Philosophy Now Forum

Mike Strand wrote:In the Monty Hall problem you posed, with the 1st person switching and the second person tossing a coin to decide between the two remaining doors, let's assume they share the prize if they both select the door with the prize.

The choices of the two people are independent of each other. 1st person chooses prize door with prob. 2/3; 2nd person chooses prize door with prob. 1/2.
Dealing with every possibility:
Probability that both win and have to share the prize: 2/3 times 1/2 = 1/3. (win-win)
Probability that 1st person wins and 2nd loses: 2/3 times 1/2 = 1/3. (win-lose)
Probability that 1st person loses and 2nd wins: 1/3 times 1/2 = 1/6. (lose-win)
Probability that both lose, and neither gets prize: 1/3 times 1/2 = 1/6. (lose-lose)

Note that the 1st person is twice as likely as the 2nd to be the sole winner, 1/3 vs. 1/6.

Note the four probabilities add up to 1.

We can use these to get other probabilities:
Probability that 1st person gets all or part of prize: 1/3 + 1/3 = 2/3
Probability that 2nd person gets all or part of prize: 1/6 + 1/3 = 1/2

In the coin tossing, if you had already tossed four heads in a row, and asked me what's the prob. of tossing another head, I would say 1/2, and this is the case, whether I know about your previous tosses or not. But at the start, before you started tossing, the probability of five heads in a row is 1/2 to the fifth power, or 1/32.

Another way of seeing this: By the time you asked me about the fifth toss, you already had a result that only had a 1/16 chance of occurring, and so the overall chance of getting all five as heads is 1/16 times 1/2 = 1/32. But the chance of that fifth toss alone being a head is still 1/2. All of these calculations reflect the independence of each toss.

Another way of checking independence of tosses: Record the results of successive tosses in successive columns. You'll find that the proportion of heads in each column tends to 1/2. If there are four tosses, with four columns, you'll also see that only 1/16 of the rows (after many, many tosses) have all heads (or all tails). Otherwise, you've got a coin with a memory!

The seeming contradictions in games of chance with independent basic outcomes (because coins, dice, cards have no memory) is a trick of the human brain, and it still tricks me until I start looking at it in a "hard-headed" manner.
____________________________________________________________________________________________________________________

Probabilities can change in other ways. Take a deck of cards with half red and half black cards, 52 in all. On the first shuffle and draw, I have a 1/2 chance of drawing a red card. But if I don't place the card back into the deck, as happens in card games, the probability of drawing another red card reduces to 25/51, and the chances of drawing a black card increases to 26/51. To analyze card games thus requires a new function (other than the binomial prob. function). It's called the hypergeometric probability function. But this still doesn't mean the cards have developed a memory -- it only means the deck from which they are being drawn has changed. In tossing a coin, it's like drawing from an infinite pot with half heads and half tails. This is like "sampling with replacement". Card games are like "sampling without replacement". But all of this may be for a new topic in this forum.

Impenitent wrote:as an aside, what kind of tosser worries about seeming contradictory?

-Imp

The real interesting thing is when we end up with a Monty Hall problem, but expand on it. If we have a second person select randomly from two doors, instead of the original three, we know that person has a 50% chance of being a winner. However, the first person has a 2/3rds chance of winning if he switches. This means that the second person makes selections that are different from the first person, (assume they have no contact and are unaware of the other's selection), yet, they should select the same cases in the long run, if independent.

But in the Monty Hall problem, they aren't independent of each other.

ForgedinHell wrote:... Therefore, if one wants to uncover the "mind of god," one studies math. Spinoza's god is alive and well. All others are dead.

Therefore, the first person and the second person must make different selections to preserve the odds of winning.

Mike Strand wrote:OK, FiH, some of your writing about the Monty Hall problem and a second person wasn't clear to me. But in one place you said:

Therefore, the first person and the second person must make different selections to preserve the odds of winning.

I think my example showed that they don't have to make different selections, and the odds can still be calculated using theory based on independent events. In my previous post, I showed what would happen if they had to make different choices at the final stage, and the resulting probabilities depended on who went first: If the rule is for player 1 to go first (switching), he still wins with prob. 2/3 and player 2 wins with prob. 1/3 (he has to choose the only remaining door). If the rule is for player 2 to go first (flipping a coin to decide), then player 2 wins with prob. 1/2 and so does player 1 (who now is the one stuck with the only remaining door).

So if they must make different selections, as you suggest, the odds of winning actually change, but only for one or the other player, depending on who goes first in the final step.

I posed several variations involving two persons in my last post. Here's another, and you can tell me whether this is the one you meant.

Monty goes through the exercise with each person separately.

Case 1. Two sets of three doors, and the two players get separate sets. This gives player 2 (who flips a coin to decide between the two remaining doors) a 1/2 chance of winning, and player 1 (the switcher) the 2/3 chance. There are two prizes, and it's like two games on separate episodes of the program. Both winning their prizes has probability 2/3 times 1/2 = 1/3. Player 1 wins solo with prob. 2/3 times 1/2 = 1/3. Person 2 wins solo with prob. 1/3 times 1/2 = 1/6. Both losing has prob. 1/3 times 1/2 = 1/6. The litmus test of this analysis succeeds: The probabilities add up to 1.

Further, to verify the consistency of this with the overall probabilities, person 1 wins (regardless of whether 2 does) with prob. 1/3 (both win) + 1/3 (only player 1 wins) = 2/3 (that's good). And person 2 wins (regardless of whether 1 does) with prob. 1/3 (both win) + 1/6 (only player 2 wins) = 1/2 (good again).

Case 2. There is only one set of three doors. The two players are isolated from each other. Monty keeps track of the game on paper. Player 1 makes an initial choice of one of the three, then Monty shows him on paper which of the remaining doors is empty, and Player 1 decides to switch. Player 2 makes an initial choice of one of the three (could be different from the initial choice of player 1), Monty shows him on paper which of the other two doors is empty, and Player 2 flips a coin to decide between the remaining closed two doors.

Is Case 2 the one you were thinking of? If so, I invite you and other readers to analyze the probabilities of each player winning (solely or shared). Note that there still may have to be sharing, since there is only one prize and both may happen to select the prize door. Check your work against this: The probability of both winning in this Case, and sharing the prize, is 1/3; player 1 is the sole winner with prob. 1/3, and player 2 is the sole winner with prob. 1/6. Finally, both players lose (both go home with no prize) with prob. 1/6.

Arising_uk wrote:

ForgedinHell wrote:... Therefore, if one wants to uncover the "mind of god," one studies math. Spinoza's god is alive and well. All others are dead.

I thought it more that one studies Physics? As Maths can vary with the Physics. If you really want to know the mind of 'God' and you don't suit Maths then Formal(Symbolic) Logic is a pretty good subject as well. Even 'God' can't fuck with Logic.

Mike Strand wrote:OK, FiH, some of your writing about the Monty Hall problem and a second person wasn't clear to me. But in one place you said:

Therefore, the first person and the second person must make different selections to preserve the odds of winning.

I think my example showed that they don't have to make different selections, and the odds can still be calculated using theory based on independent events. In my previous post, I showed what would happen if they had to make different choices at the final stage, and the resulting probabilities depended on who went first: If the rule is for player 1 to go first (switching), he still wins with prob. 2/3 and player 2 wins with prob. 1/3 (he has to choose the only remaining door). If the rule is for player 2 to go first (flipping a coin to decide), then player 2 wins with prob. 1/2 and so does player 1 (who now is the one stuck with the only remaining door).

So if they must make different selections, as you suggest, the odds of winning actually change, but only for one or the other player, depending on who goes first in the final step.

I posed several variations involving two persons in my last post. Here's another, and you can tell me whether this is the one you meant.

Monty goes through the exercise with each person separately.

Case 1. Two sets of three doors, and the two players get separate sets. This gives player 2 (who flips a coin to decide between the two remaining doors) a 1/2 chance of winning, and player 1 (the switcher) the 2/3 chance. There are two prizes, and it's like two games on separate episodes of the program. Both winning their prizes has probability 2/3 times 1/2 = 1/3. Player 1 wins solo with prob. 2/3 times 1/2 = 1/3. Person 2 wins solo with prob. 1/3 times 1/2 = 1/6. Both losing has prob. 1/3 times 1/2 = 1/6. The litmus test of this analysis succeeds: The probabilities add up to 1.

Further, to verify the consistency of this with the overall probabilities, person 1 wins (regardless of whether 2 does) with prob. 1/3 (both win) + 1/3 (only player 1 wins) = 2/3 (that's good). And person 2 wins (regardless of whether 1 does) with prob. 1/3 (both win) + 1/6 (only player 2 wins) = 1/2 (good again).

Case 2. There is only one set of three doors. The two players are isolated from each other. Monty keeps track of the game on paper. Player 1 makes an initial choice of one of the three, then Monty shows him on paper which of the remaining doors is empty, and Player 1 decides to switch. Player 2 makes an initial choice of one of the three (could be different from the initial choice of player 1), Monty shows him on paper which of the other two doors is empty, and Player 2 flips a coin to decide between the remaining closed two doors.

Is Case 2 the one you were thinking of? If so, I invite you and other readers to analyze the probabilities of each player winning (solely or shared). Note that there still may have to be sharing, since there is only one prize and both may happen to select the prize door. Check your work against this: The probability of both winning in this Case, and sharing the prize, is 1/3; player 1 is the sole winner with prob. 1/3, and player 2 is the sole winner with prob. 1/6. Finally, both players lose (both go home with no prize) with prob. 1/6.