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Ask The Wizard #376
The wine dispenses from a box at a rate that corresponds to the amount of wine remaining inside. For example, if a 3-liter box is one-third full, the wine flows out at a speed of 0.01 liters per second.
You have a completely full 3-liter box of wine. How much time will it take to pour out 2.9 liters?
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
At the starting point, when time t equals 0, our initial volume v is 3. Substitute these values into the equation we established earlier to solve for the integration constant.
-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 need to determine the value of t when only 0.1 liters of wine remains in the bag.
t = 100*ln(3/0.1) = 100*ln(30) = 340.119738 seconds = 5 minutes, 40 seconds.
This topic was raised and deliberated upon in my online forum at Wizard of Vegas
If I place a $20 bet on both 4 and 10 while wagering $30 on the numbers 5, 6, 8, and 9, what will my house edge be? Please consider that the commission on 4 and 10 is only taken from winning bets. Calculate it in two scenarios:
- Leave the bets up for one roll only
- Keep the bets active until a noteworthy event occurs (any roll that lands between 4 and 10).
- Maintain the bets until all outcomes are resolved.
The first table presents my evaluation for keeping the bets active for just one roll. The return rate is determined by multiplying the win by its probability and dividing by the total amount wagered. The bottom-right cell indicates 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 details my analysis for keeping the bets active until one of them resolves. In essence, rolling again after a total of 2, 3, 11, or 12. The return is computed in the same manner, with the lower right showing 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 includes my review for maintaining the bets active until all outcomes are determined. The return computation follows the previously mentioned formula, yielding a house edge of 2.44% in the lower right cell.
Analysis of rolling until all bets have been resolved.
| 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 |
Remember from your statistics class that the probability of an event x NOT occurring can be expressed as exp(-x). Thus, it's straightforward to state that the likelihood it has occurred at least once is 1-exp(-x). The subsequent list demonstrates the probability for any duration x of the specified points appearing. After that, integration is performed across all time intervals from 0 to infinity. I prefer using the integral calculator available at www.integral-calculator.com/ . Lastly, keep in mind that these probabilities should be adjusted based on similar events. For instance, the probability of rolling a 4 is equivalent to that of rolling a 10.
- 4 or 10 -- \t(1-exp(-3x/36))*exp(-3x/36)*exp(-4x/36)^2*exp(-5x/36)^2*exp(-x/6)/6
- 5 or 9 -- \t(1-exp(-x/9))*exp(-5x/36)^2*exp(-3x/36)^2*exp(-x/9)*exp(-x/6)/6
- 6 or 8 -- \t(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 -- \t(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 -- \t(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 -- \t(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 -- \t(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 -- \t(1-exp(-3x/36))^2*(1-exp(-4x/36))*exp(-5x/36)^2*exp(-4x/36)*exp(-x/6)/6
- 4,6,10 -- \t(1-exp(-3x/36))^2*(1-exp(-5x/36))*exp(-4x/36)^2*exp(-5x/36)*exp(-x/6)/6
- 4,5,9 -- \t(1-exp(-4x/36))^2*(1-exp(-3x/36))*exp(-5x/36)^2*exp(-3x/36)*exp(-x/6)/6
- 5,6,9 -- \t(1-exp(-4x/36))^2*(1-exp(-5x/36))*exp(-3x/36)^2*exp(-5x/36)*exp(-x/6)/6
- 4,6,8 -- \t(1-exp(-3x/36))^1*exp(-3x/36)*exp(-4x/36)^2*(1-exp(-5x/36))^2*exp(-x/6)/6
- 5,6,8 -- \t(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 -- \t(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 -- \t(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 -- \t(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 -- \t(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 -- \t(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 -- \t(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 -- \t(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 -- \t(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 -- \t(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 approximately 365.24217 days, when rounded to five decimal places. As many of you may know, the criterion for determining a leap year is as follows:
- A year that can be evenly divided by 4 qualifies as a leap year, with exceptions...
- If a year can be evenly divided by 100, it is not considered a leap year unless...
- If a year is divisible by 400, it will be recognized as a leap year.
These rules yield an average of 365.2425 days annually, which is quite close to the actual value of 365.24217, differing by just 0.00033.
I pose the question: Is there a more precise method for defining leap years in a shorter cycle than the existing 400-year period?
Yes!
Selecting 85 leap years over a span of 351 years results in an average of 0.242165 days per year. This falls short of the target by a mere 0.000005 days.
A feasible approach to determine if a year is a leap year might be stated as follows:
- A year that can be evenly divided by 4 qualifies as a leap year, with exceptions...
- If a year can evenly be divided by 31, it will not be classified as a leap year.
This topic was raised and deliberated upon in my online forum at Wizard of Vegas . The original source is 538 .
Can you explain how the magic trick in this YouTube video This method is questionable. I've attempted it numerous times without success. Am I incorrect in my approach, or is it merely a trick?
It's a hoax!
For those who missed the video, here's how Jason, the illusionist, claims it operates:
- Utilize a complete deck of 52 playing cards without any jokers.
- Pick a rank from ace to 10.
- Deal the cards sequentially until you reveal the third card of your designated rank. Make a note of the total number of cards that have been dealt up to that moment.
- The fourth card matching the chosen rank will emerge the same number of cards down from the top of the remaining deck as it took to unveil the first three.
This entire scenario is a practical joke. He relies on a predetermined deck that is crafted to align with the rank he selects. While it seems he's shuffling, he actually possesses remarkable card manipulation skills, capable of executing a deceptive shuffle.
On YouTube, he can filter comments prior to sharing them, only displaying responses from individuals who falsely profess that the trick works for them. Ultimately, it is an elaborate ruse to deceive the audience.
I get into it in even more detail in my December 22, 2022 newsletter .