I may have to lay down a couple of bucks myself.

[jclalor 12]

You have to like these odds.

[Croc 0]

I don't know anything about betting. What do those odds mean in plain English?

"Here's a good specimen of my own wisdom. Something is so, except when it isn't so."

Charles Fort, commenting on the many contradictions of astronomy

[quade 4]

Croc

I don't know anything about betting. What do those odds mean in plain English?

The lower the number, the more people have placed a be on that person to win. The odds aren't being set by individual betting companies, but by the math of how many people have placed bets.

Again, lower number = more people have betted thinking that person is going to win.

The numbers can change over time too.

https://en.wikipedia.org/wiki/Parimutuel_betting

quade -
The World's Most Boring Skydiver

[jcd11235 0]

Croc

I don't know anything about betting. What do those odds mean in plain English?

For example, 1/3, i.e., one to three against, means you pay $3 for a chance to win $1 (plus the return of your $3) if Clinton wins. In other words, the odds suggest Clinton has a 75% chance of winning in November (disregarding the bookie's rake).

Math tutoring available. Only $6! per hour! First lesson: Factorials!

[Coreeece 2]

Croc

I don't know anything about betting. What do those odds mean in plain English?

If you bet $100 on Rubio, you could win $94,900.

I take cash, check, paypal and all major credit cards.

Never was there an answer....not without listening, without seeing - Gilmour

[turtlespeed 226]

Coreeece

***I don't know anything about betting. What do those odds mean in plain English?

If you bet $100 on Rubio, you could win $94,900.

I take cash, check, paypal and all major credit cards.

Funny how Cruz wasn't on there - but Rubio was.

I'm not usually into the whole 3-way thing, but you got me a little excited with that. - Skymama
BTR #1 / OTB^5 Official #2 / Hellfish #408 / VSCR #108/Tortuga/Orfun

[SkyDekker 1,465]

jcd11235

***I don't know anything about betting. What do those odds mean in plain English?

For example, 1/3, i.e., one to three against, means you pay $3 for a chance to win $1 (plus the return of your $3) if Clinton wins. In other words, the odds suggest Clinton has a 75% chance of winning in November (disregarding the bookie's rake).

Just a reminder that the bolded part is false. Odds are set based on expected betting as well as actual betting. They have very little to do with the actual possibility of an outcome.

[jcd11235 0]

SkyDekker

******I don't know anything about betting. What do those odds mean in plain English?

For example, 1/3, i.e., one to three against, means you pay $3 for a chance to win $1 (plus the return of your $3) if Clinton wins. In other words, the odds suggest Clinton has a 75% chance of winning in November (disregarding the bookie's rake).

Just a reminder that the bolded part is false. Odds are set based on expected betting as well as actual betting. They have very little to do with the actual possibility of an outcome.

No, it was correct as I wrote it. Probability can be directly obtained from odds and vice versa.

ETA: http://www.had2know.com/academics/calculate-odds-probability.html

Math tutoring available. Only $6! per hour! First lesson: Factorials!

[quade 4]

jcd11235

In other words, the odds suggest Clinton has a 75% chance of winning in November (disregarding the bookie's rake).

As determined by;

1) People who place bets.
2) People NOT US citizens (since it's illegal for them to place those bets).

None of that is particularly scientific in terms of determining who will win.

quade -
The World's Most Boring Skydiver

[jcd11235 0]

quade

***In other words, the odds suggest Clinton has a 75% chance of winning in November (disregarding the bookie's rake).

As determined by;

1) People who place bets.
2) People NOT US citizens (since it's illegal for them to place those bets).

None of that is particularly scientific in terms of determining who will win.

Actually, it's determined by the oddsmakers who work for the bookies. It's a safe bet (so to speak) that their methodology is more scientific than nearly all of the individual bettors.

The line can move, and often does in response to more people placing bets on one side than the other (among other reasons), but it's still fixed odds betting, not parimutuel betting. The bettors don't directly set the odds collectively. The odds are fixed at time of betting. If you place a bet at 1:3 on Clinton, you'll win $1 for each $3 you lay, even if the odds change to 1:4 later (at which point you'd need to lay $4 to win the same $1).

Math tutoring available. Only $6! per hour! First lesson: Factorials!

[quade 4]

In that case, then it's not the actually odds of the person winning, but the odds at which the companies think they can remain profitable, which is still not the same as the odds of the person winning the election.

quade -
The World's Most Boring Skydiver

[jcd11235 0]

quade

In that case, then it's not the actually odds of the person winning, but the odds at which the companies think they can remain profitable, which is still not the same as the odds of the person winning the election.

That's a meaningless distinction. The true probability, and hence the true odds, cannot be calculated and are not known. It's not like selecting red and white ball from urns in which we know the proportion of each color. It's much more like trying to calculate the probability of the Buffalo Bills beating the Seattle Seahawks on Monday, November 7, 2016. In the real world (as opposed to ball and urn problems), we rarely know true population parameters. We have to rely on estimates.

The best estimates we have of the true probability are the probability provided by sportsbooks and other statistical analysts, e.g., Nate Silver. They've put significant time and energy (and, in most cases, money) into making sure their estimators as accurate and unbiased as possible.

In the case of the probability provided by oddsmakers via the sportsbooks, if a bettor believes that the given probability that, for example, Clinton wins, is too high, then they place a wager on Trump. If they think it's too low, then they place a wager on Clinton.

Or, perhaps a bettor feels like the probability of a Clinton win is going to approach 0.5 as voters get more information about the candidates and their policies. In that case, they might place a wager on Trump now and a similar sized wager on Clinton in early November. If Trump wins, their winnings will more than cover the loss from their wager on Clinton, and if Clinton wins, they'll still (approximately) break even.

Math tutoring available. Only $6! per hour! First lesson: Factorials!

[SkyDekker 1,465]

Quote

That's a meaningless distinction.

No it isn't. A $10 billion bet, if accepted at a bettinghouse, would dramatically change the payout ratios. It won't drastically change the mind of voters and hence the odds of election don't actually change.

[jcd11235 0]

SkyDekker

Quote

That's a meaningless distinction.

No it isn't. A $10 billion bet, if accepted at a bettinghouse, would dramatically change the payout ratios. It won't drastically change the mind of voters and hence the odds of election don't actually change.

No, it wouldn't. Not directly, anyway. This isn't parimutuel betting. It's fixed odds betting. The odds are changed manually and only apply to future wagers.

Such a bet as you propose would not be accepted, because the bookmaker knows that there's either a 75% probability of losing $3.33 billion or a 25% probability of losing $30 billion (depending only upon which candidate the bettor bets). It's too much risk. It's no different from an insurance company electing to not sell a $10 billion life insurance policy on a single individual. Selling the policy or not doesn't affect the probability of the individuals death (by a covered cause) during the coverage period. It just represents too much financial risk.

It's a meaningless distinction between true probabilities and probability estimates because most real world probabilities are based on parameter estimates; true parameter values are unknown and, largely, unknowable, since the entire population cannot be measured. That doesn't mean the estimates are inaccurate or biased. Quite the opposite, they are usually quite good. Consider Nate Silver's electoral college predictions for the last two presidential elections. Do you really believe he just flipped a coin for each state and got lucky?

Math tutoring available. Only $6! per hour! First lesson: Factorials!

[SkyDekker 1,465]

Quote

Such a bet as you propose would not be accepted, because the bookmaker knows that there's either a 75% probability of losing $3.33 billion or a 25% probability of losing $30 billion (depending only upon which candidate the bettor bets). It's too much risk.

Or on a smaller scale, the odds would be changed to attract more betting on the other party to mitigate the risk.

Quote

true parameter values are unknown and, largely, unknowable, since the entire population cannot be measured. That doesn't mean the estimates are inaccurate or biased. Quite the opposite, they are usually quite good.

Which would apply to closing odds, not opening odds which was what OP posted.

[jcd11235 0]

SkyDekker

Or on a smaller scale, the odds would be changed to attract more betting on the other party to mitigate the risk.

That depends on how much confidence the bookmaker has in the estimator. It also depends on whether previous wagers are part of the model. As more information is fed into the model, the resulting estimate can change.

Quote

***true parameter values are unknown and, largely, unknowable, since the entire population cannot be measured. That doesn't mean the estimates are inaccurate or biased. Quite the opposite, they are usually quite good.

Which would apply to closing odds, not opening odds which was what OP posted.

You're talking about two different probabilities as though they should be the same. The probability that Clinton wins the election given the information publicly available in May is different from the probability that Clinton is elected given the information publicly available in early November. The two values do not have to be similar to both be accurate.

Math tutoring available. Only $6! per hour! First lesson: Factorials!

[SkyDekker 1,465]

Quote

That depends on how much confidence the bookmaker has in the estimator. It also depends on whether previous wagers are part of the model. As more information is fed into the model, the resulting estimate can change.

Which shows it is separate from the odds of something happening.

[kallend 2,175]

If the true probability of an event happening were the same as what people believe to be the probability of that event happening, Las Vegas NV would be a ghost town.

...

The only sure way to survive a canopy collision is not to have one.

[jcd11235 0]

SkyDekker

Which shows it is separate from the odds of something happening.

No, it absolutely does not. If it did, then we could simply wait until after the election and then claim that the probability of [insert winning candidate's name] winning the election was exactly 1 all along. That would be absurd. The true probability changes as the available information changes.

Math tutoring available. Only $6! per hour! First lesson: Factorials!

[jcd11235 0]

kallend

If the true probability of an event happening were the same as what people believe to be the probability of that event happening, Las Vegas NV would be a ghost town.

No it wouldn't. Bet on a coin toss in Vegas, and you'll have to lay down $110 for a chance to win $100. That extra $10 is the vigorish. The vigorish is the primary source of gambling profits. Without it, Vegas would be a ghost town because oddsmakers tend to be pretty good at developing accurate estimators. That's their goal (in the majority of cases, where they don't possess information not available to the general public), because bettors are allowed to wager on either side.

Math tutoring available. Only $6! per hour! First lesson: Factorials!

[kallend 2,175]

The lavish hotels on the strip were paid for by gamblers who thought they would win.

...

The only sure way to survive a canopy collision is not to have one.

[jcd11235 0]

kallend

The lavish hotels on the strip were paid for by gamblers who thought they would win.

Yes, because the loser pays the vigorish. That's an indication of the effectiveness of the models developed by the oddsmakers. If the models weren't reasonably accurate and unbiased, the sportsbooks in Las Vegas could make no money, because most bettors would wager on the side that provides the highest expected value.

Math tutoring available. Only $6! per hour! First lesson: Factorials!

[jcd11235 0]

Note: I'm replying to this post, but it seemed more topical to this thread.

quade

******I can think of no better way to commit political suicide than accept the VP slot on a Trump ticket. It's nothing but downside.

You don't think there's upside if he wins? Don't get me wrong; I don't want to see him in the Oval Office. But I also think the probability of him winning is non-zero.

Infinitesimal is technically non-zero. Sure.

So, yes, I'm open to the possibility of Trump possibly winning on some fluke of nature, but I can not in all seriousness see it happening without something catastrophic happening first.

That said, my statement stands, I see nothing but downside for anyone who accepts the VP offer.

You provided a link to http://www.270towin.com. I decided to run the numbers. In the interest of reproducibility:

I accessed the Webpage Tuesday, May 10, 2016.

First, I selected the "2016 Toss-Up Map" from the drop down menu (located below the Gulf Coast).

Next, I clicked on Map Options, and selected "Safe, Likely, Leaning, Tossup". Note that this did not change the shade of any state color, despite trying in three browsers (Safari, Chrome, Firefox). This might indicate individual states strongly favor one candidate or another, or it might indicate poor Web site coding.

The map indicates ten states are toss-up states, Florida, Pennsylvania, Ohio, North Carolina, Virginia, Wisconsin, Colorado, Iowa, Nevada, and New Hampshire.

I did two separate calculations for a Trump win probability. For both, I assumed that states already assigned to one candidate or another will ultimately be won by that candidate.

I wrote my code in:

R version 3.3.0 (2016-05-03) -- "Supposedly Educational"

Copyright (C) 2016 The R Foundation for Statistical Computing

Platform: x86_64-apple-darwin13.4.0 (64-bit)

I commented extensively so that the code can (hopefully) be followed by those not familiar with R. The # symbol indicates everything that follows on that line is a comment.

For the first calculation I assumed that each candidate is equally likely to win each of the toss-up states. This is not exactly correct, but is pretty close.

results = matrix( nrow = 1024, ncol = 10)

# Initialize 1024 x 10 matrix


colnames( results) = c( "FL", "PA", "OH", "NC", "VA", "WI", "CO", "IA", "NV", "NH")

# Each column represents tossup state


stateElectoralVotes = c( 29, 20, 18, 15, 13, 10, 9, 6, 6, 4)

# Electoral votes for each state FL, PA, OH, etc. (see column names)


for (i in 1:10) {


results[ , i] = rep( c( 1, 0), each = 2 ^ (10 - i), length = 1024)


}

# Create all 2^10 possible results for 10 tossup states

# Note that results are mutually exclusive.


trumpElectoralVotes =

apply( results, 1, function( X) {191 + sum( X * stateElectoralVotes)})

# Assign electoral votes for tossup victories, total, and add to 191


trumpWinProb = length( trumpElectoralVotes[ trumpElectoralVotes > 269]) / 1024

# Each of 1024 possible tossup state results is equally likely. so

# find number of outcomes in which Trump wins at least 270 electoral votes

# and divide by 1024 possible outcomes to obtain probability of Trump

# victory.


# display probability value

trumpWinProb


[1] 0.2929688

With our equally likely assumptions, we have obtain a probability of a Trump general election victory of just over 29 percent.

Next, I wanted to see what happened if I assigned Clinton a win probability of 55 percent for each toss-up state. Any higher than that, and the state should fall into one of the "leaning" or "likely" categories, so this value should provide a reasonable lower bound for Trump's general election victory probability.

Since each outcome is no longer equally probable with this assumption, I elected to use simulations. I simulated the election in the toss-up states 100,000 times.

# Simulate 100000 election results for tossup states, giving Clinton a 55% 

# probability of winning each state.


set.seed( 11235)

# set seed so that random number generator results can be reproduced


stateElectoralVotes = c( 29, 20, 18, 15, 13, 10, 9, 6, 6, 4)

# Electoral votes for each state FL, PA, OH, etc.


victoryCount = 0

# set counter to zero


for (i in 1:100000) {


trial = sample( c( 1, 0), size = 10, replace = TRUE, prob = c( .45, .55))


if (191 + sum( trial * stateElectoralVotes) > 269) {


victoryCount = victoryCount + 1


}


}

# Run 10000 simulations, keeping count of number of Trump wins


trumpWinProb2 = victoryCount / 100000

# Calculate percentage of simulations Trump wins general election


# display probability value

trumpWinProb2


[1] 0.2054

With our assumption that Clinton has a 55 percent probability of winning each toss-up state, we find that Trump has about a 20.5 percent chance of winning the general election.

We can see that the 1:3 odds, implying a 75 percent chance of a Clinton victory, are reasonable. Our calculations suggest 71-80 percent.

Further, it is clear that Trump's chance of victory is not infinitesimal, as claimed by Paul.

Math tutoring available. Only $6! per hour! First lesson: Factorials!

[quade 4]

I think you're making assumptions not in evidence.

An example would be this comment;

Quote

# Each of 1024 possible tossup state results is equally likely.

The odds of every toss up state going for Trump is considerably less than equal to half of them going for Trump. There's probably a bell shaped curve in there.

quade -
The World's Most Boring Skydiver

[jcd11235 0]

quade

I think you're making assumptions not in evidence.

An example would be this comment;

Quote

# Each of 1024 possible tossup state results is equally likely.

The odds of every toss up state going for Trump is considerably less than equal to half of them going for Trump. There's probably a bell shaped curve in there.

It appears you understand neither math nor probability.

The assumption is that the probability of winning a toss-up state is 50 percent, which is reasonable, hence why the state is categorized as a "toss-up" state (think coin toss) by your source site. Given that, each of the 1024 possible outcomes for those states is equally likely. Of course, not every toss-up state is exactly 50/50. In some, Clinton has a slight edge, in others, Trump has a slight edge.

In the second calculation, I addressed your concern, and gave Clinton a fairly large edge in every toss-up state. That's not likely, but it does provide us with a reasonable worst case for Trump's chances. Clinton still only wins about 80 percent of the time.

While I didn't include it in the post, I also ran the simulation giving Clinton an unrealistic probability of 2/3 for winning each toss-up state. Trump still wins over 6.7 percent of the time, which gives him a win probability many orders of magnitude more than infinitesimal.

Math tutoring available. Only $6! per hour! First lesson: Factorials!