# Ask the Wizard #348

Two cities, Fauntleroy and Southworth, lay directly across a channel. Two ferries go back and forth all day long between the two cities. The ferries travel at different speeds. At the same time, they both start out, one from each city.

The first time they cross is 5 miles from Southworth. The second time they cross is 3 miles from Fauntleroy. Assume no time to load and unload, but both make an immediate u-turn. Also assume they go in a straight line.

How far apart are the two cities?

"Anonymous" .

_{1}= time until first crossing

Let t

_{2}= time until second crossing

r = ratio of speed of ferry initially leaving Fauntleroy to the speed of the ferry initially leaving Southworth.

c = Distance of channel between two cities.

We are given that the first time they cross is 5 miles from Southworth. To express this in formulas:

c-5 = r*t_{1}

5 = t_{1}

Equating t_{1}, we get:

c-5 = 5r, or r = (c-5)/5

We are also given that second time they cross is 3 miles from Fauntleroy. To express this in formulas:

3c - 3 = r*t_{2}

c+3 = t_{2}

Equating t_{2}, 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 be the probability of winning?

"Anonymous" .

As a reminder, in Crapless Craps the 2, 3, 11 and 12 do not immediately resolve a pass line bet, but are considered points, much like the 4, 5, 6, 8, 9, and 10.

The first step in my solution requires calculating the probability of any given outcome of the passline 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 | 1/36 |

Point 5 win | (4/36)*(4/10) | 0.044444 | 2/45 |

Point 6 win | (5/36)*(5/11) | 0.063131 | 25/396 |

Point 8 win | (5/36)*(5/11) | 0.063131 | 25/396 |

Point 9 win | (4/36)*(4/10) | 0.044444 | 2/45 |

Point 10 win | (3/36)*(3/9) | 0.027778 | 1/36 |

Point 11 win | (2/36)*(2/8) | 0.013889 | 1/72 |

Point 12 win | (1/36)*(1/7) | 0.003968 | 1/252 |

Point 2 loss | (1/36)*(6/7) | 0.023810 | 1/42 |

Point 3 loss | (2/36)*(6/8) | 0.041667 | 1/24 |

Point 4 loss | (3/36)*(6/9) | 0.055556 | 1/18 |

Point 5 loss | (4/36)*(6/10) | 0.066667 | 1/15 |

Point 6 loss | (5/36)*(6/11) | 0.075758 | 5/66 |

Point 8 loss | (5/36)*(6/11) | 0.075758 | 5/66 |

Point 9 loss | (4/36)*(6/10) | 0.066667 | 1/15 |

Point 10 loss | (3/36)*(6/9) | 0.055556 | 1/18 |

Point 11 loss | (2/36)*(6/8) | 0.041667 | 1/24 |

Point 12 loss | (1/36)*(6/7) | 0.023810 | 1/42 |

If you add all the ways to lose, you get 7303/13860 = apx. 0.526912.

The next step in my solution to this problem uses calculus. It relies on the fact that the answer would be the same if there was a random period of time between pass line bets resolving. Let's call the mean time between bet resolutions 1 and distributed by the exponential distribution, meaning it has a memoryless property.

Let x represent the time since the shooter started his turn.

The probability the shooter did not get a point 2 win is exp(-x/252). Thus, the probability that get got at least one point-2 wins is 1-exp(-x/252).

The probability the shooter did not get a point 3 win is exp(-x/72). Thus, the probability that get got at least one point-3 wins is 1-exp(-x/72).

The probability the shooter did not get a point 4 win is exp(-x/36). Thus, the probability that get got at least one point-4 wins is 1-exp(-x/36).

The probability the shooter did not get a point 5 win is exp(-2x/45). Thus, the probability that get got at least one point-5 wins is 1-exp(-2x/45).

The probability the shooter did not get a point 6 win is exp(-2x/45). Thus, the probability that get got at least one point-6 wins is 1-exp(-x/72).

Note these probabilities are the same for 8 to 12, so we can square them to show they have been achieved twice each.

The probability the shooter did not lose is exp(-7303x/13860).

The probability of losing is 7303/13860.

We can solve for the problem by integrating from t = 0 to infinity of the probability of the product of all winning requirements have been met, the losing outcome has not been met, and the probability of losing given a bet has been resolved.

The function being integrated is exp(-7303x/13860)*(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).

Put that into an integral calculator like the one at integral-calculator.com. Remember to put in the limits from 0 to infinity. The answer will be what is expressed as the answer above.

Thank you for your analysis of must-hit-by progressives. My question is does your formula for the hit point to play assume an immediate player advantage or a situation that may be slightly negative at first, but will shortly turn positive as the player contributes to the meter?

"Anonymous" .

Good question. It previously gave a formula for a "short term" player, where the jackpot must be positive on bet one.

However, for the long-term player, who can afford to play until the jackpot hits, the hit point is less. I have updated the page to include formulas for both types of players. Briefly, the two formulas are:

j (short term) = m × (1-f)/(1-f+r)

j (long term) = m × (1-f-r)/(1-f+r)

Where:

j = Breakeven jackpot size (with 0% house edge)

f = Value of all fixed wins plus slot club points and incentives.

m = Maximum jackpot (the must-hit-by point)

n = Minimum jackpot (the reseed point)

r = Rate of meter rise

You wish to play a game that requires an ordinary six-sided die. Unfortunately, you lost the die. However, you have four index cards, which you may mark any way you like. The player must choose two cards randomly from the four, without replacement, and take the sum of the two cards.

How can you number the cards so that the sum of two different cards represents the roll of a die?

Gialmere

Number them 0, 1, 2, and 4.

There are six ways to draw two out of four cards, as follows.

- 0+1 = 1
- 0+2 = 2
- 1+2 = 3
- 0+4 = 4
- 1+4 = 5
- 2+4 = 6

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