If a point is established on the come out roll, then
on average,
how many additional rolls of the dice will
be needed for the
shooter's line bet to be resolved ?
| suggestion: |
Split into 3 cases; and,
for each case, find the probability that consecutive rolls of the dice will never yield the point or a 7. |
Case 1 The shooter's point is 4 or 10
To be specific, let's assume that the shooter's point is 4.
Suppose the shooter needs to roll the dice N additional times
after the come out roll
before his line bet is decided.
If a roll of the dice resolves the line bet, we will say that
the roll was a "success";
otherwise, we will call the roll a "failure".
In this particular case, success means rolling 4 or 7, and failure means rolling anything else.
We can easily compute the probability of getting N consecutive failures.
On each roll we have:
P( failure ) = 1 -
[ P(4) + P(7) ] =
1 - [ (3/36) + (6/36) ] = 3 / 4
Since the dice rolls are independent events, the probability of not getting a 4 or a 7 in two consecutive rolls is ( 3 / 4 )2 . The probability for three consecutive failures is ( 3 / 4 )3 , etc.
Let's make a table showing the probability for never getting a 4 or a 7 in N consecutive rolls of the dice.
| Rolls 1 thru N | Probability of not getting 4 or 7 in N consecutive rolls |
|---|---|
| N = 1 | 0.751 = 0.7500 |
| N = 2 | 0.752 = 0.5625 |
| N = 3 | 0.753 ≈ 0.4219 |
| N = 4 | 0.754 ≈ 0.3164 |
| N = 5 | 0.755 ≈ 0.2373 |
Scan down the 2nd column of the above table, and notice that there is a better than 50-50 chance that we will not get a 4 or 7 during the first two rolls. But by the 3rd roll this turns around. The probability of 3 consecutive failures is less than one half ( only 42.19% ).
Since 1 - 0.4219 = 0.5781, there is a 57.81% chance that we will not get 3 consecutive failures.
So, when the shooter's point is 4 or 10, there is a 57.8% probability that at most 3 rolls will be needed to decide the outcome of his line bet.
Case 2 The shooter's point is 5 or 9
To be specific, let's assume that the shooter's point is 5.
The probability of not getting a 5 or a 7 in any roll is
1 - [ P(5) + P(7) ] = 1 - [ (4/36) + (6/36) ] = 26 / 36
≈
0.7222.
| Rolls 1 thru N | Probability of not getting 5 or 7 in N consecutive rolls |
|---|---|
| N = 1 | 0.72221 = 0.7222 |
| N = 2 | 0.72222 ≈ 0.5216 |
| N = 3 | 0.72223 ≈ 0.3767 |
| N = 4 | 0.72224 ≈ 2721 |
In the 2nd column of the table the first entry less than one half occurs at N = 3. So it is unlikely to get 3 consecutive failures.
Since 1 - 0.3767 = 0.6233, there is a 62.3% chance that a 5 or 7 will occur within the first 3 rolls after the shooter has established a point of 5.
So, when the point is 5 or 9, there is a 62.3% probability that 3 or fewer rolls will be needed to decide the outcome of the shooter's line bet.
Case 3 The shooter's point is 6 or 8
To be specific, let's assume that the shooter's point is 6.
The probability of not getting a 6 or a 7 in any roll is
1 - [ P(6) + P(7) ] = 1 - [ (5/36) + (6/36) ] = 25 / 36
≈
0.6944.
| Rolls 1 thru N | Probability of not getting 6 or 7 in N consecutive rolls |
|---|---|
| N = 1 | 0.69441 = 0.6944 |
| N = 2 | 0.69442 ≈ 0.4823 |
| N = 3 | 0.69443 ≈ 0.3349 |
| N = 4 | 0.69444 ≈ 0.2326 |
Since 1 - 0.4823 = 0.5177, there is a 51.8% chance that we will not have 2 consecutive failures.
There is a 51.8% chance that a 6 or 7 will occur within the first 2 rolls after the shooter has established a point of 6 ( or 8 ).
In other words, when the point is 6 or 8, there is a 51.8% probability that 2 or fewer additional rolls after the comeout will be needed to decide the outcome of the shooter's line bet.
Summary:
| point is 4 or 10 | → | usually [ about 58% of the time ] need no more than 3 additional rolls | |
| point is 5 or 9 | → | usually [ about 62% of the time ] need no more than 3 additional rolls | |
| point is 6 or 8 | → | usually [ about 52% of the time ] need no more than 2 additional rolls |