How much change in strategy with 50 or 100 play?

Because of the number of iterations/hands involved, does anyone have any numbers to support playing more aggressively for a royal on basic JB? I mean breaking from optimal 1-line strategy, such as keeping 2 suited face cards instead of a low pair or keeping 3 suited face cards instead of a high pair? It sounds slightly foolish, but with max coins, a royal is obviously what you want.

Similarly, on the bonus games, it seems more advantageous to split up 2 pair and play for the high bonus (as with As or 2s, 3, and 4s on many of the bonus games), especially where 2 pair is paying even money.

Any thoughts, anecdotal evidence or theories?

Thanks
Seviay

Quote: seviay

How much change in strategy with 50 or 100 play?

None.

The Wiz gave the short answer. Here's the slightly longer answer:

Just because you're playing whatever scenario 100 times all at once, it's no different than getting it 100 times on a single-play machine.

So whatever strategy applies to the single-play, applied to the multi-play.

Thank you Wiz and DJ. Allow me to change my focus a little, because I think the original motivation for the question was trying to develop strategy on games that are not addressed on the original wizardofodds site.

I am typically just a straight-up 9/6JB player, but I am intrigued by the possibilities in the multi-line bonus games. For example, I have played the 50-play and 100-play super double double bonus games on videopoker.com with great success, but I am unsure whether I am always making the mathematically correct play.

Is there a program (similar to the VP analyzer on the original wizard site) where one could input the payouts for any given game and the optimal strategy would be output? Or is this something one would have to do the long-hand math on in order to develop? I haven't dealt with statistics in 10+ years, and I am not yet proficient enough with excel to get there without a little nudging or formula help.

Again, thank you.

Products that do that create video poker strategies for any given game and pay table are:

Video Poker for Winners
Frugal Video Poker
Video Poker Strategy Master

This site's own JB has rolled his own software to do it, and is working hard optimizing it.

Quote: DJTeddyBear

Just because you're playing whatever scenario 100 times all at once, it's no different than getting it 100 times on a single-play machine.

So whatever strategy applies to the single-play, applied to the multi-play.

I'm puzzled. In 100 play, the cards you hold are held in all 100 hands. Therefore if you're playing Deuces Wild and are dealt: 2, J, Q, K, A on suit, you have a royal with deuces, which let's say pays 125. times 100 that is a payoff of 12,500 credits. A royal without deuces pays 4,000 you'd need 4 royals out of 100 hands to beat the cards you were dealt.

So what are the odds that 4 of the 100 hands will get the suited 10 you need?

That is ho I'd see that extremely unlikely deal in multiplay. In single pay I'd hold all cards on the principle that a royal w/deuces in hand is worth two royals without deuces in the VP machine's innards.

I don't know the answer to the question I asked, but I think the odds are lousy and I'd be better off holding all the cards.

Quote:

I'd be better off holding all the cards.

Ding, ding, ding. Correct answer.


If you toss the deuce, hoping to get that one card for a natural royal, you've got a 1 in 47 chance. 1/47 = .0213

In 100 hands, statistically, you'll hit it 2.13 times.

The value of that gamble is 2.13 * 4,000 = 8,510.6
The value if keeping the deuce is 100*125 = 12,500

You're better off keeping the deuce.

And it STILL doesn't matter if it's a single play, 50 play, or 100 play. All that matters is the paytable.

Only once the natural pays more than 47 times what the deuces royal pays does it statistically pay to toss the deuce.


Although, between you and me, I'd have a hard time tossing it unless the natural payed a lot more than 47 times as much.

Quote:

So what are the odds that 4 of the 100 hands will get the suited 10 you need?

Well, since the odds in a single hand are .0213, then it's .0213 * 100 / 4 = 53.25%

Hell, the odds of hitting 3, which is slightly under the break even point, is only .0213 * 100 / 3 = 71%

Keep the deuce!

As Wiz stated, there is none.

Quote: DJTeddyBear

Although, between you and me, I'd have a hard time tossing it unless the natural payed a lot more than 47 times as much.
Well, since the odds in a single hand are .0213, then it's .0213 * 100 / 4 = 53.25%
Hell, the odds of hitting 3, which is slightly under the break even point, is only .0213 * 100 / 3 = 71%

That's a rough calculation for EV, but odds differ somewhat. More precisely, the odds of hitting at least 1 are 88.4%: 1-(1-1/47)^100.
The odds of hitting more are then at (1-(1-1/47)^100)*(1-(1-1/47)^99)*... but for simplicity's sake (I'm using a pocket calculator, not Mathcad) let's take ^98 for all, for 87.85%. Note this ^100 vs ^98 business makes a big difference.
So the odds of hitting least 3 are 67.8%.
The odds of hitting at least 4 are 59.6%. Five, ~52%, six, ~45%, eight, ~35%, ten, ~25%.

However, you don't need all these probabilities. All you need to know is that a natural pays 32 times more than a wild, and your chances of hitting it are 47 times lower. Another part comes from the deuces, of which 3 are left since you toss one out, 3/47. So your EV change is 32/47-44/47, or -12/47, or about -0.24. That's a bad gamble either way. You should no more dump a wild card for a chance at a natural in a multi-play than in single play.

Quote: seviay

Is there a program (similar to the VP analyzer on the original wizard site) where one could input the payouts for any given game and the optimal strategy would be output? Or is this something one would have to do the long-hand math on in order to develop? I haven't dealt with statistics in 10+ years, and I am not yet proficient enough with excel to get there without a little nudging or formula help.

You don't even need to buy/download a program to do it. You can simply go to the excellent website www.vpgenius.com, put in any paytable you want and the tab "strategy guide" will output the optimal strategy for you as well as the return percentage achievable with this strategy. Check it out!

Similar topic, different question: when playing APDW (99.96%), 50 play, $1 per play, shooting for Diamond in a Day ($30K play through), what sort of BR should one bring along? Thanks.

I asumme you're talking about Horseshoe Hammond.

You should play one line at $5 per play. This will limit your variance. It will take about 6 hours to complete the DIAD. I would have a bankroll of at least $3,000. I would feel extremely comfortable with $6,000, though. Your risk of ruin will actually be quite low; around 5-10%. That may be too much for you, but you are going to have to gamble a little. Your fortunes will rise and fall with how often you hit the Four Deuces.

There are Risk of Ruin calculators available but you'll have to seek those out on your own.

They have a nice Diamond Lounge there. Be sure to check it out after you finish your play.

Thanks teddys. You're right about the casino. And you were right to tell me to look around for some ROR info on my own. The Wiz has some informative tables at the other site.

If you don't mind, could you substantiate your claim about variance? A five play, $1 game would have less variance then a one play, $5 game. You say a fifty play, $1 game has more. How many $1 lines must one play before the variance exceeds that of a one line, $5 game? What data or equations are you using to support this conclusion? I'm not calling you out, I'm just curious about the math.

Thanks.

Quote: DoubleDown

Thanks teddys. You're right about the casino. And you were right to tell me to look around for some ROR info on my own. The Wiz has some informative tables at the other site.

If you don't mind, could you substantiate your claim about variance? A five play, $1 game would have less variance then a one play, $5 game. You say a fifty play, $1 game has more. How many $1 lines must one play before the variance exceeds that of a one line, $5 game? What data or equations are you using to support this conclusion? I'm not calling you out, I'm just curious about the math.

Thanks.

The general rule of thumb, as I've seen it written, is that the variance on multi-line is about 1/3 that of an equivalent bet on single-line. So you get roughly equal variance playing .25 Triple Play as you do playing the same game as .25 single line, even though in the former game, you are betting three times as much as in the latter.

Quote: DoubleDown

If you don't mind, could you substantiate your claim about variance? A five play, $1 game would have less variance then a one play, $5 game. You say a fifty play, $1 game has more. How many $1 lines must one play before the variance exceeds that of a one line, $5 game? What data or equations are you using to support this conclusion? I'm not calling you out, I'm just curious about the math.

A five play, $1 game has less variance than a one play, $5 game. A fifty play $1 game ($250 a spin) has more variance just because you are betting way more. But you can play just one line on the fifty play ($5 a spin). Once you play more than 5 lines on the fifty play $1 ($25 a spin), you would be getting a higher variance than the $5 single line.

Since you have the time, you might as well play the lowest amount you can. Why expose yourself to more risk than necessary?

A good rule of thumb, like mkl said, is that the highest single line denomination you are comfortable with should be how much you bet on multi-line games. I ignore this rule quite often, though.

Zero is the correct answer.

Yeah, everyone has kinda figured it out two months ago.