Which is the best bet?

This is a spin-off of the recent discussion in the thread, Wizard's changes to Craps webpage. For the question at hand, consider the following casino game. The player has three bet choices, as follows:

A) Pay a $1 fee, then bet $50 to have a (1/3) chance to win $100, otherwise lose.
B) Pay a $1 fee, then bet $100 to have a (2/3) chance to win $50, otherwise lose.
C) Pay a $1 fee and get 99.9¢ back, with no subsequent wager available.

The question for the poll is which is the best bet?

In other words, bet $50 on the 1st dozen, or $50 each on the 1st and 2nd dozen.
If you only get one bet, C is the only sure thing.

Quote: EvenBob

In other words, bet $50 on the 1st dozen, or $50 each on the 1st and 2nd dozen.
If you only get one bet, C is the only sure thing.

C is an absurd bet. I am supposed to trek all the way to the aluminum can recycling dealer and get my one dollar then trek all the way to the casino and they will then some how give me 99 and a half cents then throw me out into the alley? This seems strange but that half-cent piece might be worth something on ebay.

C. is the answer.

For those of you choosing C, may I ask why? Would you make this "bet"?

Its the only sure thing if you only get one bet.

Quote: Wizard

For those of you choosing C, may I ask why? Would you make this "bet"?

I didn't say I would make the bet, the question was what is the best bet. I wouldn't make any of them.

Since the outcome is certain "C" is not really a bet at all. Its a trek to give someone a tenth of a cent with no chance at all of any other outcome. Certainty is 1.0. Its not a wager merely because it has some trappings of one and is made in a casino. And it still seems a foolish option. Someone would really have to enjoy seeing pole dancers to go all the way to a casino to make one bet and leave.

To clarify, I never said this was a one bet thing. You can do it over and over, as much as you like. Assume no other benefits to making the bets, like free drinks or pole dancers.

Quote: Wizard

To clarify, I never said this was a one bet thing. You can do it over and over, as much as you like. Assume no other benefits to making the bets, like free drinks or pole dancers.

Ah, I was going to defend C by saying if this were a slot machine at Caesars, you could put in $1000 and get $999 back while earning 200 tier points. You could earn DIAD for like 15-20 bucks, or Diamond the old fashioned way for $55, plus get several hundred in Comps.

Let's keep it simple and keep out the fancy stuff. I'm trying to get at the issue of how we should quantify what is a "good" bet.

As a homework assignment, what would be the house edge of each bet? Is that an acceptable measure of how "good" each bet is?

I selected the 4th choice.

Option C reminds me of a TV show I once saw (Twilight Zone?). A gambler dies and goes to heaven. Every bet he makes, he wins. No matter what he does, he just can't lose. At the end, he realized that when he died, he went to hell.

Options A and B both give you the chance to walk out with more than you walked in with, while also giving you the chance to go broke. No house edge, after the original $1.

Option C is nothing more than a guarantee that you get nothing more and nothing less than a very slight hit. Why bother?

Quote: DJTeddyBear

Option C reminds me of a TV show I once saw (Twilight Zone?). A gambler dies and goes to heaven. Every bet he makes, he wins. No matter what he does, he just can't lose. At the end, he realized that when he died, he went to hell.

Twilight Zone, definitely.

It was a gangster, as I recall, and everything worked just fine. If he shot pool, every break sank all the balls. Things like that.

Quote: Nareed

Twilight Zone, definitely.

It was a gangster, as I recall, and everything worked just fine. If he shot pool, every break sank all the balls. Things like that.

The episode was called A Nice Place To Visit, its a classic.

Mathematically, C would be the best choice. However, it would not provide any fun or excitement, especally because it is a one-time shot and the player will still lose a small amount without any of the positive aspects of making a wager - so I would eliminate C right off the bat.

In the long run, both A and B would be the same:

In the case of A, I would expect to lose an average of $1 per play. Over 3 bets the wagers and fees would amount to $153. Two bets of $50 each would be lost and one bet of $50 would be returned with an additional amount of $100 for an net of -$3.

In the case of B I would expect to lose an average of $1 per play. Over 3 bets the wagers and fees would amount to $303. One bet of $100 would be lost and two bets of $100 each would be returned with and additional $50 each for a net of -$3.

When the loss is expressed as a percentage of the amount bet then choice B would be preferred because 3/300 is smaller than 3/150. Although, I am not sure that this is a very useful comparison because I think that cost per trial is the key number while the amount bet is more like a threshhold amount and is not necessarily relevant.

More practically, we need to determine what the ultimate goal is. If the goal is to leave with the most money in the smallest number of plays (because each play costs us money), choice A provides the scenario where this could occur. The reason for this is the ratio of the possible win amount to the amount of the fee paid for playing. For example, with choice A, you could win the first play for a net win of $99. The best you could do is $98 with choice B because it would take you at least 2 plays to win what you could win with one play of choice A.

However, if the goal is to provide fun and excitement by playing the game repeatedly over a longer period of time, choice B would be the best because the frequency of the wins is greater. For me, I prefer frequent wins to larger wins -provided the the wins are still of sufficient size to make me feel like I have won something. The best case scenario is lots of small and medium sized wins and an occaisional big win. (I can't be the only one either - I think this is why slot machines are so fun and sometimes addicting.)

The biggest downside I see to choice B is that I would expect to need a larger bankroll to cover downside streaks. I would probably tend to get caught up in the excitement of winning and keep playing until I lost my bankroll too.

The best bet for me (a bet that could actually result with me walking away with money in my pocket, while limiting my downside risk) would probably be choice A. I would make fewer wagers and either walk away with nothing at all or a sizeable win. And if I lost, I would probably limit my loss to fewer trials. The key is being able to walk away while I am up, which is much less likely if I get caught up in the excitement of the game.

Put me down for A.

a: EV = -1 + 100*1/3 - 50*2/3 = -1. HA = -1/51
b: EV = -1 + 50*2/3 - 100*1/3 = -1. HA = -1/101
c: EV = -1+.999 = -.001. HA = -.001

c is the best bet. It has both the highest EV and lowest HA, but has no variance.
a has the most variance, while b has lower variance.

Over 1,000 bets you would definitely lose $1 on bet C, while on Bets a and b you would probably lose $1000 +/- the variance.

I'd still pick bet C, ogle the women, and lose the $1. Of course it isn't gambling. In fact bet C isn't a bet at all. A "bet" is a wager with a an expectation >0 of winning the bet. So wager c is not a wager at all.

Therefore I change my answer to b, since it has the lowest house edge.

is there a reason for the fee? sorry but i wouldn't make any bet. i wouldn't pay to risk my money. it's like paying to go into a store to spend your money. i even hate paying to go to things like swap meets and product shows. but if the entry fee is low enough i'll pay in hopes of making it up with a better deal than i would get in a retail store. i sometimes go to the indoor swap meet in vegas to find unusual items and found things that i thought were a good enough deal. (one thing i bought was a tibetan singing bowl which i'm glad i got, and went back to look for a gong but the booth was no longer there.) i went to the outdoor swap meet once but i don't think i'll go there again.

sorry to get off track but the point is, a fee has to be warranted. if i had a choice to go somewhere else i wouldn't make any of these bets.

.

There can be no one correct answer. Although C can never be a good bet, by any definition, it can always be better than the other two if you are strictly using EV. A gives you more variance than B, so if you are seeking more variance then A is better than B, And vice-versa. EV can never be the sole determining factor, of course. I have $5 million, enough to retire.(Hypothetically) Me betting that $5 million on a special one roll even EV craps bet (They give me the 4 at two to one) is a far worse bet for me than $1000 on rolling a snake eyes at even money. The pure math EV of the first bet is 0, the second bet -$970. So for me, if I had to make one of those bets, I would pay the penny, then find a casino which offerred games I like to play, rather than pay the $1 in EV for a game I do not like to play. I voted 'other'.

Quote: Wizard

Let's keep it simple and keep out the fancy stuff. I'm trying to get at the issue of how we should quantify what is a "good" bet.

Quote: Wizard's Signature

It's not whether you win or lose; it's whether or not you had a good bet.

Well, it YOU don't know how to define a "good" bet...

Quote: boymimbo

a: EV = -1 + 100*1/3 - 50*2/3 = -1. HA = -1/51
b: EV = -1 + 50*2/3 - 100*1/3 = -1. HA = -1/101
c: EV = -1+.999 = -.001. HA = -.001

For the record, when I originally read the question, I thought the $1 fee was a one-time thing, like an entrance fee. That sort of scenario has been discussed here before, albeit not recently. I maintain that if a casino had a small entrance fee, followed by nothing but fair bets, the house would still win. Enough people would either go bust or quit while behind to make of for those that quit while ahead. Those that are ahead will get greedy. Those that are behind have no bankroll to work with.

Anyway...

If it's $1 per bet, then, assuming boymombo's math is right, option B is the best bet.

While option C has the lowest house edge, there is no gamble there. Nothing to get your blood boiling. Nothing to give you the "rush".

So the choice of the best bet becomes A or B. -1/101 is better than -1/51.

>So the choice of the best bet becomes A or B. -1/101 is better than -1/51.

I guess I have to go along with that. voted wrong tho.

What's the connection with the original thread?

First, the $1 fee is every single bet. This is not a ridiculous concept. In some player-banked casinos, such as in the card rooms of Los Angeles and San Francisco, the player must pay a 1% of his bet as a fee, rounded up.

This is supposed to be an analogy to craps. Perhaps I'll explain the comparison in a later post, I don't want to confuse the issue at this point.

Quote: DJTeddyBear

So the choice of the best bet becomes A or B. -1/101 is better than -1/51.

I agree, the house edge of A is 1/51=1.96% and B is 1/101=0.99%. Is that the end of the subject when it comes to evaluating how "good" each bet is. Let me submit for your consideration that the variance is the same at 5,000 for both bets. Both of them have a 1/3 chance of a bankroll change of $101, and 2/3 of $51.

So, couldn't it be said that both have the same excitement factor? I think if we asked what is the reason people knowingly make negative EV bets it would be for the excitement. It seems to me they both offer the same excitement element and same expected loss. So, why wouldn't A=B?

I see people bet B at roulette often, betting 2 dozens at the same time.
Many of them do quite well because the other dozen is sleeping and they
sometimes win 20 out of 25 bets. But they never know when to stop and
they keep going till their BR is gone.

Quote: Wizard

A) Pay a $1 fee, then bet $50 to have a (1/3) chance to win $100, otherwise lose.
B) Pay a $1 fee, then bet $100 to have a (2/3) chance to win $50, otherwise lose.
C) Pay a $1 fee and get 99.9¢ back, with no subsequent wager available.

It appears that making TWO of the "A" bets costs you an extra dollar but you are betting 100.00 dollars which is the amount that is required to make bet "B". So the question then becomes what is two separate one-in-three chances of winning 100 versus having one two-in-three chances of winning 50 dollars?

I still don't see any "bet" involved in C. It is simply pay 1.00 and receive 99.9 cents in change. No wager is made, no outcome is dependent upon any event taking place. Or it might be considered a bet on an event of probability of 1.0. Pay one dollar and "win" 99.9 cents.

Well, I can't figure out whether A is better than B or not.

Quote: FleaStiff

Well, I can't figure out whether A is better than B or not.

Its the same bet, neither is better. Risking 50 to win 100,
or risking 100 to win 50. In the end its equal.

Quote: FleaStiff

I still don't see any "bet" involved in C. It is simply pay 1.00 and receive 99.9 cents in change. No wager is made, no outcome is dependent upon any event taking place. Or it might be considered a bet on an event of probability of 1.0. Pay one dollar and "win" 99.9 cents.

Option C is a red herring. It is there to make a point that just having the lowest house edge doesn't necessarily make something the best "bet."

Quote: EvenBob

Its the same bet, neither is better. Risking 50 to win 100, or risking 100 to win 50. In the end its equal.

So, B has the lower house edge, but both bets are equally good? Care to elaborate? How do you think the degree of a goodness of a bet should be defined if the house edge does not cut it alone?

Quote: Wizard

but both bets are equally good? Care to elaborate?

How can they not be equal? If I bet 50 to win 100
999 times, I'll win 333 times and lose 666 times, and
make $0. If I bet $100 to make $50 999 times, I
make $0. Why would the $1 fee make any difference,
you pay it win or lose. In the end I'd be out $999
for the fee's, the bets are a wash.

In fact, I constantly play the betting two dozens bet
in practice. Its a good bet but its too volatile. All you
need is a couple of zeros and a couple of wrong
bets and you're down 8 units. You should never bet
more than you're going to get back, it will always bite
you on the ass eventually. Thats whats wrong with
progressions too, the volatility will kill you.

Quote: Wizard

First, the $1 fee is every single bet. This is not a ridiculous concept. In some player-banked casinos, such as in the card rooms of Los Angeles and San Francisco, the player must pay a 1% of his bet as a fee, rounded up.

oh, ok. i thought i was missing something with the fee concept. thanks.

.

I think A is the best bet. On negative expectation games you should always go for maximum variance to maximize the chances to end up ahead. Variance is the only way to "overcome" house edge in the short run.

Game B seems to have a smaller house edge but it's only because the wager is twice as large, and in fact you pay the same fixed amount of house edge ($1) per round on both games A and B. But in B you stand to win less for every dollar you pay, so game A is the best bet for majority of players.

How about this game: 98 times out of 100, the result is a push. 1 times out of 100 you lose 10 units and 1 times out of 100 you win 1 unit. The house edge of this game is only 0.9%, but you risk 10 units for an equal chance to win only 1 unit, so the "effective" house edge (with all pushes ignored) is 45%. So in some cases the traditional definition of house edge doesn't really tell you how good bet the game is.

Quote: Wizard

to make a point that just having the lowest house edge doesn't necessarily make something the best "bet."

I would imagine that there are always other factors to be considered anyway. The person betting on the pass line is paying 1.414 percent to make the bet but he is also paying to listen to the stickman trying his damnedest to get him to make far worse bets as well.

Of course I have a suspicion that although Lowest House Edge does not necessarily make something the best bet, it is most likely to be a pretty good indicator most of the time.

Quote: FleaStiff

Of course I have a suspicion that although Lowest House Edge does not necessarily make something the best bet, it is most likely to be a pretty good indicator most of the time.

This Heresy on the Wizard's own site!? tsk tsk.

the Wizard can speak for himself but I really think he rarely wavers on saying that HE is to be your total guide on this; even the difference on P/DP in Craps [edited]

actually, do I adhere to the idea that lowest HE is always the best bet? I don't have the guts to say it here! [vbg]

Quote: odiousgambit

This Heresy on the Wizard's own site!? tsk tsk.

I'm not one to just go with the flow if I don't agree with something. For example, I've been saying for years that the house edge unfairly makes games that involve a lot of raising look like worse bets than they really are. Take the case of roulette and Caribbean Stud. Double-zero roulette has a house edge of 5.26%, and Caribbean Stud is almost the same at 5.22%.

However, if you factor in all the raises in Caribbean Stud, the ratio of expected loss to average bet is 2.56%. In my opinion, that makes Caribbean Stud a much better bet, by a factor of 2. As far as I know, I'm the only gambling writer to have embraced this concept.

The example that Jufo81 gave a few posts back is another good that shows a low house edge bet, that I would call a lousy bet. It goes to show that the house edge is not the end of the story. The house edge is appropriate for estimating losses over time. For example, if you know your average initial bet, the house edge, and hands per hour, you can know how much you can expect to lose over time, which is important.

However, if you're gambling for reasons of excitement, or the chance of winning, then I think one should consider variance, and embrace it as a good thing. In fact, I would hold that that for excitement-seeking recreational gamblers that a good measure of how "good" a bet is would be the same as the Kelly Criteria, the house edge divided by the variance.

Let's look at how the pass and don't pass, with full odds, at 3-4-5X odds looks under this measure of how "good" each bet is.

Pass

Statistic	Value
Expected loss	-0.014141
Combined house edge	-0.003743
Exp. bet	3.777778
Total variance	24.163436
Variance per unit bet	6.396204
House edge / variance	-0.002211

Don't Pass

Statistic	Value
Expected loss	-0.013636
Combined house edge	-0.002727
Exp. bet	5.000000
Total variance	24.135673
Variance per unit bet	4.827135
House edge / variance	-0.002825

So, by this definition, the pass side is the "better" bet. Although the house edge is higher, there is more volatility, and I would say a greater chance of winning. Factoring those things in, the "good" factor, in the bottom of each table, is better (lower) on the pass side.

What do you think?

Quote: odiousgambit

House Edge is to be your total guide on this; even the difference on P/DP in Craps...

Not quite. DontPass does have a lower house edge but to play it properly it takes a larger bankroll to lay maximum odds so for someone whose bankroll is extremely limited (or atleast his courage is limited) it is indeed better to be on the PassLine and Take Odds. There are probably other situations wherein a slavish fixation on House Edge is not necessarily wise.

However, I do stand by my belief that Lowest House Edge would be Damon Runyon's criteria as well. You know: The Race is ... but that's the way to bet.

Many a man has gone to Vegas to play Blackjack at a razor thin House Edge and given his girlfriend some Go Get Lost money so she can play that ultra dumb wheel at the casino entrance while he does some serious gambling. And to his chagrin, he loses his entire bankroll while the girlfriend comes back with enough to pay for the entire trip.

How do you measure "greater chance of winning"? As you approach infinite trials, your results approaches EV. Are you saying that the better bet needs to take variance into account, and that it's a value decision of whether you like volatility or not.

Quote: boymimbo

How do you measure "greater chance of winning"? As you approach infinite trials, your results approaches EV. Are you saying that the better bet needs to take variance into account, and that it's a value decision of whether you like volatility or not.

What I'm saying is that you have to examine your reason for gambling when choosing a game. For some people, namely those who crave excitement, or have a certain win target, then they will have a better chance with a high variance game. Others care more about bankroll longevity, in which case they should seek lower variance.

Yes, variance doesn't matter as your trials approach infinity. However, most people don't think like that. They have certain goals for THIS trip, and they don't care about past or future trips.

Good gamblers do think of maximizing their win, or minimizing loss over their whole lifetime. However, most people are not that patient. For those of you who also have short term goals, variance should be taken under consideration.

In the last posts Wiz pretty much nailed my thoughts and why I chose game A as the best option.

A typical real-life gambling scenario would be: I walk with $250 in my pocket to the casino and try to end up with $1000 total and then I will walk away. I expect this to happen slightly less than once every four times, but not too much less. Out of the options given, Game A, even though with seemingly the highest house edge, would maximize the probability of attaining this goal, so it would be the best game to play.

I always thought that there has been missing a mathematical measure for a "recreational gambler's good bet" (as Wiz put it), which the traditional house edge or element of risk (house edge per unit wagered) didn't cover well enough. In games with pushes, simplest approach would be to use the house edge with pushes ignored, which increases the "effective" house edge of BJ games (that's why I prefer Pontoon as it has no pushes). The proposed "good bet measure" of house edge divided by variance seems like a good one so maybe it could be used more as a measure of "goodness of a bet" along with traditional house edge.

removed
silly

Great question Wiz.

If I choose to play this game, I would play 1 bet only. Based on this scenario, I choose B. The fewer the # of rolls played, probability of outcome has greater significance. Therefore, I choose B. This does NOT mean I believe scenario B is the best bet. I simply prefer (enjoy) scenario B over the others.

If pressed to play multiple bets, I would abstain from this game.

Quote: Wizard

However, if you're gambling for reasons of excitement, or the chance of winning, then I think one should consider variance, and embrace it as a good thing. In fact, I would hold that that for excitement-seeking recreational gamblers that a good measure of how "good" a bet is would be the same as the Kelly Criteria, the house edge divided by the variance.
What do you think?

Consider dividing the square of the house edge by the variance.

Since the house edge is an expected value and an expected value is the mean--if one calculated the square root of mean^2/ variance, which is then mean/std deviation, you have calculated the signal to noise ratio as used in engineering and electronic analysis. But for our purposes it is important to note that the reciprocal of that statistic multiplied by 100 is the statistical measure called the coefficient of variation which is used as a standardized statistic to compare variation in different distributions. i.e. the standard deviation as a percentage of the mean. So CV = std dev/house edge.

Quote: EvenBob

How can they not be equal? If I bet 50 to win 100 999 times, I'll win 333 times and lose 666 times, and make $0. If I bet $100 to make $50 999 times, I make $0. Why would the $1 fee make any difference, you pay it win or lose. In the end I'd be out $999 for the fee's, the bets are a wash.

They are equal in terms of expected loss. However, option B allows the player to bet more.

Consider two blackjack games. One has a house edge of 0.5%, and the other 2.0%. Would you make the argument they are equally good, as long as you bet 75% less on the second table?

Quote: matilda

Consider dividing the square of the house edge by the variance.

I was afraid somebody would ask why not divide by the standard deviation, which is the metric I usually use when discussing volatility. The reason is the Kelly Criteria is based on advantage/variance. For the same reason casinos fear high variance games, the risk-seeking player should embrace them. However, I know I'm not make a very good case. To do so I think would get very math heavy.

How about this for a question. If the player is playing a negative EV game, is confined to flat betting, and has some winning goal to win w units, with a starting bankroll of b. Is there some statistic based on house edge and variance (or std dev) we could use to estimate the probability of success? I bet a pair of old socks that it would be proportional to variance/house edge.

Quote: Wizard

They are equal in terms of expected loss. However, option B allows the player to bet more.

But its more volatile in the short term. In the long term they're equal,
but you can get under water easily on option B with just a few losses.
You really have to work at it to get under water with option A. The best
bet is always the even chance bet, but thats not included here.

Quote: Wizard

How about this for a question. If the player is playing a negative EV game, is confined to flat betting, and has some winning goal to win w units, with a starting bankroll of b. Is there some statistic based on house edge and variance (or std dev) we could use to estimate the probability of success? I bet a pair of old socks that it would be proportional to variance/house edge.

Alan Krigman has an excel worksheet that does just that. 3 formulas by the way.

Here is the webpage that has the link and explains his math.

Edge, Volatility, and Risk of Ruin in Gambling

Quote: Wizard

To do so I think would get very math heavy.

Is there some statistic based on house edge and variance (or std dev) we could use to estimate the probability of success?

Well, the density function of the CV is known so at least I don't have to derive it.

Just found this thread again. This is the subject most interesting to me at the moment, outside of the NFL season starting.

Quote: Wizard

I'm not one to just go with the flow if I don't agree with something. For example, I've been saying for years that the house edge unfairly makes games that involve a lot of raising look like worse bets than they really are. Take the case of roulette and Caribbean Stud. Double-zero roulette has a house edge of 5.26%, and Caribbean Stud is almost the same at 5.22%.

However, if you factor in all the raises in Caribbean Stud, the ratio of expected loss to average bet is 2.56%. In my opinion, that makes Caribbean Stud a much better bet, by a factor of 2. As far as I know, I'm the only gambling writer to have embraced this concept.

You know, I had forgotten that you have been calculation this "element of risk" and touting that.

Quote:

However, if you're gambling for reasons of excitement, or the chance of winning, then I think one should consider variance, and embrace it as a good thing.

Agree; it is all about that assuming Player Advantage is not there. To me it is this question, why would anyone ever play a game with negative expectation unless he is hoping for variance? Therefore it does seem that the more variance vis a vis the same or similar HE has to be something to seek out.

Quote:

In fact, I would hold that that for excitement-seeking recreational gamblers that a good measure of how "good" a bet is would be the same as the Kelly Criteria, the house edge divided by the variance.

Interesting. Using this webpage here is some quick game comparisons. The closer the number to zero the better the bet with the theory that more variance is better with similar HE? Craps PL with odds figures taken from the Wizard post this is replying to. As you might suspect that game with odds towers over the others in this perspective.

Bet	HE/Variance
Craps PL	-0.0141
Pai Gow Poker	-0.02596
Baccarat Banker	-0.012256
Baccarat Player	-0.01374
Any Craps	-0.0176
Pass w 3x4x5x	-0.002211
Don't Pass w 3x4x5x	-0.002825

I will do more unless someone points out I am making some mistake in my thinking or something.

Mustangsally: your prof. lost me with saying one player could ruin it for everybody.

Quote: Wizard

Let's look at how the pass and don't pass, with full odds, at 3-4-5X odds looks under this measure of how "good" each bet is.

Pass

Statistic	Value
Expected loss	-0.014141
Combined house edge	-0.003743
Exp. bet	3.777778
Total variance	24.163436
Variance per unit bet	6.396204
House edge / variance	-0.002211

[...]

So, by this definition, the pass side is the "better" bet. Although the house edge is higher, there is more volatility, and I would say a greater chance of winning. Factoring those things in, the "good" factor, in the bottom of each table, is better (lower) on the pass side.

What do you think?

I looked at the above Craps Pass Line odds closer and there is something I don't understand about how you derived the "good bet" measure (house edge divided by variance).

In the above table you have two values for both house edge and variance. The first values are the house edge and variance relative to the initial pass line bet, and the latter values are the house edge and variance relative to unit wagered for the whole bet (pass line bet plus maximum Odds bet). But when calculating house edge / variance, you have divided house edge per initial bet with variance per unit wagered. I think you should either divide house edge per initial bet by variance per initial bet OR house edge per unit wagered by variance per unit wagered, and not mix the two up. With doing so I get a value of: 0.014141/24.163436 = 0.003743/6.396204 = 5.85E-4 for pass line.

I may not completely understand something here so I would like to hear a reasoning why you chose to divide initial bet house edge by unit variance and not unit house edge by unit variance?

Wizard,
This is the most interesting thread I have come across in this forum and touches on the only reason, I would imagine, it makes sense to gamble -- having a brush with Lady Luck. Or to be slightly more technical, as I am only a layman, benefiting from enough volatility that you overcome the house edge.

What would be most interesting to me, and I imagine a good deal of gamblers, would be a chart that shows for all games the probable range of losses to wins with a certain percent of certainty, say 68%, in a typical gambler's session (as determined by standard deviation, etc). Something like 3 hours of play, betting the minimum or slightly more, etc.

Alan Krigman starts to do this in the article guido111 linked, but 100 hands and 10,000 hands aren't typical. See below:

A) After 100 hands at $10 each, the player's expectation is to lose 0.5% of the $1,000 handle or $5. The standard deviation for the session is the standard deviation per unit bet, times the size of the bet, times the square root of 100 hands or 1.12 x $10 x 10 = $112. The so-called "empirical rule" of statistics says that for wide ranges of situations like this, approximately 68% of all players will be within one standard deviation of the mean. In this case, after 100 hands, the range is therefore from a loss of $117 to a win of $107. The $5 represented by edge is buried in the $112 standard deviation.

B) After 10,000 hands at $10, the player's expectation is to lose 0.5% of the $100,000 handle or $500. The standard deviation is $1,120. About 68% of all players can now expect to be from $1,620 behind to $620 ahead. The $500 biases the range of anticipated results, but volatility is still dominant.

You could figure out what a typical visit to the casino would be in terms of hours, bankroll, bet amount, etc. EDIT: and/or a reasonable win goal, like 25% of bankroll.

It seems like this is what you are getting at and would be the next level of analysis beyond your "House Edge, Element of Risk, and Standard Deviation of casino games compared" Chart. If not, please consider it. You could call it the Shackleford Quotient or something catchy like that...

Quote: Wizard

For the same reason casinos fear high variance games, the risk-seeking player should embrace them.

If you are playing roulette and you bet

(a) $5 on red, and $5 on black
(b) $10 on red
(c) $10 on 1-12
(d) $10 on #14

All four bets have the same EV. Well it is very customary to say that roulette has a very high HA, I always felt that it was a pretty low HA for a bet that pays 36:1.

I've always felt that bet (b) was a sucker bet because you could just as easily go to the craps table and make several different bets for a much lower HA.

I would call bet (d) the "best bet" of the four.