Standard deviation question

Hey guys, I mostly like to play the pass line backed with as many odds as possible. If I bring say 1k, play for 8 hours (let's say 800 rolls), and bet a 10 dollar pass line, what is the most I can max out my odds and still have a 95 percent certainty I will not run through my entire bankroll in those 800 rolls

Quote: grex123

Hey guys, I mostly like to play the pass line backed with as many odds as possible. ...bet a 10 dollar pass line, what is the most I can max out my odds...?

Well, if you go to a casino that allows 100x odds, place a $10 pass line bet, set a point with the come-out roll, put the rest of your 1k behind the line, then roll a 7 ... I guess the bankroll only lasts a few seconds. I don't think I want to sweat the statistics to find the max odds bet that will give a 95% chance of survival for 800 rolls.

Quote: grex123

Hey guys, I mostly like to play the pass line backed with as many odds as possible. If I bring say 1k, play for 8 hours (let's say 800 rolls), and bet a 10 dollar pass line, what is the most I can max out my odds and still have a 95 percent certainty I will not run through my entire bankroll in those 800 rolls

this can be calculated exactly using a Markov chain, but that is not understood by the majority and takes time to setup properly.

using a formula, as in ror from Don Schlesinger, we get errors as Blackjack is way different from free odds and I am too lazy (I tried once) to find what could work.

a simulation is very simple and easy (at least for me)
100,000 sims
this had the same flat bet each time ($10)

pass odds taken	bust
1x	0.086%
2x	2.595%
2.5x	3.74%
3x	9.666%
345x	19.149%
5x	26.638%
10x	52.687%
20x	75.85%
100x	94.8%

I would say $25 odds on the 6&8 should be closer (and below) 5%
the 2x sim used $20 odds and most casinos allow $25 with a $10 flat bet on the 6&8
(called 2.5 times odds - sim shows 3.74% bust)

Using 10x odds as an example, that means your average total bet is $10 + 2/3 * $100 = $77, giving you a bankroll of $1000 / $77 = 13 units.

The simulation shows a 53% RoR after 800 bets. I’d think your RoR would be 53% after more like 100 bets. $1000 is a quite small bankroll for $10 plus 10x odds...I’d think you’d be lucky to last 1 hour.

Quote: Ace2

The simulation shows a 53% RoR after 800 bets. I’d think your RoR would be 53% after more like 100 bets. $1000 is a quite small bankroll for $10 plus 10x odds...I’d think you’d be lucky to last 1 hour.

800 rolls and not 800 bets. there is a BIG difference there
1 million simulations of 800 rolls taking 10x odds
Bankroll was busted . . = 52.572% of the time

This is awesome! Thanks so much!

Quote: grex123

This is awesome!

agree
knowing something (even about Craps) has to be better than knowing nothing.

Just remember the $10 was made every time.
Increase that bet, say after a win, and increase the Bust rate.

Quote: 7craps

800 rolls and not 800 bets. there is a BIG difference there
1 million simulations of 800 rolls taking 10x odds
Bankroll was busted . . = 52.572% of the time

Oops, I was thinking 800 bets. 800 rolls would be an average of 237 pass line decisions.

Here’s a formulaic way to estimate the RoR.

The standard deviation is approximately the odds plus 0.8. So, for instance, for 5x odds the SD is about 5.8, close enough to the actual figure of 5.824.

237 decisions x 1.41% edge on flat bet is an expected loss of 3.4 units. Bankroll is $1000 / $10 = 100 so we expect to end the session with 96.6 units. SD is 5.8 * 237^.5 = 89.3.

The probability of ending the session 96.6/89.3 = 1.082 deviations to the left of expectations is 13.96 %. The chance of being there at any time (risk of ruin) is about double that or 27.9%.

You can use this method for any odds amount and it’s pretty close (within a couple percent) to the simulation.