Something about probablility

**ting** · 03-08-2008

now that we resolve this, let's try something more challenging:

1) there are now five doors with two prices
2) the player select one door, but do not open,
3) the host opens the door with a price behind it
4) the player now choose to stay or open one of the remaining door

let's work out this one mentally first, with predicate logic and probability, before attacking it with brute force algorithm.

--TING

**robwhit** · 03-08-2008

I'd keep it. The money's right there.

**whiteflags** · 03-08-2008

I believe the odds are 4 : 1 that you don't get a prize. This degrades into the Monty Hall problem, depending on how the game operates and what the host knows. I really don't see a benefit in switching either, in this case. The odds seem overwhelming even if we do get a peek behind another door.

**tabstop** · 03-08-2008

Originally Posted by ting

now that we resolve this, let's try something more challenging:

1) there are now five doors with two prices
2) the player select one door, but do not open,
3) the host opens the door with a price behind it
4) the player now choose to stay or open one of the remaining door

let's work out this one mentally first, with predicate logic and probability, before attacking it with brute force algorithm.

--TING

In the 40% chance that you have a good door, you must lose when you switch. In the 60% chance that you have a bad door, you now have a 1-in-3 chance of finding a good door on a switch -- but you don't lose anything if you don't. So it would appear that switching is a bad thing 40% of the time and a good thing 20% of the time.

**CornedBee** · 03-09-2008

Oh, great, now I can think of a good, readable explanation for the Monty Hall problem.

The initial choice is not between three doors. The choice is between two categories: good (33% chance) and bad (66% chance).
The second choice is not between the two remaining doors. The choice is between two strategies: switch your category or not. (Eliminating one bad door guarantees that the switch is not another random choice, but specifically means switching category.)
Since the change of being in the good category is only 33%, switching is good.

The new problem:
Initial choice: 40% good, 60% bad
Second choice if initial was good: switch -> 0% good, no switch -> 100% good
Second choice if initial was bad: switch -> 33% good, no switch -> 0% good

Therefore:
switch to good = (0.4 * 0) + (0.6 * 1/3) = 0.2
no switch to good = (0.4 * 1) + (0.6 * 0) = 0.4

Verify:
switch to bad = (0.4 * 1) + (0.6 * 2/3) = 0.8 (+ 0.2 = 1 -> correct)
no switch to bad = (0.4 * 0) + (0.6 * 1) = 0.6 (+ 0.4 = 1 -> correct)

-> Switching leads to a win rate of 20%, no switching to 40%. No switching is superior.

**laserlight** · 03-09-2008

The initial choice is not between three doors. The choice is between two categories: good (33% chance) and bad (66% chance).
The second choice is not between the two remaining doors. The choice is between two strategies: switch your category or not. (Eliminating one bad door guarantees that the switch is not another random choice, but specifically means switching category.)
Since the change of being in the good category is only 33%, switching is good.

What do you mean by the initial "choice is between two categories: good (33% chance) and bad (66% chance)"? As I noted in post #41 of this thread, I think the initial choice simply does not matter: it is all about choosing between two strategies. So, what you call the first and the second choice seem like one single choice to me.

EDIT:
Ah, okay, looking at your analysis again I see what you mean. Perhaps it is not so much of a choice between two categories, but that there are two categories for analysis. The actual choice is only between strategies.

**CornedBee** · 03-09-2008

The strategy defines your winning probability, but your initial choice defines whether you actually win or lose a round.

**laserlight** · 03-09-2008

The strategy defines your winning probability, but your initial choice defines whether you actually win or lose a round.

hmm... are you saying that the initial choice is between being rational (choose the good category) and irrational (choose the bad category)?

Or are you talking about initial chance instead of choice?

**CornedBee** · 03-09-2008

"Chance" is indeed a better word.

**vart** · 03-09-2008

I think the initial choice simply does not matter

It does matter - It changes probabilities of the second event (what door is opend by the host)
If we missed - The probabilities will be 1/2,1/2,0,0
If we selected right door - The probabilities will be 1,0,0,0
That's why conditional probabilities of the second choice will be different for different results of the original event...

So the original event should be taken into consideration exactly as CornedBee does it. It is simple Full Probabilities formula P(B) = P(A)*P(B|A)+P(~A)*P(B|~A)
~A - is notA event, P(B|A) is conditional probability of B when A occured

**laserlight** · 03-09-2008

It does matter - It changes probabilities of the second event (what door is opend by the host)

Ah, but that there exists a choice before the second choice is what matters. The initial choice itself (whichever door is chosen) does not matter.

EDIT:

That's why conditional probabilities of the second choice will be different for different results of the original event...

I think that holds true only in analysis, where we know what is the result of the original event. If we look at it from the point of view of the player, we can only consider strategies.

**Mario F.** · 03-09-2008

Another simplistic way of trying to explain it to the common folks (read, me) is:

By eliminating a door, the host is not increasing my chances to 50%. He is essentially giving me another door. In short, I was given 2 doors out of 3. By making sure I switch, I'll grab those odds.

My mistake all along was refusing to observe that the initial problem couldn't just go away. There were 3 doors initially, and there were 3 doors at the end.

Despite everything I still find the whole bloody thing fascinating. All I need is to think of something else, to be assaulted again by doubt. A veridical paradox indeed.

**vart** · 03-09-2008

I think that holds true only in analysis, where we know what is the result of the original event.

No, it is not like that. The formula of the Full Probability - gives the resulting probability of the event B even if we do not know what occured - event A or event ~A, becuse it incorporates both possibilities. But we need to know what is the probability of the event B when event A occured, and when event ~A occured (conditional probabilities)

**laserlight** · 03-09-2008

No, it is not like that. The formula of the Full Probability - gives the resulting probability of the event B even if we do not know what occured - event A or event ~A, becuse it incorporates both possibilities. But we need to know what is the probability of the event B when event A occured, and when event ~A occured (conditional probabilities)

I agree. In retrospect, my statement is misleading: what I mean by "the first choice does not matter" is that while the player deliberates the first choice, each option has an equal chance of holding a prize. My line of thought was in response to Mario F.'s assertion that the probability for the door initially chosen to contain the prize could change from 1/3 to 1/2.

**vart** · 03-09-2008

Originally Posted by laserlight

My line of thought was in response to Mario F.'s assertion that the probability for the door initially chosen to contain the prize could change from 1/3 to 1/2.

Actually - It could...
Because we are talking here about 2 different probability spaces - one is unconditional space P(.), second is conditional space p(.|C) where C is the known event that the door opened by host does not contain price... so probabilities in these two spaces can be different...

For example.

You take a 52 cards stack and choose one card - randomly. Put it before you closed. What is the probability to have a King? 4/52 = 1/13

Now I take a card from the stack and open it - suppose - it is a King
No, what is the probability, that the card that you have choosen before, but haven't open yet is a King?

Thread: Something about probablility

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