Philosophy Now Forum

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

-Imp

A coin tosser or card shuffler or dealer who is trying to control the outcome, I would think. If he fouls up, the result contradicts his desires.

forgedinhell wrote:
Take a coin, evenly weighted. It will have a 50% chance landing heads and a 50% chance landing tails. This means if we toss it a million times, we should expect about have the results to be heads, and the other half to be tails. Otherwise, we wouldn't have 50/50 probability. That being said, what is the probability that a coin toss yields any of the following results?

(a) H, H, T, T, H, T, H, T, H, H

(b) T,T, T, T, T, T, T, T, T, T

(c) H, H, H, H, H, H, H, H, H, H

Which result is more likely to occur and how does that answer confirm that we should expect 500,000 heads to come up in a million tosses? Is there a paradox here?

You have to consider if the order is important.

Consider the simpler scenario of only 2 coin tosses C1 and C2

There are 4 possible outcomes (O1, O2, O3, O4) as per the following table:


O1 O2 O3 O4

C1 T H T H
C2 H T T H


The probability of an outcome is the product of the probability of its components results.

Thus, the outcome O1 is 0.5 x 0.5 = 0.25. In fact, they are all the same which is what seems initially puzzling.

However, if the order of the results is unimportant, then the probability of getting a heads and a tails is the sum of the two outcomes O1 and O2. That is 0.25 + 0.25 = 0.5. Wheras the probability of two heads or two tails is still 0.25.


There seems to be a residual uneasiness as one might think that a mixed specific ordered pattern of heads and tails seems to be more likely than a straight pattern of heads or a straight pattern of tails. However, I think this is mereley due to the ease with which we can grasp a straight repetitive pattern than grasping one that seems more random.

In reality both patterns have an equal chance of being realised. Imagine a four sided dice where each side represents an outcome of the 2 coin tosses.

ForgedinHell wrote:

mickthinks wrote:There is no paradox. There is no contradiction.

Then explain how flipping a coin with 50% odds of landing heads or tails, leads to less than a 50% chance of an even distribution of heads and tails being tossed after 4 tosses? Explain it?

Explain what? Throwing 1 head with one toss, and throwing 2 heads and 2 tails with 4 tosses, are different events with different probabilities.

It seems to me that the only thing that needs explaining, Einstein, is why you would expect them to be the same.

MGL wrote:

forgedinhell wrote:
Take a coin, evenly weighted. It will have a 50% chance landing heads and a 50% chance landing tails. This means if we toss it a million times, we should expect about have the results to be heads, and the other half to be tails. Otherwise, we wouldn't have 50/50 probability. That being said, what is the probability that a coin toss yields any of the following results?

(a) H, H, T, T, H, T, H, T, H, H

(b) T,T, T, T, T, T, T, T, T, T

(c) H, H, H, H, H, H, H, H, H, H

Which result is more likely to occur and how does that answer confirm that we should expect 500,000 heads to come up in a million tosses? Is there a paradox here?

You have to consider if the order is important.

Consider the simpler scenario of only 2 coin tosses C1 and C2

There are 4 possible outcomes (O1, O2, O3, O4) as per the following table:


O1 O2 O3 O4

C1 T H T H
C2 H T T H


The probability of an outcome is the product of the probability of its components results.

Thus, the outcome O1 is 0.5 x 0.5 = 0.25. In fact, they are all the same which is what seems initially puzzling.

However, if the order of the results is unimportant, then the probability of getting a heads and a tails is the sum of the two outcomes O1 and O2. That is 0.25 + 0.25 = 0.5. Wheras the probability of two heads or two tails is still 0.25.


There seems to be a residual uneasiness as one might think that a mixed specific ordered pattern of heads and tails seems to be more likely than a straight pattern of heads or a straight pattern of tails. However, I think this is mereley due to the ease with which we can grasp a straight repetitive pattern than grasping one that seems more random.

In reality both patterns have an equal chance of being realised. Imagine a four sided dice where each side represents an outcome of the 2 coin tosses.

The order is important, because it is how we establish the true sample space. If you ignore the order, you will be committing an error.

mickthinks wrote:

ForgedinHell wrote:

mickthinks wrote:There is no paradox. There is no contradiction.

Then explain how flipping a coin with 50% odds of landing heads or tails, leads to less than a 50% chance of an even distribution of heads and tails being tossed after 4 tosses? Explain it?

Explain what? Throwing 1 head with one toss, and throwing 2 heads and 2 tails with 4 tosses, are different events with different probabilities.

It seems to me that the only thing that needs explaining, Einstein, is why you would expect them to be the same.

You are too stupid to even comprehend the issue.

LOL NO U!

Forgedinhell wrote:
I forgot to add something. If we just tossed the coin twice, then we have the following possibilities: H H; H T; T H; T T. Each possibility is equally likely, 1/4 chance of any of the four coming up. And 2 of the four possibilities have one head and one tail, so when we toss the coin twice, we do have 50% odds of heads coming up. So, why is it that the odds shifted down to 6/16 or 3/8 when we toss the coin two additional times? Shouldn't the higher number of tosses make it more likely that we end up with 50% heads, since we are eliminating the randomness associated with small sample sizes? Yet, just the opposite happened when we went from 2 to 4 tosses.

You make an interesting point here, which I have to admit I did not expect. That is to say, my first intuition was to think that the probability that a series of coin tosses that result in an exactly even distribution of heads and tails should certainly not decrease as the number of coin tosses increase. However, surely this just means my intuition about probability is wrong. You also asked where our expectation of 50% probability for heads ( or tails ) comes from, if the probability of getting an exact even distribution of heads and tails is so low. Surely this is just because the distribution of probabilities always cluster around the even distribution which is always going to have the highest probability in any number of coin tosses. That is to say the probability of 50% heads will always be higher than 49/51% which in turn will be higher than 48/52% etc.

It may help to refer you to the binomial probability function, which deals with the general problem. This function appears in spreadsheet programs (e.g. EXCEL) and in tables of values (so you don't have to run through the involved arithmetic) found in standard textbooks in probability theory or statistics. If you're ambitious and have the background, you can derive the formula from scratch.

Here is an application of the binomial probability function: Given an event with probability p of success, the function gives you the probability of x successes in n trials of the event, where the successes can occur in any order. Let's take the event of tossing a coin. A success is heads showing (or tails showing, you choose which), and p is 0.5, the probability of success. Then, for example, throwing 7 coins at once (or one coin 7 times in succession) constitutes n=7 trials of the event. Taking the example further, the probability of x=3 heads (and thus 4 tails) in 7 tosses is 0.2734 from the binomial function. By symmetry, this is also the probability of 4 heads and 3 tails in 7 tosses. The order of the heads and tails is of no concern here, consistent with the way games of chance are usually played. If you specify an order, the probability is 0.5 raised to the seventh power, or 0.0078.

The binomial probability function is the basis for analyzing games of chance (cards, dice, etc.), and the gaming houses use it, along with the betting mechanisms, to make profits.

FiH wrote:

If we just tossed the coin twice, then we have the following possibilities: H H; H T; T H; T T. Each possibility is equally likely, 1/4 chance of any of the four coming up. And 2 of the four possibilities have one head and one tail, so when we toss the coin twice, we do have 50% odds of heads coming up. So, why is it that the odds shifted down to 6/16 or 3/8 when we toss the coin two additional times? Shouldn't the higher number of tosses make it more likely that we end up with 50% heads, since we are eliminating the randomness associated with small sample sizes? Yet, just the opposite happened when we went from 2 to 4 tosses.

If I understand, you are concerned with the probability of exactly two heads in four tosses, in any order. Before appealing to the binomial function, let's go from scratch: hhtt, htht, htth, thht, thth, tthh are the only outcomes with 50% heads. Each of these six outcomes has probability 0.5 to the fourth power, or 0.0625. Six times this is 0.3750 (or 3/8, as you wrote). The probability of exactly half of the tosses being heads has gone down because the total number of possible outcomes relative to those with exactly 50% heads, each with probability .0625, has gone up -- in fact there are 16 possible orderings now: one with all heads and one with all tails, four with 1 head, four with one tail. Plus the six with two heads and two tails makes 16 in all.

If you toss a coin 20 times, the probability of exactly 50% heads goes down to .1762.

However, the probabilities of outcomes that are much different from 50% are also going down. With four tosses, the probability is 0.1250 that there is 25% (1) or less heads or 75% (3) or more heads. But with twenty tosses, (check me with the binomial function), the probability is only 0.0414 that there is 25% (5) or less heads or 75% (15) or more heads. With increasing sample size, the percentage of heads begins to "cluster" tighter around 50%!

Even with a huge sample size, the probability of exactly 50% heads is low, but the other possible outcomes are so tightly clustered around 50% (or 50.001%, if the coin is slightly unbalanced in favor of heads) that you have a good idea of what the actual probability of heads is.

I think this is what MGL was getting at.

By the way, thanks, Grendel, for earlier listing explicitly the 16 possible outcomes of tossing a coin four times, and thus making clear the 6/16 probability of getting exactly 50% heads. And I think I'm just repeating, in another way, what you said here:

It's very simply, it's to do with distribution, 499,999 tails to 500,001 heads is nearly 50/50 and so on. More of these nearly 50/50 probabilities are distrubuted closer to the mean than the higher and lower ends of the range, so the result will always be closer to 50/50.

That is, the higher the number of tosses, the more the various possible outcomes cluster around the 50% outcome.

Mike Strand wrote:FiH wrote:

If we just tossed the coin twice, then we have the following possibilities: H H; H T; T H; T T. Each possibility is equally likely, 1/4 chance of any of the four coming up. And 2 of the four possibilities have one head and one tail, so when we toss the coin twice, we do have 50% odds of heads coming up. So, why is it that the odds shifted down to 6/16 or 3/8 when we toss the coin two additional times? Shouldn't the higher number of tosses make it more likely that we end up with 50% heads, since we are eliminating the randomness associated with small sample sizes? Yet, just the opposite happened when we went from 2 to 4 tosses.

If I understand, you are concerned with the probability of exactly two heads in four tosses, in any order. Before appealing to the binomial function, let's go from scratch: hhtt, htht, htth, thht, thth, tthh are the only outcomes with 50% heads. Each of these six outcomes has probability 0.5 to the fourth power, or 0.0625. Six times this is 0.3750 (or 3/8, as you wrote). The probability of exactly half of the tosses being heads has gone down because the total number of possible outcomes relative to those with exactly 50% heads, each with probability .0625, has gone up -- in fact there are 16 possible orderings now: one with all heads and one with all tails, four with 1 head, four with one tail. Plus the six with two heads and two tails makes 16 in all.

If you toss a coin 20 times, the probability of exactly 50% heads goes down to .1762.

However, the probabilities of outcomes that are much different from 50% are also going down. With four tosses, the probability is 0.1250 that there is 25% (1) or less heads or 75% (3) or more heads. But with twenty tosses, (check me with the binomial function), the probability is only 0.0414 that there is 25% (5) or less heads or 75% (15) or more heads. With increasing sample size, the percentage of heads begins to "cluster" tighter around 50%!

Even with a huge sample size, the probability of exactly 50% heads is low, but the other possible outcomes are so tightly clustered around 50% (or 50.001%, if the coin is slightly unbalanced in favor of heads) that you have a good idea of what the actual probability of heads is.

I think this is what MGL was getting at.

By the way, thanks, Grendel, for earlier listing explicitly the 16 possible outcomes of tossing a coin four times, and thus making clear the 6/16 probability of getting exactly 50% heads. And I think I'm just repeating, in another way, what you said here:

It's very simply, it's to do with distribution, 499,999 tails to 500,001 heads is nearly 50/50 and so on. More of these nearly 50/50 probabilities are distrubuted closer to the mean than the higher and lower ends of the range, so the result will always be closer to 50/50.

That is, the higher the number of tosses, the more the various possible outcomes cluster around the 50% outcome.

But since the percentage went down from two tosses to four, it is not correct that the higher the volume the more we get to 50/50. The order is important, and cannot be ignored in setting up the sample space. We can easily see this if we want to know what the probability is of tossing a head and a tail in two tosses. If we ignore the order, then our sample space would only consist of 3 possibilities, (H,H); (T,T); (H,T). So, we would compute the odds of having one heads and one tails as 1/3rd, since there would be one way for this to occur out of three sample spaces. However, when we look at order, we get (T,T); (T,H); (H,T); (H,H). This gives us four possibilities, with two of the four involving one head and one tail, so the odds then become 2/4 or 50%.

I think to get an intuitive grasp of this coin-toss thing requires getting familiar with a probability distribution drawn as a graph. The graph consists of bars of height corresponding to the probabilities of getting 0 heads, 1 head, 2 heads, and so on up to N heads, where N is the number of tosses. These heights add up to 1.00, no matter how big N is -- no matter how many tosses.

The entire graph looks sort of like a symmetrical mountain, with peak corresponding to one bar at N/2 if N is even, and two bars at (N-1)/2 and (N+1)/2 if N is odd. That is, if you toss a coin 5 times, you can never get 50% heads, but you can get 2 heads (40%) or 3 (60%) heads.

Imagine, you can get 50% heads if you throw 2 coins or 4 coins or 6, but you can never get 50% heads if you throw 1 coin or 3 coins or 5, etc. -- that alone suggested to me that more than my earlier intuitions was involved.

The bar(s) corresponding to the peak may gradually get lower with increasing N, but the sides of the mountain become steeper, indicating that more and more of the total probability (1.00) is becoming concentrated around the peak and closer to the bar(s) defining the peak. Less and less of the total probability is found out in the foothills of the mountain, further away from the middle (at around 50% heads).

By the way, in computing all these bar heights (which can be accomplished with the binomial probability function), all possible outcomes (four in the case of tossing 2 coins; 16 in the case of tossing 4 coins) have to be accounted for and comprise, as you suggest, the "sample space". But the probabilities of individual outcomes yielding a given number of heads are combined to draw the probability distribution graph. For example, in tossing 4 coins, there were six outcomes that yeild two heads, and that becomes the middle bar of height 6/16, or 0.375. So in the graph for 2 tosses there are three bars corresponding to no heads, 1 head, 2 heads. And in the graph for 4 tosses there are five bars corresponding to 0 heads, 1 head, 2, 3, and 4.

Mike Strand wrote:I think to get an intuitive grasp of this coin-toss thing requires getting familiar with a probability distribution drawn as a graph. The graph consists of bars of height corresponding to the probabilities of getting 0 heads, 1 head, 2 heads, and so on up to N heads, where N is the number of tosses. These heights add up to 1.00, no matter how big N is -- no matter how many tosses.

The entire graph looks sort of like a symmetrical mountain, with peak corresponding to one bar at N/2 if N is even, and two bars at (N-1)/2 and (N+1)/2 if N is odd. That is, if you toss a coin 5 times, you can never get 50% heads, but you can get 2 heads (40%) or 3 (60%) heads.

Imagine, you can get 50% heads if you throw 2 coins or 4 coins or 6, but you can never get 50% heads if you throw 1 coin or 3 coins or 5, etc. -- that alone suggested to me that more than my earlier intuitions was involved.

The bar(s) corresponding to the peak may gradually get lower with increasing N, but the sides of the mountain become steeper, indicating that more and more of the total probability (1.00) is becoming concentrated around the peak and closer to the bar(s) defining the peak. Less and less of the total probability is found out in the foothills of the mountain, further away from the middle (at around 50% heads).

By the way, in computing all these bar heights (which can be accomplished with the binomial probability function), all possible outcomes (four in the case of tossing 2 coins; 16 in the case of tossing 4 coins) have to be accounted for and comprise, as you suggest, the "sample space". But the probabilities of individual outcomes yielding a given number of heads are combined to draw the probability distribution graph. For example, in tossing 4 coins, there were six outcomes that yeild two heads, and that becomes the middle bar of height 6/16, or 0.375. So in the graph for 4 tosses there are three bars corresponding to no heads, 1 head, 2 heads. And in the graph for 4 tosses there are five bars corresponding to 0 heads, 1 head, 2, 3, and 4.

I agree that the graph idea is a good way to allow us to see what is going on. And it seems to me that even in the case of two tosses, there is no logical reason to expect one head and one tail, because at most this happens only 50% of the time.

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. So, not only does the coin toss seem to be affected by prior tosses, when we look at four tosses, we can expand on this and see how it affects human behavior. Literally, this shows an underlying mathematical logic to the cosmos. One that governs us all. Therefore, if one wants to uncover the "mind of god," one studies math. Spinoza's god is alive and well. All others are dead.

Thanks, ForgedinHell, I'm glad the graph idea helped. There are pictures of such graphs in books and on the internet, or they can be generated using software such as Excel.

Correction -- in my last post, second-to-last sentence. It should read "So in the graph for 2 tosses there are three bars corresponding to no heads, 1 head, 2 heads." In general, for N tosses there will be N+1 bars or spikes in the graph. From left to right, the first spike height is the probability of throwing 0 heads in N tosses; the second spike height is the prob. of throwing 1 head in N tosses, and so on to the last spike on the right, the probability of throwing N heads in N tosses.

If an event has probability of success other than 0.5, a graph can still be drawn, and it will have a peak to one side or the other, but the shape won't be symmetrical.

If the horizontal range is kept the same (N+1 bars all within a page-width, say 10 inches, no matter how large N is), as N gets bigger and bigger, the graph mountains get steeper and steeper until the picture approaches a vertical line. The line is in the middle (at 5 inches) for a graph depicting a balanced coin toss. For a graph that depicts rolling a single fair die with a success being a particular face, such as the boxcar (6 showing on the die), the line would be at 10/6 = 1 and 2/3 inches from the left.

Regarding the Monty Hall problem -- if the two people both pick the winning door, let them sell the prize and share the proceeds. Otherwise let the winner keep the whole prize. The first person will select the winning door more often, and so has a better chance of getting the whole prize. I guess the probabilities could be worked out, but not here.

In the theory behind the coin toss, the assumption is made that the tosses are independent -- that the probability of getting heads on a given toss does not affect the probability on any subsequent toss (tosses are "without memory"). Any toss in a sequence of tosses has a probability of heads of 0.5 for a balanced coin. Otherwise, the theory becomes more complex. For two tosses, when we calculate the probability of HT, it is 0.5 times 0.5 = 0.25. This holds also for TH, TT, and HH. This assumes independent tosses. If it were, say, more likely for a tail to appear after a head were tossed, or visa-versa, these fundamental probabilities would change, and the going would get tougher, and the binomial probability function could not be applied to get the distribution graphs.

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.