House Edge On Simplest Of "Strategies"

This method of play is so simple, I hesitate to even call it a strategy.

Anyway, Caesars Windsor allows 5x odds on all numbers across the board. So I put a $5 pass line bet and I put $25 odds down no matter what the point is. The only exception to this rule is the 5/9 which require an even number for the odds for the math to work right. At CW you are allowed to round up by $1, so on a 5 or 9, you'd bet $26 in odds on a $5 PL bet.

This is literally the only bet I do in this strategy. No further come bets, no place bets, etc. Just PL bets backed with max odds.

What does the overall (average) house edge work out to with this strategy? I know the PL bet alone is 1.41% but the odds brings that amount down. I imagine the amount that 5x odds affect the H.E. would vary depending on what the point is. The 5/9 scenario I mentioned above further complicates things.

Is there a mathematical formula that can be used to calculate the vig on this simplest of all strategies?

Yes. It's simply a 0.326% house edge on total money wagered.

Source: https://wizardofodds.com/games/craps/.

The odds bet has no house edge, so if you're always putting down 5x odds, you're effectively just spreading the 1.41% HE on the passline bet across more money. I'm getting about 0.325% as the house edge you're facing.

edit: fixed transposition error.

Make sure to always work your odds for the best overall edge!!! It's all random so you're no more or less likely to get a seven on the comeout than any other roll!!!

Quote: etr102

Is there a mathematical formula that can be used to calculate the vig on this simplest of all strategies?

Yes there are a few.
HE = EV/Avg$Bet
It is the Avg$Bet that can get messed up.

ev=(-7/495)*$5
Avg$Bet= (1/3*$5) + (2/3*$30) = 5/3 + 60/3 = 65/3 = $21 and 2/3
=0.32634% (-7/2145)

or another formula without the average $ bet
=(-7/495)*1/(1+2/3*OddsFactor)

so your extra $1 on the 5and9 will be a tad lower because your average bet is a tad higher at $21 and 8/9 (21.89)
math left for the OP
-0.3230272%
(-7/2167)

for uneven odds, like 3,4,5
https://wizardofodds.com/games/craps/appendix/1/

(-7 / 495) / [ 1 + ((5x + 4y + 3z) / 18) ]

you can use the Wizard's formula for your 5.2X odds on the 5&9 (26/5)
=(-7 / 495) / (1 + ((5*5 + 4*5.2 + 3*5) / 18))
-0.3230272%
(-7/2167)

Good Luck

Quote: etr102

...simple...strategy....$5 pass line bet and...$25 odds...a mathematical formula?

The simplest way to think of it is that you've taken the option to make the odds bet (a sensible choice) but the house edge on the $5 bet is about 1.41% per "game" (note not roll). Thus every "game" (the the shooter makes or fails) your average loss will be roughly 7 cents. This average is unaffected by your odds bet; the only difference in making odds bets are your wins will be bigger and your losses will be bigger, but still average out at a loss of 7 cents per "game".

Now the discussion above goes along the thought, your average bet is $30 (let's assume they use 50c chips) and you only lose 7 cents per "game". Thus for every $100 you bet, ignoring in some games the odds won't apply, the loss is about 23c.

Reference https://wizardofodds.com/games/craps/appendix/2/

If you want to consider the cost per roll, the a $5 pass-line bet usually takes 3.38 rolls to resolve and costs 7c per "game", i.e. about 2c per roll. So if there were 100 rolls in an hour (to make the maths easier), the average cost would be about $2.07. Taking the odds would still leave the average loss, but your actual balance at the end of the hour would have a higher variance.

Compare this to a $6 bet on 6 (assuming it pays 7/6 as in the UK). First it does not work on the come out roll, so in an hour there are 100/3.38 "games" of which the first roll is ignored, so only about 70 active rolls. Each 36 rolls cost you $1 (you win $7 five times, and lose $6 six times). So in an hour you lose just under $2.

This reminds me, and we should all know, the non free bets are 0.40% to 0.42% per roll. The unit of bet event is a roll IMO, that's just how I look at it to keep it simple.

I appreciate the information, guys. 0.326% HE on average with this "strategy" seems pretty darn good to me. Makes me wonder why people even bother with place bets. At CW, it seems that very few people play with full odds.

Here's another question. Suppose I play with the strategy described above, but I add 1 come bet (with full 5x odds no matter what point it moves to), what affect does that have on the HE? Does the HE maintain a consistent 0.326%, or does the fact that it hedges the PL bet while the bet is still in the Come area have an overall effect?

In general making multiple wagers generally tends to decrease your variance (because you are often betting against yourself). This tends to allow you to play longer with a reduced chance to win for a given amount of play.

The most significant reason to do this instead of just increasing your bet size is if you just like seeing more "action" or chips being moved around and get bored too easily.

Otherwise, I believe, the rule is not to hedge.

When you bet a pass line followed by a come bet, you are, for one roll, removing the bet you have against rolling a seven on the passline by hedging it with a win on the come that will occur with the same chance of rolling that seven. This causes the edge to peek out at you more often because your chance to lose or win with those two units (one on the pass and one on the come) has been removed.

It's easier to see this effect when you think about betting red and black on roulette because there is zero chance to win in that case.

But any time you reduce your chance of winning or losing, you are decreasing your variance, allowing yourself to play longer on average (assuming the same amount of money is being risked) at the cost of more likely just losing than if you were to be okay with just having fewer bets and allowing the exposure to enable a win.

Quote: etr102

does the fact that it hedges the PL bet while the bet is still in the Come area have an overall effect?

this has been asked before and it seems clear the HE stays the same while the momentary correlation has no relevance. You should get the same results [on average] as betting the same amount serially instead of making a come bet.

However, come betting usually leads to betting more overall.

Quote: odiousgambit

this has been asked before and it seems clear the HE stays the same while the momentary correlation has no relevance. You should get the same results [on average] as betting the same amount serially instead of making a come bet.

However, come betting usually leads to betting more overall.

He's right, but understand on average he means on average for millions of rolls.

On average for a hundred rolls, you're more likely to lose it all with a seven and you're less likely to get that fantastic good roll.