How Likely is a Tie in the U.S. Electoral College Vote?

Transcription

1 How Likely is a Tie in the U.S. Electoral College Vote? Linus Schrage University of Chicago Alex Orden University of Chicago Mark Wiley LINDO Systems 16 November 2000 People sometimes ask if a tie is possible in a U.S. Presidential election electoral vote outcome. In fact, not only is a tie possible, there are 107,281,775,062,476 different ways in which states can vote so that a tie can occur, given the year 2000 allocation of electoral votes to states. The next question one might ask is: If there are so many ties possible, why have we not seen a tie in the previous 100 years? The answer is that there are a lot of different outcomes possible, specifically, 16,888,498,602,639,400 different possible outcomes. The ratio of outcomes with a tie to all possible outcomes is Thus, if all outcomes are equally likely, we would expect a tie in about every 157 elections. That probability is low enough to be consistent with not having seen a tie as yet, however, it is high enough that a well designed election system should be prepared for such an eventuality. Loose Ends and Complications The states of Maine and Nebraska explicitly allow their electoral votes to be split among two or more candidates, whereas all other states vote all their electoral votes for the single candidate obtaining the highest popular vote in the state. Our calculations take into account the special behavior of Maine and Nebraska. What About Modest Alternatives of the Current System? The curious might ask what is the effect of the special flexibility available to Maine and Nebraska? The answer is that if Maine and Nebraska were all-or-nothing states, then the probability of a tie increases to So allowing Maine and Nebraska to split their electoral vote decreases the probability of a tie. Suppose we allow all states to split their vote. This would be somewhat closer simply using the popular vote to elect the president. It still gives smaller states more power because the number of electoral votes awarded to a state equals its number of representatives in the House + 2. The number of representatives in the house is approximately proportional to population. Surprisingly, allowing all states to split their electoral votes arbitrarily substantially increases the probability of a tie, to Thus, we would expect a tie about every 29 elections. What seems to be going on here is that allowing a small state, such as Maine or Nebraska, to split its votes increases the number of possible outcomes more than it increases the number of ties. Allowing a large state, such as California, to split its vote introduces lots of additional opportunities for ties.

2 Methodology The number of possible ties and the total number of possible outcomes was calculated with the LINGO model shown in the Appendix A. The basic approach is a standard one: When faced with an apparently difficult problem, solve an even bigger and more general problem, such that the solution to our specific problem falls out as a special case. The special problem is to count the number of ways in which 51 states(washington, DC counts as a state as far as the electoral college is concerned) can cast their electoral votes so that the Democrats get exactly 538/2 = 269 votes. We assume that only two parties get votes. The general problem that is solved(by dynamic programming) is: For j = 1 to 51: For i = 1 to 269: Compute WAYS( i, j) = the number of ways that the Democrats can get exactly i electoral votes using just the first j states. We are interested in the special case: i = 269, j = 51. If state j has V(j) electoral votes and state j is an all or nothing state, then WAYS( i, j) = WAYS( i, j-1) + WAYS( i-v(j), j-1). In case one is interested in generating an example vote configuration that results in a tie, the Appendix B contains a What sbest spreadsheet model that allows one to easily generate tie solutions, perhaps with additional conditions such as: If California votes Republican, is it still possible to have a tie?. Statistical Approaches When all states can split their electoral votes and each vote has probability.5 of going to the Democrats, then the number of votes for the Democrats has a binomial distribution with n = 538, p =.5. The probability of a tie is then easily computed as above to be The Normal approximation to the binomial is good in this case. The mean is 269, with a standard deviation of This gives the probability of a tie as One can also use the Normal approximation when all states vote all-or-nothing. The mean is still 269, however, the standard deviation is This gives the probability of a tie as Note, this is slightly off from the exact probability of computed above. Appendix A! LINGO model to count the number of ways that a tie can result in U.S. Presidential Election Electoral College. LINGO can be downloaded from ; SETS: STATE: VOTES; KNT/1..269/:; KXS( KNT, STATE): WAYS; ENDSETS 2

3 DATA: STATE, VOTES = Maine 4 Nebraska 5 Alabama 9 Alaska 3 Arizona 8 Arkansas 6 California 54 Colorado 8 Connecticut 8 Delaware 3 DC 3 Florida 25 Georgia 13 Hawaii 4 Idaho 4 Illinois 22 Indiana 12 Iowa 7 Kansas 6 Kentucky 8 Louisiana 9 Maryland 10 Massachusetts 12 Michigan 18 Minnesota 10 Mississippi 7 Missouri 11 Montana 3 Nevada 4 New_Hampshire 4 New_Jersey 15 New_Mexico 5 New_York 33 North_Carolina 14 North_Dakota 3 Ohio 21 Oklahoma 8 Oregon 7 Pennsylvania 23 Rhode_Island 4 South_Carolina 8 South_Dakota 3 Tennessee 11 Texas 32 Utah 5 Vermont 3 Virginia 13 Washington 11 West_Virginia 5 Wisconsin 11 Wyoming 3 ; ENDDATA!Definition: WAYS( I, J) = number of ways of get a Democrat electoral count of I, 3

4 using only the first J states;! Get started with Maine and Nebraska. They are special in that they can split their KNT( I) I #LE# VOTES( 1): WAYS( I, 1) = KNT( I) I #GT# VOTES( 1): WAYS( I, 1) = 0! If Maine is an all or nothing state, replace above by this;! WAYS( VOTES( 1), 1) = KNT( I) I #NE# VOTES( 1): WAYS( I, 1) = 0! Now do KNT( I): WAYS( I, 2) = WAYS( I, 1) KNT( K) I - K #GT# 0 #AND# K #LE# VOTES( 2): WAYS( I-K,1))! Now the rest of the STATE( J) J #GT# KNT( I) I #LT# VOTES( J): WAYS( I, J) = WAYS( I, J - 1 WAYS( VOTES( J), J) = WAYS( VOTES( J), J-1) + 1;! We can get to I,J either without this state or with KNT( I) I #GT# VOTES( J): WAYS( I, J) = WAYS( I, J - 1) + WAYS( I - VOTES( J), J - 1)! Finally, number of ways of getting 269 votes using all states; TIES = WAYS( STATE)! Number of possible vote outcomes of any kind; ALL = (VOTES(1)+ 1)* (VOTES(2)+1)*2^(@SIZE( STATE) -2! If Maine and Nebraska are all or nothing states, turn this on;! ALL = 2^@SIZE( STATE! It may be of interest to look at the numbers in a spreadsheet. Turn on the following to send results to an appropriately named spreadsheet containing appropriately pre-defined "C:\P123\ELECTORAL.XLS") = "C:\P123\ELECTORAL.XLS") = ALL;!ENDDATA; 4

5 Appendix B A What'sBest model for finding tie solutions to the Electoral College vote in the U.S. Presidential Election. Note What'sBest is downloadable from:electoral State Votes Go for Dems? Alabama 9 0 Alaska 3 0 Arizona 8 1 Arkansas 6 0 California 54 1 Colorado 8 1 Connecticut 8 1 Delaware 3 0 D.C. 3 0 Florida 25 0 Georgia 13 1 Hawaii 4 0 Idaho 4 0 Illinois 22 0 Indiana 12 0 Iowa 7 1 Kansas 6 0 Kentucky 8 1 Louisiana 9 0 Maine1 1 0 Note, Maine is special in that Maine2 1 0 it may split its 4 electoral votes. Maine3 2 0 Maryland 10 0 Massachusetts 12 0 Michigan 18 0 Minnesota 10 0 Mississippi 7 1 Missouri 11 0 Montana 3 0 Nebraska1 1 0 Note, Nebraska is special in that Nebraska2 2 0 it may split its 5 electoral votes. Nebraska3 2 0 Nevada 4 0 New Hampshire 4 0 New Jersey 15 1 New Mexico 5 0 New York

6 North Carolina 14 1 North Dakota 3 1 Ohio 21 0 Oklahoma 8 1 Oregon 7 1 Pennsylvania 23 1 Rhode Island 4 0 South Carolina 8 1 South Dakota 3 1 Tennessee 11 1 Texas 32 1 Utah 5 0 Vermont 3 1 Virginia 13 1 Washington 11 0 West Virginia 5 1 Wisconsin 11 1 Wyoming 3 0 Total= 538 For dems 269 Constrain the votes for Dems to equal 538/2----> = To tie 269 6