Monty Hall problem (again)

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phyti said:
'most likely' is not a probablity.
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Yes, it's a way of describing probability.
If something is 98.1% likely, it is "most likely" compared to something with a 1.9% chance.
Don't play word games.

phyti said:
The person with the card in hand does not have the A.
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They might.

phyti said:
You are saying there are 51 possible cards that could be the A.
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This sequence started with post #157:
"
Get a pack of cards.
Shuffle them.
Pick one.
Then think about whether you'd be better off keeping that 1 or swapping to what's effectively all of the other 51.
"

There are 52 cards in total. The one in your hand and the 51 left on the table. You then have dragged this off course by focussing on what happens with the 51 too soon.

phyti said:
Probability needs a history of card choices which allows it to be calculated.
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Not always.
To calculate the probability that you picked up the Ace of Hearts on the first action, we just need 1/52 to calculate the 1.9%

phyti said:
P=number of successful choices/all possible choices. P=1/51.
If 51 people remove a card from the table, 1 will be successful and 50 will not.
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Only if the Ace was left on the table after you took the first card. This would be 98.1% of the time.

phyti said:
50/51=the probability of any 1 card on the table not being A. 1-50/51=1/51.
The solution depends on choices, not locations.
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No, you need to include the probability that the Ace was one of those cards in the first place. There are 51 cards on the table at that point because you picked a card first.



I think we need to start with something even simpler:

phyti, do you have a deck of cards?
 
One more time...

Let's play a game, phyti...

Take three doors playing cards and lay them face down, making sure one is an Ace.

Shuffle them, then you pick one, and I'll take the other two.

You think that you're going to end up with the Ace just as often as I do, right?

We each have a 50% chance of winning, right, phyti?
 
phyti said:
"The player doesn't know if he chose the A. So, he estimates the probability of the A being on the table as 51/52."

If the A has 1/52 chance of being in the deck, then the chance of being on the table, a set 1 card smaller, is 1/51.
You are comparing 2 different size sets, 1 vs 51.
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Are you telling me that you don't understand the content of post #167?

Really?
 
pzkpfw said:
(Edit: the above seems to have been removed from the quoted post. I'll leave my reply in place.)

Yes, the Ace can only be in one place, the odds are about where it likely is.

In this case, it's most likely (98.1% vs 1.9%) to be among the 51 cards on the table, not the 1 card in your hand.

1 Card in your hand51 Cards on the Table
???????????
??????????
??????????
??????????
???????????
The Ace of Hearts might be in the left column or the right column. Yes - not both. But which is most likely?


(If you were betting on watching one person draw one card from a fair deck; would you put money on them getting the Ace of Hearts, or not getting the Ace of Hearts?)



If all 51 cards were removed from the table by 51 different people:
There's a 1.9% chance the Ace was in your hand and all of them are losers.
There's a 98.1% chance the Ace was on the table, so one of them is a winner.
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Surely it is not possible that phyti actually doesn't understand this.

If this simple stuff has him all at sea and confused, then he isn't mentally equipped to be discussing the Monty Hall problem.
 
phyti said:
If 51 people remove a card from the table, 1 will be successful and 50 will not.
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"Successful" in removing the Ace of Hearts, yes.

And if one person removes one card from the table at random and another person removes the remaining 51 cards from the table, then the person who removed the 51 cards will be 51 times more likely to hold the Ace of Hearts than the person who removed one card.

What is the probability that the person who removed 51 cards will be "successful" in ending up with the Ace of Hearts? It is 51/52.

Why are you struggling with this?
 
phyti:

You seem to have forgotten to do what I asked you to do in post #171, above.

Please do what I asked you to do. There's a good chap.
 
James R said:
"Successful" in removing the Ace of Hearts, yes.

And if one person removes one card from the table at random and another person removes the remaining 51 cards from the table, then the person who removed the 51 cards will be 51 times more likely to hold the Ace of Hearts than the person who removed one card.

What is the probability that the person who removed 51 cards will be "successful" in ending up with the Ace of Hearts? It is 51/52.

Why are you struggling with this?
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Based on a history of choosing A from a set of 52 cards, the probability is 1/52.

The probability of choosing a not-A = 1- 1/52 = 51/52.

If person #1 chose a not-A card from the set, then A must be in the remaining 51 cards.

Based on a history of choosing A from a set of 51 cards which includes A, the probability is 1/51.

You are considering all possible 51 locations for A, before a choice is made.

Probability is calculated from a history of choices, i.e. after the choices are made.

From all 51 possible outcomes, only 1 includes A.

1/51>1/52 by a small amount only because the set is large.
 
The probability tree

1785176360703.gif
A session is a sequence of actions by the player and host in playing the MH game defined as: player 1st choice, host choice, player 2nd choice. All numbers within cells are door ID's. Prize distribution is (c, g, g).

Left, the tree based on the Selvin and Savant interpretation with the assumption of number of sessions equals number of doors, which contradicts the tree having 4 different outcomes. There are 4 sessions of player chooses a door, host opens a door, noted as (1,2), (1,3), (2,3), and (3,2). Their solution to get 4 different results from 3 doors was to play session 1 half the time with host opening door 2 and half the time with host opening door 3. That forms a bias with half as many car wins as goat wins. That allows the strategy of switching to beat the system. The probability is formed in reverse order, left to right.

Right, the method of forming probability values based on the frequency of occurrence of all possible sessions, right to left.
The tree structures are identical, only the frequencies differ.

For a fair game with no deception and offering the same opportunity to all players, the game show producers must offer the sessions with equal frequencies on average.
 
Two quotes come to mind:

"Arguing with an idiot is like playing chess with a pigeon: it'll knock over the pieces, crap on the board, and strut around as if it's won." - someone.

"Don't argue with fools. They will drag you down to their level and beat you with experience." - Twain
 
Given, a shuffled deck of 52 cards face down on a table. Each card is unique.
One card u is selected as the prize winner for a new car.
A person removes 1 card from the top of the deck until they remove u.
Probability of success is the ratio of (successful events)/(all possible events)=s/a.
Initially there is 1 successful event and 52 possible events.
event...prob
1......... 1/52
2......... 1/51
3......... 1/50
4......... 1/49
5......... 0/48
The person wins on the 4th try.
The numerator s is always 1 until a win.
The denominator a is always the remaining set.
 
Phyti, you're still focussing on independent events. Monty Hall is not a matter of independent events.
If a person comes in from the blue and sees the two doors left closed and knows a car is behind one of them, then for that person the choice would be 50:50 as to which door. Because that would be an independent event to what has gone before.
The person playing, who picked 1 from 3, and then saw Monty open one of the other 2 doors, the choice to swap is not independent from what has gone before. When you argue that it is a 50/50 choice, you are looking at the probability of someone coming in from the blue, not the initial player.

You have been told of your mistakes almost ad nauseam. Yet, here you still are, peddling your nonsense.

So, another try from me:
Notice in your post #189 your "corrected" frequency diagram has door 1 picked 50% of the time. Yet we know that in a fair game the player will pick each door 33.33% of the time. You argue that there are 4 "sessions" but you need to fit those within the starting probabilities of each door being picked 1/3 of the time. You can't alter that just to satisfy the conclusion you want.

Do you accept that, in a fair game, each door has equal probability of being picked by the player? If so, that initial probability MUST be 1/3 for each door. Not the 1/2 for door 1 and 1/4 for the other 2.

For some reason you seem resistant to learning.
 
phyti said:
Success is removal of the u card and there is only 1 of those.
If 52 people each removed a card from the deck, there would be 1 winner and 51 losers.
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So you seem to know you have a 1 in 52 chance of picking the Ace of Hearts when taking the first card.

If that doesn't turn out to be the Ace, and you know it, you are correct that the chance the next (single) card taken is the Ace is 1 in 51.

But that's not the relevant question for this thread. It is:

After taking the first card - but without looking at it - what is the chance the Ace is still one of the 51 cards left?

Hint: the Ace exists, the total odds of where the card is should add to 1 (or 100% if you prefer).
 
phyti said:
Success is removal of the u card and there is only 1 of those.
If 52 people each removed a card from the deck, there would be 1 winner and 51 losers.
Click to expand...
Right.

Now, if person A removed one card, and person B took the other 51 - who is more likely to have the Ace of Hearts in their hand?
 
phyti said:
That's a different game.
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I know, it's a comparison. Let's see if you can understand it.

If person A removed one card from a shuffled deck, and person B took the other 51 cards - who is more likely to have the Ace of Hearts in their hand?
 
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