Activity for DNB
| Type | On... | Excerpt | Status | Date |
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Why isn't MathJax table appearing? $\begin{array} {|r|r|}\hline \text{If I already earned a} & \text{then I must roll another} \\ \hline 9 & 1 \\ \hline 8 & 2 \\ \hline 7 & 3 \\ \hline 6 & 4 \\ \hline 5 & 5 \\ \hline 4 & 6 \\ \hline 3 & \text{Impossible: I need $\ge 2$ more rolls to gain a 10.} \\ \hline 2 & Merge with cell above. \\... (more) |
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| Edit | Post #285432 |
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| Comment | Post #285596 |
Thanks. Can you pls elaborate why this hitch doesn't happen on Stack Exchange, as far as I know? (more) |
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| Suggested Edit | Post #285596 |
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helpful | almost 3 years ago |
| Edit | Post #285594 |
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| Edit | Post #285594 |
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| Edit | Post #285594 | Initial revision | — | almost 3 years ago |
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Why doesn't \$ work? Why isn't $1 rendering? Both `\$1` and `$\$$1` failed! (more) |
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| Edit | Post #285524 | Initial revision | — | almost 3 years ago |
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How's it possible to arrange 0 objects? How can 0! = 1? 1. I don't understand the following explanation. Isn't it physically impossible to "arrange 0 objects", or nothing? It's senseless to raise the concept of arrangement when you have nothing, like in a vacuum! 2. Division by zero is undefined because you can't divide something (e.g. cookies) by not... (more) |
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| Edit | Post #285431 |
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| Comment | Post #285443 |
Completed. I just edited my post. (more) |
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| Edit | Post #285443 | Initial revision | — | about 3 years ago |
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Why's Pr(the running total of a fair dice rolled repeatedly = n) = $\frac16 (p_{n-1} + p_{n-2} + p_{n-3} + p_{n-4} + p_{n-5} + p_{n-6})$? Where did $\color{red}{pn = \frac16 (p{n-1} + p{n-2} + p{n-3} + p{n-4} + p{n-5} + p{n-6})}$ spring from? The solution doesn't expatiate, and makes it appear out of the blue? Can you please familiarize and naturalize me with this? I've tried to contemplate the ways that rolling can sum to $n$. For ... (more) |
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| Edit | Post #285434 |
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| Edit | Post #285434 | Initial revision | — | about 3 years ago |
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"the people who move have a percentage of Democrats which is between these two values" — How does this furnish intuition for Simpson's Paradox? Why does the phrase colored in red below ($\color{red}{\text{the people who move have a percentage of Democrats which is between these two values}}$) matter? How does it assist with intuiting the peaceful coexistence of $P{new}(D|B) > P{old}(D|B)$ and $P{new}(D|B^C) > P{old}(D|B^C)$? >59. The ... (more) |
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| Edit | Post #285433 | Initial revision | — | about 3 years ago |
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Why's a 2-player game with "win by two" rule = Gambler's Ruin where each player starts with $2? Why N = 4? 1. Why can this problem "be thought of as a gambler's ruin where each player starts out with $2"? 2. Please see my red arrow. Why is this exponent 4? I quote op. cit. p 73. >Example 2.7.3 (Gambler's ruin). Two gamblers, A and B, make a sequence of \\$1 bets. In each bet, gambler A has probabi... (more) |
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| Edit | Post #285429 |
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| Edit | Post #285432 |
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| Edit | Post #285432 | Initial revision | — | about 3 years ago |
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How to intuit p = Calvin's probability of winning each game independently = $1/2 \implies$ P(Calvin wins the match) = 1/2? Please see the sentence beside my red line. The notion of a "sanity check" suggests that these resultant integers should be obvious, without calculation or contemplation. But why's it plain and intuitive that $p = 1/2 \implies P(C) = 1/2$? Indubitably, a game differs from a match. Just because ... (more) |
— | about 3 years ago |
| Edit | Post #285431 | Initial revision | — | about 3 years ago |
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If C = Calvin wins the match, and $X \thicksim Bin(2, p) =$ how many of the first 2 games he wins — then why P(C|X = 1) = P(C)? The author's solution doesn't expatiate why $\color{red}{P(C|X = 1) = P(C)}$? This similar question on Math Stack Exchange has 0 answers, as at 4 January 2022. >50. Calvin and Hobbes play a match consisting of a series of games, where Calvin has probability p of winning each game (independently).... (more) |
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| Edit | Post #285430 | Initial revision | — | about 3 years ago |
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Intuitively, why does $p$ vary inversely with $P(C_3 \mid D_2)$? But directly with $P(C_2 \mid D_3)$? I'm seeking merely intuition here — NOT about the algebra that I know how to execute. Please see the fractions colored in red and orange at the bottom. 1. $P(C3 \mid D2) = \color{red}{\dfrac1{ 1 + p}}$ means that $p$ is inversely related with $P(C3 \mid D2)$. How can you intuit this inverse re... (more) |
— | about 3 years ago |
| Edit | Post #285429 | Initial revision | — | about 3 years ago |
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For (a variation) of the Monty Hall problem, what permits you to repaint doors 2 and 3? What permits you to replace p with $1 - p$? Please see the text colored in red at the bottom. 1. Why can you haughtily just — and what legitimizes you to — "[i]magine repainting doors 2 and 3, reversing which is called which"? This is completely unworkable, half-baked because you can't reverse doors in the real game! 2. What legitimate... (more) |
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| Edit | Post #285365 |
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| Edit | Post #285371 | Initial revision | — | about 3 years ago |
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How does $P(C > D \mid C = 2) \neq P(C > D \mid C \neq 2)$ prove that B > C depends on C > D? I grok that $\color{limegreen}{P(C > D \mid C = 2) = P(D = 1 \mid C = 6) = 1/2}$, and $\color{red}{P(C > D \mid C \neq 2) = P(C > D \mid C = 6) = 1}$. But I don't grok the last sentence in the quotation below, colored in blue. How do these two probabilities prove that B > C DEPENDS ON C > D? >3... (more) |
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| Edit | Post #285364 |
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| Edit | Post #285365 |
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| Edit | Post #285365 | Initial revision | — | about 3 years ago |
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Acceptable, usual to write $\ge 2$ pipes simultaneously? I'm NOT asking for the solution to this exercise that's publicly accessible. Rather, pls see the green and red underlines. If I apply the author's green definition to the red underline, then $\tilde P({\color{red}{L \mid M2}}) \equiv P(\color{red}{L \mid M2} \quad \color{limegreen}{\mid M1})$. Is... (more) |
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| Edit | Post #285364 |
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| Edit | Post #285364 | Initial revision | — | about 3 years ago |
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In general, does $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$? In this question, $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$. But the author's solution didn't annunciate this, and doesn't unfurl how to compute $\color{red}{P(B|M)}$. Can I simply subtract as $\color{red}{P(B|M)} = 1 - \color{forestgreen}{P(A|M)}$? I hanker to avoid calculating $\color... (more) |
— | about 3 years ago |
| Edit | Post #284738 | Initial revision | — | about 3 years ago |
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Can the bijection for the Lost Boarding Pass Probability Problem, be formulated or pictured? The two proofs below moot "bijection", but I don't see a formula. 1. Does the bijection have an explicit formula? 2. Can this bijection be pictorialized? My 16 y.o. kid doesn't understand the abstract, Daedalian phrasing below. hunter's answer dated Dec. 4 2013 >Claim 3: There is a biject... (more) |
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