## Poll

19 votes (48.71%) | |||

13 votes (33.33%) | |||

6 votes (15.38%) | |||

2 votes (5.12%) | |||

12 votes (30.76%) | |||

3 votes (7.69%) | |||

6 votes (15.38%) | |||

5 votes (12.82%) | |||

11 votes (28.2%) | |||

8 votes (20.51%) |

**39 members have voted**

You guys probably already know this…Quote:Wizard

Where φ is the golden ratio of 1.618, the long side (A) of the large rectangle is: 2*(φ^-2 + 1)^-0.5 = 1.7013. The short side (B) is: A / (2φ) = 0.52573.

The long side of the small rectangle is: A / φ = 1.0515. The short side is: B / φ = 0.32492

- Voters rank their choices from first to last.
- If any candidate gets over 50% of the first place vote, they win.
- Otherwise, the candidate with the least votes is eliminated. The second choice votes on those ballots are assigned to those candidates as first place votes.
- There is a second round of ballot counting, based on the reallocated votes from the first round. If any candidate gets over 50% of the first place vote, they win.
- Otherwise, the candidate with the least votes in the second round is eliminated. The second place votes on those ballots are assigned to those candidates as first place votes. However, if any of these second choice votes were to the candidate who lost in the first round, then the third choice votes are assigned to those candidates as first place votes.
- There is a third round of ballot counting, based on the reallocated votes from the second round. If any candidate gets over 50% of the first place vote, they win.
- Otherwise, the candidate with the least votes in the third round is eliminated. The second place votes on those ballots are assigned to those candidates as first place votes. However, if any of these second choice votes were to the candidate who lost in the first or second rounds, then the third choice votes are assigned to those candidates as first place votes. If the second and third choices were the two candidates who got eliminated in the first and second rounds, then the fourth choice votes are assigned to those candidates as first place votes.
- This process keeps repeating until a winner is found.

Here was the voting in Alaska for congress.

Source: https://www.elections.alaska.gov/results/22PRIM/ElectionSummaryReportRPT.pdf (see page 2).

You can see there were 22 total candidates and 152,174 votes cast. However, not every precinct has reported at the time of this writing, so let's say the total votes are 200,000.

Assuming the same 22 candidates and 200,000 votes, what is the fewest first-choice votes a candidate can get and still win?

Let's ignore ties for the winner, to keep things simple. Let's assume if there is a tie for loser, then all the next choice votes among the candidates in a tie for last get redistributed at the same time.

Quote:Wizard...Assuming the same 22 candidates and 200,000 votes, what is the fewest first-choice votes a candidate can get and still win?...

This is single tranferable vote under another name. Let's assume the winner has everyone's second vote. Working backwards the winner needs to land up with 100001 votes to 99999. Thus candidate #2 had 99999 votes.

Prior to that round the winner needed 50001 and candidate #3 had 50000 votes.

The previous round winner needed 25001 and candidate #4 had 25000 votes.

Continuing this process (nearly halving each round) enables the winner to have started with 2 votes. Roughly the winner votes goes 2x-1 on each round, picking up all the votes of the lowest candidate. For simplicity assume some cadndidates had no votes (otherwise have them having 1 each taken from #2.)

The starting values for each candidate could have been something like 99999, 50000, 25000, 12500, 6250, 3125, 1562, 781, 391, 195, 98, 49, 24, 12, 6, 3, 2, 1, with the winner starting with 2. So the winner wins each round 2-1, 3-2, 5-3, 8-6 etc.

There is a sidebet that is based on the Blackjack total of three cards (probably from one standard deck given the base game!) using the Blackjack rule that all Aces count as 1, except the first Ace counts as 11 if the total would have 11 or less. Using this rule which total is most common. (Hint: this landed up being the sidebet!)

Quote:charliepatrickSimple problem based on an interesting side bet I stumbled across while looking for something. I've assumed infinite decks but suspect the answer's the same for finite ones.

There is a sidebet that is based on the Blackjack total of three cards (probably from one standard deck given the base game!) using the Blackjack rule that all Aces count as 1, except the first Ace counts as 11 if the total would have 11 or less. Using this rule which total is most common. (Hint: this landed up being the sidebet!)

link to original post

Quote:charliepatrickJust two votes!

This is single tranferable vote under another name. Let's assume the winner has everyone's second vote. Working backwards the winner needs to land up with 100001 votes to 99999. Thus candidate #2 had 99999 votes.

Prior to that round the winner needed 50001 and candidate #3 had 50000 votes.

The previous round winner needed 25001 and candidate #4 had 25000 votes.

Continuing this process (nearly halving each round) enables the winner to have started with 2 votes. Roughly the winner votes goes 2x-1 on each round, picking up all the votes of the lowest candidate. For simplicity assume some cadndidates had no votes (otherwise have them having 1 each taken from #2.)

The starting values for each candidate could have been something like 99999, 50000, 25000, 12500, 6250, 3125, 1562, 781, 391, 195, 98, 49, 24, 12, 6, 3, 2, 1, with the winner starting with 2. So the winner wins each round 2-1, 3-2, 5-3, 8-6 etc.

link to original post

I agree!

Quote:charliepatrickSimple problem based on an interesting side bet I stumbled across while looking for something. I've assumed infinite decks but suspect the answer's the same for finite ones.

There is a sidebet that is based on the Blackjack total of three cards (probably from one standard deck given the base game!) using the Blackjack rule that all Aces count as 1, except the first Ace counts as 11 if the total would have 11 or less. Using this rule which total is most common. (Hint: this landed up being the sidebet!)

link to original post

Doing a brute force number-crunching search, I get 21 for any number of decks in the shoe from 1 to 100, with 20 being the second most likely result each time.

Note that the result is 22, with 23 the second most likely, if the first Ace counts as 11 even if the total would be more than 11 if all of the Aces counted as 1.