| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1103.1 | | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Mon Jul 31 1989 22:32 | 12 |
| If you keep the 2346 and roll the remaining die, rolling
it again if it isn't a 5, then you have a probability of
11/36 of ending up with 23456. Likewise, if you keep the
1234 and roll the remaining die, rolling it again if it
isn't a 5, then you have the same probability, 11/36, of
ending up with 12345.
Is there a strategy that has a higher probability of getting
a five-straight? What if you keep the 234 and roll the
remaining two dice?
Dan
|
| 1103.2 | I'll keep four, thanks | NIZIAK::YARBROUGH | I PREFER PI | Wed Aug 02 1989 15:22 | 14 |
| If you keep only the 234, then on the second roll:
1) In 4 cases you will hit 12345 or 23456
2) In 9 cases you will improve to 1234
3) In 7 cases you will improve to 2345
4) In 7 cases you will improve to 2346
5) In 9 cases you will remain with 234
The improvement chances for cases 2-5 are 1/6,1/3,1/6, and 1/9, so the
total probablility of success is
4/36 + 9/(36*6) + 7/(36*3) + 7/(36*6) + 9/(36*9)
=.27777...
which is less than 11/36 = .30555..., so the best strategy appears to be to
try to get a 5 in two rolls.
Lynn
|
| 1103.3 | yahtzeee | ESCROW::MUNZER | | Wed Aug 02 1989 17:27 | 5 |
| Lynn,
I agree. Suppose you have N rolls. What's best?
John
|
| 1103.4 | You lose on doubles | AKQJ10::YARBROUGH | I prefer Pi | Wed Aug 02 1989 20:15 | 7 |
| I think you always do better by trying to roll one number with one die than
to roll any n numbers with n dice. The odds calculations are a bit hairy,
but it resembles the calculations for the carnival game called
chuck-a-luck, which favors the house because you lose ground when you roll
multiples. The difference between .2777 and .3055 in .-2 is 1/36, and that
is the chance of rolling two 5's on your second roll; you must have one 5
to win, but two doesn't buy you anything.
|
| 1103.5 | N matters to me | ESCROW::MUNZER | | Thu Aug 03 1989 20:38 | 5 |
| Lynn,
Now, I disagree. but I can't follow your reasoning.
John
|
| 1103.6 | | KOBAL::GILBERT | Don't Worry, Be Bobby McFerrin | Sat Aug 05 1989 15:25 | 11 |
| If you keep the 2346 and roll the remaining die, rolling
it again if it isn't a 5, for up to N times, then the
probability of ending up with 23456 is:
( 1 - (5/6)^N ) / 36
Likewise, you have the same probability if you keep the 1234.
If you keep only the 234, the calculation of the probability
of success is tedious.
|
| 1103.7 | | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Sun Aug 06 1989 02:20 | 5 |
| re .6
That's just 1 - (5/6)^N; not divided by 36.
Dan
|