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

Quote:RHK6446Question: How many come out rolls can the player expect given the initial parameters?

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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.

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.

Quote:ThatDonGuyQuote:RHK6446Question: How many come out rolls can the player expect given the initial parameters?

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

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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...

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 !

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 rolledQuote:ChumpChangeSo 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.

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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

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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)?

Good catch. I had transposed the EV/SD ratio. Fixed the original postQuote:RHK6446Quote: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

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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)?

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Now that you’ve acknowledged that erroneous assumption, perhaps it’s time to rethink your “hot vs cold table” and “tourist vs experienced shooter” assumptionsQuote:ChumpChangeSo nearly 30% of rolls will resolve a PL bet, not 1/6th as I dumbly figured

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if you are actually looking to come out ahead, don't tipQuote:Ace2In this case you only need to beat expectations by about 0.6 deviations, which is much more doable

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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