Here is a semi-fictitious promotion.
Any quad gets to play a dice game with a 1 in 10 chance to win.
The prize is a progressive jackpot that starts at 100 coins.
If you fail to win the jackpot, the prize goes up 100 coins.
To simplify things, let's assume the quad cycle for the best
game in the casino is 500 hands.
Here is my analysis.
PART 1. Assume 1 machine that I and only I play. Assume dollar coin
play. Assume I sit down at 100 coins on the jackpot.
Since I am the only player and since I can certainly make sure I don't
quit if the jackpot is above 100 coins, I can guarantee that I will
get the initial 100 coin jackpot and all added funds for the times
I miss.
If I play until I hit, the results are....
win 100 if it takes one spin OR
win 200 if it takes 2 spins OR
win 300 if it takes 3 spins....
So as long as I finish the cycle, I always win 100 coins per spin,
but instead of getting 100 every time, I get lumps for the times
I win the progressive. And so my added ER is simple to figute.
It is simply 100 / 2500 = 4% And my EV is simply the ER times the
average cycle to hit, which is 25,000 dollars. So the EV is
1000 dollars.
Now I know that I get the same money in the long run as I do if
I got 100 coins each time. Note that this makes total sense if we
just look at the game from the houses perspective. They put in
100 coins each time a quad it hit. So THEIR EV is to lose 100
coins on each quad. and to pay out 1000 coins per side game win.
PART 2. Now let's assume that the jackpot starts at 500 coins.
I look at this by breaking it into 2 pieces. 100 coins and 400
coins. Ignoring the 400 coins means we have case 1. Now I can
just analyze a flat 400 coins to get that piece. This stuff is all
linear, so I can just add the ERs. Since I only hit the jackpot
1 in 10 times I get quads, My avg coin in for the jackpot is 25,000.
And therefore that piece produces an ER of 400 / 25000 = 1.6%
A formula based on the start level of the jackpot can now be
inferred.
ER = 4% + 0.4%*((Start - 100) / 100)
In this example, the ER is 5.6% The cycle for winning the dice
game is still 5000 hands which is $25,000. This makes the total
EV = 1400.
PART 3. What if another player plays with you?
At first glance it looks like1 each player splits up the ER.
Assuming we each play at the same speed and skill level, the
ER from the point of view of the house is unchanged. But then
we realize that there is twice as much coin in now. So IF the
two of us were to play per the rules of PART 1 for a long time,
it is pretty clear we would each get the same ER as before. The
only difference would be some variance. One of us might win a
lot of spins or win the big jackpots and come out ahead, but
combined we would take the same 100 coins per spin.
What if the 400 coin, non-progressive jackpot was the only
bonus and suppose it was not going to be replaced?
I think the ER is STILL unchanged, but variance goes up
if there is one other player. In this case, if the other player
hits the bonus, it is equivalent to having the machine pay table
suddenly changed by management.
I do think my EV is impacted in this case. Assuming the base game
is break even, my EV is now simply the 400 coins plus the
probability I will win them. When I am the only player, I can
play long enough to ensure that this probability is infinitesimally
close to 1. With 2 players, that probability drops to 0.5
So the total EV of the game is divided equally among the players.
The good news here is that the time required to hit the jackpot
is cut in half, so my HOURLY EV remains the same.
Again I think viewing this as a flat plus a progressive jackpot
simplifies things. If myself and one other player both start play
with a jackpot of 500 coins, the flat part gets, in essence, split
between us. But the progressive part grows twice as fast now, so
his play helps me as much as it hurts me. If we both leave when
it gets hit, between the two of us we will have won the same
jackpot that one of us would have won on his own. If this is '
confusing, just imagine one player playing 2 machines and imagine
he can actually double his hands per hour in that way. Viewed in
that way, we see the ER as a constant, but the Hourly EV doubled.
But since each of us is actually independent, we must split the
doubled hourly EV and remain with the EV we had when we played alone.
Part 4. What if I leave?
If, for example, I am forced to abandon the play when it reaches
2000 coins, what happens? If I am playing alone and start at a
100 coin jackpot and never quit in the middle of a cycle, my EV
for a cycle is 1000 coins. My wins will be scattered from 100
up to several thousand coins, but the average win will be 1000
coins. The probability that I fail to win the jackpot 20
straight times is low, but the effect on EV will be high.
The math here gets messy. I see no way to avoid finding the exact
distribution of all winning results. i.e. probability win in 1
quad = 1/10 Probability win in exactly 2 quads = 9/10 * 1/10
Probability win in exactly 3 quads = .9 * .9 * .1 Etc.
Then each of these must be multiplied by the jackpot for that
result. This sum is technically an endless series, but we do
see that the higher pieces get smaller. (Eventually)
I will start the equation....
EV = Terms
100 * .1 + = 10
200 * .09 + = 18.0
300 * .081 + = 24.3
400 * .0729 + = 29.16
500 * .06561 +
600 * .059049 +
700 * .0531441 + = 37.20087
800 * .04782969 +
...
1500 * .022876792... = 34.315
...
2000 * .015009 = 30.01
...
2900 * .005233476... = 11.5
...
5700 * .00027389... = 1.56
etc.
And now we see that the higher terms will, in fact, approach
zero, which must be true if the game is to have a finite EV.
And although I did not do a complete check, the sum of these
terms should be 1000. And, of course, the sum of the probabilities
should be 1.
What we are seeing is that EVENTUALLY the value attached to the
situations where we miss the 1 in 10 chance become a smaller
part of the overall EV, but it takes awhile. And it is not
unusual to go a long time without winning. We will fail on
our first 19 tries almost exactly 15% of the time.
Adding up and guessing at the values of the missing terms, I
found that for the first 20 terms only, the EV is around 700
coins.
Simply stated, being unable to guarantee you will finish and
win the jackpot costs a big chunck of your EV. Quitting after
the jackpot reaches 2000 would cost you about 300 coins, on
average. This could be considered analagous to getting too
tired, drunk, etc to play.
Part 5. Tying it all together.
My original equation for the ER is...
ER = 4% + 0.4%*((Start - 100) / 100)
Bringing in other players will not hurt our ER, but they will
hurt our total EV, though not our HOURLY EV. I only looked at 2
players, but the idea would apply to any number of additional
players.
Example. If 2 players start with a jackpot of 500 coins, the
EV is given by that big messy formula above, but add 400 to each
payoff term in the equation. That EV equation is correct for one
player. If two people play, each of us gets half the EV, but
in return it is far less likely that either of us will have to'
walk away. Also, our individual hourly EV remains more or less
constant, since we will, on average, hit the jackpot twice as fast.
The EV (if we know we will finish it) is 1000 + 400 divided by
the number of players.
Since the cost of forced leaving is quite high, as well as
fairly common if we were to play alone, bringing in other
players can be viewed as a good thing, since it greatly reduces
the chance you will have to walk away from a highly favorable
situation. But once in awhile some guy will walk in and "swipe"
a jackpot you spent time building and that won't feel like a good
thing.
Does anybody see any flaws in this analysis???
Quad Zilla