## Wizard Recommends

Wine flows from a box of wine at a rate in proportion to the wine left in the box. When a 3-liter box is 1/3 full, wine flows at a rate of 0.01 liters per second.

You have a full 3-liter box of wine. How long will it take to pour 2.9 liters?

"Anonymous" .

100*ln(30) =~ 340.119738 seconds

Let:
v = volume of wine in the box
t = time
c = constant of integration

We're given dv/dt = -0.01v

Rearrange to dv = -0.01v dt

-100/v dv = dt

Integrate both sides:

-100*ln(v) = t + c

We're given when t=0, v=3. Put those in our equation above to find the constant of integration.

-100*ln(3) = c

Now our equation is:

-100*ln(v) = t -100*ln(3)

t = 100*ln(3) - 100*ln(v)

t = 100*(ln(3)-ln(v))

t = 100*ln(3/v)

We're asked what is t when the wine left in the bag is 0.1.

t = 100*ln(3/0.1) = 100*ln(30) =~ 340.119738 seconds =~ 5 minutes, 40 seconds.

This question is asked and discussed in my forum at Wizard of Vegas

If I make a buy bet for \$20 on the 4 and 10 and place \$30 on the 5, 6, 8, and 9, what is my house edge? Please assume the commission on the 4 and 10 is paid on wins only. Please calculate it whether I:

• Leave the bets up for one roll only
• Leave the bets up until some significant event happens (any roll between 4 and 10)
• Leave the bets up until they are all resolved.

John Cokos

The first table shows my analysis for leaving the bets up for one roll only. The return column is calculated as the win*probability/(total bet). The lower right cell shows a house edge of 0.69%.

### One Roll Analysis

Roll Bet Net Win Combinations Probability Return
2 0 0 1 0.027778 0.000000
3 0 0 2 0.055556 0.000000
4 20 39 3 0.083333 0.020313
5 30 42 4 0.111111 0.029167
6 30 35 5 0.138889 0.030382
7 0 -160 6 0.166667 -0.166667
8 30 35 5 0.138889 0.030382
9 30 42 4 0.111111 0.029167
10 20 39 3 0.083333 0.020313
11 0 0 2 0.055556 0.000000
12 0 0 1 0.027778 0.000000
160 36 1.000000 -0.006944

The second table shows my analysis for leaving the bets up until a bet is resolved. In other words, rolling again after a total of 2, 3, 11, or 12. The return column is calculated as the win*probability/(total bet). The lower right cell shows a house edge of 0.83%.

### One Significant Roll Analysis

Roll Bet Net Win Combinations Probability Return
4 20 39 3 0.100000 0.024375
5 30 42 4 0.133333 0.035000
6 30 35 5 0.166667 0.036458
7 0 -160 6 0.200000 -0.200000
8 30 35 5 0.166667 0.036458
9 30 42 4 0.133333 0.035000
10 20 39 3 0.100000 0.024375
Total 160 30 1.000000 -0.008333

The third table shows my analysis for leaving the bets up until all are resolved. The return column is calculated as the win*probability/(total bet). The lower right cell shows a house edge of 2.44%.

### Roll Until All Bets Resolved Analysis

Win 4,10
Rolled
5,9
Rolled
6,8
Rolled
Combinations Probability Return
-160 1 0 0 2,677,114,440 0.200000 -0.200000
-101 0 1 0 594,914,320 0.044444 -0.028056
-88 0 0 1 823,727,520 0.061538 -0.033846
-95 2 0 0 1,070,845,776 0.080000 -0.047500
-42 0 2 0 74,364,290 0.005556 -0.001458
-16 0 0 2 149,768,640 0.011189 -0.001119
-30 1 1 0 267,711,444 0.020000 -0.003750
-29 1 0 1 421,812,160 0.031512 -0.005712
-36 0 1 1 562,464,448 0.042020 -0.009455
-23 1 1 1 800,192,448 0.059780 -0.008593
36 2 1 0 751,055,104 0.056109 0.012625
30 2 0 1 93,017,540 0.006949 0.001303
23 1 2 0 127,949,276 0.009559 0.001374
43 0 2 1 136,097,920 0.010168 0.002733
49 1 0 2 276,379,776 0.020648 0.006323
29 0 1 2 259,917,112 0.019418 0.003519
42 2 1 1 383,915,862 0.028681 0.007529
95 1 2 1 280,463,688 0.020953 0.012441
108 1 1 2 430,248,448 0.032143 0.021696
101 2 2 0 626,008,276 0.046767 0.029522
102 2 0 2 48,772,745 0.003644 0.002323
88 0 2 2 101,392,694 0.007575 0.004166
114 2 2 1 243,130,194 0.018164 0.012942
167 2 1 2 263,665,646 0.019698 0.020560
160 1 2 2 409,147,802 0.030566 0.030566
173 2 2 2 679,339,612 0.050752 0.054875
232 0 0 0 832,156,379 0.062168 0.090144
Total 13,385,573,560 1.000000 -0.024848

I used integral calculus. The key is the odds are the same whether there is one unit of time between rolls or the length of time follows an exponential distribution with a mean of 1.

Recall from your statistics call the probability of an event x NOT happening is exp(-x). It is then easy to say the probability it has happened at least once is 1-exp(-x). The following list shows the probability for any length of time x the given points have been rolled. Then, integration over all periods of time x from 0 to infinity. I prefer the integral calculator at www.integral-calculator.com/. Finally, remember to weight these probabilities by similar events. For example, the probability of rolling a 4 is the same as rolling a 10.

• 4 or 10 -- (1-exp(-3x/36))*exp(-3x/36)*exp(-4x/36)^2*exp(-5x/36)^2*exp(-x/6)/6
• 5 or 9 -- (1-exp(-x/9))*exp(-5x/36)^2*exp(-3x/36)^2*exp(-x/9)exp(-x/6)/6
• 6 or 8 -- (1-exp(-5x/36))*exp(-4x/36)^2*exp(-3x/36)^2*exp(-5x/36)exp(-x/6)/6
• 4 and 10 -- (1-exp(-3x/36))^2*exp(-4x/36)^2*exp(-5x/36)^2*exp(-x/6)/6
• 5 and 9 -- (1-exp(-4x/36))^2*exp(-5x/36)^2*exp(-3x/36)^2*exp(-x/6)/6
• 6 and 8 -- (1-exp(-5x/36))^2*exp(-4x/36)^2*exp(-3x/36)^2*exp(-x/6)/6
• 4 and 5 -- (1-exp(-3x/36))*(1-exp(-4x/36))*exp(-5x/36)^2*exp(-4x/36)*exp(-3x/36)*exp(-x/6)/6
• 4 and 6 -- (1-exp(-3x/36))*(1-exp(-5x/36))*exp(-4x/36)^2*exp(-5x/36)*exp(-3x/36)*exp(-x/6)/6
• 5 and 6 -- (1-exp(-4x/36))*(1-exp(-5x/36))*exp(-3x/36)^2*exp(-5x/36)*exp(-4x/36)*exp(-x/6)/6
• 4,5,6 -- (1-exp(-3x/36))^1*exp(-3x/36)^1*exp(-4x/36)^1*(1-exp(-4x/36))^1*(1-exp(-5x/36))^1*exp(-5x/36)^1*exp(-x/6)/6
• 4,5,10 -- (1-exp(-3x/36))^2*(1-exp(-4x/36))*exp(-5x/36)^2*exp(-4x/36)*exp(-x/6)/6
• 4,6,10 -- (1-exp(-3x/36))^2*(1-exp(-5x/36))*exp(-4x/36)^2*exp(-5x/36)*exp(-x/6)/6
• 4,5,9 -- (1-exp(-4x/36))^2*(1-exp(-3x/36))*exp(-5x/36)^2*exp(-3x/36)*exp(-x/6)/6
• 5,6,9 -- (1-exp(-4x/36))^2*(1-exp(-5x/36))*exp(-3x/36)^2*exp(-5x/36)*exp(-x/6)/6
• 4,6,8 -- (1-exp(-3x/36))^1*exp(-3x/36)*exp(-4x/36)^2*(1-exp(-5x/36))^2*exp(-x/6)/6
• 5,6,8 -- (1-exp(-3x/36))^0*exp(-3x/36)^2*exp(-4x/36)^1*(1-exp(-4x/36))*(1-exp(-5x/36))^2*exp(-5x/36)^0*exp(-x/6)/6
• 4,5,6,10 -- (1-exp(-3x/36))^2*exp(-4x/36)^1*(1-exp(-4x/36))^1*(1-exp(-5x/36))^1*exp(-5x/36)^1*exp(-x/6)/6
• 4,5,6,9 -- (1-exp(-3x/36))^1*exp(-3x/36)^1*exp(-4x/36)^0*(1-exp(-4x/36))^2*(1-exp(-5x/36))^1*exp(-5x/36)^1*exp(-x/6)/6
• 4,5,6,8 -- (1-exp(-3x/36))^1*exp(-3x/36)^1*exp(-4x/36)^1*(1-exp(-4x/36))^1*(1-exp(-5x/36))^2*exp(-5x/36)^0*exp(-x/6)/6
• 4,5,9,10 -- (1-exp(-3x/36))^2*exp(-3x/36)^0*(1-exp(-4x/36))^2*(1-exp(-5x/36))^0*exp(-5x/36)^2*exp(-x/6)/6
• 4,6,8,10 -- (1-exp(-3x/36))^2*exp(-3x/36)^0*(exp(-4x/36))^2*(1-exp(-5x/36))^2*exp(-5x/36)^0*exp(-x/6)/6
• 5,6,8,9 -- (1-exp(-3x/36))^0*exp(-3x/36)^2*(1-exp(-4x/36))^2*(1-exp(-5x/36))^2*exp(-5x/36)^0*exp(-x/6)*exp(-x/6)/6
• 4,5,6,9,10 -- (1-exp(-3x/36))^2*exp(-3x/36)^0*(1-exp(-4x/36))^2*(1-exp(-5x/36))^1*exp(-5x/36)^1*exp(-x/6)/6
• 4,5,6,8,10 -- (1-exp(-3x/36))^2*(1-exp(-4x/36))^1*exp(-4x/36)*(1-exp(-5x/36))^2*exp(-x/6)/6
• 4,5,6,8,9 -- (1-exp(-3x/36))^1*exp(-3x/36)^1*(1-exp(-4x/36))^2*(1-exp(-5x/36))^2*exp(-x/6)/6
• 4,5,6,8,9,10 -- (1-exp(-3x/36))^2*(1-exp(-4x/36))^2*(1-exp(-5x/36))^2*exp(-x/6)/6

A year consists of 365.24217 days, to five decimal places. As you probably know, the test for whether a year is a leap year goes as follows:

• If a year is evenly divisible by 4 it is a leap year, except ...
• If a year is evenly divisible by 100 it is not a leap year, except ...
• If a year is evenly divisible by 400 it is a leap year.

The above rules result in 356.2425 days per year. Pretty close to the correct 365.24217, off by 0.00033.

My question is whether there is a more accurate way to choose leap years with a shorter cycle than 400 years?

"Anonymous" .

Yes!

If we choose 85 leap years out of a cycle of 351 years, we get a an average year of 0.242165. That is off the 0.24217 target by only 0.000005 days.

A way to test for whether a year is a leap year would be as follows:

• If a year is evenly divisible by 4 it is a leap year, except ...
• If a year is evenly divisible by 31 it is not a leap year.

This question is asked and discussed in my forum at Wizard of Vegas. The original source is 538.

Can you explain how the magic trick in this YouTube video is possible. I've tried it many times and it doesn't work for me. Am I doing it wrong or is the whole thing a hoax?

"Anonymous" .

It's a hoax!

For those who didn't watch the video, here is how Jason, the magician says it works:

• Use a complete 52-card deck with no jokers.
• Pick a rank from ace to 10.
• Deal cards one at a time until you get to the third card of the chosen rank. Take note of the total cards dealt to that point.
• The fourth card of the chosen rank will appear the same number of cards from the top of the remaining cards as it took to find the first three.

The whole thing is a practical joke. He uses a prearranged deck, made to work for the rank he picks. It appears he shuffling, but he is a very good card mechanic with a fake shuffle.

On YouTube you can pre-screen comments to show and he only shows comments from his fans that falsely claim it works for them. It's all a big hoax to gaslight the audience.

I get into it in even more detail in my December 22, 2022 newsletter.