the choice would be 50:50, however the chance of getting to the two door position in this scenario would be small: 2×98!/(99!)=2%.

Edited by blinkythefish on Tuesday 9th April 07:23

It's equally unlikely that out of the initial 100 doors, it was behind the other remaining door.

What if, I choose one door out of 100 - 98 have been opened and 2 are left.

We then go outside and get a stranger to come in - say you have a choice of two doors, one has a car behind it and one has a goat. It is absolutely not 99% more likely that if he chooses the "other" door, he's going to get a car and not a goat. He has a 50/50 chance because what happened before is not relevant to what happens with the remaining two doors.

[foxbody-87]

Ok so here’s another one. I have roughly a 1 in 45 million chance in winning the lottery. So if I play every week and find a way to live over ~865k years I’m pretty much guaranteed a win right?

[Roger Irrelevant]

davek_964 said:

blinkythefish said:

Id say it becomes more obvious with more doors. At the start you picked a random door. The host then opens 98 he knows are empty, leaving one goat and one car. It is unlikely(1%) you picked correctly the first time, meaning the prize is almost certainly(99%) behind the other door.

The key is the fact the host knows which doors to open. It makes the 99 door probability collapse onto the last door.

If the host didn't know and was randomly guessing, then at the two door position the choice would be 50:50, however the chance of getting to the two door position in this scenario would be small: 2×98!/(99!)=2%.

Edited by blinkythefish on Tuesday 9th April 07:23

It's equally unlikely that out of the initial 100 doors, it was behind the other remaining door.

What if, I choose one door out of 100 - 98 have been opened and 2 are left.

We then go outside and get a stranger to come in - say you have a choice of two doors, one has a car behind it and one has a goat. It is absolutely not 99% more likely that if he chooses the "other" door, he's going to get a car and not a goat. He has a 50/50 chance because what happened before is not relevant to what happens with the remaining two doors.

It is relevant if the host knows where the car is. Because the host knows where the car is, the choice you have when he asks you whether you want to switch after he's opened 98 other doors is essentially the same as if he'd offered you the choice of having all 99 other doors before opening the 98. He knows where the car is, so he's always going to show you 98 empty doors, the chance that you guessed correctly in the first place is still 1/100 and so the other 99/100 chance has 'collapsed' into the one remaining door. If you get a stranger to come in at this point he's still have a 1/100 chance of winning if he chooses the door you originally chose, it's just that (if he's unaware of what's gone before) he won't know that.