| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Anyone who's played YAHTZEE is familiar with rolling dice to get the
highest possible score for 'chance'. In YAHTZEE, the player may toss
any number of the five dice for as many as three times.
Consider a slightly simplified game involving only one die. We wish
the best strategy when the maximum number of tosses varies.
It's obvious that if the maximum number of tosses is two, then the
player should roll again if the result of his first roll is 1, 2, or 3
(since these are all less than the average toss of 3.5).
If the maximum number of tosses is three (as in YAHTZEE), he should
roll again if the result of his first roll is 1, 2, 3, or 4 (proof is
left to the student). A third roll should be made if the result of the
second roll was 1, 2, or 3 (previous paragraph).
What's the winning strategy if the maximum number of tosses is four?
What's the winning strategy if the maximum number of tosses is not
known to the player, but he is told after each roll whether or not he
may roll again?
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 662.1 | Remember 623.*! | VINO::JMUNZER | Mon Feb 09 1987 09:42 | 9 | |
Dwayne:
What makes a strategy best? Beating someone else? Striving for
a high expectation value? Other?
Is the existence of a best strategy known, or is existence part of
the problem?
John
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| 662.2 | Best Strategy | COMET::ROBERTS | Dwayne Roberts | Mon Feb 09 1987 10:30 | 8 |
Sorry, I wasn't specific enough. Let's define 'best strategy' here
as that strategy which has the highest expected value.
The first part of the problem (where maximum number of rolls is
known) should be quite easy. When the maximum number of rolls is
unknown, however, I'm not sure if the problem is easy or hard.
The existence of a best strategy here may be a good question.
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| 662.3 | VINO::JMUNZER | Thu Feb 12 1987 13:14 | 9 | ||
Let's try to make the question more definite. Suppose you always
have the choice of stopping with your current roll, or asking for
another roll. If you ask for another roll, it's probability p that
you get it, and probability (1-p) that you don't. What's best to
choose if your current roll is 1? 2? 3? 4? 5? 6?
I think this is another nice problem for non-regular contributors.
John
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| 662.4 | CLT::GILBERT | eager like a child | Thu Feb 12 1987 15:14 | 11 | |
The answer to the 'probabilistic stop' problem takes a simple form.
If your current roll is: Then:
1 ask for another roll
2 ask for another roll
3 ask for another roll
4 ask for another roll if p > ?
5 ask for another roll if p > ?
6 don't ask for another roll
All that's required is solving for the two '?'s, and this can be
done easily by hand with a little algebra.
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| 662.5 | CLT::GILBERT | eager like a child | Thu Feb 12 1987 23:35 | 9 | |
Spoiler follows.
For an M-sized die with sides numbered 1 thru M, and a 'current roll'
of n, the expectation is maximized by asking for another roll if n=1,
or if
M 2n-M-1
p > --- --------.
n n-1
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| 662.6 | CLT::GILBERT | eager like a child | Sun Mar 22 1987 19:58 | 1 | |
But what about the case when the maximum number of rolls is 4? | |||||
| 662.7 | Re: .6 | ZFC::DERAMO | Daniel V. D'Eramo | Tue Dec 01 1987 18:59 | 23 |
For one roll, each of 1,...,6 is equally likely, and the
expected value is E1=3+1/2. For two rolls, roll again if
your first roll is less than E1. Your expected value is
1/6(4+5+6+3*E1)
because you stop with 4,5,6 and play the one roll game on
the other three initial rolls. So E2=4+1/4. For the three
roll game, stop at a 5 or 6 and otherwise roll again
[playing the two roll game.] E3 = 4+2/3. Continuing in
this manner you get E4=4+17/18 and E5=5+7/54. So the
strategy this leads to is:
After you roll,
if you have 0 rolls left, stop
if you have 1 roll left, stop at 4,5,6
if you have 2,3,4 rolls left, stop at 5,6
if you have 5 or more rolls left, stop at 6
otherwise, roll again.
Dan
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