This is some math I don’t know how to do. In a 98.0% video poker game what is the chance of breaking even in a one million hand sample space? Thanks for any help.
Mickey
A Probability Question
Insufficient information. However, provide game variance and permit the assumption that game results are “near-normally” distributed for values central to the mean, and it’s possible to calculate the answer you are looking for (… which I would leave as an exercise for NOTI, whose much more facile with the math).
Variance is key here for a simple reason … you’re looking for results that exceed the EV by a given margin. The high the game variance, the greater the proportion of results that will exceed that margin.
—In vpF…@…com, <mickeycrimm@…> wrote :
This is some math I don’t know how to do. In a 98.0% video poker game what is the chance of breaking even in a one million hand sample space? Thanks for any help.
Mickey
Okay, vp_Wiz. So lets say 7/5 Bonus Poker which has a 98.01% payback and a 20.75 variance. Thanks.
Mickie,
Standard deviation for a million hands of 8/5 BP is .00456. So this is more than a 4 Sigma event or less than 3 in 10,000.
It might not be quite
this simple, but I believe the standard deviation is the square root of the number
of trials times the variance, which would be 4555. I assume you want to know the chance of breaking
even or better, not just exactly breaking even, so you’d have to fluctuate at
least 20,000 units, or about 4.4 standard deviations. According to the table at the following link,
that chance is around 1 in 100,000.
https://en.wikipedia.org/wiki/Standard_deviation#Confidence_interval_of_a_sampled_standard_deviation
A double check on this is
that the standard deviation of the number of jackpots is roughly the square
root of the number of cycles. If you
assume that the entire fluctuation in 1 million hands is due to how the number
of royal flushes has deviated, in 25 cycles, a 5 jackpot variation is 1 standard
deviation. To fluctuate by 25 royals
would be 5 standard deviations.
···
On Sun, Feb 12, 2017 at 5:14 AM, mickeycr…@…com [vpFREE] <vpF…@…com> wrote:
Okay, vp_Wiz. So lets say 7/5 Bonus Poker which has a 98.01% payback and a 20.75 variance. Thanks.
Very interesting. kcace, thanks for the response.
If you want the math behind it, check out wikipedia and the article titled “Z Table”. Otherwise, Nzero is useful for negative expectation games as well, it’s just that, well, the expectation is negative, i.e. your opponent has the edge. So, in this case the Nzero is variance/edge/edge = about 20/(-2%)/(-2%) = 50,000 hands. Meaning, at this point, about 84% of your possible results are “negative”, leaving just 16% positive. A million hands would be 20Nzero or about 4.5 SD which basically means almost all of your possible results would be negative.
I wrote a little program to do this type of computation some time ago. If I entered all the numbers correctly the probability of being even or better after 1 million hands of 7/5 BP are about 3.3 x 10^-5, or in other words the odds against are a little worse than 30,000:1. That’s with perfect play.
Mike
Out of curiosity, can you add in cash equivalent promotions (making the game “playable”) … say 2.5%, in the case of 7/5 bp? (equivalent to asking the p that play return > 97.5% after 1 mil hands))
—In vpF…@…com, <mpeck1@…> wrote :
I wrote a little program to do this type of computation some time ago. If I entered all the numbers correctly the probability of being even or better after 1 million hands of 7/5 BP are about 3.3 x 10^-5, or in other words the odds against are a little worse than 30,000:1. That’s with perfect play.
Mike
Thanks for the responses, guys.
This is easier than it may first seem. Suppose 1M hands at $5 a spin. 2.5% of that is $125,000.
So now you will just want to calculate the probability of losing $125,000 or less. With a flat game like 7/5 BP, you can pretty safely go with expected value after a million hands. That game drops roughly 2%, so your expected value is about $25,000. To lose would be to run just awful. Assuming everything else besides royals runs about average (which it almost has to be on this game) you’d have to be 6.25 royals below average.
I have forgotten the math, but there is a fairly accurate and simple formula that will tell you the odds of getting fewer than 17 events when the expected number of events is 23. So, for example, if you were flipping a coin 46 times, what is the odds of only getting 16 heads (or less). The answer to that question will be very close to the correct answer for this situation.
If this is a real life game, you’ve got a decent return, but nothing special. If you crank out 1000 hands an hour, you make roughly $25 an hour. As for bankroll… This game will be about as mellow as can be and I’m thinking 20K should be viable, though that assumes you have other money for things like eating.
Much easier that doing the math is to run a bankroll analysis on one of the many VP programs out there.
That will, in effect, put your actual break-even point at a loss of $25,000 on the game. It should tell you P(losing 25K or less), which will also be the answer to your question. The VP program I have allows you to include cashback, which then allows you to simply look to see the answer with no need to do any further analysis to understand what the output means. What you should see, if you have done it right, is an “average” result of +25,000 for a million hands and the frequency of results that deviate by more than $25,000 will be very low (and only half of those will be losses) If the software is good, it will also give you the probability curves for all the outcomes.
QZ
From What I remember from my math days, I think it would be closer to this. Take the expected value and multiply by pi. Then take the variance and square it. The expected loss will be close to a squared x b squared x c squared. Don’t forget that this is only an approximation and could vary by 10 to the power of 6.
mikepeck5440, I have a follow up question. Some people think that 10,000 deals on a hundred play would be the equivalent of playing 1,000,000 deals on a single line game. But I don’t think so as far as chances of success go. I would think that 10,000 deals would have a much better chance of being at breakeven or ahead than 1,000,000 deals. Could you tell me what your programs says about the chance of being at breakeven or ahead at 7/5 bonus poker for 10,000 deals?
Mickey, I can’t give you an exact answer for 100-play, but you’re correct in thinking that you’d have a much better chance of at least breaking even than you would with the same coin-in playing single line.
With my program I can compare 1,000,000 deals of a single line game to 100,000 deals of 10-line, and even with that you can see the difference. I did each case 10,000 times.
With single line, none of the 10,000 trials resulted in a breakeven or positive result. In fact, only 65 out or 10,000 times did the player lose less than 40,000 coins. (i.e., lose less than $10K for a 25c game)
With 10-line, there were 28 cases out of the 10,000 trials that ended with a positive result, and 346 other cases when the player lost less than 40,000 units.
Of course, the reverse is also true; namely, big losses are more likely with 10-line. In a 25c game, the chance of losing more than $40K playing single line is 0.25%. But with the same coin-in on 10-line, a $40K loss is 6 times more likely.
All these effects would be amplified again with 100-line.
–Dunbar
—In vpF…@…com, <mickeycrimm@…> wrote :
mikepeck5440, I have a follow up question. Some people think that 10,000 deals on a hundred play would be the equivalent of playing 1,000,000 deals on a single line game. But I don’t think so as far as chances of success go. I would think that 10,000 deals would have a much better chance of being at breakeven or ahead than 1,000,000 deals. Could you tell me what your programs says about the chance of being at breakeven or ahead at 7/5 bonus poker for 10,000 deals?
Thanks, Dunbar. I ordered the risk analyzer a few days ago. I should have had it a long time ago.
With the multiplays, you add some big cycle times, like the dealt royal (649,740), dealt aces (54,145), dealt deuces (same), dealt aces with a kicker in games where that matters (216,580), and so on. On a three play these hardly make a difference, but on a 10 play or 100 play or 250 play they make a huge difference because these are clumps of big payoff hands that you either get lucky and get or you get unlucky and don’t get (which puts you in a hole). For example on a 10 play, the dealt royal is a 40,000 coin hole. If you play 649,740 dealt hands, roughly a third of the time you will not get a dealt royal and be in the hole, a third of the time you will get one dealt royal, and a third of the time you will get more than one dealt royal.
NOTI, it’s actually not that bad. After 650,000 rounds (1 round = 100 “hands”), say $0.25 100 play, with a half percent edge your EV is $406,250. A dealt RF is $100,000. Although that’s a decent chunk of your EV, about 25% of it, it doesn’t mean you’ll be “in the hole” (negative) without a dealt royal.
Of course, the problem here isn’t whether you do or don’t get dealt big hands and all that, it’s the minuscule half percent edge you’re playing with.
With something like a 4% edge, after 650,000 rounds of $0.25 100 play, your EV is $3.25 million. That $100,000 dealt royal doesn’t even make a dent in your overall EV. (Not saying you can play 650k rounds on such a play.)
···
Sent from my iPhone
On Feb 16, 2017, at 2:09 PM, nightoftheiguana2…@…com [vpFREE] <vpF…@…com> wrote:
With the multiplays, you add some big cycle times, like the dealt royal (649,740), dealt aces (54,145), dealt deuces (same), dealt aces with a kicker in games where that matters (216,580), and so on. On a three play these hardly make a difference, but on a 10 play or 100 play or 250 play they make a huge difference because these are clumps of big payoff hands that you either get lucky and get or you get unlucky and don’t get (which puts you in a hole). For example on a 10 play, the dealt royal is a 40,000 coin hole. If you play 649,740 dealt hands, roughly a third of the time you will not get a dealt royal and be in the hole, a third of the time you will get one dealt royal, and a third of the time you will get more than one dealt royal.
The math behind it is pretty straightforward. In a 52 card deck, the dealt royal cycle is fixed, at 649,740. The total royal cycle depends on the drawing strategy used, lets assume it’s 40,000. So, in a dealt royal cycle on a single play, the total royals (on average) would be 649,740/40,000 = about 16. The standard deviation is the square root of the average, so sqrt(16) = 4. On a single play, the dealt royal represents one royal, and one is much less than the SD at 4, so it’s basically lost in the noise. On a four play, you get four times as many royals per dealt royal cycle, so 64, and sqrt(64) = 8. The dealt royal represents four of those royals, or half the SD, now it’s getting significant. On a 100 play, you get 100 times as many royals per dealt royal cycle, so 1600, and sqrt(1600) = 40. The dealt royal now represents 100 of those royals, or 2.5 times the SD. The dealt royal is now the more dominant noise factor over the standard deviation. This effect isn’t limited to royals, but applies to all hands.
Mickey’s posts always spur long threads, and this one about DEALT RFs on Multi-lines got me to dreaming about DEALT Royals. I had a chance to REVISE my play plan so that I played a LOT more multi-lines on my latest trip (to RENO).
Result: check out my latest photo.
First time for me on the mutliLines.
So, just wanted to mention this to give some credit to the Probabilitizers and to those who write off threads like these with all its Math Probababble. Sometimes, good things come out of vpFree chit chat. Thanks for the inspiration!
Trip report on vpFree_Reno later this week…
~Mark my words
Very awesome hit!!
—In vpF…@…com, <mark_my_words_again@…> wrote :
Mickey’s posts always spur long threads, and this one about DEALT RFs on Multi-lines got me to dreaming about DEALT Royals. I had a chance to REVISE my play plan so that I played a LOT more multi-lines on my latest trip (to RENO).
Result: check out my latest photo.
First time for me on the mutliLines.
So, just wanted to mention this to give some credit to the Probabilitizers and to those who write off threads like these with all its Math Probababble. Sometimes, good things come out of vpFree chit chat. Thanks for the inspiration!
Trip report on vpFree_Reno later this week…
~Mark my words