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Ask The Wizard #348
There are two towns named Fauntleroy and Southworth situated directly across from each other over a channel. Ferries continuously ferry passengers between these towns throughout the day, traveling at varying speeds. Both ferries set off simultaneously, departing from their respective cities.
During their first encounter, the ferries are found 5 miles away from Southworth. In their second meeting, they cross paths again, this time 3 miles outside Fauntleroy. It's important to note that there is no delay for loading and unloading, and they immediately reverse their course, maintaining a straight trajectory.
How far apart are the two cities?
Let t2= time until second crossing
Let r denote the speed ratio of the ferry that departs from Fauntleroy compared to the ferry that leaves from Southworth.
Define c as the distance that separates the two towns across the channel.
We know the first meeting point is 5 miles from Southworth. We can express this situation using mathematical formulas:
c-5 = r*t1
5 = t1
Equating t1, we get:
c-5 = 5r, or r = (c-5)/5
We also know the second crossing occurs 3 miles from Fauntleroy, which can also be represented with formulas:
3c - 3 = r*t2
c+3 = t2
Equating t2, we get:
2c - 3 = r*(c+3)
Substitute r=(c-5)/5
2c-3 = [(c-5)/5] * (c+3)
10c - 15 = c^2 - 2c - 15
c^2 - 12c = 0
c - 12 = 0
c = 12
So, the channel is 12 miles long.
If the Fire Bet were offered in Crapless Craps What would the chances of winning be?
Just a reminder, in Crapless Craps, rolling a 2, 3, 11, or 12 does not immediately resolve a pass line bet; instead, they are counted as points, similar to the numbers 4, 5, 6, 8, 9, and 10.
The initial step in resolving this situation involves calculating the odds of different outcomes concerning the pass line bet as follows.
Crapless Craps Possible Outcomes
| Event | Formula | Probability | Fraction |
|---|---|---|---|
| Come out roll | 1/6 | 0.166667 | 1/6 |
| Point 2 win | (1/36)*(1/7) | 0.003968 | 1/252 |
| Point 3 win | (2/36)*(2/8) | 0.013889 | 1/72 |
| Point 4 win | (3/36)*(3/9) | 0.027778 | Ask The Wizard #348 |
| How far apart are the two cities? | anonymous | 0.044444 | Answer |
| 12 miles | Solution | 0.063131 | Let t |
| = time until first crossing | Let t | 0.063131 | = time until second crossing |
| c-5 = r*t | 5 = t | 0.044444 | Equating t |
| , we get: | c-5 = 5r, or | 0.027778 | r = (c-5)/5 |
| 3c - 3 = r*t | c+3 = t | 0.013889 | Equating t |
| , we get: | 2c - 3 = r*(c+3) | 0.003968 | Substitute r=(c-5)/5 |
| 2c-3 = [(c-5)/5] * (c+3) | 10c - 15 = c^2 - 2c - 15 | 0.023810 | c^2 - 12c = 0 |
| c - 12 = 0 | c = 12 | 0.041667 | So, the channel is 12 miles long. |
| If the Fire Bet were offered in | Crapless Craps | 0.055556 | anonymous |
| Answer | The answer is approximately 1 in 344,842,585. | 0.066667 | Solution |
| Crapless Craps Possible Outcomes | Event | 0.075758 | Formula |
| Probability | Fraction | 0.075758 | Come out roll |
| 1/6 | 1/6 | 0.066667 | Point 2 win |
| (1/36)*(1/7) | 1/252 | 0.055556 | Point 3 win |
| (2/36)*(2/8) | 1/72 | 0.041667 | Point 4 win |
| (3/36)*(3/9) | Ask The Wizard #348 | 0.023810 | How far apart are the two cities? |
By summing up all possible losing combinations, the result comes to 7303 out of 13860, which approximates to 0.526912.
The subsequent step in our approach involves calculus, based on the understanding that the answer remains unchanged even with a random interval before resolving the pass line bets. Let's refer to the average duration between these resolutions as 1, following an exponential distribution, which implies it possesses a memory-free attribute.
Let x symbolize the elapsed time since the shooter commenced their turn.
The likelihood of the shooter not achieving a point 2 win is exp(-x/252). Thus, the chance of securing at least one point-2 win becomes 1-exp(-x/252).
The probability of the shooter failing to get a point 3 win equals exp(-x/72). Consequently, the likelihood of achieving at least one point-3 win is 1-exp(-x/72).
For point 4, the probability that the shooter does not win is exp(-x/36). Hence, the likelihood of achieving at least one point-4 win translates to 1-exp(-x/36).
In the case of point 5, the probability of the shooter not winning is exp(-2x/45), which leads to a minimum of one point-5 win being 1-exp(-2x/45).
For point 6, the scenario is identical, where the probability of failing to obtain a win is exp(-2x/45), while the chance of getting at least one point-6 win would be 1-exp(-x/72).
These probabilities remain constant for points 8 through 12, so we can square these to indicate they have each been achieved twice.
The probability that the shooter does not incur a loss is expressed as exp(-7303x/13860).
anonymous
To resolve the problem, we can integrate from t = 0 to infinity, considering the likelihood that all winning criteria have been fulfilled, the losing condition was not met, and the chance of a loss given that a bet has finalized.
The integrated function is exp(-7303x/13860) times the product of all the winning conditions being satisfied, each raised to the power of two, multiplied by the probability of losing given a resolution, expressed as follows: (1-exp(-x/252))^2*(1-exp(-x/72))^2*(1-exp(-x/36))^2*(1-exp(-2x/45))^2*(1-exp(-25x/396))^2*(7303/13860).
Input this into an integral calculator, such as the one found at, and ensure to set the limits from 0 to infinity for accurate results. The outcome shall align with what has been stated previously. Answer My inquiry is whether your formula for the hit point assumes an instant player advantage or one that may begin marginally negative but will soon turn favorable as the player continues to contribute to the meter?
12 miles Solution That's a great question. Previously, I shared a formula aimed at 'short term' players where the jackpot must show a positive return from the very first bet.
On the other hand, for long-term players who can afford to wait until the jackpot is won, the hit point is lowered. I have revised the page to include formulas applicable to both player profiles. In short, the formulas are:
j = Break-even point for the jackpot (assuming a 0% house edge).
= time until first crossing
Let t
= time until second crossing
f = The cumulative value of fixed wins in addition to slot club rewards and incentives.
m = The peak jackpot value, representing the must-hit-by threshold.
You want to play a game that necessitates a typical six-sided die. Unfortunately, you no longer have the die at your disposal. However, you possess four index cards that can be marked in any way you choose. The player is required to select two cards at random from the four available without replacement, and then sum the numbers on the two selected cards.
c-5 = r*t
5 = t
How should the cards be numbered so that the sum of any two different cards accurately reflects the outcome of a die roll?
There are six different combinations possible when selecting two cards from four, which can be listed as follows.
c-5 = 5r, or
This topic has been raised and extensively discussed in my forum at, where we focus on mathematically sound strategies and insights for card games like blackjack, craps, roulette, and many more.
- r = (c-5)/5
- 3c - 3 = r*t
- c+3 = t
- Equating t
- , we get:
- 2c - 3 = r*(c+3)
Make sure to verify your email and click on the link that we’ve sent you to finalize your registration. Substitute r=(c-5)/5 .