Estimated Bankroll Question

Hi Friends. Once a year I come to Vegas. I stay for three nights and I only play craps. My favorite tables for me are $5 tables with 10X odds. There are 3 casinos I visited in the past, and I would visit others with those odds.

I play straight pass/come line max odds (5/50) and try to keep three numbers covered. I don’t bet any other bets on the table. I just don’t seem to have good luck laying 6’s and 8’s. And I don’t have much fun playing wrong way. (Even though I wish I knew when to switch to keep winning).

I play 2 hours in the afternoon and 4 hours at night for 3 days.

Covering 3 cards can bring some wild bankroll swings. That’s cool. But, I want to make sure I can cover everything barring a catastrophic loss trip. Given the above, how much of a bankroll would you bring for Craps?

Asking for a friend...

MrCraps550 plays a straightforward Molly at 10x Odds at $5 games. Excluding trivial losses to crap numbers, his worst-case loss per shooter is (3)($5 + $50) = ($165). Although four consecutive such losses to four miserable shooters is unlikely, he should be prepared to endure and survive. He should buy in for $700 or $800. Two such table stakes should suffice for each afternoon and three should suffice for each evening.

A useful benchmark for recreational gamblers is to begin with a bankroll that is eight to ten table stakes. For professional gamblers, the bankroll should be double that standard. For MrCraps550, a Craps bankroll of $8000 will see him through any three-day trip that is beset by only moderately bad luck.

Quote: MrCraps550

Given the above, how much of a bankroll would you bring for Craps?

$6000
that is from old data I have that looks like this
The rolls could be too high for full crowded tables.
enjoy your play!

10x your betting method per shooter per session to survive table swings.
$165 x 10 = $1650

Also 3 losses in a row, or 3 out of 4 you’re done with that session.

Quote: tconley19

10x your betting method per shooter per session to survive table swings.
$165 x 10 = $1650

this has already been shown to be way off (very wrong) in other threads because it refuses to take into consideration the 'time' factor for each session.


it (the 10X gig) assumes
craps players are ALL so very foolish (is lacking good sense or judgement)
and foolishly stupid (is lacking in intelligence or exhibiting the quality of having been done by someone lacking in intelligence.)

nice try rehashing old trash, imo (not really)

Quote: tconley19

Also 3 losses in a row, or 3 out of 4 you’re done with that session.

imo, another system for losers

+1 for sharing your opinions here

How about 90 times your base bet with odds? There's lots of point 7-outs at the table.
$55 x 90 = $4950 ~ $5,000

Here’s a rough estimate.

Assuming average point value of 5/9, in 45 bets your expectation would be -2 % of flat bets made, comprised by:

5 losses of 1 unit
18 losses of 11 units
10 wins of 1 unit
12 wins of 16 units

Which is a standard deviation of 10.8.

Assuming you play 20 hours at 50 resolutions per hour is 1,000 resolutions. So your expectation is -20 units +/- 341.

The chance of you losing more than 680 units (about 2 deviations below expectations) is about 2.5 %.

The chance of you busting a bankroll of 700 units = $3,500 is about double that : 5%.

I'd have $450 for 3 $150 sessions of $5 3 Point Molly with no odds. If I pull ahead, I'll add odds.

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