Lotto Probability

[BillyJoe]

There are two distinct scenarios presented. One where there are 3 choice options followed by one where there are 2 choice options. One of the 2 choice options was made from the odds of the previous 3 choice option.

In the second scenario, we have a carry over from the first scenario, namely that the contestant has already chosen one of those doors. This changes things for the second scenario.
So, perhaps it is better to call it "one evolving scenario" rather than "two distinct scenarios".

 

[BillyJoe]

You are playing with your evil twin brother.

You decide to never change doors.
(Are we agreed that means you will win a car in 1/3 of the cases?)

Your evil twin brother decides to always start out with the door you pick, but he will always change.

So in the 2/3 of cases that you lose, he will win and vice versa. So if you agree that not changing means you win the car in 1/3 of all cases, changing has got to let you win in all of the remaining cases.

That is excellent, Rasmus, I must remember that one.
What do you think, ynot?

 

[ynot]

In the second scenario, we have a carry over from the first scenario, namely that the contestant has already chosen one of those doors. This changes things for the second scenario.
So, perhaps it is better to call it "one evolving scenario" rather than "two distinct scenarios".

Yes - "one evolving scenario" is better. Not only is the first choice carried over from the first part of the scenario to the second part, but one of the alternate choices has also been removed.

 

[BillyJoe]

Yes - "one evolving scenario" is better. Not only is the first choice carried over from the first part of the scenario to the second part, and one of the alternate choices has also been removed.

Indeed it has.

So, what are your thoughts on posts #419 and #420?

 

[ynot]

Indeed it has.

So, what are your thoughts on posts #419 and #420?

Working on it - Wach this space.

 

[Bjorn]

<monty python>Yes, it is!</monty python>

Changing the doors means the player will win in 2/3 of all games. (Unless I am just missing something very subtle in this definition and this is clear to everyone involved. Just ignore me if that should be so...)

Allow me to confuse things further by offering a different explanation:

You are playing with your evil twin brother.

You decide to never change doors. (Are we agreed that means you will win a car in 1/3 of the cases?)

Your evil twin brother decides to always start out with the door you pick, but he will always change.

So in the 2/3 of cases that you lose, he will win and vice versa. So if you agree that not changing means you win the car in 1/3 of all cases, changing has got to let you win in all of the remaining cases.

That must be the best explanation I ever saw. I'll be the evil twin brother every time.

 

[BillyJoe]

Working on it - Watch this space.

The problem is that it is counterintuitive.

But Rasmus example shows that it must be correct.
However, we are not finished until we understand why it is correct.

Sounds like you are working on that part.

 

[ynot]

Exactly.

So, we have the bit you agreed with:
The odds that the prize is behind the door chosen by the contestant remains 1 in 3 (1/3)

Which means that....
The odds that the prize is behind the other door must be 2 in 3 (2/3)

The odds must add up to 3 in 3 (3/3)
So, if the odds of the chosen door remains 1 in 3 (1/3), the odds of the other door must be 2 in 3 (2/3).


Does this make sense?

BJ

I guess I’m arguing against the wording used, not the math. I don’t believe it is valid to keep calling the odds 1 in 3 when they are applied to a to a 2 choice option. The odds would be more accurately expressed as 1/3 of 2 to 2 when the first choice is applied to the second choice scenario, and 2/3 of 2 to 2 for the alternate choice. This means that choosing the alternative option doubles your chances of being right.

Does this make sense?

 

[Rasmus]

I guess I’m arguing against the wording used, not the math. I don’t believe it is valid to keep calling the odds 1 in 3 when they are applied to a to a 2 choice option.
[...]
Does this make sense?

Little.

x in n just says you will win x games for every n games plaid. It doesn't necessarily have to do with the number of choices you can make.

I agree that it might not be very intuitive - but what you said (and which I snipped before i realized I would comment on it and i am not going back, so there!) isn't intuitive at all. There isn't a point where I have anything like 1 in 2 chance.

 

[Rasmus]

The problem is that it is counterintuitive.

But Rasmus example shows that it must be correct.
However, we are not finished until we understand why it is correct.

Sounds like you are working on that part.

Yes, it doesn't help you to understand it at all. But once you actually see that it does work - if you don't have an evil twin brother, I would recommend to just try it - you can make progress towards understanding it.

As far as I remember, I kept falling back into the "but it can't be true" line of thought before I first got it ....

 

[BillyJoe]

I guess I’m arguing against the wording used, not the math. I don’t believe it is valid to keep calling the odds 1 in 3 when they are applied to a to a 2 choice option.

I see. Now I understand what you meant when you said.....

The chance the prize is now behind the remaining door is not 2 in 3. The third choice has been revealed, removed, exposed, negated. There are now only 2 choices remaining.

It doesn't matter though, the odds are still 2 in 3 (2/3)

The odds would be more accurately expressed as 1/3 of 2 to 2 when the first choice is applied to the second choice scenario, and 2/3 of 2 to 2 for the alternate choice. Does that make sense?

Not really. I don't understand what you mean by 2 to 2.

This means that choosing the alternative option doubles your chances of being right.

At least we get the same answer.
...but, I think your wording is.....idiosyncratic? (in a friendly way)

regards,
BillyJoe

 

[ynot]

Little.

x in n just says you will win x games for every n games plaid. It doesn't necessarily have to do with the number of choices you can make.

I agree that it might not be very intuitive - but what you said (and which I snipped before i realized I would comment on it and i am not going back, so there!) isn't intuitive at all. There isn't a point where I have anything like 1 in 2 chance.

I have never said that there is every a chance of having 1 in 2 odds.

What I have said is .
. .
In the first choice scenario (3 choice options) the odds of being right are 1 in 3.

In the second choice scenario (2 choice options) the odds can never be 1 in 2.

In the second choice scenario the odds of first choice remain unchanged, but they should not be expressed as being 1 in 3 as they are being applied to a 2 choice option.

Given that 1 in 3 is essentially the same as 1/3, it would be more accurate (and less confusing) to express the odds of the first choice being right (when applied to the second scenario) as being 1/3 of 2 to 2.

 

[ynot]

I see. Now I understand what you meant when you said.....

It doesn't matter though, the odds are still 2 in 3 (2/3)

Not really. I don't understand what you mean by 2 to 2.

At least we get the same answer.
...but, I think your wording is.....idiosyncratic? (in a friendly way)

regards,
BillyJoe

Ok . . . so I’m not a mafmatishun.

Hmmmmm . . . TC = idiosyncratic - people must be warned. Ynot = idiosyncratic - people must be warned?

 

[BillyJoe]

Consider this scenario:

We will label the three doors, 1, 2, and 3.
There are two contestants, A and B.
Contestant A chooses door 1.
The odds that the prize is behind door 1 is 1/3.
Contestant B chooses door 2.
The odds that the prize is behind door 2 is 1/3.
The host opens door 3 and reveals it to be empty.
The host allows contestant A and B to switch.
They both switch to increase their odds to 2/3.



BillyJoe

 

[BillyJoe]

Ok . . . so I’m not a mafmatishun.

I thought your maths was okay.
But, about your terminology......what does 2 to 2 mean?

Hmmmmm . . . TC = idiosyncratic - people must be warned. Ynot = idiosyncratic - people must be warned?

Using different terminology can make communication difficult (though not impossible as we have demonstrated ).

I thought JT (I think you meant JT) was thinking along the same lines as myself but using different terminology. But he wasn't. he was using different terminology and thinking along different lines!

I thought you were thinking along different lines to myself but using different terminology. But you weren't. You were using different terminology but thinking along the same lines.

 

[ynot]

I thought your maths was okay.
But, about your terminology......what does 2 to 2 mean?

Means I'm very tired (1:48?). I just meant 1/3 of 2 and 2/3 of 2.

Using different terminology can make communication difficult (though not impossible as we have demonstrated ).

I thought JT (I think you meant JT) was thinking along the same lines as myself but using different terminology. But he wasn't. he was using different terminology and thinking along different lines!

Yup - I meant JT

I thought you were thinking along different lines to myself but using different terminology. But you weren't. You were using different terminology but thinking along the same lines.

“Thanks” for post #434 - now I won’t get any sleep!

 

[Rasmus]

I have never said that there is every a chance of having 1 in 2 odds.

I didn't mean to say that. But your 2 to 2 would imply that to me if I wasn't aware of the context. (If it would imply anything to me at all, which i am not too sure about.)

I meant to say that whilst saying 1 in 3 when you have two options might be somewhat counter-intuitive, saying 2 to 2 or any such thing when describing the probability wouldn't be helpful at all.

And I think it is perfectly okay to say 1 in 3, regardless of the number of doors. If you had six doors, with two prizes total, and were allowed one pick, your chances of winning could be described as 1 in 3. I think it carries more and better information than saying 2 in 6.

Telling you the odds should happen independent of the game, so that they can e.g. be compared to the odds in other games easily. So bringing it down the most easy number seems right to me. Doing that also tells you how many games you can expect to play before you win (yes, I know the math is off here ...)

 

[Just thinking]

Is everyone on the same page with this now?

If not, I have a game to play with you. I have 3 identical shoe boxes labeled 1, 2 and 3; two boxes will be empty and one will have a red scarf within. I offer you to choose one box -- then I will show you that one of the non-chosen boxes is empty. You will now be given the option of switching your box with the other unpened one, but be warned. In the end, the person with the red scarf must pay the other person $100. Anyone that believes switching is a 50/50 deal can play with me as often as they like, but must always switch.

 

[69dodge]

Consider this scenario:

We will label the three doors, 1, 2, and 3.
There are two contestants, A and B.
Contestant A chooses door 1.
The odds that the prize is behind door 1 is 1/3.
Contestant B chooses door 2.
The odds that the prize is behind door 2 is 1/3.
The host opens door 3 and reveals it to be empty.
The host allows contestant A and B to switch.
They both switch to increase their odds to 2/3.



BillyJoe

troublemaker...

If there is only a single contestant, the host can always open an empty door. If there are two contestants and they pick both empty doors, the host will be stuck. In your case, by discovering that he's not in fact stuck, the contestants get new information that they didn't have before, which changes the probability. Initially, the probability was 1/3 that each picked the car; now, it's 1/2.

 

[BillyJoe]

spoilsport

I suppose another way of explaining it is as follows:

In 1 out of every 3 games, the prize will be behind the unchosen door and, when the host opens that door, the game is over and both contestants lose. In the remaining 2 out of every 3 games, there are equal odds (1 in 2) of the prize being behind either chosen door. So they will both win 50% of the time regardless of whether or not they switch.