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Ask the Wizard #312
According to BookMaker, the Washington Post, who keep a count political false statements in general, would be used as the source of number of lies. According to that source, Trump averaged 15 false statement per day during 2018. The next question to be answered in analyzing this bet is how much time does Trump spend making public statements a day? Between tweets, interviews, and off-the-cuff statements, 20 minutes seems like a reasonable estimate to me. A nice round number at least. Simple division gives us 15/20 = 0.75 false statements per minute, or one every 80 seconds.
The address was estimated to last six to eight minutes by the media before it started. Let's split the difference and go with seven minutes. Seven minutes at 0.75 false statements a minute gives us an estimated 5.25 false statements. So, I would have set the over/under at 5.5.
By the way, if we assume 5.25 to be the mean number of false statements, then the probability of three or less false statements is 23.17%, if we assume the total is distributed according to the Poisson distribution, which I think is a reasonable assumption.
By the way, in the end, the number of false statements was scored at six.
This question is asked and discussed in the very long thread on Trump at Wizard of Vegas, but discussion of this specific topic starts here.
In single-zero roulette, what is the mean and median number of spins required for every number to appear at least once?
Answering the mean is much easier, so we'll start with that. Let's go through it step by step:
- The first spin is definitely going to be a new number.
- The second spin will have a probability of 36/37 of being a new number. If an event has a probability of p, then the expected number of trials for it to occur is 1/p. In this case, the expected number of trials to get the second number is 37/36 = 1.0278.
- After two numbers have been observed, the probability that the next spin will result in a new number is 35/37. Thus, the expected number of spins after the second number to see the third is 37/35 = 1.0571.
- Following this logic, the average number of spins to see every number is 1 + 37/36 + 37/35 + 37/34 + ... + 37/2 + 37/1 = 155.458690.
The median is much more complicated. To find the exact answer, as opposed to using a random simulation, one needs to use a lot of matrix Algebra. I've discussed how to solve similar problems in other Ask the Wizard questions, so I won't go through the details again. One example of a similar question is the one on getting a 6-6 pair in the hole three times in a row, as discussed in Ask the Wizard #311. Suffice it to say that the probability of seeing every number in 145 spins is 0.49161779, and in 146 spins is 0.501522154. Thus, the median is 146.
This question is asked and discussed in my forum at Wizard of Vegas.
Suppose you have 12 six-sided dice. You roll them and may set aside any dice you wish. You then re-roll the other dice. What is the probability of getting a 12-of-a-kind in the two rolls?
There are 58 different types of sequences on the initial roll. The way I identify each is the number of the face in majority, then the number of dice of the face second in total, and so on. For example, a roll of of 3,3,3,3,6,6,6,5,5,2 would be signified as 4-3-2-1. The following table shows the number of combinations of each sequence, the probability of rolling it, the probability of completing a 12 of a kind in the second roll, and the product of the two. For the probability on the second roll, I assume the player holds the dice that have the greatest total on the initial roll. The lower right cell shows an overall probability of 0.0000037953, which equals 1 in 263,486.
12 Dice Question
Sam from Fountain Valley CA
As measured by square feet of gambling space, here they are. This comes as a surprise to me, as I've barely heard of the two Oklahoma casinos in the top five.