I recently ran across an interesting promotion idea.
Here is the premise. A casino offers to cover 10% of all your
losses on a given day. This sounds pretty cool. It looks like it
is worth 5% if the game is a break even game. However, it is really
quite a bit trickier. Here is my analysis. If I have erred, please
correct me.
Lets start by assuming the game is a coin toss.
Now suppose we have $60. How do we maximize the return. The answer
is to simply bet all the money on our first bet. We win 60 or lose
54 The Expected value is simply (60 - 54) / 120 = $3, which is the
5% I mentioned earlier. Assuming the game is favorable to the
house, this represents an upper bound on the EV we can get.
However, we can surely do worse. Suppose we get clever and decide
to make 6 $10 wagers. Not so clever. When we consider that there
are 64 WLWLLL sequences and sum the products of the overall win or
loss and their respective probabilites, the answer is the EV for
this scheme. What we get is about 94 cents.
So it seems we are stuck with a single fairly large wager that will
have a nice EV, but we can't really make much money unless we make a
very large wager.
Can we do better?
I think so. The key is to seek out a game where the return is not 2
for 1. Let's look at an imaginary roulette game where there are no
zeros. Now let's imagine having $100 to wager. Suppose we bet it
all at once. Our expected value is
35*(-100 + 10) + 3500 / 36 = $9.72
This is a very interesting result. We have broken through the 5%
barrier by choosing a game with a high payoff. That 5% upper bound
was only valid for games that had even money bets. What about
single zero roulette. Now the equation changes a bit.
36*(-100 + 10) + 3500 / 37 = $7.02
A nasty hit, but we are still far above 5%.
What about sequences of wagers. Let's consider betting $100 twice,
but quitting if we win the first wager. Probability of winning the
first wager is 1/37. Probability of making and winning winning 2nd
wager is 36/37 * 1/37. Probability of making and losing 2nd wager
is 36/37 * 36/37. These give wins of $3500, $3400, and -$180.
Summing the products give $94.595 + $89.408 - $170.40 = $13.603.
This is a lower EV than we would get by betting the $200 all at once.
In that case, we would simply double the EV from the single $100
wager and arrive at $14.04.
Even so, we see that we are doing a better job of reducing risk
without doing so much damage to our return.
Lets revert to the no zero version and lengthen the sequence. Our
absolute EV will continue to rise (but by smaller sizes) until we
reach a loss of $3500. At that point, we gain nothing by playing
further. If we win, we lose the 350 bonus that we have coming. The
equations becomes 35*(-90) + 3500 - 350 / 36 = 0.
For this imaginary no zero roulette game, we can see that $100
wagers will continue to yield over 5% return until we have made
roughly 17 wagers. The math is really mess now, but it seems that
our average sequence about 14 wagers retruning an average of 7% or
so. Our EV for this scheme is roughly $98. Of course, we will lose
the entire $1400 quite often. I am guessing this will happen about
60% of the time. As before, our EV is best if we just bet it all to
start. If we just bet $1400 to start, we get an EV of $136.08.
When we consider other imaginary games with various returns, we see
a common theme. We can reduce the damage caused from splitting our
wager into pieces. However, we will always face the prospect of
losing until playing gives no benefit. For games like roulette,
this is just the point where a winning bet no longer gets us above
even.
Now let's change gears and look at fixed wagers size games. Suppose
we had a no zero roulette game with a table maximum of $10. Now we
cannot simply wager our $350 on the first turn. Instead, we would
be forced to employ a sequence of wagers. As shown above, we will
maximize our EV by playing as long as we have not yet lost 35 units.
What is surprising is that this also applies to the winning side.
Suppose we win the 21st game. We now have a tidy profit of $150.
Should we quit? Not necessarily. If we bet again, our wager just
has an ER of 100%. However, there is a non-zero probability that we
will fall back down to even. Once that happens, we are once again
able to start making wagers with a large advantage. Of course, if
we are fortunate enough to win again, we will have a much lower
chance of falling back to even. Since the game is break even, we
may never again fall back to even! So, we should quit when
retruning to even is not very likely and will probably take quite
awhile. This is subjective. Even if you are ahead 100 units, there
is some small advantage to be gained by further play. My guess is
that quitting somewhere between 17 and 35 units is about right.
If we consider our earnings per unit time, we certainly want to set
some sort of stop win so that we don't pursue a tiny advantage.
Now finally, what about VP. VP is tricker, since it has lots of
payouts. However, let's consider Jacks or better. Let's take a
major liberty and just focus on quads. That hand is worth 25 units.
The above analysis seems to indicate that we probably do very well
if we play until we lose 25 units or until we get ahead about 15
units, though playing until you are ahead 25 units may work well too.
Just how much this adds to the overall return is unclear. A rough
approximation of your advantage at a given time is to take %10 and
divide it by 25. Now subtract that number from 10% for each bet we
lose. This number is .004. What we see is that after we are 24
bets down, the ER is now almost precisely 100%, since the game
itself carries an ER of 99.55% or so. If you seek to play only with
an advantage of 1% or better, you probably want to quit after losing
about 22 units.
One interesting thing about VP in this scenario is that the ebb and
flow of lesser hands will tend to lengthen our play (and therefore
our EV). But the fact that quads are much harder to hit than their
payofff will tend to reduce the lenght of our play.
If our game was 10/7 DB, we would get much more volatility, but we
would get to play until we lost roughly 48 units. With the reduced
pays on other hands, that may or may not translate to more coin in.
An analysis would be very difficult, but it could be that the JOB
game create a larger average session wager sum and end up returning
a higher EV, though the hourly EV probably lands in favor of the
10/7 DB game.
Just what this adds to your EV is not clear, but it clearly gives
you the chance to play with a high advantage for a decent hourly
return. On JOB, you start with well over a 5% edge (since there
are payouts are higher than even money). You eventually will wander
too far from break even and find that playing is no longer
profitable enough. If you simply lose 22 hands in a row, you will
have had an EV of roughly $3 on dollar JOB. However, the average
coin in for this scenario is probably quite a bit higher. I'd
guess you should average at least 60 hands with an average return
somewhere around 3%. That is still only about a $9 overall EV.
This shows that even now the game is a marginal affair. Sure, you
get a great hourly EV, but you only play for five minutes.
Now as I write, I have finally figured out the best way to handle
this, though I don't know if you could pull it off. Suppose the
casino offered this to two players. Each player places a $1000
wager on pass/don't pass. If they are risk averse, they hedge with
a $67 bet on hard 12(sic). If 12 comes, they win $10. If not, they
win $100 less the $67 hedge. This is free money with no risk.
In fact, you can do even better with more players. Suppose you have
6 players bet 6 numbers on a single zero roulette game. Now you
hedge the zero. For a $1000 each, you lose 5 * $900 and win $5000.
You do get hit with a big charge for the hedge. $180 gives you a
$300 win on zero. Of course, I have no clue how one could actually
get away with this, but it is a good thing to keep in the back of
your mind.
Jim Morgan
