Speeves
Speeves
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May 5th, 2013 at 1:09:20 PM permalink
Can anyone help me with the house edge in playing the following:

If I have equal dollar amounts on the 6 and the 8 for only 5 rolls what are the odds of rolling a 6 or an 8 before a 7?


Thanks for the help!

To follow up.....I was reading a book called Make Your Living Playing Craps by Larry Edell. In this book he offers a 6 and 8 strategy. It works like this:

Wait for a 7 to be rolled
then
Wait for a 4,5,9,10
Then place the 6 and the 8 for $30 a piece
If a 6 or 8 is rolled, take both down and repeat the process.
If there is no "decision" after 5 rolls, take down both and start again.

I am trying to figure out what the house edge is in this strategy.
odiousgambit
odiousgambit
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May 5th, 2013 at 1:44:37 PM permalink
each roll: 5 ways to roll a 6, 5 ways to roll an 8, 6 ways to roll a 7, so your chances of winning one or the other exceed the chances of rolling a 7 = 10 to 6. Each roll.

Doing it for 5 rolls only does not increase the chances for one way or the other, but increases the chances that *something* happens [over one roll and then taking them down- edit]. It makes me wonder if you are trying to trick the dice [g]
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
Speeves
Speeves
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May 5th, 2013 at 2:06:28 PM permalink
I'm confused. If the odds are in my favor.....where is the house edge? Does it have to do with the payout amounts and the amount wagered?
cclub79
cclub79
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May 5th, 2013 at 2:16:56 PM permalink
Quote: Speeves

I'm confused. If the odds are in my favor.....where is the house edge? Does it have to do with the payout amounts and the amount wagered?



You are risking $12 to win $7, so yes, you'll win 7 a bit more often, but when you lose, you lose 12.
cowboy
cowboy
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May 5th, 2013 at 2:20:23 PM permalink
I'm assuming this is a Place bet.

You'll win about 59.25% of the time and lose about 36.44% of the time. The remainder of the time you won't win or lose but will take the bet down. Let's say you are betting $6. Then overall you'll win $4.15 and lose $4.25. (i.e. winning $7 59.25% of the time and losing $12 36.44% of the time.

So you lose 10 cents on a $6 bet, giving a house edge of 1.67%. This is with rounding to the nearest cent.

I ran your scenario over a million iterations and got a house edge of 1.755%, actually.

Edit: all the stuff about waiting for certain rolls is to just set up the table so 6 and 8 are not the current point.
Speeves
Speeves
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May 5th, 2013 at 2:27:06 PM permalink
Is it possible to offset the house edge by increasing your wager......it just seems that if you could win 59.25% of the time you could increase your wager when you lost to offset the house edge. Any thoughts?
cowboy
cowboy
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May 5th, 2013 at 2:31:27 PM permalink
Quote: Speeves

Is it possible to offset the house edge by increasing your wager......it just seems that if you could win 59.25% of the time you could increase your wager when you lost to offset the house edge. Any thoughts?



No because these multiples are always the same. You'll always be betting (and risking) 2N where N is the wager and you can only win 7/6 * N.

One thing though, I did not include the possibility of 6 or 8 being hit multiple times in those 5 rolls. What does the strategy say to do if you win? Likely it says take the bet down.
Speeves
Speeves
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May 5th, 2013 at 2:37:38 PM permalink
Yes. It does say take it down if you win. If you started with two $6 bets and doubled up each time you lost? can you run a scenario like that?
cowboy
cowboy
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May 5th, 2013 at 2:48:55 PM permalink
Good thing it says take it down, because the house edge <Edit: stays at about 1.6%> if you leave it up.

There are several discussions here about doubling up when you lose. The problem is eventually you hit a streak so bad that you'd exceed the table limit by doubling your bet. FREX if you doubled your bet 4 times you'd have to be willing to bet $192, so that you could win $112 and you've already lost $180.
dicesitter
dicesitter
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May 7th, 2013 at 6:41:29 AM permalink
Speeves



Since there is no way to determine ahead of time when a 7 will appear, the house edge is what ever it is
when you place that bet. For each bet you place that edge applies.

There are all kinds of systems and trend betting ideas, however they do not change the house
advantage. For instance my simulation indicates about 14.9 % seven outs..... now that would seem to say
you can bet alot more on the first roll and have a 85.1% chance of not being a 7. that is the basis of the
iron cross, we have so many chances to win and so few to lose. However we have all seen 5-6 7's in
a row and 5-6 of the same number in a row, back to back to back hardways etc.

The house edge always applies.

dicesitter.
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