what are the odds?

Quote: Stoney

i've been successful recently with a simple craps betting method, and i'm here because i can't seem to find anywhere on the net that can tell me how to figure the odds to see if i'm good or lucky.

i just make equal bets pass line and field(come out only) and take full odds after the point. yes, there are a lot of break-even rolls with the 3 and 7, but the 2, 11(x2), 12(x2) are pure wins. and yes, the 5,6,8 are initial losses with a chance to break even, but the 4,9,10 are initial wins with a chance to double up. can anyone do the math for me on this?

Quote: Stoney

math breakdowns on this would be greatly appreciated.

Quote: Stoney

math breakdowns on this would be greatly appreciated.

Quote: Stoney

the field pays 3x on the twelve. i just stated that the win is x2 since i lost 1 on the pass line. and the 11 is a double win since i win both bets, but the 2 is only a 1x win since the pass line lost.

as far as the math side goes, i can calculate the return on the field side + the 7 (which is positive.... i think). but i don't know how to calculate the return of the point numbers (4, 9,10 pay double and the 5,6,8 net 0) on a win vs loss (wich would be +0 and -2). and i take the 3-4-5x odds. and, how does this affect the return rate of the field bet.

just started playing this way vs various wagers based on my confidence in the shooter. yesterday, i started with $100 and was up $80 in less than an hour when my buddy went broke playing three card poker. and 2 weeks ago i started with $100 was up $225 in about 3 hours when my wife nagged me to leave.

so, for the computer, $500 to start, $1000 win goal, $0 stop loss.

Quote: Stoney

thank you for this. i didn't know how to figure the pass line edge post-point. and i didn't know if the edge on the 2 bets were cumulative or averaged, since they get resolved differently. but if i could pick up my pass line bet after the point, i'd have the advantage.... assuming my chart above is correct. looks like i've just hit those hot shooters.

and btw, the ones i've won the most off of are dice setters, myself included. i almost always lose on the 'shake and throw'.

Quote: 7craps

Guido111 did a slick calculation taking the 345X odds into total consideration.

Quote: Ahigh

In general, you lose everything. If you expect to win, and try to win faster, in the short run, you can win faster with high edge bets. But in the long run, the long run goes away, so there's no point in talking about the long run.

Quote: buzzpaff

I will be at the WOV G2E luncheon. Care to share a diet coke ?

Quote: buzzpaff

Bet 5 with 100 odds seems better ?

Quote: Ahigh

250/20000 = 1.25% edge for the event. Better than all on the pass. Better edge than all on the don't.
...
please correct me on anything I'm doing wrong here.

Quote: Ahigh

If you just want to play the game, play minimum bets and enjoy it knowing you're going to lose.

That's what I say. Pay to play and have fun.

If you want to have a chance at winning, I think you should be advised to minimize the number of large bets to have a better chance.

The Wizard has his scheme of how to double your money.

I say I would rather have an exactly 50/50 chance to win like this:

$5000 buy on the 4
$5000 buy on the 10

Hit a seven? Lost! Go home.
Hit a 4 or a 10? Win, bring it down.

Chance to win: 50/50

Cost to get paid: 5% of $5000 = $250.

Total haul: $10,000 - $250 - dealer tips

250/20000 = 1.25% edge for the event. Better than all on the pass. Better edge than all on the don't. Better chance of winning and easier to understand a 50/50 chance and easier to understand your cost to take your pay when it's an explicit vig.

How much lower edge and/or better chance can you get than that for having exactly a 50/50 chance to win as close to twice what you risk to win? Put the simulator and all your might behind it, and let me know.

Being shy of ten g's for that last little extra chance to win is worth it to me. Not that I would lay the four for ten grand and settle for less than 5 grand, but I'm just saying, it's cheap to play if you just want to play, and it's expensive to get paid big money if you just look at the raw costs of big play, it's expensive compared to what other things you could do with that money.

This was precisely one of the areas where I think a couple of people blew a gasket saying I had the wrong math, so please correct me on anything I'm doing wrong here.

But as far as the details of the simulator for the OP, I'm not sure they are sophisticated enough to understand all that crap.

Even if the experts have a hard time with my math, most newcomers understand you have exactly a 50/50 chance to win, and the house keeps $250 when they pay you $10,000 .. so it's really $9,750.

It's possible you guys get so technical that you forget that there are people who just want to win without talking about complicated stuff.

> br = 10000      # Starting Bankroll
> goal = 14750  # Target bankroll
> odds = 1   # Payoff Odds
> p = 244/495    # P(win bet)
> 
> prob = 0
> p_fail = 1
> count = 0
> 
> while ( (br >= 1) & (br < goal) & (count < 1000) ) {
+ 
+ # Compute probabity of completing parlay up
+ 
+   consec = 0 
+   while (br + br*odds <= goal) {
+     br = br + br*odds
+     consec = consec + 1  
+   }
+ 
+   p_parlay = p^consec
+ 
+ # Compute probabilty of reaching goal
+ 
+   bets = 0
+   if (br < goal) {
+   while ( br >= ceiling((goal-br)/odds) ) {
+     br = br - ceiling((goal-br)/odds)
+     bets = bets + 1
+   }
+ }
+ 
+   p_goal = ifelse(bets, 1-(1-p)^bets, 1)
+ 
+   delta = p_fail*p_parlay*p_goal
+   prob = prob + delta
+   p_fail = p_fail*p_parlay*(1-p)^bets
+   count = count + 1
+ }
> 
> prob   # Probability of reaching goal
[1] 0.6686613
> delta
[1] 0
> count
[1] 1000
> 

Quote: Ahigh

250/20000 = 1.25% edge for the event. Better than all on the pass. Better edge than all on the don't.
...
please correct me on anything I'm doing wrong here.

Quote: MathExtremist

Your denominator is incorrectly inflated. You shouldn't include the action for the unresolved bets.
50% of the time you win 9750 on action of 5000, and 50% of the time you lose 10000 on action of 10000. EV = (0.5*9750 - 0.5*10000) / (0.5*5000 + 0.5*10000) = -125/7500 = -1.667%.

Quote: guido111

This is the best way to hijack a thread. This is how threads get way out of whack and become useless posts and way too difficult to read and follow.

All your above words have nothing to do with the OPs original questions.

Start by moving this to a new thread and then I will destroy (nicely) some of your conclusions.
Here is one to start

You take $10,000 make two buy bets.
Risking $10,000 to win $4750.
Your probability to hit a $4750 win is exactly 50%

Quote: Ahigh

I say I would rather have an exactly 50/50 chance to win like this:

$5000 buy on the 4
$5000 buy on the 10

Hit a seven? Lost! Go home.
Hit a 4 or a 10? Win, bring it down.

Chance to win: 50/50

Cost to get paid: 5% of $5000 = $250.

Total haul: $10,000 - $250 - dealer tips

250/20000 = 1.25% edge for the event.

Quote: clacam

Hi everyone, I am a new member and I have a probobilities question.
Can anyone help me to write the formula to calculate on a 37 numbers roulette wheel
the correct odds of the expected repetitions of numbers at any time between spins 10 and 37.

ie: at spin 17 there should be 4.22 repetitions.
I consider a second repetition a number that has come up three times.

I thank you for your help.

Quote: ThatDonGuy

Somebody correct me if I'm wrong, but don't you have to pay the 5% up front when you buy a number (i.e. you're betting $5250 against the casino's $10,000 when you buy a 4 or 10)? In that case, here's what I get:

Buy the 4 and 10 for $5000 each, plus $250 commission on each number; total bet is $10,500
1/2 of the time, 4 or 10 comes up before 7; result = +10,000
1/2 of the time, 7 comes up before 4 or 10; result = -10,500
EV = (1/2 x (+10,000) + 1/2 x (-10,500)) / 10,500 = -250 / 10,500 = -2.38%.
Am I missing something?

Quote: MathExtremist

Your denominator is incorrectly inflated. You shouldn't include the action for the unresolved bets.
50% of the time you win 9750 on action of 5000, and 50% of the time you lose 10000 on action of 10000. EV = (0.5*9750 - 0.5*10000) / (0.5*5000 + 0.5*10000) = -125/7500 = -1.667%.

Quote: guido111

You take $10,000 make two buy bets.
Risking $10,000 to win $4750.
Your probability to hit a $4750 win is exactly 50%

All on the Pass or Don't is still way better with NO Odds bet.
It is called Bold Play. Mathematically proven.

IF I have the same bankroll Risking $10,000 to win $4750 and I lose the first time, I am still in the game,
you lose the first time you are done.
For the Pass I have a [1] 66.86613% chance of success to hit my $4750.
Bold Play, most times, kicks ass to win the most the fastest.
added:
[1] 0.6571952 chance of winning exactly $5000. Better than your score
[1] 0.5621962 chance of winning exactly $7500. Again, Better than your score

added: I see ME stepped in to correct you on your house edge calculations.
I doubt it will do any good.

Quote: ThatDonGuy

Somebody correct me if I'm wrong, but don't you have to pay the 5% up front when you buy a number (i.e. you're betting $5250 against the casino's $10,000 when you buy a 4 or 10)? In that case, here's what I get:

Buy the 4 and 10 for $5000 each, plus $250 commission on each number; total bet is $10,500
1/2 of the time, 4 or 10 comes up before 7; result = +10,000
1/2 of the time, 7 comes up before 4 or 10; result = -10,500
EV = (1/2 x (+10,000) + 1/2 x (-10,500)) / 10,500 = -250 / 10,500 = -2.38%.
Am I missing something?

Quote: 7craps

Man, where is the $5000 Buy 4 and Buy 10 that only pays $4750 on a win?
Stay away from that table!
I guess you know where your mistake is?

Bold play, $10,000 to win $9750
pass: 0.4987689
Dpass: 0.4988305

Not that much less than AHigh's 2 bet wager.

Quote: Ahigh

You are correct that the lower edge depends on 5% commission on the win. Every casino that I know of in Vegas does buy bets this way, but I do understand that many other casinos don't. The Riviera charged commissions up front when they went to 1000x odds, for example. When I asked them about it, they said it was because they had to make money somehow.

But in general, here in Vegas (this is wizardofvegas.com) the commission is on the win. I apologize if anyone else was led astray to thinking that this works with commission up front. It's a horrible deal with commission up front to do this.

Quote: ThatDonGuy

I'm a little confused as to what "commission on the win" means.

Suppose I buy the 10 for $100, and I win - what profit do I make? $195 (200 minus 5% of the bet), or $190 (200 minus 5% of the 200 that I won)?
If it's 5% of the bet, then, if I buy just the 10 for $100:
1/3 of the time, I win $195 (200 minus 5% of the 100)
2/3 of the time, I lose $100
EV = (1/3 x (+195) + 2/3 x (-100)) / 100 = (-5/3) / 100 = 1.6667%
And if I buy the 4 and 10 for $100 each:
1/2 of the time, I win $195 (200 minus 5% of the 100, with the other number being taken down)
1/2 of the time, I lose $200
EV = (1/2 x (+195) + 1/2 x (-200)) / 200 = (-5/2) / 200 = 1.25%
Since you'd lose the whole 200 on a 7, you're "betting 200" each time.
(If it's 5% of the amount won, this becomes 2.5% - but then, they might as well be asking for the 5% up front.)

Something tells me I'm still missing something...

Quote: Ahigh

The action is $10,000 the way that I constructed the bet. The reason why is that I intend to take both bets down on resolution of any of the 12 possible resolving rolls and be done.

Quote: MathExtremist

Except you don't do that -- you keep betting, as demonstrated by the charts for your progression system that you posted in another thread. In the final analysis, you are still expected to lose 1.67% of the total amount you wager on every resolved buy 4 wager, plus 1.67% of the total amount you wager on every resolved buy 10. If you inflate the denominator by considering the 4 bet resolved when a 10 rolls, or vice versa, then of course your computed edge will drop. But I don't think that's a valuable statistic; in fact, I think it's misleading, and therefore, less valuable than other, more typical ways of looking at the bet. Quoting the combined edge as 1.25%, without any further information, would imply that a single person who makes both $100 bets simultaneously would somehow have a lower expected loss than the total loss of two different players making $100 bets independently. Of course that's not true: the expected loss from the combination is exactly the same as the sum of the loss from each component.

Quote: MathExtremist

Except you don't do that -- you keep betting, as demonstrated by the charts for your progression system that you posted in another thread. In the final analysis, you are still expected to lose 1.67% of the total amount you wager on every resolved buy 4 wager, plus 1.67% of the total amount you wager on every resolved buy 10. If you inflate the denominator by considering the 4 bet resolved when a 10 rolls, or vice versa, then of course your computed edge will drop. But I don't think that's a valuable statistic; in fact, I think it's misleading, and therefore, less valuable than other, more typical ways of looking at the bet. Quoting the combined edge as 1.25%, without any further information, would imply that a single person who makes both $100 bets simultaneously would somehow have a lower expected loss than the total loss of two different players making $100 bets independently. Of course that's not true: the expected loss from the combination is exactly the same as the sum of the loss from each component.

Quote: ThatDonGuy

I think I see MathExtremist's point.

Suppose there was a bet that was always a push. If you buy the 10 for $100 and then bet $9900 on the "always a push" bet, 1/3 of the time you win 195, and 2/3 of the time you lose 100, for an EV of -5/30,000, or -0.016667%. In fact, there is such a bet; it's called "keeping $9900 off the table."

Then again, suppose there was a bet that paid 0.975-1 (equivalent to 1.95-1 on half the bet and a push on the other half) if a 4 or 10 was rolled before a 7; on a $200 bet, 1/2 of the time you would win $195, and 1/2 of the time you would lose $200, for an EV of -5/4000, or -1.25%. How is this different from Ahigh's bet?

Something tells me it comes down to a question of whether or not you include pushes in EV - and I seem to recall there being a rather heated debate about this in mathematical/statistical circles.

Quote: ThatDonGuy

Something tells me it comes down to a question of whether or not you include pushes in EV - and I seem to recall there being a rather heated debate about this in mathematical/statistical circles.

Quote: Ahigh

And one last final thing, and I will repeat this again, all of this is simply confusion, and nothing else. If you turn everything into edge per roll, all of this becomes way less confusing.

Quote: 7craps

But your idea of just using the HE per roll for all bets plays into the Enron Accounting tricks book.
And is never always the proper way to calculate and compare EV to different bets.....

Quote: RaleighCraps

Excellent post 7craps! For me, one of your best posts ever.

Your comparison of the field to the Place 5 bet, using HE, then edge per roll, and total bet action finally helped me understand this a bit more. A working example I could easily relate to.

I still can't explain it, but now I have this post to refer back to, and with the other reference links you have provided, I have some studying to do.

Quote: AHigh

Just coming up with a bunch of complicated math results and believing to one's self that you are right because you understand things that most people don't understand is not doing yourself any favors. Neither by using the results of that complicated math to benefit from playing the game nor by having a better life as the result of being an unpleasant person to tell other players how stupid they are.