Ask The Wizard #263

My apologies for the dated reply on this. I wrote the following before the election.

First, I’m going to get rid of that other 6%, who are either undecided or will waste their votes on a third party candidate or "none of the above," which is an option in Nevada. Some may disagree with this assumption. To be honest, another reason for ignoring them is the math gets more complicated with more than two candidates. So, after rounding, that would leave us with 306 votes for Angle, and 281 votes for Reid, for a total of 587 in the sample.

I’m going to use the standard normal approximation to answer this question. If I were going to be a perfectionist, then I would use the T distribution, because the actual mean and variance are not known. However, in my opinion, a sample size of 587 is perfectly fine for the normal distribution.

Sample size = 306+281 = 587.

Angle sample mean is 306/587 = 0.521295.

The estimated standard deviation of the mean is (0.521295 × 0.478705 / (587-1))^0.5 = 0.0206361.

Angle’s share above 50% is (0.521295-0.5)/0.0206361 = 1.031917 standard deviations.

According to the normal distribution, the probability of Reid finishing 1.031917 standard deviations above expectations is 0.151055. This can be found in Excel with the function NORMSDIST(-1.031917). So Angle’s probability of winning is 1-0.151268 = 84.89%.

To create a 95% confidence interval, note that the 2.5% point on either side of the Gaussian curve is at 1.959964 standard deviations from the mean. This can be found in Excel with the function NORMSINV(0.975). As already noted, the estimated standard deviation of the sample mean is 0.0206361. So there is a 95% chance that either candidate will get within 0.0206361×1.959964 = 0.040446 standard deviations of the poll results. So Angle’s 95% confidence interval is 0.521295 +/- 0.040446 = 48.08% to 56.17%.

I’m told it would be mathematically incorrect to phrase that as "Angle’s share of all Angle/Reid votes has a 95% chance of falling between 48.08% and 56.17%." That was how I originally phrased my answer, but two statisticians recoiled in horror at my wording. To paraphrase their response, they said I had to use the passive voice, and say that "48.08% and 56.17% will surround Angle’s share with 95% probability." To be honest with you, it sounds the same to me. However, they stressed that the confidence interval is random and Angle’s share is immutable, and that my original wording implied the opposite. Anyway, I hope the frequentist statisticians out there will be satisfied with the second wording.

The "margin of error" is half the difference between the two ends of the 95% confidence interval. In this case (56.17% - 48.08%)/2 = 4.04%.

As a follow-up, here are the actual results:

Reid: 361,655
Angle: 320,996
Other: 21,979

So, not counting the "other" votes, Reid got 53.0% and Angle 47.0%. That is a comfortable 6% win for Reid. It begs the question of why the poll was so far off. Was it chance? Did voters change their minds? Or was it a bad poll to begin with? I leave those questions to the reader (I hate it when textbooks say that).

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

There are eight ways two win: three rows, three columns, and two diagonals. There are combin(9,3)=84 ways to pick 3 squares out of 9. So the probability of winning is 8/84 = 9.52%.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

I agree, that does seem like a good advantage play. At the 5dimes live casino game they offer the following rules:

• 6 decks.
• One burn card at the beginning of the shoe, and one with every dealer change, which occur every 30 minutes. If the supervisor has to step in, which I’ve seen happen twice, then three cards are burned.
• Dealer stands on soft 17.
• Early surrender, except against an ace.
• Double after split allowed.
• Split to two hands only.
• No hole card, but player loses original bet only to dealer blackjack.
• Bet ranges: $5-$250, $10-$250, or $25-$500

The house edge under these rules is 0.24%.

The rules say the penetration is 75%. Twice I counted the number of cards seen per shoe, and I got 211 both times, which is 68%. Shortow, a member of my forum, placed two videos of the cut card placement on YouTube: Video 1 Video 2. It looks to me like the penetration was a little deeper in video 2. I would say it ranges from 60% to 70%, depending on the dealer.

Not only could the player easily count this game, but he could use a calculator on a separate screen to make perfect decisions. The Blackjack Real Time Analzyer is one product that does this. However, I have no idea what kind of game protection they employ. With every card scanned, it would not be difficult to determine the correlation between count and bet size and raise a red flag if the correlation was sufficiently positive.

If you give it a try, I'd be interested to know what happens.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Outstanding bet, like taking candy from a baby. It’s like getting 2 to 1 on a coin flip. Even better, since you would win on a tie.

At the time of this writing, 95 games have been played out of 256 in the regular season. Using the binomial distribution, I show your probability of winning is 99.87%. A fair settlement price would be $998.02.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.