Question #11: Mission's 2023 Top Hits—Part 1 of 2

 Greetings, LCB Readers!

It’s time for two more installments of, “Ask Mission.” These two installments will contain my best answers, on our WizardofVegas online game site, for 2023. I was also going to include some Facebook responses I gave, but that would be a lot of DM’s to go through.

In any event, if you have any gambling questions, then I would encourage you to check out our WizardofVegas site. Just complete the easy process to sign up for the Forums, start a thread and ask your question. In addition to myself, many other forum participants are skilled at gambling math, so you’ll be sure to get an answer from someone. 

Some questions are so complex as to require highly precise computer simulations, which I can’t do; if that ends up being the case, I’ll be sure to tell you.

I always try to check new threads every couple of days, but if I somehow miss your question, please send a PM to, ‘Mission146,’ along with a link to the thread. Generally, I’ll take a look within 24 hours. You can also find me on Facebook at “Brandon James,” request me as a friend, and I’d answer any questions you want to DM me on there.

To be fair, I’d much prefer you ask questions on WizardofVegas; not only is it one of our sites, but trying to type long answers into a Facebook DM is an absolutely miserable process.

Without further ado, let’s get into my 2023 Best Of:

CRAPS ODDS DO NOT CHANGE EXPECTATION

The first question that we will cover comes from WoV user, Payner 12, who asked:

Payner12: Hi, everyone. Im new to craps. What is mathematically better for me. Lets sayI i bet $100 on the pass line and the point has been established. Is it better to back it up with another $100 or do the max 2x with $200? Its less loss in the long run backing it up with 2x? THANK YOU!

Mission146: Let's start with the basic premise that you are absolutely determined to bet $100 on the Pass Line for each trial.

The notion that it is, "Less loss in the long run backing it up with 2x," is not correct. Your expected loss will be $1.41 for every Pass Line bet you make, regardless of whether or not you back those bets up with Odds.

Odds bets have an expected value of $0.00, unless they count towards your comps (varies by House), in which case they have a slightly positive expected value, but only on the back end. Because Odds have an expected result of $0.00, in the long run, they do not change your expected loss whatsoever.

Okay, so what do Odds change?

They change your expected loss relative to your total action, which is an important distinction. Isolating one specific decision, if you bet $100 on the Pass Line and there is a Point Number (2/3 to occur) and you put $200 in Odds behind it, then your expected loss on the bet will be $1.41 relative to $300 in total bet as opposed to $100 in total bet.

Now, the probability of there being a point number is .6666666667, which means that the expected amount of Odds, per Come Out roll, is:

200 * .666666666667 = $133.33

In other words, for every $100 Pass Line bet you would make, you would expect to have $233.33 in total action. $133.33 of this expected action would have no House Edge and the other $100 has an expected loss of $1.41.

1.41/233.33 = 0.00604294347

Or 0.604294347%, approximately, becomes your new Expected Effective House Edge. Your actual Effective House Edge for those bets which do, in fact, get Odds is:

1.41/300 = .0047

Or 0.47%, approximately, but this does not always happen.

Either way, none of that changes your expected loss in $$$ terms as it will all relate back to the amount bet on the Pass Line.

Basically, the short answer is that Taking Odds on the Pass Line, or laying Odds on the Don’t Pass, does not change your Expected Loss on the Pass Line bet whatsoever. The reason why is because these are totally different Craps bets; line bets have a House Edge working against you and Odds Bets do not.

Once upon a time, a handful of casinos would Triple both the Two and Twelve on Field Bets, which resulted in there not being a House Edge on those; I don’t believe that’s the case anywhere now. In addition to that, the Santa Ana Star is a Native American casino in New Mexico that used to allow Buy Bets on the four and ten that paid true odds, and therefore, had no House Edge.

Naturally, you can’t make an Odds bet without having a Pass Line, or Don’t Pass, bet. That being the case, I don’t believe that there are presently any casinos offering any Craps Bets that can be made without a House Edge that don’t force you to also make a bet that does have a House Edge.

In any event, Odds Bets do not change your Expected Loss ($1.41 on a $100 bet) because they don’t change the Pass Line bet in any way whatsoever. They do change your expected loss percentage relative to your total action, which is an important distinction to make that some players might consider significant. Of course, taking/laying odds also increases your variance significantly.

HOME POKER VARIANT QUESTIONS

The first thing that I want to say is that I absolutely love home poker games, and variant games, because just understanding how the math impacts the game can give you a significant advantage over your opponents. On WoV, Linksjunkie asks about drawing odds for a unique five-card home poker game:

LinksJunkie: Need a little help with a new 5 card poker variant we’ve recently added to our home game

Everyone gets a 5 card poker hand to start. We frequently play Jacks or better to open. After someone opens everyone that stays in gets an additional 3 cards - no cards are discarded. So basically you have 8 cards to make your best 5 card hand.

My question is - what is the probability that a two pair hand improves to a full house? My gut tells me probably somewhere between 20-25% but not versed enough in statistics to confirm.

All input appreciated.

Mission146: Greetings!

This should be a pretty easy question to answer.

You have Two Pair and are specifically looking to improve to a Full House, I'd prefer quads, myself, but Full Houses are also good.

Cards: 52

Known Cards: 5

Unknown Cards: 47

Cards to be Drawn: 3

Since we already have a four-of-a-kind on the brain, let's get that knocked out! We're going to make your hand:

J-J-Q-Q-5

Jacks Remaining: 2

Fives Remaining: 3

Queens Remaining: 2

4OAK

The first thing we should do is look at our easiest ways to make quads. The easiest ways would be to get the other two Jacks or Queens:

calculator

When it comes to getting the other two Jacks or Queens, we really don't care what else comes out, so we will treat it all as just the other 45 cards.

nCr(2,2)*nCr(45,1)/nCr(47,3) = 0.0027752081406105

We want to multiply this by two because it can either be Queens or Jacks, so:

0.0027752081406105 * 2 = 0.00555041628

1/0.00555041628 = 180.166666706

Which means that you are about 1 in 180.16666671 to get a Four-of-a-Kind of either Jacks or Queens. You can't get a Four-of-a-Kind in both, because you would then need to be drawing four cards.

The second you of Four-of-a-Kind you can get is fives, which would require that you draw all three other fives:

nCr(3,3)*nCr(44,0)/nCr(47,3) = 0.0000616712920136

As you can see, that is much less likely and only occurs 1/0.0000616712920136 = 1 in 16,215 attempts in this scenario. That's okay; quad jacks and queens are better, anyway.

Let's add that result to our result for Jacks and Queens to get:

1/(0.00555041628 + 0.0000616712920136) = 1 in 178.186813226

Full Houses

Full Houses are slightly trickier than they normally would be because you don't discard anything. For that reason, we will start with a Full House, "Queens Full of Anything," because that is the best ranking Full House possible. The good news is there will always be a best-ranking Full House possible, at least, of cards we already have in hand; though I will admit that the math is going to end up being slightly different if the kicker to the Two Pair is HIGHER than the rank of either of the Two Pair, or both of the Two Pair, but, in this case, Queen is also the highest ranking card.

If you have a hand like AQQJJ, however, Aces become the highest ranking Full House. That won't change things very much (in terms of overall probability) but it does change things a little.

The reason it would change things is because, in this hand, if we get ONE queen, then we could get one or two other fives and it wouldn't make a difference that our best hand is Queens Full of Jacks. In the AQQJJ hand, our best Full House would be Aces Full of Queens, in the event we got two aces.

All of the math in the Four of a Kind section would apply to anything.

Anyway, your Mission, should you choose to accept it, is to do the math for Full Houses on the AQQJJ hand; you should be able to do it by doing something similar to what I am about to perform!

FULL HOUSE-QUEENS FULL OF?

Okay, so what we are going to do here is we want to get a Queen, but not both, and we don't want to get the other two Jacks; getting one Jack doesn't change anything, so let's do the math for both zero jacks and one jack:

ZERO JACKS:

nCr(2,1)*nCr(2,0)nCr(43,2)/nCr(47,3) = 0.1113783533765032

ONE JACK:

nCr(2,1)*nCr(2,1)nCr(43,1)/nCr(47,3) = 0.0106074622263336

With that, we will add these two things together to determine the probability of getting a Full House in Queens. Remember, we have already done the math for Four of a Kind Jacks, so we really don't care if the third card (to the Jacks) is another Queen because we have 4OaK already.

0.1113783533765032 + 0.0106074622263336 = 0.1219858156 or 1/0.1219858156 = 1 in 8.1976744188

Of course, we could also do better than the Jacks and get a Full House Queens full of Kings, or Aces, but we really don't care for these purposes. We're just going to do all Full Houses Queens Full of Anything, which this covers. The Four of a Kind in Fives is irrelevant because, if we got three fives, then we did not get a Queen anyway.

FULL HOUSE IN JACKS

For the Full House in Jacks, once again, the fives are immaterial as Jacks Full of Queens (that we already have) is better anyway. However, we can't draw a Queen as one of the other two cards, otherwise, we have a Full House Queens Full of Jacks, Kings or Aces, depending how the other two cards go.

Okay, so we want no Queens whatsoever:

nCr(2,1)*nCr(2,0)*nCr(43,2)/nCr(47,3) = 0.1113783533765032 or 1/0.1113783533765032 = 1 in 8.97840531561

As you can see, this will be the only probability we have to deal with. We can't possibly get Quad Fives because we have a Jack and two cards is fewer than three. We could get a Jack with two Kings, or Aces, but that's immaterial to us getting a Jack, but not a Queen.

FULL HOUSE IN FIVES

Finally, we could get Fives Full of something, but that would only happen if we got Two Fives, No Jacks and No Queens, so let's do that:

nCr(3,2)*nCr(2,0)*nCr(2,0)*nCr(40,1)/nCr(47,3) = 0.0074005550416281 or 1/0.0074005550416281 = 1 in 135.125

The left hand numbers inside of all parenthesis on the left side of the divisor add up to 47, and add up to three on the right, so I think we have this right.

Once again, this is the probability of getting two fives, but not three fives and the third card being neither a Jack or a Queen.

FULL HOUSE IN ANYTHING ELSE

What, you thought we were done? No way! We have ten ranks remaining and, from all ten of them, we could conceivably draw three of the four available cards. Fortunately, since that is what would be required, no card can be anything else, which makes the math easy:

nCr(4,3)*nCr(44,0)/nCr(47,3) = 0.0002466851680543 * 10 = 0.002466851680543 or 1/0.002466851680543 = 1 in 405.375

I think we're about ready to wrap it up, so here we go:

FINAL HAND PROBABILITIES

Four of a Kind (Fives): 0.0000616712920136 or 1 in 16,215---->(0.0061671292%)

Four of a Kind (Queens): 0.0027752081406105 or 1 in 360.3333333333---->0.27752081406%

Four of a Kind (Jacks): 0.0027752081406105 or 1 in 360.333333---->0.27752081406%

Four of a Kind (ANY): 1/(0.00555041628 + 0.0000616712920136) or 1 in 178.186813226 (Approximate, remember, we rounded off a little on Queens + Jacks)---->0.561208757%

Full House (Queens Full): 0.1219858156 or 1 in 8.1976744188---->12.19858156%

Full House (Jacks Full): 0.1113783533765032 or 1 in 8.97840531561---->11.13783533765032%

Full House (Fives Full): 0.0074005550416281 = 1 in 135.125---->0.74005550416281%

Full House (A/K/10/9/8/7/6/4/3/2 Full) = 0.002466851680543 or 1 in 405.375---->0.2466851680543%

With that, we must sum up the potential Full Houses:

0.1219858156+0.1113783533765032+0.0074005550416281+0.002466851680543 = 0.24323157569 or 1 in 4.11130831662 or 24.323157569%

Full House ANY: 0.24323157569 or 1 in 4.11130831662---->24.323157569%

ANY Full House OR Quads: (0.00561208757+0.24323157569) = 0.24884366326 or 1 in 4.01858736083 or 24.88436626%

With that, you're close enough to 1 in 4 to get either a Full House or Quads that you'll likely not play enough hands to notice the difference.

You probably tried to catch me forgetting to include the other ten Full Houses; I almost did---you're a sly one!

If you want to check out that thread, then you can find it here. In addition to the work above, I also determined the probability of hitting either a Full House, or Quads, if the player starts off holding a Three-of-a-Kind.

There was additional discussion in that thread, which is definitely worth reading, through which I determined that a final hand straight is actually REALLY BAD under these rules, if you would believe such a thing, and that flushes with a face card are probably the worst hands that could still be considered playable.

With that, if you were playing at a table with players who thought straights are good, then you should absolutely clean up, in the long run, just by knowing how weak they are and how likely Full Houses, or better, are to get under these rules. Drawing three cards to Two Pair, without the need to discard, as opposed to discarding one and drawing one, is obviously quite powerful when it comes to final hand strength.

Once again, just taking deep dives into any sort of novelty Home Poker rules, and understanding the underlying math and probabilities, is sure to give you a tremendous advantage compared to most of your competition!

U.S. LAND CASINOS AND TAXES

Isn’t it crazy to think that, with ever-increasing inflation, the IRS wanted to reduce the W-2G threshold for gambling winnings (mandatory tax reporting) from $1,200 to $600?

Fortunately, that hasn’t gone through because, if nothing else, the W-2G threshold should…well, it SHOULD be abolished, but it definitely should be increased if it’s going to do anything.

You can actually find a quite old thread on this site addressing that issue.

“Mental” pointed out that amassing large amounts of W-2G’s isn’t unusual as it is.

Mental: Nice article!

I want to emphasize an important aspect of the proposed IRS change. While they would drop the reporting threshold to $600, they would also net your W/L for the day. I calculated that this would have reduced my W-2G total for the year by half. So, a gambler like me would benefit greatly from the proposed change that everyone is bellyaching about.

I once received 36 W-2Gs totaling $91.378.48 in a single losing session. I would have received zero W-2Gs under the proposed change to IRS regulations! Sadly, gamblers and the casino industry fought this proposal.

It is congress that is mostly to blame for the current situation. Congress passes tax legislation and the IRS does its best to implement the spirit of the law. Scalia was the gamblers friend on the Supreme Court, but he is long gone.

I think your estimates for numbers of W-2Gs filed by a casino in a month are way too low.

Mission146: Thank you very much for saying so!

That's an excellent addition to the drop to $600 that escaped my attention. That said, it does somewhat strengthen my position that the IRS has come to depend on the actual profit/loss for gamblers not being reported for the year as that would likely result in less revenue. I tend to think a huge chunk of their revenue comes from gamblers who don't know how to write it off with their losses, don't care to take the time to do so or from people to whom the Standard Deduction would be superior anyway.

Of course, with the Standard Deduction, that person could just have a completely truthful and accurate gambling log that also includes winning days, just with losing days to more than cover those winning days as well as any W2-G such that the amount now exceeds the standard deduction. The optimal decision would certainly depend on the exact numbers involved and starting income (presumably from wages) could also be a factor in it.

It's difficult to argue with the casino industry fighting the proposal, from their perspective, because $1,200 is already ridiculous as it is. There's no denying that the W2-G mechanism increases available staff needed, wait times, paperwork, reporting, accounting and, perhaps most importantly to them, decreases the player's time on device.

By a major casino? Do you think it could be tens of thousands? I suppose some months it could. I guess I should have taken into account how many HL players there would be compared to other areas.

Even with the net W/L, perhaps the problem is that could be rife for deliberate malfeasance, and even absent that, would be extremely difficult to strictly track anyway. It would end up having to go to an all electronic system, where---if not every single bet---every buy in and cash out is accounted for as belonging to a specific individual.

In any event, since net gambling losses for a year (other than to offset winnings) do not constitute a loss of income, for recreational gamblers, I don't think net wins should constitute an increase in income, either. That failing, they should just make it 5k.

Anyway, given that you're talking about 91.3k+ in a single day, I assume you keep pretty detailed logs anyway.

Of course, while the IRS counts gambling winnings as income, they don’t count gambling losses as a loss of income to any extent beyond offsetting winnings.

Naturally, while I don’t recommend filing anything but a perfectly accurate tax return, I will emphasize that there are still plenty of forms of gambling upon which a gambler might put losses in a log book that the IRS would have no way to verify. One example would be Table Games, where buy-ins wouldn’t be tracked if you weren’t using your players club card; besides that, casino win/loss tracking is generally considered inaccurate and is often not even accepted by the IRS.

Additionally, some states have Limited Video Lottery such that wins and losses, other than W-2G aren’t tracked at all, so any losses that go into your logbook from those sorts of activities could be used to offset gambling winnings.

In theory, sports betting also could, but it’s probably best to keep losing (and winning) sports tickets if you’re going to try to use that. The IRS might get suspicious if you claim a huge amount in sports betting losses, but don’t have any tickets to prove that you actually made the bets.

Technically, you’re supposed to report ALL winnings to the IRS, not just those that result in mandatory reporting. For that reason, what I would recommend doing is also including winning sessions in your logbook if you plan to try to offset any winnings that have been mandatorily reported, as it might look odd if you report that is the only day you ever left ahead.

I also suspect that the IRS is going to be working more closely to monitoring player accounts on regulated in-state online casinos, in the United States, which could also come with some tax implications.

Naturally, since most gambling is done at a negative expected value, most players who do any substantial amount of gambling finish the year with a loss. You definitely don’t want to end up paying taxes for winning money gambling when you’ve actually lost overall, so it’s very important to realize, if you file an itemized return (rather than a simple return) that gambling losses can offset gambling winnings; you’re kind of losing twice if you don’t avail yourself of your right to do that.

DO THE MATH CORRECTLY

It’s definitely beneficial to players to understand concepts such as Expected Value, House Edge and Standard Deviation, but when you’re trying to take the mystery out of your results, it’s also important to avoid making mistakes.

“Daypay,” made a perfectly reasonable mistake that anyone could make trying to determine possible ranges for their Blackjack results. I immediately saw the problem and was able to help him, believe it or not, without actually doing very much.

Daypay: Thank you to all of vegas of wizard users. I was got lots of useful information from this forum.

And sorry about my wreck of English. I'm not native.

Anyway. As I know, I can measure range of profit and loss using variance.

Assume below condition.

Bet same amount of money = unit

number of play hands = hands

ignore house edge

No side bet.

Follow basic rule

At a rough level. Sqrt (hands)*1.21*unit is range of profit and loss.

So if I play 100 hands. I will suffer 12.1 units up and down.

And standard Normal Distribution chart show over 20 units of profit and loss probability under 10%.

But, my game result history show. Over 20 units volatility about 20~25%.

Sample size, about 20k played hands.

Additionally, I play 5 hands against one dealer's hand.

I mean, 1unit $20 is mean my bet $4 each hole.

That makes me confuse.

So, question Is my volatility formula is right? Or something wrong.

If I'm misses anything to measure risk. Please let me know.

Thank you.

Mission146: First, Blackjack is not really my strong suit.

Second, it's very possible that I misunderstood your post.

That said, I don't know what basis you are using to calculate what you're doing, in terms of variance/SD. 1,000 initial deals playing five-handed at $4/hand is going to have more variance than 5,000 initial deals playing one-handed at $4/hand because so much of it comes down to the dealer's final hand.

Anyway, perhaps I did misunderstand what you're asking. If I didn't, then it seems like you're using the SD on $20 as if you were betting $20 playing one-handed, possibly.

As it turned out, Daypay didn’t need my help for the math itself and was able to do that after I told them what the problem was. After correcting that the problem needed to be done differently when playing a multi-handed game (with the same bet per hand) as opposed to treating it as playing that number of single-handed games, Daypay was well on his way to getting the answer that he was trying to get.

The key here is the variance of the multi-handed game is greater because there are fewer dealer hands; additionally, there is a greater amount of money being bet (which means greater amounts of money to be won or lost) per dealer final hand; therefore, the standard deviation is going to be much greater than single-hand because the range of possible results is much greater, which means greater amounts that results can deviate from the mean.

UTH COUNT OUTS—21 NO GOOD?

OdiousGambit made a long opening post on WoV in which he indicated that he thinks counting outs, on Ultimate Texas Hold ‘Em, is not the best way to make the final decision to Raise or Fold. The thread can be found here, but again, it’s a very long opening post, so while I suggest you hop over and read the thread, I’m not going to quote the whole thing.

Ultimately, I gave an example of why the 21 Out count is something that I find useful, and instead of trying to, “Reinvent the Wheel,” I found it more practical just to find ways to add more outs to count when it comes to making marginal 1x Raise decisions. I’ll quote my post below:

Mission146: I was asked to check out this thread and have happily done so.

Personally, I have never had a problem with the 21 outs count, so I wouldn't really have any great need for any type of chart where any variations to counting 21 outs would result in a negligible save to EV, though I could see where someone who plays a lot might need an alternative or someone who has trouble grasping 21 outs. 21 outs can also get tricky with the prospect of the dealer needing two cards for a straight or flush.

These charts actually seem more complicated to me than counting the outs; while I am a very average poker player, I have probably played more than some people, so just counting the outs comes pretty naturally to me. Take a board like:

Ah Qd 9c 5s 3h

If I have:

Jd 6c

Okay, so any card on the board being matched beats me, that's 15. Kings beat me, so that's another four cards. That's nineteen outs and the dealer can also have a jack with a better kicker. I would just count that as one out, so that gets me to twenty; I should call.

If we have a board with a high pair:

Ac Qd Qs 8h 5d

I have:

10s 7d

Three Aces, Four Kings, Two Queens, Four Jacks, Three Eights, Three Fives. No straights or flushes possible, 19 outs, easy call. I wouldn't add an out for a ten with a better kicker because that would only be 10x9x without the dealer already beating me some other way.

If you use the calculator (on WizardofOdds):

You can compare the EV of -1.716 for the bottom hand to -1.91 for the top hand. Superficially, both hands would have nineteen outs for the dealer, but that's why I went ahead and added an out due to the top hand possibly having a Jack that outkicks me with the second kicker as opposed to beating me some other way. Jx10x, Jx8x and Jx7x all beat me for the top hand. In the bottom hand, any 10 other than 10x9x (except cards that individually beat you anyway) is also using the eight as the final card.

I also count dealer two cards against an open-ended straight (think of a hand like Kc Qs 4d 5c 3h) or dealer needing two cards to make a flush as one out. If both are applicable, then I count that as two outs. A hand with the above and player has:

Jd6d

Becomes a call with the board above, but a fold if you change any of the non-club cards to a club.

Aces-Four

Kings-Three

Queens-Three

Fours-Three

Fives-Three

Threes-Three

Jack (Better Kicker) +1

Two for a Straight +1 (Makes 21, but is actually a very close call---that's likely because the better jack shouldn't be a full +1 and we're blocking some straights...also, dealer Ax2x beats us by the ace anyway)***

Two for a Straight + Two for a Flush (changing a non-club to a club) +2 (Definite fold)

I've played on Wizard's game and this method seems fairly accurate and works with the fact that I prefer just counting the outs anyway.

***When I say close, the Raise has an EV of -1.99495. I don't mind folding something like this and being wrong. If our hand were instead Jd2d, then it's definitely a fold as there are more ways to outkick our jack by having Jx, dealer having Jx6x now beats us, for example.

As you can see, I simply find it more useful to try to make subtle improvements to a method that generally works as opposed to throwing the entire method away. Given that the math generally holds up with the 21 Out Count, and the times it doesn’t are usually in favor of folding, I simply decided to try to find different ways of adding to the number of outs.

While I’m not actually capable of writing a full-on computer simulation, I have tested my Out Count method against more than a hundred applicable hands and find that my method generally gets increasing the Out Count right, when it comes to making the final decision. Even when my Adjusted Out Count would favor folding (when the proper EV decision is, slightly, to make the Raise) the occasions where my Count Method does fold are enough to be overall better than strictly using the 21 Outs.

My method for adding to Outs makes the game more complicated, but it’s an excellent game that’s worth learning and has a very low House Edge relative to most Table Games. It’s also, arguably, one of the only Table Games that has a low House Edge with the potential for HUGE wins, which in my book, makes Ultimate Texas Hold ‘Em the best Table Game that exists right now.

I’d recommend checking out the strategy page on WizardofOdds, practicing on the free to play game and getting the basic strategy down, which is mostly just memorization of a few concepts. When you have that all squared away, then you can implement my Count Adding strategy to the 21 count for final decisions, and it’ll get you that much closer to that low theoretical House Edge.

Also, many online casinos seem not to restrict Ultimate Texas Hold ‘Em, even live (sometimes) from promotions where they do exclude other low House Edge games such as Baccarat, Craps and Blackjack. Putting UTH in your repertoire, and learning the game well, could very well result in more opportunities to play profitable promotions that you might otherwise dismiss because other Table Games, such as those mentioned, are not allowed.

LOOK FOR THE SLOTS THEY MISS! (SLOT ADVANTAGE PLAY)

One undervalued must-hit play of old was on WMS machines, which were largely overlooked by other advantage players because the must-hit amount increased by a percentage of your winnings on any given spin, with a losing spin not increasing the meter at all. Some players might have thought this increased the variance to an unacceptable level, others might not have understood how it worked in the first place, others might not have known how to calculate expected meter increases and some flatly just didn’t like them.

In any event, RSActuary and I exchanged some words about how great these opportunities were, mostly, because we didn’t have much competition.

RSActuary: Comments:

Sahara Gold is a game under the Lightning Link brand, I wouldn't specifically list it.

Your MHB section is reasonable, although after my last post, I thought about the old WMS G+ machines with MHB jackpots. Instead of the value going up by some percentage of the amount bet, the jackpots when up by some percentage of the amounts won. So on those machines, you did need a winning spin to advance the meters and hopefully cross over the secret target amount. But it wasn't symbol related... any win advanced the meter. There's a thread on these machines someone on here.

Mission146: Those were fantastic. A ton of variance, but you could find good numbers on those a lot because even people in the know didn't like them being win based, even though it would often make increases faster as opposed to slower, by expectation. Let's see how that works out vs. the $0.01 per $6.67 (just for one example, some are faster) that we see on other machines:

Of course, on games such as Zeus (whichever number Zeus) and the Cowboy one, actual meter increase was high variance and would depend on hitting free games and the like, but it was still faster at the end of the day.

Had a machine configured at 0.00582 times the money won, but even then:

6.67 * .8 * .00582 = 0.03105552/6.67 = 0.004656 or 0.4656% based on an extremely conservative (way too low) assumption of 80% RTP on the reels. Compare to a penny for $6.67 bet (0.15%) or a penny per $5.00 (0.2%).

With that, even some competition thought that they had to be way high because they never really calculated the meter move relative to the win amount. At the same time, some casual players would think they could run the thing up just by hitting good bonus games, line pays or what have you.

Also, many players would make some of the higher bet amounts, so the meters would sometimes go up pretty quickly. At that point, you could pop on and bet fairly low, if you wanted to, to keep variance under control as you waited for decent pays to advance the meter.

For one reason or another, some casinos must have decided that they didn't want them or they lost popularity because I don't see this variation of the must-hit concept around as much anymore, but you could find some fantastic numbers to play when it was, in my experience.

It’s honestly a shame that they’re not as prevalent as they used to be because they once represented quite an opportunity that others players either didn’t like, or flatly ignored.

Another type of slot machine advantage play I’ve seen get widely ignored are slots that only have limited upside, but little to no downside. Bottom line: If I know there’s even a potential play on a machine, and the chair is free, I’m checking it.

STRAIGHT FLUSH WITH FEWER CARDS

I’d love to have known Topdog’s reason for asking (because it could be advantage play related), but I don’t think he’s talking. In any event, he wanted to know the odds of a Straight Flush in a modified deck; I would presume this is for some sort of Progressive Jackpot.

TopDog: I've searched a few sites and maybe I'm just an idiot but can't seem to find a site that can answer my question. I know what the odds are for hitting a straight flush with 52 cards but what I'm curious about is what are the odds of hitting a straight flush with only using the 2s,3s,4s,5s,6s,7s,8s and 9s??? If someone could assist me that would be great and IF you can could you also assist me in letting me know the calculations needs to determine the odds??

Thank you

Mission: Right. The first thing to do is figure out how many straight flushes there are:

2-3-4-5-6 (1)

3-4-5-6-7 (2)

4-5-6-7-8 (3)

5-6-7-8-9 (4)

4 * 4 = 16

The next thing we have to do is figure out how many total cards there are:

2 3 4 5 6 7 8 9 ---> 8 * 4 = 32

The next thing we have to do is figure out how many ways we can pick five of 32 cards:

calculator

nCr(32,5) = 201,376

The next thing we have to do is divide the number of combinations that give us a straight flush by the total possible number of combinations:

16/201376 = 0.0000794533608772 or 0.00794533608772%--->1 in 12,586 to be dealt a straight flush.

Significantly more likely than with a full deck.

Basically, this is similar to how you’d figure out the same problem for a 52 card deck, except that’s widely known. You just determine the number of different ways to make a straight flush, multiply by the four suits and divide that by the number of combinations. You should try it for yourself! If you do a deck with just deuces through sevens, that should be an easy and quick one.

CAESARS LONG-SHOT SPORTS BETTING PROMOTION

Soopoo, over on WizardofVegas, has been doing some pretty good work with sports betting, these days, but did have a question about a rather complicated Caesars promotion.

In general, it would also be good to know that United States regulated (by individual states) have been tightening the belt on promotions lately. While there are still promotions with an expected profit to be found, I would say the majority of promotions are small losers, barely winners or, in the case of this one, pure enticement to make bad bets, as you will see below:

Soopoo: This will be an easy one. If I flip a coin 14 times, how often will heads appear 10 or more times?

I’m asking because Caesars has promo. Bet 14 NFL pointspread games at $25 each. Essentially all will pay at -110.

If you win 10 or more you get $150 ‘free bet’ bonus.

I think the EV on the 14 bets is -$17.50. And the EV on the $150 free bet is around +$100.

So I think the overall EV of this promo is still slightly negative. My brain guesstimated I hit 10 or more 8% of the time.

They had similar promo last year which I think was better AND last year I’d get rewards credits and tier credits that were way more valuable last year. Unless someone corrects me I’m skipping this offer.

link 

Mission146: Greetings!

Interestingly, I was asked about this very promo last night and roughly (though a little less roughly than your guestimate) figured it out; it is pretty easy.

ASSUMPTIONS

The first thing that we are going to do is start with the following assumptions:

1.) We are going to assume that every spread bet is 50/50 to be a winner. If a person considers himself/herself to be an excellent handicapper, then perhaps they think they can do better than 50/50 long-term, but for our purposes, it's 50/50.

2.) We are going to make pushes (ex. a +3 line and the team loses by three) impossible for these purposes. Of course, they aren't impossible, but since the promotion is terrible anyway, there is no reason to consider the probability of a push.

3.) We have 28 cards. On the picture side of the cards, 14 are black and 14 are red; this is a good mental visual aid and also means we can use combinatorics to get to the cumulative probability.

THE MATH

The first thing that we will consider is the probability of hitting 10+ wins:

nCr(14,10)*nCr(14,4)/nCr(28,14) = 0.0249772164141527

nCr(14,11)*nCr(14,3)/nCr(28,14) = 0.0033027724184004

nCr(14,12)*nCr(14,2)/nCr(28,14) = 0.00020642327615

nCr(14,13)*nCr(14,1)/nCr(28,14) = 0.0000048857580154

nCr(14,14)*nCr(14,0)/nCr(28,14) = 0.0000000249273368

We sum these probabilities:

0.0000000249273368 + 0.0000048857580154 + 0.00020642327615 + 0.0033027724184004 + 0.0249772164141527 = 0.0284913227940553

This is so fundamental as to not really beg for proof, but we will prove it anyway. What we are going to do next is nine wins and eight wins:

nCr(14,9)*nCr(14,5)/nCr(28,14) = 0.099908865656611 * 2 = 0.199817731313222 (The *2 is to account for five wins and nine losses)

nCr(14,8)*nCr(14,6)/nCr(28,14) = 0.2247949477273747 * 2 = 0.4495898954547494 (The *2 is to account for six wins and eight losses)

Next, we will look at seven wins and seven losses, which can only happen one way***:

nCr(14,7)*nCr(14,7)/nCr(28,14) = 0.293609727643918

***Technically, this can happen many ways, but that's what the combinatorics is for. For our purposes, 7-7 can only happen one way.

We sum these probabilities and double the 10-14 cumulative probability to account for the inverse in losses:

(0.0284913227940553 * 2) + 0.293609727643918 + 0.4495898954547494 + 0.199817731313222 = 1

Every probability has been accounted for and our logic is proven. Of course, it's a simple problem, so we had no doubt that would happen.

THE JUICE

The next thing that we have to do is account for the vig on our main bets. As I was presented this promotion, you would get $1,500 in free bets if you made a $250 spread bet for every game:

(250 * (100/110) * .5)) - (250 * .5) = -11.3636363636363636

In other words, we expect to lose $11.36 each game, which checks out as:

11.36/250 = 0.04544 This represents the implied juice of 4.544%. Keep in mind we are ignoring the possibility of pushes for these purposes; the reason why is that we will prove the promotion is so bad that pushes won't matter.

$11.36 * 14 = $159.04

With that, we expect to lose $159.04 on our bets.

KICKBACK

For the purposes of the kickback, we are going to pretend that the kickback is actual CASH, as opposed to Free Bets, just to prove how bad this is:

1500 * 0.0284913227940553 = 42.73698419108295

If this were straight cash, which it isn't, it would represent additional EV of about $42.74, which:

-159.04 + 42.74 = ($116.30)

Would still leave us with an expected loss of $116.30.

Even then, still ignoring pushes and assuming it pays cash, rather than free bets:

116.30/(250*14) = 0.0332285714285714

The implied juice would be reduced to 3.3223%, which is still quite bad, and that's with the favorable assumption of ties being impossible AND if the promotion paid cash, rather than Free Bets, which have a lower value than cash.

Obviously, this promotion is enticement to make fourteen bets that, individually and collectively, would all still be bad bets. In the extreme case that a person bets the applicable amount on EVERY SINGLE SPREAD anyway, then this adds a little value; otherwise, it is simply unplayable.

What I have found is that Caesars’ promotions are notoriously bad these days. I’d also encourage anyone looking to take advantage of strong new player promos in the sports betting realm, when there are any such promos to begin with, to understand that Caesars is careful to look out for offset betting with a confederate and will void the promotions pretty aggressively.

If you bet on something that gets a lot of action, such as an NFL point spread, it might fly under the radar. I think what got an associate of mine and his confederate nailed was offset betting on a random early season Major League Baseball game that almost nobody cared about and probably took very few bets. They also bet a few minutes, or less, apart from one another, which isn’t ideal.

DO NOT TRUST CHATGPT (OR ANY OTHER AI) WITH COMPLEX GAMBLING MATH!

Our final question for this first of two will involve a discourse between myself and OdiousGambit in which he wondered about the conclusion reached by the AI Bot called, “ChatGPT.” As you will see, I’d suggest NOBODY use AI for gambling math for two reasons:

1.) If you don’t phrase your question perfectly, then it might misunderstand and just try to give you the best answer as it understands you. A human, on the other hand, would ask for clarification.

2.) Even if ChatGPT understands your question, if it’s not something for which it can find an answer online, then it will try to do it itself and probably get the answer wrong, as you will see:

OdiousGambit: Chatgpt is not coming up with the correct answer for the question I was trying to get it to do, which is: I can make a wager where I wager $100 with a 40% chance of winning and get paid 4:3. Only If I lose, I get a second chance where I make the same $100 bet, do not get the wager returned, and receive only the winnings. Is this a good bet?

It's kind of helped me, though, in that the thing to do is figure out first what the EV of the second chance is, which is 0.40* 133.33 = 53.33. So if I lose the first bet, I lose the amount of the wager minus the EV of the second chance? I would lose $100, normally, but on average now I lose 100-53.33 = 46.67

it led me here and this may be incorrect but I get

EV = (0.40 * 133.33) - (0.60 * 46.67) = 25.33

The below is what Chatgpt gets,

CHATGPT: I apologize for the oversight. [I keep having to correct that same error] Let's calculate the corrected expected value (EV) using the formula you provided:

EV = (Probability of winning initial bet * Expected value of winning initial bet) - (Probability of losing initial bet * Expected value of winning second bet)

Probability of winning the initial bet = 40% (0.40)

If you win, you receive 4/3 times your wager ($100 * 4/3 ≈ $133.33).

So, the expected value of winning the initial bet = 0.40 * $133.33 ≈ $53.33.

Probability of losing the initial bet = 60% (0.60)

If you lose, you get a second chance with a $100 bet.

Now, the probability of winning the second bet is 40% (0.40), and if you win, you receive $100 in winnings.

So, the expected value of winning the second bet after losing the initial bet = 0.60 * 0.40 * $100 = $24.00.

Now, let's calculate the overall expected value using the corrected formula:

EV = (0.40 * $53.33) - (0.60 * $24.00)

EV = $21.33 - $14.40

EV = $6.93

The corrected expected value (EV) of this bet is approximately $6.93. This means that, on average, you can expect to win $6.93 in the long run for every $100 wagered.

Based on this corrected calculation, it still appears that this is a favorable bet in the long run, as the expected value is positive. However, as mentioned earlier, individual outcomes can still vary, and there is always a risk of losing money in any single instance of the bet.

And…the correct answer:

Mission: Assuming I understand the question correctly, there are ultimately three possibilities.

Possibility 1: You win the initial $100 bet that pays 4:3, thereby resulting in a return of $233.33 of which $133.33 is profit.

Possibility 2: You lose the initial $100 bet, but win the follow up bet that pays to the extent of winnings only, thus resulting a return of $133.33 of which $33.33 is profit.

Possibility 3: You lose both bets, therefore losing $100.

The probability of winning either of the two bets, individually, is 40%. We start with the 40% probability of winning the first bet to get its EV:

$133.33 * .4 = $53.33

The second possibility is to lose the first bet, but win the second, which nets a profit of $33.33:

(.6 * .4) * 33.33 = $8

The third possibility is to lose BOTH bets, which results in a loss of $100:

(.6 * .6) * -100 = -$36

The overall probabilities have been included in the EV calculations, but for proof, they are 40%, 24% and 36%, in order. This totals 100%.

At this point, we simply do some addition and subtraction:

$53.33 + $8 - $36 = $25.33

Therefore, the expected outcome is that you will see a profit of $25.33, ergo, your advantage on an initial investment of $100 is 25.33%.

This is actually earlier in the same thread that SOOPOO’s question was, on the same page, in fact. What a great day for me!

In any event, I would later go on to explain how and why ChatGPT made errors, which I won’t include here, but would recommend jumping over and looking at that thread.

The thing to understand about the way these Chatbots work is that they generally trawl the internet for information, and often, just give the closest thing they find to the correct answer. It would almost be more appropriate to call them, “Searchbots,” with the only meaningful differences being that they, ‘Learn,’ and that they answer your queries in a conversational way.

The fact is, movies are great, but Artificial Intelligence has a long way to go before it gets to the level seen in films. Basically, if the AI can’t find the information that you’re asking for by doing an internet search, especially if it’s the sort of question that would required independently coming up with a unique process to solve a problem, humans are going to outperform…for now.

CONCLUSION

That’s it for Part 1 of my Best of 2023 answers. Please stay tuned for Part 2, which should be released fairly shortly. If you have any questions for me, then please hop over to our partner site at WizardofVegas dot com, create an account, start a thread and let it rip! Earlier, I said to shoot me a PM if you don’t get an answer from me within a few days, but you’re also free to PM right away.

Even if I’m extremely busy, I’ll make sure to respond to your PM within 24 hours and let you know roughly when I can get around to your question, but I’ll generally answer most within 24 hours, provided I have access to a proper computer in that time.