Lotto Probability

[macgyver]

It's much easier to see why it's better to switch if instead of 3 doors (where you initially pick 1) we use 20.

Thanks, that was a good way to describe it!

Basically, is it because of eliminating the other doors as potential "wins" that you've changed the game, and therefore the odds?

 

[BillyJoe]

It was referring to the Monty Hall TV gameshow where three doors are hiding potential prizes. The participant chooses a door, then another door is revealed that shows nothing behind it. The participant is then given the opportunity to stick with their first choice, or switch to the other unopened door.

It is critically important to explain the setup fully, otherwise you risk opening up the discussion to a variety of interpretations each with different solutions.




For example:

You do not state explicity whether the host decides whether to open one of the other doors only after the contestant makes his choice or whether the host always opens one of the other doors. You also do not state whether the contestant knows this.

If it is the latter, and the contestant knows this, he should switch (as JT explained).
If it is the former, and the the contestant knows this, he would be inclined to think that the host is trying to get him to switch - because he has chosen the door with he prize - and therefore he would be inclined to stay with his original choice.

If the contestant does not know whether the host decides whether to open one of the other doors only after the contestant makes his choice or whether the host always opens one of the other doors, he has no basis on which to decide on whether he should switch or not.

Yeah, a pandora's box!


Another, for example:

You do not state whether the host knows which door contains the prize.
If not and if he opens the door with the prize, the contestant loses. If he opens one of the other doors and it does not contain the prize, then what the contestant should do will depend on whether he knows whether the host does or does not know which door contains the prize.



BJ

 

[ynot]

Haven’t ever seen a simple explanation of the Monty Hall problem so here’s my attempt . . .

You are given three choices, and are told one is right and two are wrong.
Whatever choice you make therefore is twice as likely to be wrong than right. (odds = 2 to 1 against)

After you have made a choice, a wrong choice is removed from the two other choices (the ones you didn‘t choose).

You are then given the option of retaining your original choice or changing to the other remaining choice.

You are better off changing your choice - Here’s why . . .

Your original choice has odds of 2 to 1 against (removing a wrong choice from the two other choices doesn’t change this fact).

The other choice has worst odds of 50/50. (they’re actually better than 50/50 given that the odds of the original choice is less than 50/50 and there are now two choices)

Your choice therefore is . . . 2 to 1 against or better than 50/50

 

[BillyJoe]

Well, let's state this scenario correctly first....

We have a game show host, a contestant, and three doors - one of which contains a prize.
The host knows, from the outset, which door contains the prize.
The contestant chooses one door.
The host opens one of the remaining doors according to the following rule:
- if only one of them is empty he will open that door.
- if both are empty, he will randomly choose one and open it.
The host then makes an offer to the contestant that he can either:
- stick with his original choice.
- switch to the other unopened door.

The contestant knows from the outset:
- that the host knows which door contains the prize
- that he will be asked to choose one of the three doors.
- that the host will then open one of the remaining doors.
- that the door the host opens will be empty.
- that the host will then make him an offer to either stick with his original choice or switch to the other unopened door.

What, rationally, should he do to maximise his chance of winning the prize?


BJ

 

[Rasmus]

What, rationally, should he do to maximise his chance of winning the prize?

The contestant should turn to the host; posing to all nearby cameras in a way that would communicate thoughtfulness, excitement and a trace of desperation, and whisper under his breath to the host:

"I have in my wallet several pictures showing what you did with the goat behind the door that is now open and presumably the goat that is behind one of the hidden doors; unless you have more than two goats total for the show. Now, if the car is behind the door I already chose, I want you to clear your throat and blink; if, on the other hand the car is behind the door I can switch to, I want you to loudly ask me what my choice is again."

The contestant should then smile, wait a moment, and finally announce his choice to a nearby camera, with a voice of confidence but a sympathetic trace of anxiety.

 

[BillyJoe]

Your original choice has odds of 2 to 1 against (removing a wrong choice from the two other choices doesn’t change this fact).

The other choice has worst odds of 50/50. (they’re actually better than 50/50 given that the odds of the original choice is less than 50/50 and there are now two choices)

Your choice therefore is . . . 2 to 1 against or better than 50/50

This is not actually correct. Consider...

After the contestant makes his original choice:
- the chance that the prize is behind the door he chose is 1 in 3.
- the chance that the prize is not behind the door he chose (ie behind one of the other two doors) is 2 in 3.

After the host opens one other two doors and reveals it to be empty, the chance the prize is behind the remaining door is 2 in 3.

If the contestant sticks with the original choice, the chance of winning the prize remains 1 in 3.
If he switches to the remaining door, the chance of winning the prize is 2 in 3.

He therefore doubles his chance of winning by switching.


BJ

 

[BillyJoe]

? In the case with only 3 doors the odds are not so extreme, but you go from 33% to 67% by switching.

Close enough!

 

[Just thinking]

Thanks, that was a good way to describe it!

Basically, is it because of eliminating the other doors as potential "wins" that you've changed the game, and therefore the odds?

Actually, what the game host does is offer you to either A: keep your choice of door, or B: take what's behind all the other doors (be it 2 or 19, as I described). What difference does it make if the host shows you all but one of the other doors? -- you're essentially getting them all and what's behind them if you switch, including the one that remains unopened. (We're assuming of course that all the non-winning doors have nothing behind them.) The greater the number of doors, the better it is for you to switch, because at the onset your odds for winning were 1 to whatever the number of doors were against.

For the sake of both clarity and risking further lottery discussions , let's assume that someone offered you to keep your single lotto ticket or switch it for all the other possible lottery ticket combinations -- before the drawing. Would you keep your single ticket, or switch? You know only 1 ticket will win -- but do you really think you have it against all the others?

 

[ynot]

This is not actually correct. Consider...

After the contestant makes his original choice:
- the chance that the prize is behind the door he chose is 1 in 3.
- the chance that the prize is not behind the door he chose (ie behind one of the other two doors) is 2 in 3.

Semantics - Saying “2 to 1 against” is essentially the same as “1 chance in 3 of being right” or “2 chances in 3 of being wrong“.

After the host opens one other two doors and reveals it to be empty, the chance the prize is behind the remaining door is 2 in 3.

If the contestant sticks with the original choice, the chance of winning the prize remains 1 in 3.
If he switches to the remaining door, the chance of winning the prize is 2 in 3.

He therefore doubles his chance of winning by switching.


BJ

I think you meant to say “After the host opens one of the other two doors and reveals it to be empty”?

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.

 

[BillyJoe]

Semantics - Saying “2 to 1 against” is essentially the same as “1 chance in 3 of being right” or “2 chances in 3 of being wrong“.

Yes, I wasn't finding fault here, just putting it more clearly

I think you meant to say “After the host opens one of the other two doors and reveals it to be empty”?

Yes. I think quicker than I type sometimes, sorry.

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.

This isn't correct, but I'm not sure how to explain it for you. Try this....

There are three doors with only one containing a prize.
The host opens one of them to revealing it to be empty.
The contestant chooses one of the remaining doors.
The odds that the prize is behind the door chosen by the contestant is 1 in 2
Okay so far?


Now try this:

There are three doors with only one containing a prize.
The contestant chooses one of the doors.
The odds that the prize is behind the door chosen by the contestant is 1 in 3
The host opens one of the other two doors revealing it to be empty.

Question: Do you think that the odds that the prize is behind the door chosen by the contestant has changed?


....to be continued

BJ

 

[Just thinking]

I think you meant to say “After the host opens one of the other two doors and reveals it to be empty”?

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.

BJ is correct ... the chances (probability, odds, whatever) that the contestant chose the correct door at the onset are fixed. If it's 3 doors, it's 1:3 -- 8 doors, it's 1:8 -- with n doors it becomes 1:n. Now, when the host asks if the contestant wishes to switch for the other unopend door, it actually doesn't matter if he (the host) opens all but one of the other non-chosen doors or not -- he is essentially giving all of them to the contestant if the contestant switches. Why? -- because the contestant already knows that only at most one of them contains the prize. (The only other option is that the contestant was lucky enough to have picked the winning door right off. But that is against the odds with 3 or more total doors.) So when the host opens all but one of the non-chosen doors he is giving the contestant no new information probability-wise. Hence, switching is in the contestant's best interests.

NOTE: This scenario is different from the recent game show DEAL or NO DEAL where the contestant picks off cases one at a time hoping that the big prize is the case initially chosen. Here, if it gets down to 2 cases then it's 50/50 that you've got the big one -- but when the host picks all but one door for you to see you can be sure it's because he knows which one has the prize.

Comparison with the Monty Hall Problem

When only two cases remain, Deal or No Deal might appear to be a version of the Monty Hall problem. Consider a game with three cases (similar to the three doors in the Monty Hall problem). The player chooses one case. Then, the host chooses a case to open. Finally, the player is given the option to trade his or her case for the one unopened case remaining. The Monty Hall problem gives the player a 2/3 chance of winning with a switch and a 1/3 chance of winning by keeping his or her case. However, statistical testing has shown that there is no advantage in switching in the Deal or No Deal situation. The player has a 50/50 chance of increasing his winnings by either switching or keeping his case.

The reason that the Monty Hall problem gives the player an advantage is because the host purposely exposes a losing door. In other words, there is no randomness to the door exposed. It is purely based on the player's first choice. In the Deal or No Deal situation, the player chooses one case and then exposes another. The one exposed may very well be the winning case, leaving two lesser cases still in play. Therefore, there is no relation between the player's first choice and the case that is exposed. That breaks the Monty Hall advantage in switching.

 

[ynot]

There are three doors with only one containing a prize.
The host opens one of them to revealing it to be empty.
The contestant chooses one of the remaining doors.
The odds that the prize is behind the door chosen by the contestant is 1 in 2
Okay so far?

Agreed

There are three doors with only one containing a prize.
The contestant chooses one of the doors.
The odds that the prize is behind the door chosen by the contestant is 1 in 3
The host opens one of the other two doors revealing it to be empty.

Question: Do you think that the odds that the prize is behind the door chosen by the contestant has changed?

No - The odds of the original choice being right remain 1 in 3 (as I have made clear earlier).

....to be continued

BJ

Can't wait!

 

[ynot]

BJ is correct ... the chances (probability, odds, whatever) that the contestant chose the correct door at the onset are fixed. If it's 3 doors, it's 1:3 -- 8 doors, it's 1:8 -- with n doors it becomes 1:n.

Lets keep it simple and use just 3 choices (doors or whatever).

Now, when the host asks if the contestant wishes to switch for the other unopend door, it actually doesn't matter if he (the host) opens all but one of the other non-chosen doors or not -- he is essentially giving all of them to the contestant if the contestant switches.

He is not “giving” the contestant anything other than better overall odds by reducing 2 alternative choices to 1 alternate choice.

Why? -- because the contestant already knows that only at most one of them contains the prize. (The only other option is that the contestant was lucky enough to have picked the winning door right off. But that is against the odds with 3 or more total doors.) So when the host opens all but one of the non-chosen doors he is giving the contestant no new information probability-wise. Hence, switching is in the contestant's best interests.

No new information as to which choice is the right one but definitely new information probability-wise. Are you saying that the odds of making a correct alternate choice are the same regardless of whether there are 1 or 2 alternate choices?

 

[BillyJoe]

Sorry to be pedantic but (revealing them to be empty)

Not sure what you mean. The host opens only one door - revealing it to be empty.
(I think JT is confusing you with those twenty doors of his )

No - The odds of the original choice being right remain 1 in 3.

Good. So lets add in the bit you agree with and add in two new bits:

There are three doors with only one containing a prize.
The contestant chooses one of the doors.
The odds that the prize is behind the door chosen by the contestant is 1 in 3
The host opens one of the other two doors revealing it to be empty.

THE BIT YOU AGREE WITH:
1) The odds that the prize is behind the door chosen by the contestant remains 1 in 3 (1/3)

THE NEW BITS:
2) The odds that the prize is behind the door opened by the host is 0 (0/3)
3) Therefore the odds that the prize is behind the remaining door must be 2 in 3 (2/3)

It is counterintuitive but true.


BJ

 

[BillyJoe]

Be warned, ynot, JT will tie you up in knots, in that idiosyncratic way of his.

 

[ynot]

Not sure what you mean. The host opens only one door - revealing it to be empty.
(I think JT is confusing you with those twenty doors of his )


BJ

Sorry - My mistake - edited it out (not quickly enough)

 

[ynot]

Not sure what you mean. The host opens only one door - revealing it to be empty.
(I think JT is confusing you with those twenty doors of his )

Good. So lets add in the bit you agree with and add in two new bits:

There are three doors with only one containing a prize.
The contestant chooses one of the doors.
The odds that the prize is behind the door chosen by the contestant is 1 in 3
The host opens one of the other two doors revealing it to be empty.

THE BIT YOU AGREE WITH:
1) The odds that the prize is behind the door chosen by the contestant remains 1 in 3 (1/3)

THE NEW BITS:
2) The odds that the prize is behind the door opened by the host is 0 (0/3)
3) Therefore the odds that the prize is behind the remaining door must be 2 in 3 (2/3)

It is counterintuitive but true.


BJ

The odds that the prize is behind the door opened by the host is 0 (0/0). It not longer exists as a choice so is not a factor in the alternative choice option.

 

[ynot]

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.

 

[BillyJoe]

The odds that the prize is behind the door opened by the host is 0 (0/0). It not longer exists as a choice so is not a factor in the alternative choice option.

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

 

[Rasmus]

The chance the prize is now behind the remaining door is not 2 in 3.

<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.