Ask The Wizard #414
Someone challenged me to the following bet. We would roll a pair of six-sided dice until one of the following two events happened:
- Two totals of seven appeared.
- At least one six and one eight appeared.
I would win if the two sevens appeared first at even money. Don't I have the advantage since a total of seven is the mostly likely. However, I suspect a trick. Who had the advantage?
The other side had the advantage. Here is the probability of rolling each total involved:
- 6 = 5/36
- 7 = 6/36 = 1/6
- 8 = 5/36
The average waiting time to achieve an event of probability p is 1/p.
The probability of rolling a total of 7 is 1/6. So, on average, it takes 6 rolls to get a total of 7. Two get two of them would require 12 rolls, on average.
The probability of rolling a total of 6 or 8 is (5/36) + (5/36) = 10/36. Note that the 6 and 8 can happen in either order. The probability of achieving either the 6 or 8 is thus 1/(10/36) = 36/10 = 3.6.
Once the first total is achieved between the 6 and 8, the probability of getting the other one is 5/36. The waiting time for that second event is 1/(5/36) = 36/5 = 7.2 rolls.
Thus, the expected rolls to get both a 6 and 8, in either order is 3.6 + 7.2 = 10.8. This is less than the 12 for two sevens. Thus, taking the 6 and 8 is the better side of that bet.
An icosahedron (20-sided die) is rolled. The player may either keep the number of dollars from the roll of the die or pay $1 to roll again. The player may keep doing this an unlimited number of times. What is the the correct strategy and fair price to play this game?
Let's say the lowest roll a player will accept is r.
Once that goal is achieved, the average outcome will be (20+r)/2.
The probability the goal is achieved on any one roll is (21-r)/20. Thus the expected rolls to achieve the goal is the inverse, or 20/(21-r).
For a goal of rolling r, the expected win is thus (20+r)/2 - 20/(21-r). Here are some expected wins for plausible values of r.
- 14: $15.14
- 15: $15.17
- 16: $15.00
So, we see the expected win maximized at $15.17 with a goal of rolling a 15 or higher.
This question was adapted from puzzle 22 in Math Puzzles Volume 3 by Presh Talwalkar. In his book, a 100-sided die is used.
In tennis, assume the probability the server wins any given point is p. What is the probability the server wins the game if the score is Ad-Out, Deuce, or Ad-In?
For the benefit of other readers, in tennis a player must win by 2 points to win a game. Being a point behind is called Ad Out and being a point ahead is Ad In.
Let's create some terminology.
- a = Probability winning the game at Ad Out.
- b = Probability winning the game at Deuce.
- c = Probability winning the game at Ad In.
From here we can form a Markov Chain, as follows:
- a = pb
- b = pc + (1-p)a
- c = p + (1-p)b
Let's try to solve for b, plugging the first and third equations above into the second:
b = p(p + (1-p)b) + (1-p)pb
b = p2 + pb - p2b + pb - p2b
Some simple algebra leads to...
b = p2/(1-2p+2p2)
From there it is easy to use the first and third formulas to find a and c.
The following table shows the probabilities at the three possible stages for various values of p.
| p | Ad Out | Deuce | Ad In |
|---|---|---|---|
| 0.1 | 0.001220 | 0.012195 | 0.110976 |
| 0.2 | 0.011765 | 0.058824 | 0.247059 |
| 0.3 | 0.046552 | 0.155172 | 0.408621 |
| 0.4 | 0.123077 | 0.307692 | 0.584615 |
| 0.5 | 0.250000 | 0.500000 | 0.750000 |
| 0.6 | 0.415385 | 0.692308 | 0.876923 |
| 0.7 | 0.591379 | 0.844828 | 0.953448 |
| 0.8 | 0.752941 | 0.941176 | 0.988235 |
| 0.9 | 0.889024 | 0.987805 | 0.998780 |