RHK6446
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May 11th, 2024 at 8:57:15 AM permalink
Dear Odds Makers:

Imagine a 40 year old craps player with an life expectancy of 40 more years (asking for a friend :)
This player visits the casino 4 times per year and each visit lasts 8 hours.
The player plays at a craps table with an average of 120 rolls per hour.
The expected number of rolls over the remainder of his life would be:
40 years x 4 visits per year x 8 hours per visit x 120 rolls per hour = 153,600

If the player made 153,600 bets of $25 each and the house edge is 1.41% the expected value is - $0.35250 per bet then the players lifetime EV: 153,600 rolls x - $0.35250 EV per roll = -54,144

However, given the original parameters, not all of the rolls would be come out rolls.
Question: How many come out rolls can the player expect given the initial parameters?
Can the standard deviation be estimated?

*edited for clarity - and I also didn't have automatic recalculation enabled in excel. doh.

Thanks
Bob
Last edited by: RHK6446 on May 11, 2024
ThatDonGuy
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May 11th, 2024 at 10:27:12 AM permalink
Quote: RHK6446

Question: How many come out rolls can the player expect given the initial parameters?
link to original post


If I am crunching my numbers correctly today, the average number of rolls per come-out is 4717 / 990, or about 4.765.

There should be a way to calculate an exact SD, but it may take a little grinding, considering that, in theory, any comeout can have any positive length.
ChumpChange
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May 11th, 2024 at 10:58:28 AM permalink
That looks like $25 line bets with no odds. For 160 X 8 hour nights you get 153,600 rolls @ 120 rolls/hr. The EV is 1.41% on the PL and 1.36% on the DP.
The thing about craps is the the table is typically cold much longer than it is hot. You'll get a hot shooter (20+ rolls) much more infrequently than a series of cold shooters. Of course there are times where table limits are higher and only the better shooters bother getting to the tables and you'll have a better crop of throws. The tourists who don't know the game and don't have the beginner's luck going for them really skew the averages down. So for the tourists, bet the DP and the regulars who seem to throw a good game, bet the PL. For the ones I don't know just bet the 6 & 8 so I don't go on endless losing streaks on their come-out rolls when they are trying to roll horn numbers anyway.

Now if you are trying to figure out how many times the PL bet gets resolved in 153,600 rolls so you can figure out your EV, I'm just going to divide by 6 and get 25,600 7's and multiply that by $25 X 1.41% = $9,024, or down 361 units.

Surely there must be a way to modify your betting so you win 250 bets ahead a few times in that number of rolls. But if the rolls don't come, they don't come.
ThatDonGuy
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May 12th, 2024 at 1:24:46 PM permalink
Quote: ThatDonGuy

Quote: RHK6446

Question: How many come out rolls can the player expect given the initial parameters?
link to original post


If I am crunching my numbers correctly today, the average number of rolls per come-out is 4717 / 990, or about 4.765.
link to original post


Slight mistake - the actual mean number of rolls in a comeout, confirmed in simulation, is 557 / 165, or about 3.376.

Still working on the SD...
Ace2
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May 12th, 2024 at 5:07:07 PM permalink
No need to split hairs over this. There will be variance in both the number of rolls per hour and the number of rolls per bet resolved, but assuming 153,600 rolls the number of passline bets resolved will be very close to that times 165/557 =~ 45,500. So an expected loss of about 652 units +/- 213 or $16,300 +/- $5.325. To beat it you’d need to be about three deviations above expectations which is quite unlikely but not that remote.

But let’s say there’s a $5 electronic craps table that makes payouts to the penny. You could bet $7 on the passline plus 3-4-5 odds and have about the same amount of total action ($26.44 average total bet) and your expected loss would be $4,491 +/- 7,361. In this case you only need to beat expectations by about 0.6 deviations, which is much more doable

Bottom line, about 27% of players betting the line plus max odds will beat craps over a lifetime of playing (assuming this bet volume). For comparison purposes, about zero percent of all gamblers beat the casino over a lifetime…or even over one year !
Last edited by: Ace2 on May 12, 2024
It’s all about making that GTA
ChumpChange
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May 12th, 2024 at 6:20:59 PM permalink
So nearly 30% of rolls will resolve a PL bet, not 1/6th as I dumbly figured, OK. He does make a case to get on the stick with the odds bets though.
Ace2
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May 12th, 2024 at 6:36:09 PM permalink
Quote: ChumpChange

So nearly 30% of rolls will resolve a PL bet, not 1/6th as I dumbly figured, OK. He does make a case to get on the stick with the odds bets though.
link to original post

The chance of the next roll resolving a PL bet is always between 1/4 and 1/3. Seems like you’re only thinking in terms of sevens rolled
It’s all about making that GTA
RHK6446
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May 12th, 2024 at 7:12:50 PM permalink
Quote: Ace2



But let’s say there’s a $5 electronic craps table that makes payouts to the penny. You could bet $7 on the passline plus 3-4-5 odds and have about the same amount of total action ($26.44 average total bet) and your expected loss would be $4,491 +/- 7,361. In this case you only need to beat expectations by about 1.6 deviations, which is much more doable

link to original post



Am I reading that correctly?
If the EV is -4,491 and the SD is +/- 7,361 then you should only need to beat the EV by +0.61 SD (not 1.61 SD)?
Ace2
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May 12th, 2024 at 7:29:25 PM permalink
Quote: RHK6446

Quote: Ace2



But let’s say there’s a $5 electronic craps table that makes payouts to the penny. You could bet $7 on the passline plus 3-4-5 odds and have about the same amount of total action ($26.44 average total bet) and your expected loss would be $4,491 +/- 7,361. In this case you only need to beat expectations by about 1.6 deviations, which is much more doable

link to original post



Am I reading that correctly?
If the EV is -4,491 and the SD is +/- 7,361 then you should only need to beat the EV by +0.61 SD (not 1.61 SD)?
link to original post

Good catch. I had transposed the EV/SD ratio. Fixed the original post
It’s all about making that GTA
Ace2
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May 12th, 2024 at 10:34:43 PM permalink
Quote: ChumpChange

So nearly 30% of rolls will resolve a PL bet, not 1/6th as I dumbly figured
link to original post

Now that you’ve acknowledged that erroneous assumption, perhaps it’s time to rethink your “hot vs cold table” and “tourist vs experienced shooter” assumptions
It’s all about making that GTA
ChumpChange
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May 13th, 2024 at 1:50:16 AM permalink
I'm gonna go with cursed and non-cursed shooters.
odiousgambit
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May 13th, 2024 at 6:14:31 AM permalink
Quote: Ace2

In this case you only need to beat expectations by about 0.6 deviations, which is much more doable
link to original post

if you are actually looking to come out ahead, don't tip

if you want freeplay, do tip

I've figured for those times when the freeplay was nice, at over-comped level, I've come out at least about even... after tips
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
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