Suppose all 435 voting members of the U.S. House of Representatives join in on the same zoom call, which is scheduled to go from 9 a.m. to 10 a.m.. However, it is not required to attend the entire call, just any part of it. Each member randomly picks an exact moment to enter and leave the call somewhere within that one-hour range. What is the probability at least one Representative overlaps every other Representative on the call? In other words, seeing the face of every other member during his/her time on the call, not necessarily all at the same time.

anonymous

Click the button below for the answer.

2/3

Here is my solution (PDF).

This problem was asked and discussed in my forum at Wizard of Vegas.

It was adapted from the puzzle, Can You Join the World’s Biggest Zoom Call? at FiveThirtyEight.

Your basic strategy charts don't address what to do with a pair of aces, if the player has reached a re-splitting limit, and drawing to split aces is allowed.

J.R. from Las Vegas

It is extremely unlikely to find a blackjack game that allows drawing to split aces, to be dealt a pair of aces and then reach the splitting limit. Nevertheless, I endeavor to address the most obscure situations and admit my basic strategy tables at the time of this question didn’t address what to do in this situation.

The answer is to hit, except double if:

• The dealer has a six up (with any number of decks)
• The dealer has a five up with one or two decks.

Here is the expected value of this situation under various such situations.

### Expected Value of Hitting and Doubling Soft 12

Decks Stand
Soft 17
Dealer
Up Card
Hit
EV
Double
EV
Best
Play
1 Stand 5 0.182014 0.215727 Double
1 Hit 5 0.182058 0.215933 Double
1 Stand 6 0.199607 0.247914 Double
1 Hit 6 0.201887 0.258415 Double
2 Stand 5 0.169241 0.170637 Double
2 Hit 5 0.169339 0.171311 Double
2 Stand 6 0.192311 0.213109 Double
2 Hit 6 0.194397 0.227011 Double
4 Stand 5 0.162849 0.148228 Hit
4 Hit 5 0.162955 0.149183 Hit
4 Stand 6 0.18902 0.196249 Double
4 Hit 6 0.19074 0.211466 Double

Expected values taken from my blackjack hand calculator.

In the upcoming 2020 presidential election, what is the smallest percentage of the popular vote a candidate can receive and still win. Please assume everybody votes and for one of two candidates only.

anonymous

The answer is a candidate can receive as little as 21.69% of the popular vote and still win.

To elaborate, the following table shows the state by state population and electoral votes by state. The population numbers are taken as of 2019 and the electoral votes the last time they were adjusted in 2010. As a reminder to my readers outside of the United States, each state also gets a bonus two electoral votes. The result is states with a small population have much more influence on elections than those with a large population. As of the 2020 election, voters in Wyoming have almost four times the influence in the presidential election as those in Texas.

Given the rules, a candidate could get 100% of the vote in Texas, Florida, California ,North Carolina, New York, Georgia, Arizona, Virginia, Ohio, Pennsylvania, New Jersey, and Missouri, plus get half the vote (less one) in every other state and secure a total of 257,085,170 popular votes. Meanwhile, the other candidate would get 71,215,374 only, and win with exactly the needed 270 electoral votes.

The following table breaks it down. It is listed in order of million population per electoral vote (least to greatest).

### Electoral College Hypothetical Scenario

State Population Electoral
Million People
per Electoral Vote
Texas 28,995,881 38 1.311 - 28,995,881
Florida 21,477,737 29 1.350 - 21,477,737
California 39,512,223 55 1.392 - 39,512,223
North Carolina 10,488,084 15 1.430 - 10,488,084
New York 19,453,561 29 1.491 - 19,453,561
Georgia 10,617,423 16 1.507 - 10,617,423
Arizona 7,278,717 11 1.511 - 7,278,717
Virginia 8,535,519 13 1.523 - 8,535,519
Ohio 11,689,100 18 1.540 - 11,689,100
Pennsylvania 12,801,989 20 1.562 - 12,801,989
Colorado 5,758,736 9 1.563 2,879,369 2,879,367
Washington 7,614,893 12 1.576 3,807,447 3,807,446
New Jersey 8,882,190 14 1.576 - 8,882,190
Illinois 12,671,821 20 1.578 6,335,911 6,335,910
Massachusetts 6,949,503 11 1.583 3,474,752 3,474,751
Michigan 9,986,857 16 1.602 4,993,429 4,993,428
Tennessee 6,833,174 11 1.610 3,416,588 3,416,586
Missouri 6,137,428 10 1.629 - 6,137,428
Indiana 6,732,219 11 1.634 3,366,110 3,366,109
Maryland 6,045,680 10 1.654 3,022,841 3,022,839
Oregon 4,217,737 7 1.660 2,108,869 2,108,868
Wisconsin 5,822,434 10 1.717 2,911,218 2,911,216
Louisiana 4,648,794 8 1.721 2,324,398 2,324,396
South Carolina 5,148,714 9 1.748 2,574,358 2,574,356
Oklahoma 3,956,971 7 1.769 1,978,486 1,978,485
Minnesota 5,639,632 10 1.773 2,819,817 2,819,815
Kentucky 4,467,673 8 1.791 2,233,837 2,233,836
Alabama 4,903,185 9 1.836 2,451,593 2,451,592
Utah 3,205,958 6 1.872 1,602,980 1,602,978
Iowa 3,155,070 6 1.902 1,577,536 1,577,534
Nevada 3,080,156 6 1.948 1,540,079 1,540,077
Connecticut 3,565,287 7 1.963 1,782,644 1,782,643
Arkansas 3,017,825 6 1.988 1,508,913 1,508,912
Mississippi 2,976,149 6 2.016 1,488,075 1,488,074
Kansas 2,913,314 6 2.060 1,456,658 1,456,656
Idaho 1,787,065 4 2.238 893,533 893,532
New Mexico 2,096,829 5 2.385 1,048,415 1,048,414
Nebraska 1,934,408 5 2.585 967,205 967,203
West Virginia 1,792,147 5 2.790 896,074 896,073
Montana 1,068,778 3 2.807 534,390 534,388
Hawaii 1,415,872 4 2.825 707,937 707,935
New Hampshire 1,359,711 4 2.942 679,856 679,855
Maine 1,344,212 4 2.976 672,107 672,105
Delaware 973,764 3 3.081 486,883 486,881
South Dakota 884,659 3 3.391 442,330 442,329
Rhode Island 1,059,361 4 3.776 529,681 529,680
North Dakota 762,062 3 3.937 381,032 381,030
Alaska 731,545 3 4.101 365,773 365,772
DC 705,749 3 4.251 352,875 352,874
Vermont 623,989 3 4.808 311,995 311,994
Wyoming 578,759 3 5.184 289,380 289,379
Total 328,300,544 538 71,215,374 257,085,170

Sources:

Assuming a seven-out didn’t cause the Fire Bet to lose, how many rolls would it take to win on all six points, on average?

anonymous

There are two ways I can think of to solve this. The first is with a Markov Chain. The following table shows the expected rolls needed from any given state of the 128 possible.

### Fire Bet — Markov Chain

Point 4
Point 5
Point 6
Point 8
Point 9
Point 10
Expected
Rolls
No No No No No No 219.149467
No No No No No Yes 183.610129
No No No No Yes No 208.636285
No No No No Yes Yes 168.484195
No No No Yes No No 215.452057
No No No Yes No Yes 177.801038
No No No Yes Yes No 203.975216
No No No Yes Yes Yes 160.639243
No No Yes No No No 215.452057
No No Yes No No Yes 177.801038
No No Yes No Yes No 203.975216
No No Yes No Yes Yes 160.639243
No No Yes Yes No No 211.272344
No No Yes Yes No Yes 170.911638
No No Yes Yes Yes No 198.520513
No No Yes Yes Yes Yes 150.740559
No Yes No No No No 208.636285
No Yes No No No Yes 168.484195
No Yes No No Yes No 196.113524
No Yes No No Yes Yes 149.383360
No Yes No Yes No No 203.975216
No Yes No Yes No Yes 160.639243
No Yes No Yes Yes No 189.938796
No Yes No Yes Yes Yes 137.865939
No Yes Yes No No No 203.975216
No Yes Yes No No Yes 160.639243
No Yes Yes No Yes No 189.938796
No Yes Yes No Yes Yes 137.865939
No Yes Yes Yes No No 198.520513
No Yes Yes Yes No Yes 150.740559
No Yes Yes Yes Yes No 182.290909
No Yes Yes Yes Yes Yes 121.527273
Yes No No No No No 183.610129
Yes No No No No Yes 136.890807
Yes No No No Yes No 168.484195
Yes No No No Yes Yes 113.177130
Yes No No Yes No No 177.801038
Yes No No Yes No Yes 126.849235
Yes No No Yes Yes No 160.639243
Yes No No Yes Yes Yes 98.046264
Yes No Yes No No No 177.801038
Yes No Yes No No Yes 126.849235
Yes No Yes No Yes No 160.639243
Yes No Yes No Yes Yes 98.046264
Yes No Yes Yes No No 170.911638
Yes No Yes Yes No Yes 113.931818
Yes No Yes Yes Yes No 150.740559
Yes No Yes Yes Yes Yes 75.954545
Yes Yes No No No No 168.484195
Yes Yes No No No Yes 113.177130
Yes Yes No No Yes No 149.383360
Yes Yes No No Yes Yes 80.208000
Yes Yes No Yes No No 160.639243
Yes Yes No Yes No Yes 98.046264
Yes Yes No Yes Yes No 137.865939
Yes Yes No Yes Yes Yes 53.472000
Yes Yes Yes No No No 160.639243
Yes Yes Yes No No Yes 98.046264
Yes Yes Yes No Yes No 137.865939
Yes Yes Yes No Yes Yes 53.472000
Yes Yes Yes Yes No No 150.740559
Yes Yes Yes Yes No Yes 75.954545
Yes Yes Yes Yes Yes No 121.527273
Yes Yes Yes Yes Yes Yes 0.000000

Briefly, the expected rolls from any given state is the expected rolls until point is either made or lost (5.063636) plus the expected number of rolls if the player advances to a further state, divided by the probability of not advancing in state.

The other method uses integral calculus. First calculate the expected rolls for each possible outcome to happen. Then take the dot product of the probability of each event and average rolls to get the average rolls to resolve a pass line bet, which the lower right corner shows is 3.375758 = 557/165.

### Fire Bet — Expected Rolls

Event Probability Average Rolls Expected Rolls
Point 4 win 0.027778 5 0.138889
pt 5 win 0.044444 4.6 0.204444
pt 6 win 0.063131 4.272727 0.269743
pt 8 win 0.063131 4.272727 0.269743
pt 9 win 0.044444 4.6 0.204444
pt 10 win 0.027778 5 0.138889
pt 4 loss 0.055556 5 0.277778
pt 5 loss 0.066667 4.6 0.306667
pt 6 loss 0.075758 4.272727273 0.323691
pt 8 loss 0.075758 4.272727273 0.323691
pt 9 loss 0.066667 4.6 0.306667
pt 10 loss 0.055556 5 0.277778
Come out roll win 0.222222 1 0.222222
Come out roll loss 0.111111 1 0.111111
Total 1.000000 3.375758

From there we can get the expected rolls between any given point winning:

• Rolls between a point of 4 winning = (3/36)*(3/9)*5*(557/165) = 6684/55 = apx 121.527273.
• Rolls between a point of 5 winning = (4/36)*(4/10)*4.6*(557/165) = 1671/21 = apx 75.954545.
• Rolls between a point of 6 winning = (5/36)*(5/11)*(47/11)*(557/165) = 6684/125 = apx 53.472.

The expected rolls for a 10, 9, and 8 point winner are the same as for 4, 5, and 6, respectively.

Let's say that instead of a point-4 winner happening on a discrete basis, it follows an exponential distribution with a mean of 6684/55. The probability such a random variable lasts x units of time without happening is exp(-x/(6684/55)) = exp(-55x/6684).

The probability it has happened within x units of time, at least once, is 1-exp(-55x/6684).

If we represent all six points as continuous variables, then the probability all six have happened within x units of time is (1-exp(-55x/6684))^2 * (1-exp(-22x/1671))^2 * (1-exp(-125x/6684))^2.

The probability at least one of the six events not happening within x units of time is 1 - (1-exp(-55x/6684))^2 * (1-exp(-22x/1671))^2 * (1-exp(-125x/6684))^2.

We can get the expected time for all six events to happen by integrating the above from 0 to infinity.

Using this integral calculator gives an answer of 8706865474775503638338329687/39730260732259873692189000 = apx 219.1494672902.

Why this works is harder to explain, so please take that part on faith.