Lotto Probability

[joe87]

I don't understand why everyone doesn't agree that after the host opens an empty door, the odds change to 1 in 2 for each remaining door. Therefore, it doesn't matter whether the player stays with his original pick or switches, he has a 50-50 chance of being right either way. The 1/3 probablilty of being right only applies to the original situation, not to the changed situation after the host opens an empty door.

 

[Rasmus]

I don't understand why everyone doesn't agree that after the host opens an empty door, the odds change to 1 in 2 for each remaining door.

Do you mean the original puzzle, or the new one with two players?

Therefore, it doesn't matter whether the player stays with his original pick or switches, he has a 50-50 chance of being right either way. The 1/3 probablilty of being right only applies to the original situation, not to the changed situation after the host opens an empty door.

If the fornmer, please just re-read the last page or so of this thread, there should be anywhere between 5 and 10 different explanantions. Ask, if you have trouble understanding anything, maybe we can come up withj even more ways of displaying this

 

[joe87]

Do you mean the original puzzle, or the new one with two players?

The same logic applies to both puzzles. There are two doors and one prize. Each door is equally likely to contain the prize. Each player has a 50-50 chance to pick the correct door, whether he switches or not. The new puzzle graphically illustrates this, because the two contestants can't both have a probablility of 2/3 of being right if they both switch, because in that case, the combined probablility is 1.5, an impossibility.

If the fornmer, please just re-read the last page or so of this thread, there should be anywhere between 5 and 10 different explanantions. Ask, if you have trouble understanding anything, maybe we can come up withj even more ways of displaying this

I have read the last several pages, which is why I made the post. Somewhere back there, everyone agreed that the odds remain 1 in 3 for the original door after the host opens the empty door. This is wrong because the host opening an empty door added information to the puzzle, thereby changing the odds. Probablilities are used only when there is less than perfect knowledge, and when additional information becomes available , the odds are changed.

edit: the combined probability is actually 4/3, or 1.333, not 1.5 as I stated above. My conclusion is still good, because the combined probablilty can't exceed 1.

 

[Just thinking]

I have read the last several pages, which is why I made the post. Somewhere back there, everyone agreed that the odds remain 1 in 3 for the original door after the host opens the empty door. This is wrong because the host opening an empty door added information to the puzzle, thereby changing the odds. Probablilities are used only when there is less than perfect knowledge, and when additional information becomes available , the odds are changed.

Actually, there is no new information. Please re-read what I posted earlier carefully ...

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.

The contestant knows that one of the non-chosen doors has nothing behind it (there is only one winning door). The host merely confirms this when he reveals the empty door(s).

What exactly is the new information given to the contestant?

 

[joe87]

Actually, there is no new information. Please re-read what I posted earlier carefully ...

The contestant knows that one of the non-chosen doors has nothing behind it (there is only one winning door). The host merely confirms this when he reveals the empty door(s).

What exactly is the new information given to the contestant?

The new information is that there is one less door that the prize could be behind. The contestant did not know that before. He only knew that all the original doors but one had no prize. Opening a single door changes the odds that each of the other doors has the prize behind it by one divided by the number of doors.

If we go back to the 20 doors and one prize scenario, to start the game the contestant picks one door and he has 1/20 chance of picking the prize door. After the host opens one door with no prize, the odds are 1/19 that any door, including the one picked by the contestant has the prize behind it. Suppose the host continues opening doors with no prize behind them. After each door is opened, the likelihood that each remaining door holds the prize increases. When there are 3 doors left, the odds that one will hold the prize is 1/3. When 2 doors are left, the odds are 1/2. If the host then opens the door that the contestant did not pick when 2 doors are left, if that door does not have the prize behind it, the probability is 1 that the contestant will get the prize. The odds change after each door is opened, because one less door remains for the prize to be behind.

If we go to the 3 door scenario, with the contestant picking a door, and the host opening one empty door, suppose there is another contestant that is now allowed to pick a door, and if he is right, he gets the prize. With 2 doors to chose from, he knows that one has a prize, so his odds are 50% that he picks the correct door. What is the difference between him and the contestant that has picked a door when there were 3 doors available? Nothing. Both contestants have equal odds of picking the right door, and neither door has a higher likelihood than the other. The game starts over with a new slate of probablilities every time a door is opened. The odds have no memory.

 

[Rasmus]

The same logic applies to both puzzles.

The new puzzle is bogus. The new puzzle doesn't work, since in 1/3 of the gazes, the game can't even continue.

There are two doors and one prize. Each door is equally likely to contain the prize.

No.

Please do not just assume this, since it is at the core of the disagreement.

Each player has a 50-50 chance to pick the correct door, whether he switches or not. The new puzzle graphically illustrates this, because the two contestants can't both have a probablility of 2/3 of being right if they both switch, because in that case, the combined probablility is 1.5, an impossibility.

The new puzzle doesn't work.

The host cannot open a door in 1/3 of the cases.

The new puzzle has nothing to do with the old puzzle, because the mechanism the old puzzle relies on is ruled out.

I have read the last several pages, which is why I made the post. Somewhere back there, everyone agreed that the odds remain 1 in 3 for the original door after the host opens the empty door. This is wrong because the host opening an empty door added information to the puzzle, thereby changing the odds.

I ask you to simply try this with a few matchboxes, a coin and a trustworthy partner. Or use some of the many online simulations, if you trust them.

Probablilities are used only when there is less than perfect knowledge, and when additional information becomes available , the odds are changed.

If the chances were 50%, then a player would win half the time, even if he never changed, right?

But if a player never changed, it wouldn't matter if he had the chance, right?

So how can the chances be 1/2, if they previously were 1/3?

 

[Just thinking]

The new information is that there is one less door that the prize could be behind. The contestant did not know that before. He only knew that all the original doors but one had no prize. Opening a single door changes the odds that each of the other doors has the prize behind it by one divided by the number of doors.

Afraid not. Using just 3 doors as in the original Monty Hall scenario, the contestant knows full well that at least one of the non-chosen doors is empty -- the host, by opening up one of the two non-chosen doors, merely confirms it. The host must know where the prize is (otherwise he risks exposing it), so we know the host will open up an empty door. The original choice the contestant made had a 1/3 chance of winning -- the prize being behind one of the other 2 doors is 2/3. This does not change by showing the contestant that one door has nothing, since this is known before any door is opened.

If we go back to the 20 doors and one prize scenario, to start the game the contestant picks one door and he has 1/20 chance of picking the prize door. After the host opens one door with no prize, the odds are 1/19 that any door, including the one picked by the contestant has the prize behind it. Suppose the host continues opening doors with no prize behind them. After each door is opened, the likelihood that each remaining door holds the prize increases. When there are 3 doors left, the odds that one will hold the prize is 1/3. When 2 doors are left, the odds are 1/2. If the host then opens the door that the contestant did not pick when 2 doors are left, if that door does not have the prize behind it, the probability is 1 that the contestant will get the prize. The odds change after each door is opened, because one less door remains for the prize to be behind.

You are assuming (by your description) that the host is opening these doors at random -- not knowing which one holds the prize. Sorry, but that is not the case. The host must know where the prize is or he risks accidently exposing it. Therefore, he will only open empty doors -- this makes the odds different.

If we go to the 3 door scenario, with the contestant picking a door, and the host opening one empty door, suppose there is another contestant that is now allowed to pick a door, and if he is right, he gets the prize. With 2 doors to chose from, he knows that one has a prize, so his odds are 50% that he picks the correct door. What is the difference between him and the contestant that has picked a door when there were 3 doors available? Nothing. Both contestants have equal odds of picking the right door, and neither door has a higher likelihood than the other. The game starts over with a new slate of probablilities every time a door is opened. The odds have no memory.

There is a big difference between your two hypothetical contestants. The first one picks his winning prize from among 3 possible doors, the second only 2. The first has odds of 1/3 -- the second 1/2. If the host never exposed anything, and let each contestant choose from among all 3, then both have a 1/3 chance of winning.

 

[Just thinking]

joe87,

Instead of doors, let's say that you buy a lotto ticket to a lottery of which I know the winning combination (but you don't). There are 10 million possible combinations of tickets, and I have all but one (the one you have). I will discard all losing combinations except for one ticket and offer you the option of switching your ticket for the one I have left (assuring you one of us holds the winning ticket); do you somehow believe that your ticket, which had a 10 million to 1 chance of winning now has a 50/50 chance of being the winner?

 

[drkitten]

The contestant knows that one of the non-chosen doors has nothing behind it (there is only one winning door). The host merely confirms this when he reveals the empty door(s).

What exactly is the new information given to the contestant?

The exact identity of one of the doors with nothing behind it. Before, he knew that at least one of A, B, and C had nothing behind it. (Actually, he knew that two of them had nothing).

Now he knows that (hypothetically) door A itself has nothing behind it.

 

[joe87]

The new puzzle is bogus. The new puzzle doesn't work, since in 1/3 of the gazes, the game can't even continue.

That doesn't change the probabilities before each pick. The new puzzle is a reductio ad absurdum for the idea that the probablilities don't change after a door is opened.

Please do not just assume this, since it is at the core of the disagreement.

One calculates probabilities in any given situation by dividing the number of possible winning picks by the total number of possible picks. Do you disagree?

The new puzzle doesn't work.

The host cannot open a door in 1/3 of the cases.

That doesn't affect the probabilities before the host takes action, or doesn't take action becase he can't.

The new puzzle has nothing to do with the old puzzle, because the mechanism the old puzzle relies on is ruled out.

That mechanism is spurious, if you are referring to the hypothesis that the odds don't change when a door is opened.

I ask you to simply try this with a few matchboxes, a coin and a trustworthy partner. Or use some of the many online simulations, if you trust them.

If I get some time, I'll program it up and play a few thousand cases, but that's really not necessary because the answer is clear (to me at least).

If the chances were 50%, then a player would win half the time, even if he never changed, right?

Yes, just as he would if he changed every time a door is opened.

But if a player never changed, it wouldn't matter if he had the chance, right?
So how can the chances be 1/2, if they previously were 1/3?

The probability of picking the correct door is independent of what the contestant has previously picked. When the number of doors is reduced by one, the denominator of the probability changes by one, thereby changing the odds. 20 doors, 1/20 chance of being right at that time. 3 doors, 1/3 chance. 2 doors, 1/2 chance. 1 door, 1 chance. If the host continues to open empty doors, the odds eventually go to 50% when there are 2 doors left, as you said above. Doesn't matter if the contestant stays with his previous pick or not. probability = number of possible wins divided by number of possible picks.

 

[Just thinking]

The exact identity of one of the doors with nothing behind it. Before, he knew that at least one of A, B, and C had nothing behind it. (Actually, he knew that two of them had nothing).

Now he knows that (hypothetically) door A itself has nothing behind it.

That (which you point out) doesn't matter. The host is essentially giving the contestant both of the non-chosen doors (if the contestant switches). What difference does it make if the host opens one of the two doors before or after the switch (if it occurs)? It really doesn't matter ... but let's look at one set of outcomes:

[The prize is behind door B]

Contestant picks door A.
Host opens door C -- empty.
Contestant swiches -- wins.
Contstant stays -- loses.

Contestant picks door B.
Host opens door A -- empty.
Contestant swiches -- loses.
Contstant stays -- wins.

Contestant picks door C.
Host opens door A -- empty.
Contestant swiches -- wins.
Contstant stays -- loses.

In the 3 cases where the contestant switches, 2 of them are wins.
In the 3 cases where the contestant stays, 2 of them are losses.

You can rearrange the combinations for the other two possible winning doors, but the win to lose ratio is going to remain the same.

Stay with your initial pick, chance of winning = 1/3.
Switch from your initial pick, chance of winning = 2/3.

 

[joe87]

Afraid not. Using just 3 doors as in the original Monty Hall scenario, the contestant knows full well that at least one of the non-chosen doors is empty -- the host, by opening up one of the two non-chosen doors, merely confirms it. The host must know where the prize is (otherwise he risks exposing it), so we know the host will open up an empty door. The original choice the contestant made had a 1/3 chance of winning -- the prize being behind one of the other 2 doors is 2/3. This does not change by showing the contestant that one door has nothing, since this is known before any door is opened.

The probability that the prize is behind one of the 2 other doors is 2/3 before one is opened. After the empty door is opened, the probability that the prize is behind the other door changes to 1/2. Divide the number of winning picks by the total number of available picks. That's the definition of probability (at least it was in the statistics courses I've taken).

You are assuming (by your description) that the host is opening these doors at random -- not knowing which one holds the prize. Sorry, but that is not the case. The host must know where the prize is or he risks accidently exposing it. Therefore, he will only open empty doors -- this makes the odds different.

Not at all, I'm assuming that the host knows where the prize is and only opens empty doors. The probability of interest is that of the contestant. If there are 3 doors, he has a 1/3 probability of picking the right one. One prize divided by 3 doors.

There is a big difference between your two hypothetical contestants. The first one picks his winning prize from among 3 possible doors, the second only 2. The first has odds of 1/3 -- the second 1/2. If the host never exposed anything, and let each contestant choose from among all 3, then both have a 1/3 chance of winning.

Perhaps I stated that wrong. I meant to say that after first contestant is down to 2 doors, he has a 50% chance of picking the right one, independent of what has gone before, or whether he switches or not. He has the same odds as another contestant would have if he were given 2 doors to pick from.

The point I'm trying to make here is that "maturity of chances" is a fallacy. The probablility that the contestant has picked the right door depends on how many doors there are, not on what his previous picks have been.

If the host NEVER opens a door that is not empty, the contestant is guaranteed to get down to a choice of 2 doors, no matter how many doors the game starts with. Since I've never seen the game played, I don't know how many doors the game starts with and I'm not sure whether or not the host sometimes ends the game before the contestant gets down to 2 doors. However, when the contestant gets down to 2 doors, if he has no way of determining which is best (i.e., the host has given him no clues), he has a 50% chance of picking the right door, whether that involves a switch or not.

 

[Rasmus]

This one only works in IE, sadly:
http://www.grand-illusions.com/simulator/montysim.htm
10.000 trials, without changing
wins: 3.291 or 32%
losses: 6709 or 68%

10.000 trials, with changing:
wins: 6636 or 66%
losses: 3364 or 34%

 

[drkitten]

One calculates probabilities in any given situation by dividing the number of possible winning picks by the total number of possible picks. Do you disagree?

Yes, I disagree. This is incorrect.

"Dividing the number of possible winning picks by the total number of possible picks" is only legitimate if the picks are all equally likely. As an example of otherwise, consider betting on ("picking") the next color that a roulette ball will land on. There are three possible picks :red, black, and green, but none of those picks have probability 1/3 of coming up.

Just as a correct analysis of the roulette wheel involves looking at sub-cases, so does the Monty Hall problem.

Assume without loss of generality that the prize is behind door A. The player can choose any of A, B, C, freely -- each with probability 1/3.

Case 1) The player chooses B. Monty shows C.
Case 2) The player chooses C. Monty shows B.
Case 3) The player chooses A. Monty can choose freely between two subcases
Case 3a) The player chooses A. Monty shows B.
Case 3b) The player chooses A. Monty shows C.

Notice that the probabilties of each situation are 1/3, 1/3, 1/6, and 1/6 (respectively), not a simply even split. In particular, the situations where the player wins by switching have a total probability of 2/3 -- the probabilities where the player loses by switching have a total probability of 1/3. Hence the player wins twice as often by switching.

ETA: typo fix.

 

[Just thinking]

The probability of picking the correct door is independent of what the contestant has previously picked. When the number of doors is reduced by one, the denominator of the probability changes by one, thereby changing the odds. 20 doors, 1/20 chance of being right at that time. 3 doors, 1/3 chance. 2 doors, 1/2 chance. 1 door, 1 chance. If the host continues to open empty doors, the odds eventually go to 50% when there are 2 doors left, as you said above. Doesn't matter if the contestant stays with his previous pick or not. probability = number of possible wins divided by number of possible picks.

In a polite word ... No. Please see my example above.

 

[Just thinking]

If the host NEVER opens a door that is not empty, the contestant is guaranteed to get down to a choice of 2 doors, no matter how many doors the game starts with. Since I've never seen the game played, I don't know how many doors the game starts with and I'm not sure whether or not the host sometimes ends the game before the contestant gets down to 2 doors. However, when the contestant gets down to 2 doors, if he has no way of determining which is best (i.e., the host has given him no clues), he has a 50% chance of picking the right door, whether that involves a switch or not.

What really matters here is how the contestant gets down to 2 doors. If it's achieved by random picking, then with 2 doors left it is 50/50 -- but the Monty Hall scenario is not done randomly, and that's important. The host will always open an empty door, as I've explained earlier. This is a ruse and adds nothing to the equation of probabilities. Yes, we know for sure that the prize is not behind the exposed door, but we still know that it has a 2/3 chance of being among the two non-chosen doors -- even if the empty one is opened.

As drkitten pointed out, two choices do not guarantee a 50/50 probability with weighted events.

 

[69dodge]

Assume without loss of generality that the prize is behind door A. The player can choose any of A, B, C, freely -- each with probability 1/3.

Case 1) The player chooses B. Monty shows C.
Case 2) The player chooses C. Monty shows B.
Case 3) The player chooses A. Monty can choose freely between two subcases
Case 3a) The player chooses A. Monty shows B.
Case 3b) The player chooses B. Monty shows C.

Typo.

Case 3b should be: The player chooses A. Monty shows C.

 

[69dodge]

There is a big difference between your two hypothetical contestants. The first one picks his winning prize from among 3 possible doors, the second only 2. The first has odds of 1/3 -- the second 1/2.

No. Both contestants have the same information about everything. How can their probabilities differ?

 

[joe87]

What really matters here is how the contestant gets down to 2 doors. If it's achieved by random picking, then with 2 doors left it is 50/50 -- but the Monty Hall scenario is not done randomly, and that's important. The host will always open an empty door, as I've explained earlier. This is a ruse and adds nothing to the equation of probabilities. Yes, we know for sure that the prize is not behind the exposed door, but we still know that it has a 2/3 chance of being among the two non-chosen doors -- even if the empty one is opened.

As drkitten pointed out, two choices do not guarantee a 50/50 probability with weighted events.

OK, you guys have convinced me, the contestant should always switch. The way the game is played gives information to the contestant and allows him to beat the 50-50 odds that he would have if he were presented with 2 doors without knowing how he got there from 3 doors.

But I don't think that means I should start playing the lottery.

 

[drkitten]

Typo.

Case 3b should be: The player chooses A. Monty shows C.

You are of course correct.