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119

The analysis is complicated by the prospect that the game goes into "overtime" in order to win by a margin of at least two points. (Otherwise it would be as simple as the solution shown at https://stats.stackexchange.com/a/327015/919.) I will show how to visualize the problem and use that to break it down into readily-computed contributions to the answer. ...


76

they are trying to assert that [...] if there have been 10 heads, then the next in the sequence will more likely be a tail because statistics says it will balance out in the end There's only a "balancing out" in a very particular sense. If it's a fair coin, then it's still 50-50 at every toss. The coin cannot know its past. It cannot know there was an ...


47

EDIT (after reading the paper): I've read the paper thoughtfully. Let's start off with what Google claimed in the paper: They defeated Stockfish with Monte-Carlo-Tree-Search + Deep neural networks The match was absolutely one-sided, many wins for AlphaZero but none for Stockfish They were able to do it in just four hours AlphaZero played like a human ...


31

The confusion is because he is looking at the probability from the start without looking at what else has already happened. Lets simplify things: First flip: T Now the chance of a T was 50%, so 0.5. The chance that the next flip will be T again is 0.5 TT 0.5 TF 0.5 However, what about the first flip? If we include that then: TT 0.25 TF 0.25 The ...


25

Using the binomial distribution and assuming every point is independent: The probability the $58\%$ player gets to $21$ in the first $40$ points (taking account of the fact the last point must be won) is $\sum_{n=21}^{40} {n-1 \choose 20} 0.58^{21}0.42^{n-21}$ $=\sum_{k=21}^{40} {40 \choose k} 0.58^{k}0.42^{40-k}$ $\approx 0.80695$ The probability $58\%$ ...


17

I went with a computational answer. Here is an R function that simulates a ping-pong game where the winner has to win by 2. The only argument is the probability that you win a point. It will return the final score of that game: ## data simulation function ---------------------------------------------------- sim_game <- function(pt_chance) { them <- ...


16

How about the game show "Deal or No Deal". Though not emphasized the banker is checking the probability given the number of remaining suitcases what the probability is that the 1,000,000 dollar prize is in your suitcase. Based on the odds he makes an offer that favors him. Whenever I watch this I am hoping the contestant understands expected gain and ...


16

Answers to the referenced question apply here directly: create a dictionary consisting only of the target word (and its possible wildcard spellings), compute the chance that a random rack cannot form the target, and subtract that from $1$. This computation is fast. Simulations (shown at the end) support the computed answers. Details As in the previous ...


15

Should we assume that the 58% chance of winning is fixed and that points are independent? I believe that Whuber's answer is a good one, and beautifully written and explained, when the consideration is that every point is independent from the next one. However I believe that, in practice it is only an interesting starting point (theoretic/idealized). I ...


13

Fantasy sports incentivize players to think with statistical intuition. For example, every week in fantasy football you must choose which players to start based on, e.g.: that player's career stats, the trend in the time series of that player's stats, the team he is going up against, weather, injuries, and many more. According to http://www.fsta.org/, 32 ...


13

Below are a couple very simple models. They are both deficient in at least one way, but maybe they'll provide something to build on. The second model actually does not (quite) address the OP's scenario (see remarks below), but I am leaving it in case it helps in some way. Model 1: A variant of the Bradley–Terry model Suppose we are primarily ...


13

So this is a Monte Carlo solution, that is, we are going to simulate drawing the tiles a zillion of times and then we are going to calculate how many of these simulated draws resulted in us being able to form the given word. I've written the solution in R, but you could use any other programming language, say Python or Ruby. I'm first going to describe how ...


13

The odds are still 50-50 that the next flip will be tails. Very simple explanation: The odds of flipping 10 heads + 1 tail in that order are very low. But by the time you've flipped 10 heads, you've already beaten most of the odds... you have a 50-50 chance of finishing the sequence with the next coin flip.


12

If you play games to $4$ points, where you have to win by $2$, you can assume the players play 6 points. If no player wins by $2$, then the score is tied $3-3$, and then you play pairs of points until one player wins both. This means the the chance to win a game to $4$ points, when your chance to win each point is $p$, is $$p^6 + 6p^5(1-p) + 15p^4(1-p)^2 + ...


12

A formula is requested. Unfortunately, the situation is so complicated it appears that any formula will merely be a roundabout way of enumerating all the possibilities. Instead, this answer offers an algorithm which is (a) tantamount to a formula involving sums of products of binomial coefficients and (b) can be ported to many platforms. To obtain such a ...


12

Much effort could be spent on a perfect model. But sometimes a bad model is better. And nothing says bad model like the central limit theorem -- everything is a normal curve. We'll ignore "overtime". We'll model the sum of individual points as a normal curve. We'll model playing 38 rounds and whomever has the most points win, instead of first to 20. ...


11

I think this is more about Engineering Problem Solving. Most successful engineering projects do not duplicate expert reasoning or or the expert's nature exactly. They solved the problem in a different way. For example washing machines use a different technique than humans, airplanes use different dynamics than birds. If you are duplicating Expert ...


11

Poker is a good game for learning probabilistic thinking. The game has been dominated at the highest level in recent years by "math nerds" who have spent a lot of time studying the odds and spend very little time trying to read their opponents. Check out the Time article World Series of Poker: Attack of the Math Brats for some details. Also, I can't find ...


11

You should try convincing them that if the previous results were to impact the upcoming tosses then not only the last 10 tosses should have been taken into consideration, but also every previous toss in the coin life. I think it's a more logical approach.


11

It appears that the policy network determines a probability distribution $p(a \mid s)$ over the possible moves $a$ when in game state $s$. When the program is searching the game tree it does so in a random fashion, and $p$ determines how it does this search. The hope is that this function will "guide" the program to good moves that a strong player is ...


9

You can use Sklansky's starting hand ranking to know the strength of a dealt hand from 1 to 8. Generate a random sample with the first 100 hands, another with the next 100 hands and compare them with the Wilcoxon test.


9

Perhaps the best known measure would be the Gini index. The R package ineq (See here) implements the Herfindahl and Rosenbluth concentration measures (in function conc). It also implements a number of inequality indexes (including the Gini) in function ineq -- the Gini coefficient, Ricci-Schutz coefficient (also called Pietra’s measure), Atkinson’s measure,...


9

What you really want to know is how to do this calculation quickly--preferably in your head--during game play. To that end, consider using Normal approximations to the distributions. Using Normal approximations, we can easily determine that two rolls for damage with a 2d6+2 have about a $30\%$ chance of equalling or exceeding $20$ and three rolls for ...


9

The question asks for the "mean of the sum of points." Because the target is 10 and no value exceeds 10, this is the mean of a distribution defined on the ten integers $10, 11, \ldots, 10+10-1$. It takes about the same amount of computing power to work out this distribution, exactly, as it does to perform a small simulation. Here's how. The deck is ...


8

The model you describe is known as the Bradley-Terry model. This model has been extended to include covariates as well as random effects predicting team abilities and contest specific covariates (including home advantages) available as the R-package BradleyTerry2, see this paper in Journal of Statistical Software. For dynamic extensions of the Bradley-...


8

You are effectively taking the probability for a single dice roll being larger than 4 (which is indeed $\frac{1}{3}$). But that is a different thing than the mean of several dice rolls being larger than 4 (you can express this also as the sum of $x$ dice rolls being larger than $4x$). Here I will provide an intuition behind the reason for your error by ...


7

Microsoft's TrueSkill algorithm, as used to rank players on XBox Live, can deal with team matches, but does not incorporate margin of victory. It may still be of some use to you.


7

For a simulation it's crucial to be correct as well as fast. Both these objectives suggest writing code that targets core capabilities of the programming environment as well as code that is as short and simple as possible, because simplicity lends clarity and clarity promotes correctness. Here is my attempt to achieve both in R: # # Simulate one play with ...


7

For the word "BOOT" with no wildcards: $$ p_0=\frac{\binom{n_b}{1}\binom{n_o}{2}\binom{n_t}{1}\binom{n-4}{3}}{\binom{n}{7}} $$ With wildcards, it becomes more tedious. Let $p_k$ indicate the probability of being able to play "BOOT" with $k$ wildcards: $$ \begin{eqnarray*} p_0&=&\frac{\binom{n_b}{1}\binom{n_o}{2}\binom{n_t}{1}\binom{n-4}{3}}{\binom{n}{...


7

The identity $\Pr(x \mid y,z) = \Pr(x \mid z)$ does not hold in general. It holds iff and only if $X$ is conditionally independent of $Y$ given $Z$.


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