Strategy question for a fixed bankroll

Let's say someone has a fixed video poker bankroll of $10,000. (Heck it could be $500 or one million, it doesn't matter.)

Their strategy is to play through the bankroll only ONCE. Would they be more likely to WIN MORE when they play through the entire bankroll at the SMALLEST denomination possible, or at the LARGEST denomination possible? Or does it not matter? All denominations have the same paytable and the same game is played regardless of the denomination.

I think that "mathematically" the answer will be IT DOESN'T MATTER as long as the pay tables are the same on the various denominations. But if you are able to play more hands at the lowest denomination (therefore more chances you'll have at getting royals or other big payoffs) will there be a difference in your wins after you run through the bankroll once?

The Wizard's 976-9-6 payout probability on Jacks or Better, for a Royal, for example, is .000028. I believe that is closest to 1:35,714 probability.

If you play with $10,000 on a $1.00 machine betting 5 units per hand, then you are looking at making 2,000 bets because you are just playing-through. The probability of you hitting for that Royal is 5.60000448% (2000/35,714). The Royal Flush w/bonus pays $4,000, so there is a 94.399996 percent chance of you NOT hitting for a Royal Flush at any point in time. The expected return on the Royal Flush alone is then $224.00. ($4,000 -94.39996%)

If you play with $10,000 on a $0.25 machine betting a maximum of 5 units per hand (it is often twenty) then you are going to be betting $1.25/hand which gives you 8,000 hands to play-through. The probability of you hitting a Royal Flush is 22.4001792% (8,000/35,714). The Royal Flush w/bonus still pays 4000 units, which, in this case, is $1,000. I'm going to go with units and say the expected return on a Royal Flush is (4000units - 77.59982%) or 896 units. The value of that 896 units is going to be $224.00, which is the expected return.

The difference is $0.00. Despite the probability ratios, every hand is completely independent of past results, so the extra hands provide no advantage.

It's 35,715. Rounding error. Doesn't affect anything.

thanks but there are other big hands, such as quads and straight flushes.

Is there a difference between playing the bankroll through once on a 25-cent machine vs a $100 per coin machine ($500 per hand)?

Quote: AlanMendelson

thanks but there are other big hands, such as quads and straight flushes.

Is there a difference between playing the bankroll through once on a 25-cent machine vs a $100 per coin machine ($500 per hand)?

I have already demonstrated that the other big hands do not matter. You said the paytables are the same, I gave you the expected return for one item (Royal Flush) on the paytable for both denominations. Logically, if the paytables are the same, and the expected value for one paytable item is the same, then they must all be the same.

I'm going to play it safe and use 35,714 again.

If you are playing $500 per hand and RF pays 4,000 units, then you have twenty hands before you have played-through your bankroll.

The probability of you hitting RF is 0.05600448%. (20/35,714)

The probability of you not hitting RF is 99.943999552%.

RF pays $400,000 dollars ($100/unit * 4,000 units won)

If you subtract 99.94399955% from that result, you get $224.00 as the expected value.

If the paytables are the same, it will never matter.

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If you are playing at a $0.01 machine (Does that exist?) and the max bet is $0.05 and RF pays 4,000 units, then you have 200,000 hands before you have played-through your bankroll.

The expectation of you hitting RF is 5.60004480035:1, or 5.60004480035 times during the entire playthrough. (200,000/35,714)

The probability of you never hitting a Royal Flush is irrelevant, because the expectation is you will hit at least 1.

4000 units * 5.60004480035 = $224.00, which is the expected value of the RF hit.

If the paytables are the same, it will never matter.

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The only thing that you are most likely to experience is an actual closer proximity to the expected value for the entire game (and for each payout) with the more hands that you play.

For example, if it were a $2,000 machine making $10,000 the max bet per hand and you lost, your actual payout would be $0.00.

In this example, however, you have one hand which gives you a 0.002800022% probability (1/35714) of hitting RF.

The probability of not hitting an RF is 99.997199978%.

The payout would be 4,000 credits * $2,000/per or $8,000,000

The expected value would be $8,000,000 -99.997199978% or $224.

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If you were to only play one hand, you WOULD deviate from the expected value. Given the payout structure, there is absolutely no other possibility, but the expected value of a single hand does not know or care that you are only playing one hand.

Keep in mind that, technically, you WANT to deviate from the expected value because the expected value has you losing 0.46% using optimal strategy, according to the Wizard or $10,000 - .46% = $46.00.

However, there are two ways to deviate from the expected value, one is good, one is bad, and you don't know which one you are going to get.

The, "Safest," play is the smallest bet because the actuality should bring you closer to the EV. It has no effect on the EV itself.

thanks for taking the time to write that. My original question was meant to find out playing which denomination would help someone win more. I think you hit the answer in your last line:

"The, "Safest," play is the smallest bet because the actuality should bring you closer to the EV. It has no effect on the EV itself. "

We've all been at machines when a run of bad luck has left us without a winner in many hands, so running the bankroll at the smallest possible denomination might give us the best shot.

BTW there are penny machines (usually on 100 play machines where you can play one line) but Ive never seen one without a reduced pay table.

thanks again

If by "win more" you mean "more times", then the lowest denomination is what you want - you are right, the more hands you play, the more chances you get to win.
If you mean "win more money", then it seems like a contradiction, because at the end of the day, you have lost your entire bankroll, so your (negative) "winnings" are exactly the same in each case.

Quote: AlanMendelson

thanks for taking the time to write that. My original question was meant to find out playing which denomination would help someone win more. I think you hit the answer in your last line:

"The, "Safest," play is the smallest bet because the actuality should bring you closer to the EV. It has no effect on the EV itself. "

We've all been at machines when a run of bad luck has left us without a winner in many hands, so running the bankroll at the smallest possible denomination might give us the best shot.

BTW there are penny machines (usually on 100 play machines where you can play one line) but Ive never seen one without a reduced pay table.

thanks again

You're welcome.

I guess the original question was kind of ambiguous. I didn't know if you meant, "Win more times," or, "Win more money." I assumed you meant win more money.

Playing $10,000 at once gives you both the best shot at winning more or losing more, (as in all of it) but it has no effect on the expected value. I could go into more detail, but simply stated, you have a 23.9979% chance of having a winning (i.e. profitable) hand at Jacks or Better. (1)

You have a 21.4585% chance of having an even hand. (1)

You have a 54.5435% chance of a losing hand. (1)

(1) https://wizardofodds.com/games/video-poker/analyzer/

I can back this up with some math, but I'm busy at the moment. If you want me to, fine, if you'll take my word for it, even better.

The LEAST you can possibly profit betting $10,000 is $10,000. (Two Pair pays 2:1 $20,000-$10,000=$10,000)

You can break even.

Your chances of profitting OR breaking even are 45.4564%.

If you hit for two pair, you double your money. You have no chance to approach the expected value because you are only making one play. If you bet $0.05/hand, and hit for two pair, you are up $0.05, bank $0.10 and have $9,999.95 to playthrough. You should approach the expected value over time. You can't do that with the former because it is one hand only, you can be expected to do so with the latter. In other words, with the latter, you are far less likely to maintain a 2:1 profit than you are to achieve a 2-1 profit with the former.

I would play the lowest denomination/highest hands (1c 100-play if possible)
My daily session is exactly $10k coin-in...I used to put in HOURS at it playing 25c, but have recently switched to $, because of a promotion that lasts 5 hours (therefore I need to put in the $10k in 5 hours)

If the total payout for say JoB is 0.9954, thats 1/(1-.0.9954) or about a 5 unit loss in 1087 hands. In 8000 hands at 25c * 5 coin, thats an AVERAGE loss of about 37 units.

As previously mentioned, in 8000 hands your chances of hitting a ROYAL are about 2 in 9. When you hit every other payline in statistical average, this will lower the payout from 0.9954 to about 0.975. Now recalculate your losses. 1/(1-0.975) is 1 unit in 40 plays. At 8000 plays, this will lose 200 units of $1.25.

Third step: the former will happen 22% of the time and the latter will occur 78% of the time. 0.22*37 + 0.78* 200 = about 164 unit loss on statistical average ($9795 remaining) on the 8000 unit wagers

Quote: 98Clubs

As previously mentioned, in 8000 hands your chances of hitting a ROYAL are about 2 in 9. When you hit every other payline in statistical average, this will lower the payout from 0.9954 to about 0.975. Now recalculate your losses. 1/(1-0.975) is 1 unit in 40 plays. At 8000 plays, this will lose 200 units of $1.25.

So in other words, if I only play 4000 hands, I'll only lose 100 units (if I don't hit a royal), which is $500 (at the $ level), which is acceptable, as the casino requires this many hands at this level per month and gives me $400 cashback PLUS (now) 1% in comps. So, while it may not be a money making opportunity (if I don't hit the Royal) it sure doesn't cost a lot to be VIP ;)

I think the casinos recognize this feature, and offer better odds at higher denominations. The penny machines will have the poorest returns, while the dollar (and higher) the best.

All in all, I think that exposing your bankroll to the edge for the least number of trials is the best way to capture any positive variance. Otherwise, the HE will grind you down over time.

If you hit a Royal on the first hand, will you continue through the entire $10k? I think the measure of "success" depends on your motivation for playing. Are you trying to win big, minimize risk, collect comps, kill time? Each calls for a different bankroll management philosophy.

I'm not so sure... there is an incresing EV as one plays more hands, because your chances of getting the top two pays increases, lowering the EV of the game overall.

I'm not disputing the short-run, its that the OP wants to maximize the chance of hitting a RSF with a FIXED bankroll, that presumes "no early quitting".

By reducing the play to 4000 bets, I think any straight-flush becomes an issue (I ignored it for simplicity with 8000 bets), and this will lower the average EV by 1/2% more if one does not hit any straight-flush. And 4000 bets reduces the chance of a RSF to about 1 in 9.

Quote: Mission146

The probability of you hitting RF is 0.05600448%. (20/35,714)

The probability of you not hitting RF is 99.943999552%.

This is incorrect. By that logic, playing 35,714 hands would result in a 100% chance of a royal. Use the binomial distribution.

Quote: deedubbs

This is incorrect. By that logic, playing 35,714 hands would result in a 100% chance of a royal. Use the binomial distribution.

(1-.000028)^2000 = 0.94553839457 Not hitting

1-0.94553839457 = 0.05446160543 Hitting

(1-.000028)^35714 = 0.36787723384 Not hitting

1-0.36787723384 = 0.63212276616 Hitting

You're correcting a mistake I made nearly two years ago!!!

Strangely, the original thing he wanted to know about 2000 hands is still pretty close!

Quote: Mission146

Strangely, the original thing he wanted to know about 2000 hands is still pretty close!

Yeah, for (1-x)^y where x<<<<<<1 and y is not >>>>>>1, a good approximation is 1 - x*y. Sadly I am old enough to learn this in my college chemistry class (1998) to solve those problems easier. Obviously it breaks down when y approaches 1/x.

No offense, of course. I was doing some video poker related googling and this thread came up on high on the first page, so I thought it'd be wise to chime in.