COIN FLIPPING AND VOTING PROBABILITIES

Developed by the Association for Conceptual Studies
Presented by: F Richard Singer III Edition Date 9/2005

Note: For a MS Word Version of his paper contact me at is available richardsinger3@sbcglobal.net 

Overview: An explorable cluster is a group type constructivist learning resources having an initial realm of interest that is manifest to a specified type of target group. It suggests activities or stories and various considerations for discussing them. A successor realm is one that can be explored at any time after exploring the initial realm. For more about what this means use one of the links below.

Explorable Cluster Constructivist Learning Resources

Altho this present resource is an extension of one of the realms from the Rock Paper Scissors explorable cluster, it could be taken as an alternative initial realm, as it is presented here. Sections 1 and 2 are both group-plan constructivist-learning resources followed by a dialog type resource in which imaginary students use the plan. The group-plan is for any group of four or more students who are learning about probability concepts. The dialog is primarily for their teacher, altho it could also be used by the students. Section 1 is intended both to help students appreciate some of the mathematical concepts involved in probability and to obtain a feeling for how mathematical probabilities relate to what actually happens in coins flipping trials. This section can be skipped or skimmed by anyone who has a functional mastery of the concepts involved. However, running the trials and formulating the dialog actually expanded my perspective. Section 2 presupposes basic probability concepts and applies them to simple voting situations and on how voting probabilities might relate to reasons for voting. Altho an expanded perspective will suggest more live options, I am not advocating anything other than a flexible perspective that will make for a better discussion of the reasons different people might have for voting. Of course, I not disinterested. I hope that this will result in a less simplistic attitude towards voting and a greater openness to options that might otherwise be dismissed by adopting a rigid or narrower attitude.

Dialog: The dialog in constructivist learning resource provides a way to focus on contrasting ideas. It involves four students {Bob, Jan, Kay, Roy} and their mentor Jo. These characters are imaginary and not intended as typical. For one thing, their interest in concepts and their ability to use and communicate about concepts is above average and in discussing voting they use some precise ordinary conceptual tools that Jo has previously discussed with them. However, some of what they say may be wrong, even if it is not challenged. One reason for not trying to use typical students is that different persons have different characteristics. What is typical for one may not by typical for another. However these students at least represent some of the attitudes or styles that influence learning mathematics. Another reason these students are not typical is that they are being used to communicate ideas to adults, and adults are the primary intended audience for the dialog. However altho these students may not be typical of students who receive traditional schooling, they are more typical of students who have been learning for some time with constructivist attitudes.

Altho the dialog is fictional, trial results were actually obtained while playing the role of our students.

PNDP: PNDP (Public Net for Descriptive Psychology) is an organized network of theory neutral conceptual tools that includes much more than what most people associate with the word ‘psychology’. Jo has introduced these students to some of these tools. In section 2 they briefly use the concept of a core social practice and the conceptual relation between status and eligibilities. They also use the distinction between a model and a theory. See Appendix 2 for more details.

SECTION 1: COIN-FLIPPING PROBABILITIES

Activity: Call 6 coin flips times a trial. One way in which a trial might give the same number of heads and tails is heads, heads, tails, tails, heads, tails (HHTTHT). Record your results for 4 trials. Before doing so write your initial impression of the probability of getting the same number of heads and tails in a trial.

Considerations:

(1) Compare probability estimates to what actually happened. Tell why you made your estimates. Discuss your concept of probability.

(2) Determine the following probabilities when flipping a coin 6 times. Express them both as fractions and as decimals to the nearest thousandth. Compare these probabilities to what actually happened.

exactly 4 heads

exactly 5 heads

exactly 6 heads

more heads than tails

(3) What is the probability of getting the same number of heads and tails when flipping a coin 2 times? What is the probability of getting the same number of heads and tails when flipping a coin 4 times? Estimate what the probability of getting the same number of heads and tails when flipping a coin 8 times. What about 10 times?

(4) A game consists of trials in which a coin is flipped 3 times. Player M gets 4 points each time all coins match. Player D gets 1 point if they do not. Estimate the number of points each player likely to get per trial if the play a game with a large number of trials.

Jo: I will put your results on the board.

Jo: Here are your trial results. Those having the same number of heads and tails are in color.	Bob:	THTHHT	TTHHTH	HTTHTT	HHHTHH
	Jan:	TTHTHT	THTHHH	HTTTHT	TTTTHH
	Kay:	HHHTTH	HTTHTT	HTTTTT	TTHHTT
	Roy:	THHTTT	HHTHTT	TTTTHH	TTHTTT

Bob: I thought the same number of heads and tails would be most likely so I said the probability was about 2/3. I got 3/4 but the other got almost none. Maybe they made a mistake.

Jan: I also thought the same number of heads and tails would be most likely, but not by enough to matter so I said about 1/2. As you see none of my trial did so, so I tried 4 more times, but only got 1.

Kay: I thought the same number of heads and tails would be more likely than any other single possibility, but there are lots of other possibilities so I said about 1/3. As you see I did not get any which had the same, so I also tried 4 more times, and got 3.

Roy: I said about 30%. I figured that I might get either 1 or 2 in 4 trials, so I did not do any more. If we combine our results we get 4 out of 16 trial which is 25%. I will do 4 more trials.

Bob: In 4 more trials more and only got 1 even split. So maybe my first 4 trials gave unusual results.

Jo: Here are the results of our second set of trials.	Bob:	HHHHTT	THHHTH	TTHHTT	THTHTH
	Jan:	HHTTTH	HTHHHH	TTTHTT	HHHHTT
	Kay:	TTTHHH	THTHTT	HHTTTH	HTTHHT
	Roy:	HHTHHH	TTTHTH	THHHHH	HHHTTH

Roy: Altogether we have 9 out of 32 which is about 27%. This is a little lower than my estimate but still closer than before.

Kay: Instead throwing six coins in succession it would be faster to role 6 dice at a time. We can use even for heads, and only record the number of heads.

Bob: Okay, but are you sure throwing this involves the same probabilities?

Jo: Here are the resulting number of heads you each gave me for 8 more trials using dice.	Bob:	4	5	4	4	3	4	3	1
	Jan:	2	2	2	2	2	3	5	1
	Kay:	3	3	3	4	4	1	4	2
	Roy:	3	2	3	2	3	1	2	5

Jan: I made a table showing how many times we got each number of heads. The differences are minor	Number of Heads	0	1	2	3	4	5	6
	Using Coins	0	3	10	9	6	4	0
	Using Dice	0	4	9	9	7	3	0
	Combined	0	7	19	18	13	7	0

Roy: Looking at results can give you a feeling about the probabilities, but it cannot tell you what they are. If we did another run of 32 we would probably get different results, but they would most likely be similar.

Jan: That is why they call it probability. You cannot predict exact results, but you can have some reasonable expectations.

Bob: There was the same number of trials with 1 head as there were with 1 tail. Since heads and tails are equally likely, this could be expected. In general I would expect symmetric results. However 19 of our trials gave 2 heads and only 13 of them gave 2 tails.

Jan: I would also have expected that we would get 3 heads more often than 2 heads. Using probabilities, even reasonable expectations may fail. It doesn’t take long with dice, so let us each do 16 more.

Bob:	1	1	3	3	3	4	3	3	4	3	2	5	3	2	3	3
Jan:	1	3	3	1	4	1	3	5	2	3	3	3	2	1	3	1
Kay:	5	2	5	4	2	4	3	4	3	2	3	2	5	4	3	3
Roy:	3	3	4	2	5	3	4	2	4	4	3	5	3	3	3	2

Jo: Here is a table showing how many times we got each number of heads. The difference between the first and last set and there average is small.	Number of Heads	0	1	2	3	4	5	6
	First 64 Trials	0	7	19	18	13	7	0
	Last 64 Trials	0	7	11	28	11	7	0
	Combined	0	14	30	46	24	14	0
	Average for 64 Trials	0	7	15	23	12	7	0

Bob: The last 64 trials gave symmetric results. Could my remarks about symmetric results have caused this to happen? Also we thought there should be more trials with 3 heads and we got lots more.

Roy: The dice did not here your remarks. It is just a matter of chance.

Jan: Some people claim that we may have paranormal abilities and these that these could influence the role of the dice. Also it may be possible that our expectations influence the way we actually handle the dice. Its seems at least somewhat plausible to me that someone might learn how to throw dice in a way that they are more likely to get what they expect. However these results are not far enough from expectations that we need an explanation other than chance.

Kay: There are about 10²¹ {H,T} sequences of length 64 sequences. The probability of any one of them happening is very small. Yet exactly one of them must happen.

Jan: So something highly improbable is bound to happen.

Roy: Altho the specific sequence that gave these symmetric results has small probability, there are many such sequences. I have no idea of the probability of getting a totally symmetric result.

Jo: The question of whether results are too different from expectations to be a matter of chance is subtle. In statistical applications, experiments are usually run with the expectation that something other than chance is involved. Using a 95% confidence level, we often reject chance if there is less than a 5% probability that the result is a matter of chance. As we just observed, unlikely things are bound to happen. In addition to the statistical information, we need some other reason to believe that chance is not the only factor. That is one reason that the statistical inferences involving the paranormal are not widely accepted. This use of statistical tests is a topic we could explore in more detail, but I now want you to focus on your concept of probability rather than on how probability relates to what actually happens.

Bob: Well no matter what happens, for a single flip of a normal coin, T and H are equally likely. So the probability of H is 1/2 and the probability of T is 1/2.

Roy: A probability is a number between 0 and 1. With 3 possibilities each has a probability of 1/3, with 4 possibilities then each has a probability of 1/4, with 4 possibilities then each has a probability of 1/5, etc.

Jan: For rolling a die there are 6 faces so the probability of getting any one of them is 1/6. Since 3 of them are even this gives 1/2 as the probability. This is why we can use dice instead of coins.

Kay: If we flip 2 coins then we can get 0 heads or 1 head or 2 heads. These three possibilities are not equally likely because there are 4 more basic possibilities {HH, TH, HT,TT} and 2 of them give 1 head.

Jo: Let me summarize the concept of probability you seem to be using. A probability is a number between 0 and 1. If you have n basic events and if each is taken to be equally likely then the probability of each basic event is 1/n. Any other event is a combination of some number x of basic events. Its probability is x/n. This is a satisfactory concept for now, but in the future be prepared to expand it so basic events do not all have the same probability, as when flipping thumbtacks.

Jan: So the probabilities are 1/4, 1/2, 1/4. With 3flips HHH,THH,HTH,TTH,HHT,THT,HTT,TTT. The probabilities are 1/8, 3/8, 3/8, 1/8. of course you cannot have the same number of heads and tails.

Bob: With 4 we can just add an H and a T to what Jan listed. This gives 16 possibilities. The numbers for them are 1,4,6,4,1.

Roy: I have seen that pattern before. They are the numbers in Pascal’s Triangle. For 5 flips we would have 1,5,10,10,5,1 for a total of 32. For 6 flips 1,6,15,20,15,6,1 for a total of 64. This means that the probability of the same number of heads and tails in trial is 20/64 or 5/16 or .312. My 30% estimate was pretty close, and even closer than Kay’s.

Bob: I am not so sure about what Roy said about Pascal’s Triangle. Just because you see the same numbers in two patterns you cannot be sure this will keep on going. So I made a list of the 64 sequences and his numbers are correct.

Jan: I agree with Bob, and I suspect his confirmation for 6 means it might continue. But I would like to know why. Anyway we know the probabilities for trials of 6 flips.

4H: 15/6 » .234

5H: 6/64 » .093

6H: 1/64 » .016

more H than T 22/64 » .344

Kay: We can have more tails than head or the same number of each or more heads than tails. The probabilities are close 11/32, 10/32, 11/32 or about .344, .312, .344.

Jan: What happen on our first 64 was .406 .281 .312, on the next 64 it was .281 ,438, .281. For the all 128 we obtained .344 .359 .300.

Bob: Perhaps the more we do, the closer these numbers might become closer to the probabilities.

Kay: Bob had doubts about why Pascal’s Triangle should apply. I have an alternative analysis. There is clearly only one way to 0 head. To get 1 head it must occur in one of 6 places. To get 2 head you must pick 2 places for head. You can check that there are 15 ways to pick 2 out of 6 places.

Bob: Since 4 heads is the same as 2 tails, this also gives 15 sequences with 4 heads. This gives all of Roy’s numbers except 20, which you can get by subtracting the sum of the others from 64.

Kay: The formula m(m-1)/2 tells how many ways to pick 2 out of m places. This fits the numbers Roy would get using Pascal’s Triangle. I think we could find a general formula that tells how many ways there are to pick n out of m places.

Roy: I don’t know about a formula, but I know why Pascal’s Triangle works. Think about going from 4 flips to 5 flips. Just add an H and a T to each sequence of 5. Those to which you added a T will have the same number of heads. Those to which you add an H will have one more H.

Bob: I think see what you mean, here going from 5 to 6.	Number of Heads	0	1	2	3	4	5	6
	With T Added	1	5	10	10	5	1	0
	With H Added	0	1	5	10	10	5	1
	Total	1	6	15	20	15	6	1

Jan: We could just add adjacent numbers to get the next list, which would be 1 7 21 35 35 21 7 1 and the list for 8 would be 1 8 28 56 70 56 28 8 1. By the way, my probability estimate for the same number of heads and tails with 8 flip trials is .273.

Roy: Yes, and mine for 10 flip trials is .246.

Bob: Those are not estimates. You calculate 70/256 and 252/1024.

Kay: For the game in consideration 4, I think M will average about 1 point per trial and D will average about ¾ point per trial. This is what I got in 8 trials THT HHT THT HHH HTT HTH TTT THT. Of course, 8 trials will not tell you much.

Jan: I ran 8 more trials and got THH HTT THT TTH HTH HTH HHT THT. This leaves M with 8 points for .500 points per trial and gives a total of 14 points or for .885 points per .

Bob: I got HHT THT HTT HTH HTT HTT HHH TTH. I think the game favors D.

Roy:. I think the game favors M. I got THH HHH THH HHT THT HHT HHH HHH. This is 12 to 5 in favor of M. All told, this only gives 26 to 24 favor of D and I do not think this will last.

Bob: Here we go again. You expected M to have the advantage and your expectations were met.

Roy: And I still do not think the dice knew what I expected.

Kay: In spite of these results, Roy’s expectations are in line with the mathematical probabilities. If we kept going M would pull ahead.

Jan: I agree. There are 8 basic events HHH HHT HTH HTT THH THT TTH TTT. If each occurred in 8 trials, M would get 8 points and D would get 6 points. So the most reasonable points to expect for an 8-trial game is 8 to 6 in favor of M. Ultimately, the game should run about as Kay said.

Kay: If we gave 3 instead of 4 points for each match then both would get 6 points in that game. Eventually they should about break even.

Jo: I asked about long average point per trial. This is related to what is called the expected value of an event. In any situation where some numerical worth is assigned to an event, the expected value of that event is this worth times the probability of that event.

Roy: So the expected value of a match is 1 and the expected value of a difference is ¾.

Kay: But if we change the points for a match to 3 then both have an expected value of ¾.

Bob: Let me see if I have the idea. Since there are 2 ways out of 8 to obtain a match and 6 ways out of 8 to obtain a difference, the probability of a match is ¼ and the probability of a difference is ¾. If a match is worth 4 points its expected value is 4·¼ = 1. If a match is worth 3 points its expected value is 3·¼ = ¾. Since a difference is only worth 1 point its expected value is 1·¾ = ¾.

Jan: This gives another way to explain what I said as an expectation for a game. If the expected value for a match is 1 for a single event it should be 8 for 8 events, and if the expected value for a difference is ¾ for a single event then it should be 8·¾ = 6 for 8 events.

Bob: I am still not convinced. I ran 6 more 8 trial games. The number of mixed was 2, 4, 1, 2, 0, 1. This gives 40 to 38 in favor of M. Combined with our previous results gives 64 to 64, which is even. If we had only given 3 points for a match the score would have been 48 to 64 in favor of D, which is not very even.

Roy: If one of those 0 to 8 scores had been 9 to 5 the total would have been 57 to 61.You did not run enough trials. Keep going in the 3-point version M will pull even.

Jo: Perhaps, but can anyone see why D might stay ahead in the short run? To help you think about this I will indicate the possible scores for 8 session games using 3 points for a match and their probabilities.

Possible Scores	24 to 0	21 to 1	18 to 2	15 to 3	12 to 4	9 to 5	6 to 6	3 to 7	0 to 8.
Probabilities	<	<	.007	.023	.086	.206	.311	.267	.100
Our 10 Sessions					.1	.1	.3	.3	.2

Kay: I think I see how you found these.

Jo: I used something we have not discussed, so please ignore this for now. If you really want to see how I did this look in Appendix 1. I repeat my question. Why might D stay ahead in the short run?

Bob: It is making more sense to me. The most likely scores favor D about 37% of the time. They will favor M about 29% of the time. The unlikely and very unlikely scores strongly favor M. Unless one of these occurs, D will stay ahead.

Jan: You mean D is likely to stay ahead. Just because the 38% of likely scores favor D, does not mean that 38% will actually favor D. Using probabilities to see predict what happens in the short run is tricky.

Kay: Even with expected values in M’s favor M can run behind for a long time. Suppose a trial is 7 flips and M gets 64 points for a match. The expected value for a match is 1 and the expected value for a difference is 63/64. D has over a 60% chance of staying ahead for at least 30 trials.

Jo: There are many other questions that we could explore, but instead I want to make a final remark on expected value before we go on to another session next week. The concept of expected value can be used as model (not a theory) for making a decision even when something other than probabilities are involved. Suppose the cost of a lottery ticket is $1 and the payoff is high enough to make the expected value $1.25. I am opposed to the state advertising which encourages gambling. If the cost to my ideals is over 25¢ then the expected value of my buying a ticket is less that $1, and if it is more than $2 it is negative. For someone who endorses state lotteries the expected value could be more than $1.25. The worth of the excitement could also increase expected value of buying a lottery ticket. Furthermore when huge gains and small costs are involved the use monetary units to compute expected value is questionable.
SECTION 2: VOTING PROBABILITIES AND REASONS FOR VOTING

Story: B’s club is voting tonight for two candidates H and T who are running for club president. Altho B wants H to win, B has a chance to see a free concert instead of going to the meeting. B is undecided about whether to vote. B knows that 10 other voters are almost certain to attend and vote for H and that 10 voters are almost certain to attend and vote for T. There are 6 other undecided voters.

Considerations: Call B’s vote crucial if it would result in a tie or a win for H.

(1) Suppose that each of the 6 undecided voters is equally likely to vote for either way. What is the probability that B’s vote is crucial? Discuss H’s prospects for becoming club president.

(2) In addition to casting a crucial vote, think of some reasons that B might have for voting for H. Could these other reasons actually carry more weight than wanting H to win?

(3) Suppose B finds out that there are only 3 undecided voters because the other 3 so-called undecided voters have been convinced to vote for T. What is the probability that B’s vote could be crucial?

(4) Again suppose that there are actually 6 undecided voters, but that one of the firm supporters of H is too sick to go to the meeting and vote. What is the probability that B’s vote is crucial? What is the probability that B’s vote could make H obtain win 14 to 12?

(5) Think about other possibilities. Can you think of any in which there are 6 undecided voters and for which the probability that B’s vote is crucial would be greater than 1/3?

Roy: The probability that B’ vote is crucial is .312.

Bob: This is because the only way there would be a tie is if the same number of undecided voters vote for H as vote for T. This is the same probability as we had of getting the same number of heads and tails in a trial of 6 coin flips. If B cares about the outcome, B is giving up a 31% chance of determining it.

Jan: H and T each have about a 34% chance of winning that night if B does not attend and vote. If B does attend and vote there will not be a tie and H will have about a 65% chance of winning.

Kay: H might become president even if there is a tie. In fact, if everyone votes as before then H will win a runoff vote. So the chance of H becoming president is at least 31% and at most 65%. I doubt that probability applies, unless we have information about the probability that everyone will vote as before.

Jo: Kay makes a good point. To appropriately apply probability to a situation, you must treat events as if their occurrence is a matter of chance and you must have a reasonable way to assign probabilities to basic events. We already assumed that the way undecided members initially vote is a matter of chance. Unlike coin flipping, it is less clear that probability is a good tool to apply.

Jan: We also assumed that the undecided voters do not abstain. If they all abstain and the other members vote as expected then there will be a tie. H’s prospects depend on how they can later be influenced.

Roy: H’s chance is better than 50%, because counting B he has 11 votes and T only has 10.

Kay: I agree, but I would not regard this as a probability statement. Just using percentages to talk about the chance of some event does not mean probabilities are appropriate. It can be useful to talk about the chance of something when there is no clear way to assign a probability.

Jo: There is different concept that can be used, namely plausibility. A detailed formulation of plausibility concepts is available in the conceptual papers section of our website. However I suggest being somewhat loose in applying probability. You may make up probabilities that you cannot really support as long as they are somewhat reasonable. Just remember to only treat the conclusions that follow as suggestive.

Bob: I can think of other reasons for voting than the chance of influencing who becomes club president. Suppose that voting is one club’s core social practices. If this is more important to B than whether H or T becomes the next club president then B has a stronger reason to vote than just wanting H to win. Of course, any single vote only has a small effect on maintaining club practices.

Kay: I can think of a way to use the concept of expected value, altho the number I use will be arbitrary, and it would be highly unusual for anyone to explicitly use numbers in this way. Suppose B assigns a worth of 10 to having H win on the first vote and a worth of 1000 to not undermining the practice of voting. Suppose H feels that there is more than a .005 chance that his not voting will weaken this practice. Then the expected value of his vote is 10·(.31)+1000·(.005) = 8.1, and the part from maintaining the practice of voting is higher than the part from having H win on the first vote.

Roy: Using .005 as a probability is questionable, but as Jo said, we can use probabilities that we cannot really support. However if B assigns a value of 10 to going to the concert, B will still go to the concert.

Jan: Since voting is a core social practice of this club, never voting would significantly affect B’s status and eligibilities. Altho failing to vote under these circumstances may have at most a small effect, voting might enhance B’s status and increase some of B’s eligibilities, especially because H has influence. If we assign 2 as the status worth of voting, the expected value of voting is greater than going to the concert.

Bob: Another significant aspect of voting is to help H to feel supported. If we assign a 10 to this, we can up the expected value of B voting to over twice as much as going to the concert. Supporting H could be a sufficient reason even if B did not think his vote would make a difference.

Kay: In addition, the more votes H is able to obtain the more this will enhance the H’s influence, even if H looses. This gives B another reason to vote for H, further increasing the expected value of voting.

Roy: This could go on and on, but it is irrelevant. People don’t use numbers this way to make decisions. In fact, numerical probabilities may be minor in comparison to other factors that influence B’s decision.

Kay: I agree, but we are using this as a model rather than as a theory.

Jo: As Roy said, factors other than probabilities may play a major role in a person’s decisions about voting. However we can use expected values to model this. It might first be useful to examine some of the other considerations.

Kay: Since I am interested in probability regardless of its applications, I am ready to move on.

Roy: I also want to move on. For consideration (3), the probability is .125.

Bob: This is because there are 13 votes for T. For H to win all 3 of the truly undecided must vote for H.

Jan: Assuming that each is equally likely to go either way, (.5)³ is the probability H will win.

Bob: For consideration (4), if B votes then there are 10 sure votes each way. For a tie, the undecided must split 3 to 3. The probability of this is 31%. This will result in a runoff which H is likely to win.

Kay: So if B goes to the concert, then there is about a 65% rather than a 34% chance T will win.

Roy: There is a 23% B’s vote would make H win 14 to 12, but then H would have won 13 to 12 anyway.

Kay: So this does not effect the probability that B’s vote is crucial.

Jan: Let me summarize what we know about consideration (4). If B goes to the concert, there will be 31 votes. So we cannot have a tie and H only has about a 34% chance o win. On the other hand if B votes there is a 34% chance win in which B’ vote is not crucial and a 31 % chance of a tie in which B’s vote is crucial. If there is a tie and if everyone votes as before then H will win a runoff election. However I do not think we can assign a reasonable probability to H winning a runoff election.

Bob: Turning to consideration (5), if there had been 8 undecided voters the probability of vote being crucial would have been about 27%. With 10 undecided voters, it would be about 25%. We obtained these numbers in our coin flipping session.

Jan: The more undecided voters you have, the less likely prospects of vote being crucial.

Jo: In our session on coin flipping Kay talked about a formula. When we study combinations, we will obtain a formula for selecting n out of m places. However I agree with Roy that it is better at present to focus on using Pascal’s triangle, altho this would be tedious for a large number of undecided voters. Instead of giving you the formula, I am just going to indicate a rough estimate. Suppose the number of undecided voters is x and they are evenly split and leaning equally in both directions. Then the probability of a vote being crucial is than 1/Ö(x+1). A partial analysis of this is given in Appendix 1. If you want the formula, look there. If you want to wait until we study combinations then do not look there.

Kay: Going back to our original situation, undecided voters might not be leaning equally in both directions. Suppose that most of them were leaning more towards T. Suppose that for each voter there was about a 60% chance that that voter would vote for T. The probability of a tie should be less.

Jan: I agree, but how would you find this probability?

Roy: The probability is 20·(.4)³·(.6)³ » .276.

Jo: This uses some ideas we have not discussed. Can anyone see why this works?

Kay: For each the probability of voting for H is 40% or .4. To find the probability of HHHTTT, just multiply probabilities. This gives (.4)³·(.6)³. We already know there are 20 such ways to tie.

Bob: So to find the probability of several thing happening just multiply probabilities of each. With voters leaning 90% towards T, the probability of a tie is about 2%.

Jo: What if thing are interrelated? Suppose two of the undecided voters confer and decide to vote the same way and that they flip a coin to see which way to vote.

Jan: Then things are more complicated. I am not sure how to handle this, but if 4 of them confer and decide to vote the same way then B’ vote will not be crucial. Why would they flip a coin?

Bob: Maybe they believe in something like casting lots, or maybe they just want to avoid feeling responsible for the outcome. Recall when we looked at Explorable Clusters and decided to start with this cluster. I preferred starting with the cluster entitled Established Social Practices and Chance.

Jo: I had something simpler in mind. I just wanted to stress that they were equally likely to vote either way. Let us save reasons making a random choices until we look at the cluster Bob preferred.

Roy: If three confer, and all vote the same the probability that B’s vote is crucial is 12½%.

Bob: I agree, because there are 16 possibilities and only 3HTTT and 3THHH give a tie.

Kay: The same idea shows that the probability of B’s vote being crucial if two voters confer is 25%. There are 32 possible outcomes and 8 of them result in a tie.

Bob: Another complication arises if there is a chance that undecided voters may abstain.

Kay: If there were only 2 instead of 6 undecided then the probability that B’ vote is crucial would be 1/3. Letting A mean abstain, we could have {AA, HT, TH, AH, HA, HH, AT, TA, TT}.

Bob: If B votes there is a 2/3 chance of H winning, a 2/9 chance of a tie, a 1/9 chance of T winning.

Roy: If we let the probabilities be 1/2 for A, 1/4 for H, 1/4 for T then the probability of a tie is 3/8.

Bob: I think that see how you got 3/8, namely (1/2)·(1/2)+(1/4)·(1/4)+ (1/4)·(1/4).

Jan: Roy’s probability is about 38%. This is higher than Kay’s probability of about 33%. This makes sense. The greater the probability of an abstention the greater will be that a single vote is crucial.

Roy: With 3 undecided there each of {AAA, AHT, ATH, HAT, HTA, TAH, THA} give a tie. There are 27 possible voting sequences. If all are equally likely this gives 7/27 or about a 26% chance of a tie. Using the probabilities I gave the probability of a tie is 50%.

Kay: With 6 undecided some of whom may abstain, there are 141 ways for a tie, 1 tie with 6A, 30 with 4A, 90 with 2A, 20 with 0A. So the probability of a tie is 141/729 or about 19% (if all are equally likely). Using the probabilities Roy gave the probability of a tie is about 23%.

Bob: Here is another consideration. If the decide voters are not close to evenly split the probability of a single vote being crucial should be smaller than if they are about evenly split.

Jan: I worked the probability of a tie with 28 decided voters and 8 undecided voters and with the decided voters split in various ways. I used the Pascal triangle number we gave earlier for coin flipping.

Decided Voter Split	10-18	11-17	12-16	13-15	14-14	15-13	16-12	17-11	18-10
Probability of a Tie	.004	.031	.109	.219	.273	.273	.109	.031	.004

Jo: Latter when we study combinations, you will find that with 100 fully undecided voter the probability that they will split 60 to 40 for candidate H is about 1%. Thus if candidate T has a lead of 20 decided voters, the probability of single vote being crucial is only about .01 or 1/8 what it would be if the voters were evenly split. With 10000 fully undecided voters and T having a 2% vote lead, the probability is about .001. For a 3% lead it is less than .0001. Using the calculator on my computer and the formula for combinations these probabilities, it took very little time to obtain these probabilities.

Bob: So not only does a larger the number of undecided voters decrease the probability of a vote being crucial; with an unbalanced split among decided voters also decreases this probability.

Roy: With over a 10% lead it would be 0.

Jan: Suppose we call the expected value of a vote being crucial its crucial value. Even if there was only .0001 probability of being crucial, its crucial value could be worth some effort to a candidate, altho I am not sure how to assign a numerical worth in order to this calculate a crucial value.

Bob: As a rough estimate, we could use the amount the candidate is willing to spend on a campaign. I looked up information on senate campaigns. Many spend well over 10 million dollars, and sometimes a significant amount of this is their own money. With only a .0001 probability that this single vote will be crucial, this makes the crucial value over $1000 for the candidate.

Roy: I am not sure expected value is an appropriate concept to use. For a $100 contributor the crucial value of that vote is only 1¢. This is not why contributions are made.

Bob: I agree that votes are made for many reasons more important than being crucial. Margin of victory or loss is also important to candidates. A large victory margin has many advantages. A small margin in losing can be grounds for a comeback in a future election. However, I wanted to point out that even the crucial value may be significant for the candidate.

Kay: I think we can use expected value as merely a suggestive model. Any reasonable model should indicate that the crucial value for the candidate seems to be significant. For most voters it cannot be very high, so a voter will have other more important reasons for voting.

Jan: Not necessarily. Many people say that voting for a third party candidate is a wasted vote. It would seem that they are indicating that the crucial value of a vote is more important than other considerations.

Bob: Perhaps they have something else in mind. People also say that if too many people vote for a third party candidate then this could take away votes from the better major party candidate. The focus is not on the crucial value of a single vote.

Kay: Still, for most voters the crucial value of his or her vote is small whatever other people do. So I repeat that a voter who understands will have other reasons for voting.

Roy: Returning to my $100 contributor, I must admit the worth of a win is likely to far exceed the contribution. This contribution could provide political access. Furthermore, the probability that this contribution could influence the election is greater than the probability that a single vote will break a tie. The contributor hopes that his money will be used to influence more than a single voter.

Jan: I disagree with Kay. The worth of a win to a contributor may involve something other than political access. It can involve important social goals. The same can be said for many voters. The worth assigned to these can make a vast difference to them. For anyone who feels that the worth of having a candidate win is extremely high, the crucial value can be the most significant reason for voting.

Roy: For a person who says there is little difference between the candidates the crucial value could be insignificant. Perhaps this is why so many people do not vote.

Bob: Even if a person sees little difference between candidates or when the crucial value is insignificant for any other reason, that person my still decide to vote. In a safe district the crucial value of a vote is 0. About 400 of the 435 congressional districts are safe districts. In a presidential election, there are a number of safe states. The crucial value of a vote in any of these is also 0. This is one reason why presidential candidates do not focus on some states. Yet people in these states still vote.

Jan: As Bob and I observed earlier, the act of voting in B’s club is one of its core social practices. The same can be said about voting in a democracy. I do agree that this is a more important reason for some people than the crucial value, altho given the usual turnout it may be insignificant for others.

Kay: If we think in terms of expected value, the chance that a single vote can affect winning or anything else is much smaller in a major election than it would be in club election. However the values involved may be much higher, so the expected values could be just as high or even higher. It all depends on how important various matters are to the potential voter.

Bob: I have uncle who seldom votes for a major party. He has always said that the crucial value is insignificant to him in comparison to other factors. For one thing, he wants to have the status of being an independent voter.

Kay: I can also thin of various status considerations. For one thing, voting for a candidate who cannot win is gives you a different eligibility in a political discussion than you would have if you voted for the winning candidate.

Jo: We could continue this indefinitely. There are many of reasons voting choices, and I hope your appreciation of them has increased to some extent.

APPENDIX 1 TIE VOTE PROBABILITY

Overview: Suppose the number of undecided voters is an even number x and if they are evenly split and leaning equally in both directions. I will show that then the probability p of a tie less is than 1/Ö(x+1). To show this I use the formula for C(x,y) where C(x,y) denotes the number of ways for selecting y out of x places. This is usually referred to as the number of combinations of y out of x items.

C(x,y) = x!/(y!·(x-y!)) where x! denotes x factorial, i.e. x! = x·(x-1)·¼·2·1.

To obtain the formula below for p, divide the formula for combinations by the total number 2^x of voting possibilities.

p = (x!/(y!·y!))/2^x) = x!/(y!·2^y)² where y = x/2.

With 100 undecided voters 50 to 50 is the most likely split. That seems reasonable enough, but something that many people find strange is that the odds are greater than 10 to 1 (i.e. p < 1/10) that this will not happen. Of course, this may not seem strange if you realize how many other splits are possible. Nor will it seem strange that altho 5000 to 5000 is the most likely split with 10000 undecided voters, the odds against it occurring are greater than 100 to 1.

Estimating p: That p = 100!/(50!·2⁵⁰)² » .08 for x = 100. Likewise p = 10000!/(5000!·2⁵⁰⁰⁰)² » .008 for x = 10000. When I tried to approximate p for x = 1000000 and asked for 1000000! I got a message saying that this calculation would take a very long time and I was given the option of stop or continue. Without a calculator, even these earlier approximations would be tedious to obtain in a direct manner. However it is possible to reduce such expressions to a numerical value whose size is easy to estimate. I illustrate how I made the estimate using x = 12.

Consider x = 12, so y = 6 and 6!·2⁶ = 12·10·8·6·4·2,

p = 12!/( 6!·2⁶) = (11/12)(9/10)(7/8)(5/6)(3/4)(1/2).

Letting q = (12/13)(10/11)(8/9)(6/7)(4/5)(2/3) gives pq = 12!/13! = 1/13. Pairing factors of p and q it follows that p < q, and so p² < pq = 1/13. Thus p < 1/Ö13.

For x = 100, p = (99/100)(97/98)…(3/4)(1/2). Letting q = (100/101)(98/99)…(4/5)(2/3), pq < 1/101 and p < q. Thus p² < 1/101, and hence p < 1/Ö101 < 1/10. The same idea works for any value of x.

Remark: Calculations for x = 100 and for x = 10000 each gave p » .8/Öx. In fact they both showed that .7/Öx < p < .8/Öx. A calculation shows that p » .226 for x = 12, and this is also between .7/Ö12 and .8/Ö12. In general it can be shown that .7/Öx < p < .8/Öx (see Appendix A in Combination Functions for details).

Remark: The combination formula also can be used to give probabilities for other than even splits of fully undecided voters. The probability p that 100 fully undecided voter will split 60 to 40 for candidate H can be given by as 100!/(60!·40!·2¹⁰⁰) » .011. Thus if the other candidate T has a lead of 20 in the decided voters, the probability of a tie is only about 1/8 what it would be if the voters were evenly split. In a larger election with 10000 fully undecided voters and T having a 200 vote lead then the probability of a tie is 10000!/(5100!·4900!·2¹⁰⁰⁰⁰) » .001. For a 300 vote lead the probability of a tie is less than .0001.

A Constructivist Comment: Jo refers to results of the combination formula, but since she thinks that an attempt to help them construct it at this time would be distracting, she leaves it to them whether they want to look it up. In doing so, she is operating on a choice principle for constructivist mentoring with these students. Since they are unlikely to just memorize a formula without wanting to understand why it works, she does not mind them seeing or using it before they understand a rationale. However she knows that constructing a rationale can often be enhanced by either pure or guided discovery activities. Kay will not want to see the formula until she discovers it without any guidance. However Roy will be content to see the formula and then construct a rationale for it. Bob and Jan are likely to want obtain an intuitive understanding of specific cases that they can generalize.

APPENDIX 2: PNDP

Conceptual vs Paraceptual: The fact that a red bishop cannot take any piece on a black square is independent of any state of affairs in the world of chess. This is purely conceptual information within in our conceptual net for thinking about the world of chess. That a red bishop has placed your king in check uses this net to give information about a state of affairs this net is intended to help us understand. Such information is paraceptual in relation to this net.

¨ Conceptual information is about concepts and relationships between concepts in some net. Such information is known by working within the net.

¨ Paraceptual information presupposes some net, but is about some particular state of affairs that the net is intended to help access. It includes any information that is not purely conceptual.

Theories and Models: Altho the term theory still occurs in mathematics, it is not used in a paraceptual manner. Outside of mathematics, a theory involves paraceptual claims about some realm. It is intended to explain how some feature of this realm works or what is true within this realm. A model also relates to some paraceptual considerations about realm, but only in a suggestive manner that provides a perspective on what considerations about the realm might be plausible. Altho conceptual claims may be made within the model, none of these is to be to be considered as a paraceptual claim about the intended realm of application for the model. Due to the difference between the intent of a theory and of a model, holding incompatible theories is problematic while using incompatible models may be highly useful.

A Conceptual Net: As indicated earlier PNDP is a conceptual net, and thus of theory neutral network of conceptual tools. Jo has introduced these students to some of these tools for several reasons. One of her main reasons is to expand their appreciation of a conceptual study beyond it role in mathematics. Another reason is that PNDP illustrates aspects of conceptual study that are not apparent in mathematics.

Community Parameters: A PNDP community description uses some or all of the seven parameters below to characterize a community and differentiate it from other communities.

{members, statuses, concepts, locutions, social practices, choice principles, world}

Below is a simplified account of these parameters. For more see Anthony Putnam, Communities Advances in Descriptive Psychology Vol 1 p 195 or Mary Shideler Persons, Behavior and the World or see the Descriptive Psychology section of this website.

Members: To be a member of a community normally is to identify oneself as a member and to be recognizable as such by other members of that community. The distinction between members and non-members will also normally be recognizable to non-members. Furthermore, this distinction behaviorally significant, i.e. members will be treated in some manners differently than outsiders. Membership may be awarded by a formal ceremony, such as an initiation in which an individual become a member of sorority. It may be recognized with specific criteria but without ceremony, such as being a member of the community of Chicago residents. Both recognition and criteria may casual, as when an individual is merely recognized as belonging to the community of football fans.

Statuses: Having a status is to have a certain set of relationships. For any X each of X’s statuses refer to X’s place in some world in the broadest possible senses imaginable. An eligibility for X is being able to play certain role. Statuses determine P’s eligibilities, i.e. P’s potential for behavior. A status may or may not have anything to do with a community. For instance, an adult great white shark has the status of being at the top of a food chain, and this has considerable implications about what it can do. For a community the most basic status is membership, however in number of other statuses will be available in any community. They may be explicitly recognized, such starting point guard for the Boston Celtics. The may be more casual, such a person you can rely upon in a crunch.

Concepts: To engage in deliberate action a person must be able to make conceptual distinctions. The concepts of a community are those that are essential for meaningful participation in its practices, and especially in its core practices. These concepts may or may not be recognized by non-members, and when recognized by non-members they may not be understood in the way they are understood by non-members. Members share these concepts in being able to act upon them in a similar manner. For instance, the community of boy scouts uses the concept of an Eagle Scout, and furthermore this concepts is understood in terms of its merit badge requirements. An outsider may also be able to use this concept, but many will use it more vaguely and few outsiders to the scouting community will know the requirements.

Locutions: The locutions of a community may include the language spoken, such as English or French. More important, they include the ways in which it is spoken and the concepts and conceptual distinctions this indicates. This involves the use of jargon and terminology and expressions that are intertwined with the social practices of the community. The distinction between locutions of members of a community and non-members can vary from being minor subtle to being highly pronounced. For pilots and bridge players the term ‘ace’ represents different concepts, but members of each community, as well as outsider to both communities, could probably understand the essence of the difference. Of course an insider would use this term with greater sophistication. The community of mathematicians uses the terms ‘ring’ and ‘field’ in ways that have no apparent relation to their use in ordinary language. In fact the meaning use of these terms would be difficult to even explain to most non-mathematicians. However a mathematician who did not know English could easily acquire full use of these locution, having similar locution in his own language.

Social Practices: A community is especially distinguished by the things members do as members of the community and the way in which they do these things. These are the social practice of the community, and the point of being a member is to be eligible to engage in these practices. There are optional social practices, in the sense that a member can be in good standing without engaging in the practice. The are also core social practices, i.e. those that a member must engage in to be considered a member of the community. For instance, using planting wheat might be an optional social practice in a farming community. However planting some crop would be a core social practice, since o person who never planted a crop would not be considered a farmer.

Choice Principles: The actions of members as they engage in its social practices are guided by choice principle. Choice principles include any of ways a community accepts the justification of the behavior of its members. For instance, a member may appeal to custom or principles. Choice principles are often expressed in the form of value statements, norms, policies, slogans, etc. They are often illustrated in stories or myths.

Worlds: In describing what we do and think about we elements that we think of as {objects, processes, events, states of affairs}. A world for a person P is a large interrelated set of such elements that P is willing to act on. P will have a multitude of such worlds. For instance, P might have world W of cycling. That P’s bicycle tire has a nail would be a state of affairs W. P’s tire and tire gage are object in W. Having the tire go flat is an event in W. Repairing a flat tire is a process in W. There are other persons who have similar worlds. A cohesive set of such individuals sharing a common world forms a community of cyclists. In general, members of a community share a world. This means they react to this world by manifesting values and attitudes and interests that are similar in a multitude of ways, that they make distinctions in a common manner, and that they share social practice and choice principles.