Dependant probability and Poisson distribution--a game question.

**yearzero** · February 4th 2007, 07:08 PM

First off let me thank anyone who is willing to assist me with this. Here is the scenario...

A gambler places a number of consecutive $3 dollar wagers. During each wager there is a 1 in 500 chance of initiating a three-tiered Event where a player may or may not win a up to three consecutive jackpots. However the chance to win the various jackpots during this Event (once initiated) and the amount of the wins are dependant on how many previous non-winning wagers have occurred.

As an example…
If the player initiated the Event on his first through 100th wagers there is a 50% chance of winning $10 dollars. If that is won there is an 25% chance of winning an additional $25 dollars. If that is won there is an additional 10% chance of winning an additional $100 dollars. (Dependant probability=50% chance of winning $10 dollars, 12.5% chance of winning $35 dollars, 1.25% chance of winning $135)

If the player initiated the Event on his 101st through 250th wagers there is a 70% chance of winning $20 dollars. If that is won there is a 50% chance of winning an additional $50 dollars. If that is won there is an additional 25% chance of winning an additional $200 dollars. (Dependant probability=70% chance of winning $20 dollars, 35% chance of winning $70 dollars, 3.5% chance of winning $270)

If the player initiated the Event on his 251st through 550th wagers there is a 100% chance of winning $30 dollars. If that is won there is an 80% chance of winning an additional $75 dollars. If that is won there is an additional 60% chance of winning an additional $300 dollars. (Dependant probability=100% chance of winning $30 dollars, 80% chance of winning $105 dollars, 48% chance of winning $405)

If the player initiated the Event after his 550th wager there is a 100% chance of winning all three jackpots (in this case a $100, $250 and a $500 dollar jackpot). (Dependant probability=100% chance of winning $850).

If the Event occurs (regardless of winning any jackpot(s) during the event) the non-winning wager counter resets.

How would you calculate the total odds for this game? What would the average loss to the player be per wager? Would it be necessary to use a Poisson distribution model to indicate the probability of the Event triggering occurrences over the course of 100,000,000 wagers?

I would certainly appreciate any help with this.

**CaptainBlack** · February 5th 2007, 02:33 AM

The approach to this problem that I would suggest is to start by calculating the proportion of events of each type (label them 1 through 4).

Then analyse the distribution of winnings for each event type (probably
using an contingency tree to keep track of the results). (I assume that
there are no additional stakes involved for the staged jackpot phase).

As the type of an event is independent of that of the proceeding event
we need only calculate the probability of each event type for a single
sequence of wagers ending at an event.

The probability that the event occurs on the $n$ -th wager is the probability
that it has not occured on any of the preceeding $n-1$ wagers times the
probability that it occurs on this one:

$ pr(n)=\frac{1}{500} \times \left( 1-\frac{1}{500} \right)^{n-1} $

So the probability of a type 1 event:

$ P(type1)=\frac{1}{500}\,\sum_{r=1}^{100}\left( 1-\frac{1}{500} \right)^{r-1} $ ,

which is a geometric series and so:

$ P(type1)=\frac{1}{500}\times \frac{1-(1-1/500)^{100}}{1-(1-1/500)}\approx 0.1814 $ ,

Similar arguments show that:

$ P(type2)=\frac{(1-1/500)^{100}}{500}\times \frac{1-(1-1/500)^{150}}{1-(1-1/500)}\approx 0.2123 $

$ P(type3)=\frac{(1-1/500)^{250}}{500}\times \frac{1-(1-1/500)^{300}}{1-(1-1/500)}\approx 0.2737 $

$ P(type4)=\frac{(1-1/500)^{550}}{500}\times \frac{1}{1-(1-1/500)}\approx 0.3325 $

From these and the results of the analysis of the return when an event of each type occurs
we can calculate the mean return in a long run (say N) of wagers, as there are N/500 events
and we know the amount wagered ($3N).

It would also be wise to support this type of analysis with a simulation to make sure that
no hidden invalid assumption/s have creapt in

RonL

**yearzero** · February 5th 2007, 05:45 AM

Thank you! That makes perfect sense, even with my limited math skills!

**CaptainBlack** · February 6th 2007, 09:37 PM

You will note that here we are using the Geometric distribution for the
number of wagers to the next event. This is the discrete analog of the
(Negative) Exponential distribution which would be appropriate for the time to
next event in continuous time (we are using the wager number an a proxy
for discrete time here).

This results in the number of events in a fixed number of wagers having a
Binomial distribution rather than a Poisson which would be the case with continuous time.

With a small probability of an event per wager (as we have here) the difference
between the discrete time and continuous time models is negligible, unless
very high precision is required, or the ability to change the frequency of events
to something much higher is required.

RonL

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