Bankroll for short VP rounds?

if I was going to play a short round of 9/6 jacks (say I need 5k coin in for a promotion, and I am playing a $5 machine, so $25/pull, therefor only need to play 200 hands)

how much bankroll would I need to have X% chance of making it to my goal without going bankrupt?

Is there a calculator to calculate something like this?

What if I had to play a $10 machine so I only needed 100 hands?

There's no simple formula for this, given that each hand has 10 different results.

Using the Wizard's 9/6 Jacks or Better probabilities, here's what I get for the 200 hand solution:

Here's what I get for 100 hands, but remember, you should be comparing each number to the number twice as high in the 200-hand table (for example, with a bankroll of $1000 which is 40 bets at a $5 machine and 20 at a $10 machine, you have a 13.4606% chance of failure with the $5 machine and a 28.637% chance of failure with the $10 machine.) It appears that the $5 machine and 200 bets has a much better chance of success given the conditions.

Thank you for putting that table together.

How did you do it? Did you just do a bunch of simulations (if so how many? ) or did you do something math and fancy?

There are two commercial products that calculate this kind of thing essentially instantly. Video Poker for Winners software and Dunbar's Risk Analyzer for Video Poker. The two products use different techniques --- but come up with very similar answers.

Quote: Bondy3

Thank you for putting that table together.

How did you do it? Did you just do a bunch of simulations (if so how many? ) or did you do something math and fancy?

Something math and fancy - it's usually called a Markov chain.
In this case, it works like this:
Draw a table, where each row is the number of bets made, and each column is your current bankroll. The value in each cell of the table is the probability of getting there. For the first row, which is 0 bets, the probability of the bankroll being your initial bankroll value is 1, and the probability for all other values is 0.
Starting from "row zero" and working down one row at a time, do this:
1. Set all values in that row to zero.
2. Since, if your bankroll was down to zero after the previous number of bets, it will remain zero (since you can't bet anything), copy the previous row's "zero bankroll" column's value into the current row's zero bankroll value.
3. For each column beginning with "bankroll 1", multiply the value in the previous row for that column by the probability of each of the 10 possible results (lose, jacks or better, two pair, and so on through royal flush), and add that value to the current row, in the column representing what the new bankroll would be.
4. In the case of this problem, I fudged it slightly; since we are interested only in what the probability is that the bankroll is (well, is not) zero after 200 bets, and you cannot lose more than 1 per bet, if a calculated value of the bankroll after a win is more than 200, treat it as 200. Without some sort of cap like this, there would be over 270,000 different results just after six bets.
Continue until you get to the maximum number of bets; the value in the zero bankroll column is the probability of failure.

Thank you for that guideline. After reading what you wrote as well as the Wikipedia on markov chains I think I understand what you did.

What programming language do you use for this type of problem?

Data point for short term return:
$15 Freeplay.
.25 9/6 job

Only got $5. :(

33% return

Quote: 100xOdds

Data point for short term return:
$15 Freeplay.
.25 9/6 job

Only got $5. :(

33% return

That's not unusual for a sample of twelve hands. In fact, Free Play is the only reason that you are noticing it at all, many similar short runs come during longer sessions and go unnoticed simply because they are part of a much bigger session. Figure if the probability for a hand with no return is:

0.54543467

Then you would look at the probability on a zero return of $15 coin-in based on twelve hands as:

(0.54543467)^12 = 0.00069328671

Which means you would have:

1/0.00069328671 = 1 in 1,442.4046871 odds on it happening as of a specified initial hand. The longer you play, the more initial hands you have (being any hand that you are about to play or have just played, depending on how you define the, 'Initial,' hand) and the more likely you are to see a run like that at some point all the way until it becomes a near certainty.

A binomial distribution would also help to determine the probability of losing x number of hands out of 12, but I don't know how many you lost. There are several different ways in which you could have gotten that $5.00 return. Well, not several, but:

Straight
Trips + High Pair
Two Pair x 2
Two Pair x1 + High Pair x2
High Pair x4

So, I'd have to know which of those it was.

all high pairs

The probability of a non-winning hand, assuming Optimal Strategy, is 0.54543467 whereas the probability of a pair of Jacks or Better is 0.21458503 (So, you technically ran better than expected on those!)

Using a Binomial Distribution, we can get a rough probability of the probability of having 8/12 losing hands, or more:

This sweet Binomial Distribution Calculator:

http://vassarstats.net/binomialX.html

Spits out a probability of 0.293245553423 to have eight or more losing hands in a sample size of twelve hands, with a probability of 0.165551207814 that it will be exactly eight losing hands. When it comes to exactly x losing hands, out of twelve, eight is the second most likely result to six and seven.

Of course, it is unfortunate that you did not get any 2P or better, but the probability of that happening can also be determined using a Binomial Distribution if you really want to. I don't. However, we can look at the probability of four or more Jacks or Better results:

The probability of exactly four such hands is about 0.151983443804 and the probability of four or more is 0.245516172993.

What I will do really quick is look at the probability of missing all other hands:

1-(.54543467)-(.21458503) = 0.2399803

That means that roughly 1 in 4 hands are 2P or better. The probability of hitting zero such hands in a twelve hand sample size is:

0.037144814519

That is about 1 in 26.92 in terms of the odds, so not even close to unusual, but certainly more unusual than looking at the occurrence of losing 8+ out of 12 hands (about 1 in 3.41) or winning four or more Jacks or Better pairs out of twelve hands (about 1 in 4.07 odds).

In other words, if you play enough VP, that sort of string of hands is virtually guaranteed to be something that you run into sooner or later, and it sounds like you play a good bit of VP. Furthermore, it has nothing to do with the probabilities or odds that you were playing free play, and even if it did, your results are not particularly unusual taken alone.

Quote: 100xOdds

Data point for short term return:
$15 Freeplay.
.25 9/6 job

Only got $5. :(

33% return

another data point:

$10 free play on .25 9/6 job.
only got $2.50 -> 25% return

yeah short term swing/variance