What Book Are You Reading at the Moment?

[Fife]

Negatory. (and switching increases your odds to 66.7%, not 50%)

Each cup has a 1/3 chance of being a winner and a 2/3 chance of being a loser. That doesn't change.

The host giving away one of the 2 losing cups doesn't change the chance your first choice (no matter which one of the three it was) has a 1/3 chance of having the prize; and the remaining field having a 2/3 chance in the aggregate of containing the winner.

Run the simulation for yourself.

You are not reading what I wrote.

Read it again.

My choice to refrain and DB's choice to switch end on the same cup. The probability cannot be different for the same cup based on the choices of the participants.

Once you add more than one chooser, the error is pretty obvious. If you bother to read.

I bothered to read it. You're wrong.

You changed the game when you added another player, and more importantly removed the host's knowledge of the winning location and his inability to reveal the winner. Unfortunately, math doesn't let you add in other rules and conditions and pretend the game is the same. Why not have 100 people playing all at once on the same three cups?

I didn't make this math up; I just read the book. I also have tried the simulation, and run the game IRL with three cards with my daughter a while back. Why don't you try that and see what your lyin' eyes tell you?

[Speaker to Animals]

In your example none of us know where the prize is. The odds only shift when someone who knows the prize location is forced to reveal a non-prize.

My point is, in my example, my decision to hold on cup A is the same, statistically, as your decision to change from cup B to cup A.

In the end, we both chose cup A. But it is a contradiction to claim my chances are less than your chances.

A third way to look at it is that after cup C was uncovered, we just started a new game. My decision was tantamount to choosing cup A in game #2 and Fife's decision was to choose cup B. We both made a choice.

But the contradiction I show above is the real killer of this argument. The argument violates the principle of noncontradiction.

[Speaker to Animals]

Negatory. (and switching increases your odds to 66.7%, not 50%)

Each cup has a 1/3 chance of being a winner and a 2/3 chance of being a loser. That doesn't change.

The host giving away one of the 2 losing cups doesn't change the chance your first choice (no matter which one of the three it was) has a 1/3 chance of having the prize; and the remaining field having a 2/3 chance in the aggregate of containing the winner.

Run the simulation for yourself.

You are not reading what I wrote.

Read it again.

My choice to refrain and DB's choice to switch end on the same cup. The probability cannot be different for the same cup based on the choices of the participants.

Once you add more than one chooser, the error is pretty obvious. If you bother to read.

I bothered to read it. You're wrong.

You changed the game when you added another player, and more importantly removed the host's knowledge of the winning location and his inability to reveal the winner. Unfortunately, math doesn't let you add in other rules and conditions and pretend the game is the same. Why not have 100 people playing all at once on the same three cups?

I didn't make this math up; I just read the book. I also have tried the simulation, and run the game IRL with three cards with my daughter a while back. Why don't you try that and see what your lyin' eyes tell you?

I cannot change the game by adding another player because players make no impact on the location of the prize. It does not matter how many people play. The outcome remains the same.

[SuburbanFarmer]

Jesus, guys...

You have a 33% chance, no matter what. You're assuming a random First Choice, fine. 33%.

Now you have 2 unknown cups, and Choice 2, which appears to be 50%. But it's not. No matter what you chose in Choice 1, you had a 33% chance. You could have chosen right or wrong, and now you know that one of the other choices was wrong. Those odds do not change based on staying with Choice 1, or changing it. Your odds are still 33%.

No matter what you choose in Choice 1, your odds are 33%. No matter what you do in Choice 2, you have the same chance that you started with. The guy just fucked with your head, showing you irrelevant information.

[Fife]

Your position is getting better, StA. Still hard-headed hubris; but moving up in the world.

:goteam: :drunk:

[Speaker to Animals]

Well, fuck. I still cannot believe it is true, but wrote it out myself.

Code: Select all

import random


count_first_guess_correct = 0.0
count_second_guess_correct = 0.0
count_total_games_played = 0.0


def run_game ():
    cups = {1:0, 2:0, 3:0}
    keys = list(cups.keys())



    # hide the prize in one of the three cups
    prize_key = random.choice(keys)
    cups[prize_key] = 1

    random.shuffle(keys)

    # player chooses random cup

    first_choice = keys[0]


    # Overturn a cup not containing the prize and not chosen by player
    keys.remove(first_choice)
    if prize_key != first_choice:
        keys.remove(prize_key)


    overturned_cup = keys[0]
    keys.remove(overturned_cup)

    # player chooses other cup

    keys = list(cups.keys())
    keys.remove(first_choice)
    keys.remove(overturned_cup)
    second_choice = random.choice(keys)

    #Sanity checks
    if first_choice == second_choice:
        print ("Error: first choice and second choice are the same")
    elif prize_key == overturned_cup:
        print ("Error: prize key and overturned cup are the same")
    elif second_choice == overturned_cup:
        print ("Error: second choice and overturned cup are the same")

    if prize_key == second_choice:
        return 1
    else:
        return 0





for n in range (0,10000):
    outcome = run_game()
    if outcome == 1:
        count_second_guess_correct = count_second_guess_correct + 1.0
    else:
        count_first_guess_correct = count_first_guess_correct + 1.0

    count_total_games_played = count_total_games_played + 1.0


percent_first_guess_correct = count_first_guess_correct / count_total_games_played
percent_second_guess_correct = count_second_guess_correct / count_total_games_played

print ("percent first guess correct: " + str(percent_first_guess_correct))
print ("percent second guess correct: " + str(percent_second_guess_correct))
print ("total games: " + str(count_total_games_played))

Output:

percent first guess correct: 0.3324
percent second guess correct: 0.6676
total games: 10000.0

Results are repeated numerous time. You actually have about 66% chance of getting the prize if you choose a second time.

[heydaralon]

Is this all referencing the Monty Hall problem?

[Speaker to Animals]

It's the semi-related Monty's Balls problem.

[SuburbanFarmer]

Well, fuck. I still cannot believe it is true, but wrote it out myself.

Code: Select all

import random


count_first_guess_correct = 0.0
count_second_guess_correct = 0.0
count_total_games_played = 0.0


def run_game ():
    cups = {1:0, 2:0, 3:0}
    keys = list(cups.keys())



    # hide the prize in one of the three cups
    prize_key = random.choice(keys)
    cups[prize_key] = 1

    random.shuffle(keys)

    # player chooses random cup

    first_choice = keys[0]


    # Overturn a cup not containing the prize and not chosen by player
    keys.remove(first_choice)
    if prize_key != first_choice:
        keys.remove(prize_key)


    overturned_cup = keys[0]
    keys.remove(overturned_cup)

    # player chooses other cup

    keys = list(cups.keys())
    keys.remove(first_choice)
    keys.remove(overturned_cup)
    second_choice = random.choice(keys)

    #Sanity checks
    if first_choice == second_choice:
        print ("Error: first choice and second choice are the same")
    elif prize_key == overturned_cup:
        print ("Error: prize key and overturned cup are the same")
    elif second_choice == overturned_cup:
        print ("Error: second choice and overturned cup are the same")

    if prize_key == second_choice:
        return 1
    else:
        return 0





for n in range (0,10000):
    outcome = run_game()
    if outcome == 1:
        count_second_guess_correct = count_second_guess_correct + 1.0
    else:
        count_first_guess_correct = count_first_guess_correct + 1.0

    count_total_games_played = count_total_games_played + 1.0


percent_first_guess_correct = count_first_guess_correct / count_total_games_played
percent_second_guess_correct = count_second_guess_correct / count_total_games_played

print ("percent first guess correct: " + str(percent_first_guess_correct))
print ("percent second guess correct: " + str(percent_second_guess_correct))
print ("total games: " + str(count_total_games_played))

Output:

percent first guess correct: 0.3324
percent second guess correct: 0.6676
total games: 10000.0

Results are repeated numerous time. You actually have about 66% chance of getting the prize if you choose a second time.

The answer is there in your code. You instructed the computer to run a completely random second choice.

second_choice = random.choice(keys)

Effectively, you're counting the number of times that choosing between two remaining cups is correct/incorrect.

Mandating that you switch on Choice 2 will yield an identical result to sticking with the cup chosen in Choice 1.

[C-Mag]

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