OK then, having started it, it's my turn to contribute to this thread.
About the matter of using a doubling cube with other games, I'm sure
it's a goer, for money games or longmatch games. The arguments against,
for completeinfo games of no chance, (chess etc), struck me as very weak.
Incidentally, I agree that for CINC games that are actually *played*,
i.e. for which perfect strategies are as yet unknown, there IS luck in
them, in any reasonable sense of the term. A chess master pal of mine
from long ago, summed it up well, even profoundly  "In chess there is no
such thing as BAD luck, but there IS good luck". Rather neat, I thought.
I was particularly struck by the suggestion that a doubling cube be used
for snooker matches, especially on telly  that it would reduce those long
boring nearcertain results, and also keep the likely outcome secret
from the programtimeaware viewer. :) Also, snooker is quite a high
volatility game, which would make it a natural for a doubling cube.

However, all that aside, what I really want to talk about here is the
UNDERdoubling part of the thread, for backgammonstyle games.
Underdoubling OR OVERdoubling, of course! A tripling cube might
well be an amusement, especially for the keen gambler.
There's not much in here except a pretty picture, (pretty crumby picture?),
except that I look at the general Mmultiplying cube. So standardly, M=2.
Like others, I *only* look at the very subbackgammonlike situation
where there is ZERO VOLATILITY. This is like the tugodice game, only
more so. We model the game as a diffusion, with a variable moving between
0 and 1, this being the cubeless P(WIN) for one player. This variable
starts at 0.5, and drifts continuously to and fro on the [0,1] interval,
with equal chances of drifting left or right, until an endpoint is hit,
or a cube is dropped. So in effect, moves are infinitesimally small
increments in a 2way race. This continuous movement, or zero volatility,
is a very poor model indeed of real backgammon, as it removes most of the
key everyday cubely problems. But we have to start somewhere.
With a doubling cube, it is wellknown that one should double, and also
eithertakeordrop, when your P(WIN) gets to 0.8
Now we look at the Mcube situation.
Here are two graphs of the player's game equity, as a function of that
cubeless winprobability variable; WITH the cube, (above, graph "f"),
and AGAINST the cube, (below, graph "e").
It is 180symmetric about the (.5 , 0) point, of course.
 :
 :
1       : * 
 /. /:
 f / . / :
 / ./ :
 / / :
 / /. :
 b / / . :
0 +/+/:: x = P(player wins) [cubeless]
 / / K :
 / / :
 / / :
 / / :
 / / e :
/ / :
1 *.../....:........:...
 : :
 :
0 0.5 1
K is the precise cubing point  and by nonvolatility, the take point also.
If the cuber waits just a tad longer, it's a drop;
if he cubes just a tad sooner, it's a take.
"b" is the beaver point, incidentally  the moment you get positive
equity, with the cube. Note one might always beaver soundly even (!) with
P(cubeless win) < .5 .
It is a standard result following from the linearity of expectation,
that the graphs are straight lines; by symmetry between players they
must be 180symmetric as stated; and the (0,1) point is obvious.
That just leaves the slope (and/or intercepts) to calculate.
Clearly we have f(x) = 2x/K  1 ; and e(x) =  f(1x).
For K to be the cubing point we must have M.e(K) = 1 .
Thus 1/M = e(K) =  f(1K) = 1  2(1K)/K = 3  2/K . Therefore...
K = 2M/(3M  1) .
~~~~~~~~~~~~~~~
Thus for the usual cube where M = 2, we get K = 0.8 , as stated.
For an underdoubling cube like a sesquicube, M = 1.5 , we get K = 6/7.
I found this somewhat counterintuitive; I would have expected a cheaper
cube to give rise to a lot *more* cubing, not less! But the key is,
that *taking* is a lot cheaper, so more cubes will be taken, and thus
there is no need to make early cubes, as no "market losing" will apply
until much later. Strictly, the continuous game doesn't have market
losers, of course, but similar reasoning applies.
On the other hand, with an OVERdoubling cube, drops come much earlier,
so market losing is a real issue, and cubing must thus be done earlier.
For a tripling cube, M = 3, K = 0.75 ; not so very big a difference.
There is an even more intriguing matter though. Without analysing the
exact discrete probability process for a game of many tiny increments,
let us assume that both players play a tactic of cubing just BEFORE the
point K, rather than just after. [Just after, would be very boring  it
means there will only ever be one cube turn, always dropped, no game never
gets fully played out! Bunny stuff! ;) ] But with marketloss
anxiety, as it were, the cube will always be turned just before the K
point, and thus always taken.
What now? The game is more interesting  it flows to and fro between
the K and (1K) points, being recubed (and taken) at each swing, till
it eventualy goes well on past one, and ends the game at 0 or 1.
What is the expected number of cubings till the game finishes?
P(hit on cube point  hit at other)
= P(reach K before 0  a diffusion on [0,K] starting at (1K) )
= (1K)/K (a standard result from Markov chain theory, easily
proved using constant expectation from zero drift)
= (M1)/(2M)
= p say.
Then the number of cubings is a geometric random variable with
p_1 = (1p) (p_0 = 0 as it must get to one of the Kpoints first),
p_2 = (1p)p
p_3 = (1p)p^2 etc. So the expected number of cubings is 1/(1p).
For the standard M = 2; K = 0.8 , p = 1/4, E[#] = 4/3.
Even with a tripling cube we get E[#] = 1.5 . Not very exciting?
HOWEVER! There is a BIIIIIIG surprise!
The key thing is, not the expected number of cubings, but the expected
size of the final stake. For this we get...
# cubings: 1 2 3 4 ... . .
probability: (1p) (1p)p (1p)p^2 (1p)p^3 ... . .
final stake: M M^2 M^3 M^4 ... . .
So E[final_stake] = sum[0:inf] {M(1p)(Mp)^k}
= M(1p)/(1Mp)
= (M+1)/(3M)
For the usual M = 2, E[f_s] = 3; moderate.
For the tame M = 1.5 E[f_s] = 1.67, tame indeed, (as 1.5 = min payout!)
But for the mildseeming TRIPLING cube, E[f_s] = infinite !!
~~~~~~~~~~~~~~~~~
With a tripling cube, you are already, unbeknownst, playing a StPetersburg
style game of infinite expectation, (though "only just").
And with a QUADRUPLING cube, ("automatic compulsory beavers"),
you're completely in outer space!!
Food for thought!

Bill Taylor W.Taylor@math.canterbury.ac.nz

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