Why do coin toss results seem contradictory?

[ForgedinHell]

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.
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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.

But in the Monty Hall problem, they aren't independent of each other. The first person choices, if we ran them out as million times, would allow her to win 2/3rds of the time by switching. Yet, the second person who comes along, he cannot have a greater than 50% chance of winning. Therefore, the first person and the second person must make different selections to preserve the odds of winning. If the second person chose the same case the first person did, let's assume she switches every time, then he would have a 2/3rds chance of winning, when he should only have 50%. If he choses without knowing what the first person's choice his, then we can't say the probabilities shifted by increasing his information. So, what must happen is the choices of each party must differ to preserve the odds, if we were to run a simulation, this would occur. So, the preservation of the odds, must affect even seemingly independent conscious choice. That's my point.

[Arising_uk]

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

-Imp

I thought you an American?

[Mike Strand]

Hi, FiH! I believe we are on the same page, with a few misunderstandings. In your prior post, you said:

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.

This would appear to contradict your present post statement:

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

By the way, I refer other readers to the topic, "Monty Hall" Three Doors Problem" earlier in this forum for the needed background.

Assuming the two persons have no contact and are unaware of the other's selection assumes independence, but this is achieved simply by having the 2nd person flip a coin to decide between the two remaining doors (you said "have a second person select randomly from two doors...") -- which implies flipping a coin. This gives the 2nd person a 1/2 chance of winning, since the prize has to be behind one of the remaining doors. The 1st person has already decided to switch, giving that person the 2/3 chance of winning.

So yes, I agree (and my example bears this out) that the 2nd person has no greater than a 50% chance of winning. In fact, the chance is exactly 1/2 of being the sole winner, if person 1 walks away before the end of the game.

But now I have posed a natural "new shared game", where they can share the prize if both happen to choose the prize door, or one of them wins all, or they both lose, as I have shown in my example in my previous post. In this game the 2nd person has a 1/6 chance of being the sole winner, and a 1/3 chance of sharing with the 1st person, or a 1/2 chance of "winning" overall. Very reassuring, don't you think? -- and according to intuition. The 1st person, even in this shared game, keeps the advantage by switching, as I showed in that example: a 1/3 chance of sharing and a 1/3 chance of being the sole winner, or a 2/3 chance of winning overall.

If the game is run this natural way, with a new second player who selects between the two remaining doors with a coin toss, and with the 1st player still deciding to switch from his first choice, they can be in the same room with Monty and the audience, and the game still involves independent basic events: the first person's initial choice, the second person's flipping of the coin, and the first person's decision to switch.

Now if you want to change the game and have two people at the beginning choosing a door, they might choose different doors, and then only one door would be left, and the game breaks down. If they happen to choose the same door initially, then the game may proceed as I illustrated. Or, the second person could just follow along with the 1st person and win (and share the prize) with probability 2/3 or both lose with probability 1/3. A third possibility is to have the first person make the switch and force the second person to take the other remaining door. Then person 1 wins with probability 2/3 and person 2 wins with probability 1/3. A fourth variation is to have person 2 flip the coin to choose a door and force the 1st person to take the other remaining door. This would give them both a 1/2 chance of winning.

However, I think my version of the game in the example in my prior post is more interesting.

I think this should clear it up. Or you and I can invite a third party to check this out and comment accordingly.

[Arising_uk]

... 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]

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.

[ForgedinHell]

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.

It's this simple: If they did not make different selections, then the second player would not have 50% odds. The odds reflect reality. The reason why we know the monty-hall analysis is correct is because the person does have a 2/3rds chance of winning if she switches. If the second person chose the door that reflected the 2/3rds chance of winning, every time the selection was made, then the second player would have 2/3rds chance of winning and not 50%. I am stating what the reality is over the long-haul, which is the only way to verify the odds. To preserve the different odds for the two people, they must make different selections.

[Mike Strand]

Right on, ForgedinHell!

By the way, I appreciate your icon. In my philosophy class we're studying Spinoza right now -- his Theological-Political Treatise, and I saw the book about it on Amazon.com by Nadler. Any chance you ... "know" Nadler? My professor calls Spinoza's book the "radical enlightenment". I think Thomas Jefferson may have adopted many of Spinoza's ideas.

[ForgedinHell]

... 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.

The reference wasn't to any supernatural god, mainly in today's world, a play on words. And I think there is a difference between math and formal logic, although math is a form of logic. It's just the idea that since the world may be described mathematically, and math is logic, that one can say that there is an embedded logic in the cosmos. This embedded logic, for lack of something better, I referred to as god. You could just refer to it as the laws of physics or some such thing and it would be just as useful.

Here is the thing. Math is useful in assisting us in deriving physical laws, but the opposite is true. Physics allows us to derive mathematical theorems. One can actually use a fishtank, fill it with water, and attach it to a pole to prove the pythagorean theorem. Just as one example. The point is not the specific example, just the fact that one can go back and forth between the two disciplines so easily is fascinating.

Logically, I would probably think that the total number of odd and even numbers is greater than the total number of even numbers, but I would be wrong. There are as many even numbers as there are even and odd numbers combined. That's what the math tells me. Would symbolic logic give me the same answer? I don't know, just asking.

[ForgedinHell]

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.

Yeah, that's right. The thing is that there are a lot of interesting things that follow from the problem. I'm not sure that too many people have taken the problem to any additional steps.