Hi,

Someone asked me this question:

Each time you roll a fair-sided die. You can stop at any point and
your score is the last score you scored on your die. If you roll a 1
you have to stop and you score 1. If you roll a 2, 3, 4, 5 or 6 you
can choose to roll again How to devise a strategy to maximise your
chances of getting a better score than your opponent?

I am not sure what the optimal strategy should be. But the
Expectation is 3.5, so my guess is that if I get 2 or 3, I will roll
again. If I get 4,5 or 6 I will stop. But it is, on average, making
me score higher than 3.5 only.

Please advise.

Rather than expectations, you should look at the probability of winning
(see "Game Theory"). Sensible strategies are:
(1) accept the first roll
[your score is then a discrete uniform on 1..6 ]
(2) roll again while your score is 2
(3) roll again while your score is 2 or 3
(4) roll again while your score is 2, 3 or 4
(5) roll again while your score is 2, 3, 4 or 5.

You can draw up a table of the probability of winning, depending
on the strategy you & your opponent choose.

My quick calculation - which could be wrong - is that you might
as well accept your first roll (which at least saves effort),
while your opponent should roll while he has a 2, 3 or 4.
You then have a probability of 1/3 of getting a better score
than your opponent.

In article <1173346704.788683.224920@t69g2000cwt.googlegroups.com>,
"jionglong" <jionglong@gmail.com> writes:
Hi,

Someone asked me this question:

Each time you roll a fair-sided die. You can stop at any point and
your score is the last score you scored on your die. If you roll a 1
you have to stop and you score 1. If you roll a 2, 3, 4, 5 or 6 you
can choose to roll again How to devise a strategy to maximise your
chances of getting a better score than your opponent?

I am not sure what the optimal strategy should be. But the
Expectation is 3.5, so my guess is that if I get 2 or 3, I will roll
again. If I get 4,5 or 6 I will stop. But it is, on average, making
me score higher than 3.5 only.

Please advise.


Rather than expectations, you should look at the probability of winning
(see "Game Theory"). Sensible strategies are:
(1) accept the first roll
[your score is then a discrete uniform on 1..6 ]
(2) roll again while your score is 2
(3) roll again while your score is 2 or 3
(4) roll again while your score is 2, 3 or 4
(5) roll again while your score is 2, 3, 4 or 5.

You can draw up a table of the probability of winning, depending
on the strategy you & your opponent choose.

My quick calculation - which could be wrong - is that you might
as well accept your first roll (which at least saves effort),
while your opponent should roll while he has a 2, 3 or 4.
You then have a probability of 1/3 of getting a better score
than your opponent.

[NOTE: if you have a rematch when scores are equal, that is
of course a different game, & is left as an exercise...]

Hi,

Someone asked me this question:

Each time you roll a fair-sided die. You can stop at any point and
your score is the last score you scored on your die. If you roll a 1
you have to stop and you score 1. If you roll a 2, 3, 4, 5 or 6 you
can choose to roll again How to devise a strategy to maximise your
chances of getting a better score than your opponent?

I am not sure what the optimal strategy should be. But the
Expectation is 3.5, so my guess is that if I get 2 or 3, I will roll
again. If I get 4,5 or 6 I will stop. But it is, on average, making
me score higher than 3.5 only.

Please advise.

Hi,

Someone asked me this question:

Each time you roll a fair-sided die. You can stop at any point and
your score is the last score you scored on your die. If you roll a 1
you have to stop and you score 1. If you roll a 2, 3, 4, 5 or 6 you
can choose to roll again How to devise a strategy to maximise your
chances of getting a better score than your opponent?

I am not sure what the optimal strategy should be. But the
Expectation is 3.5, so my guess is that if I get 2 or 3, I will roll
again. If I get 4,5 or 6 I will stop. But it is, on average, making
me score higher than 3.5 only.

Please advise.

Hi,

Someone asked me this question:

Each time you roll a fair-sided die. You can stop at any point and
your score is the last score you scored on your die. If you roll a 1
you have to stop and you score 1. If you roll a 2, 3, 4, 5 or 6 you
can choose to roll again How to devise a strategy to maximise your
chances of getting a better score than your opponent?

I am not sure what the optimal strategy should be. But the
Expectation is 3.5, so my guess is that if I get 2 or 3, I will roll
again. If I get 4,5 or 6 I will stop. But it is, on average, making
me score higher than 3.5 only.

Please advise.

Hi,

Someone asked me this question:

Each time you roll a fair-sided die. You can stop at any point and
your score is the last score you scored on your die. If you roll a 1
you have to stop and you score 1. If you roll a 2, 3, 4, 5 or 6 you
can choose to roll again How to devise a strategy to maximise your
chances of getting a better score than your opponent?

I am not sure what the optimal strategy should be. But the
Expectation is 3.5, so my guess is that if I get 2 or 3, I will roll
again. If I get 4,5 or 6 I will stop. But it is, on average, making
me score higher than 3.5 only.

Please advise.

On Mar 8, 3:38 am, "jionglong" <jiongl...@gmail.com> wrote:
Hi,

Someone asked me this question:

Each time you roll a fair-sided die. You can stop at any point and
your score is the last score you scored on your die. If you roll a 1
you have to stop and you score 1. If you roll a 2, 3, 4, 5 or 6 you
can choose to roll again How to devise a strategy to maximise your
chances of getting a better score than your opponent?

I am not sure what the optimal strategy should be. But the
Expectation is 3.5, so my guess is that if I get 2 or 3, I will roll
again. If I get 4,5 or 6 I will stop. But it is, on average, making
me score higher than 3.5 only.

Please advise.

Often a good place to start in figuring out the problem formulaicly is
to do some simulations.

There are 5 strategies:
stop at the first roll (if you get a 2 or higher),
stop if you get a 3 or higher (otherwise continue),
stop if you get a 4 or higher (otherwise continue),
stop if you get a 5 or higher (otherwise continue),
stop only if you get a 6 (otherwise continue).

If we simulate a random sequence of die rolls, we can determine for
each sequence whether it is uniquely better (W), is tied for the best
strategy (T), or is not the best strategy (L). The results for 2000
random sequences are shown below.


Stop at
stop at: 2 3 4 5 6
W 3.9% 1.4% 1.8% 4.3% 17.7%
T 49.6% 55.3% 58.0% 60.3% 48.8%
L 46.6% 43.4% 40.2% 35.5% 33.6%

Based on this, it would appear that it is better to keep rolling until
you get a 6.

However, if you have one opponent, the advantage of a "6" strategy is
less clear. Let's suppose we have one stopping at "6", the other at
"5". There's an advantage to the "6" strategy, but it's not huge.

Better head to head
6 17.7%
T 66.6%
5 15.8%

Furthermore, 6 versus 4 isn't much different:

Better head to head
6 26.6%
T 49.2%
4 24.3%

and these small differences are an artifact of these particular 2000
sequences.

Consider that a "stop only at 6" rule is really a "stop at 1 or 6,
whichever comes first", so half the time you end up with a 1, and half
the time with a 6.

A "stop if 5 or better" rule is a "stop at 1 or 5 or 6, whichever
comes first", so a third of the time you end up with a 1, a third of
the time a 5, and a third of the time a 6.

If we have two players playing against each other, one using a "stop
at 6" rule and the other "stop at 5", this means:

one third of the time both players will get a "1" score
one third of the time both players will get a "6" score
one third of the time one player will get a "5"
half of that (1/6) the other player will get a "1" and lose
half of that (1/6) the other player will get a "6" and win

So, half of the time each player will win.

I will leave consideration of other rule pairs to somebody who feels
like doing more typing!

"Bob Dole" <t...@mailcity.com> writes:


I will leave consideration of other rule pairs to somebody who feels like doing more typing!

Here is a complete table of the head-to-head results...

"Stephen J. Herschkorn" <sjherschko@netscape.net> writes:



In sci.stat.math, "jionglong" <jionglong@gmail.com> writes:



Each time you roll a fair-sided die. You can stop at any point and
your score is the last score you scored on your die. If you roll a 1
you have to stop and you score 1. If you roll a 2, 3, 4, 5 or 6 you
can choose to roll again How to devise a strategy to maximise your
chances of getting a better score than your opponent?

I am not sure what the optimal strategy should be. But the
Expectation is 3.5, so my guess is that if I get 2 or 3, I will roll
again. If I get 4,5 or 6 I will stop. But it is, on average, making
me score higher than 3.5 only.



If no player observes any other, you cannot come up with "a strategy
to maximise your chances of getting a better score than your
opponent," since each player can use the same strategy.



This analysis is faulty. It is true that each player can use the same
strategy, and thus the expected value of the game is to win half the
time. (You can either replay the game on ties or just count them as
half a win--the analysis is the same.) You can still solve for the
strategy that achieves this maximum value. I posted my results earlier.
The strategy "stop on 6" has an expected value of 1/2 against any
strategy the opponent chooses. However, the strategy "stop on 5" has
a chance of winning which is greater than 1/2, should the opponent choose
a strategy which stops on 2, 3 or 4.

I also analyzed the asymmetric game where "player one" really wants to
beat his opponent, and ties are counted the same as losses. In this
case, the best he can do is to win 1/3 of the time.



Perhaps you are looking for a strategy to maximize your expected
score. Intuitively, the strategy will be of the sort, pick a number
n between 2 and 6; stop as soon as you get 2 or higher, if you
can. Note that there are (6 - n + 1) + 1 = 8-n numbers you will
possibly stop on. With probability
(7-n) / (8-n), you wil get a conditionally expected score of (n+6) /
2. Thus, the expected score is
1 / (8-n) + (7-n) / (8-n) (n+6) / 2 = (44 + n - n^2) / [2 (8-n)].
By inspection, this value is maximized when n = 4 or 5; the maximal
expected score is 4.

Or perhaps you were interested in a two-player game where the second
player observes the first player's score. This is an interesting
question. There are several possible scenarios.



I agree that this is interesting.



1) A tie ends the game, and each player wants to maxmize his/her
chance of winning.



This formulation is a little unclear to me. If a tie ends the game,
then the possibility of ending in a tie has to be considered. How
does it count? Do both players lose? If so, the game is non-zero
sum.