How to calculate expected loss
I want to know if it is possible to calculate what the expected loss would be on a betting strategy in craps. I'd like to know the math on how to do it, so that I could then use that math for any betting strategy.
I think I understand the concept behind a single roll of the dice. If, for example, the betting strategy is $6 on Place 6, and $6 on Place 8, then there are 10 ways to win $7 and 6 ways to lose $12. So, (10/16)*$7 - (6/16)*$12 = -$0.125. Each roll of the dice, you should expect to lose 12 1/2 cents. Right?
And I can take that number and multiply it by a number of rolls, and expect to have lost that much money over that many rolls. So, over 100 rolls, I should have lost $125 dollars. Obviously, I could have rolled a 6 each and every single roll, and I could have rolled a 7 each and every single roll, so there is some variance possible. But the expected is to lose the $125 dollars.
But my question is more of what would be the expected loss if you played the betting system up until it lost. In other words, if you walked up to a craps table, placed the bets, and rolled a 7, you would have lost $12. If you rolled one 6 or 8, and then a 7, you would have lost $5, and if you rolled two 6's or 8's, you would have won $2. (Rolls of other numbers are being ignored since they would have no impact on this betting strategy. You could roll 100 4's, and then a 7, and the strategy would still have resulted in a loss of $12.) (Also, I'm obviously assuming that bets are working on the come out all the time, etc, etc.)
So, would I take the probability of rolling a 7 and losing $12 multiplied by the probability of rolling one hit and then a 7 and losing $6 multiplied by the probability of rolling two hits and then a 7 and winning $2 multiplied by the probability of rolling three hits and then a 7 and winning $9, etc, etc? And, how do I calculate this since there are obviously an infinite number of possibilities to multiply together.
I remember some things from calculus in college, and I'm sure this is going to involve sum of infinite possibilities with limits approaching zero and maybe other stuff in that realm. If someone can try and explain it to me, I can try to understand it, but I may ask for several points of clarification. :) Thanks!
The $125 you calculated IS the expected loss. It already takes into account that you'll sometimes win and sometimes lose. The only part you got wrong is that your calculation is for 100 resolutions, not rolls. I.E. It calculates 100 rolls of 6, 7 or 8 and ignores other rolls. To calculate for all rolls, the formula is (10/36)*7 + (6/36)*-12 = -0.05555 for a 5.56¢ expected loss per roll or $5.56 loss for 100 rolls.
I think the part that has you confused is that there is no system to your system. As far as I see it, you've got $6 on the 6 and 8 for every roll. A typical 'system' would start increasing the bet after a win or two. Once you add that to the mix, it gets a lot harder to calculate - too hard for me this early in the morning.
Are you asking, if you walk up with $12 and bet it, how long can you expect to stand there until you're broke?Quote: konceptumBut my question is more of what would be the expected loss if you played the betting system up until it lost.
You could hit a 7 right away, so it's one roll. Or you could get lucky and hit a few 6s and 8s and make a day of it.
Or you could hit a single 6 or 8, then a 7. You lost $5 and now have $7. You're neither broke, nor able to keep the bet up. What do you do?
Try this: You walk up with $23, and keep playing until you've lost AT LEAST $12.
In that scenario, the calculation is easy: 12 / 0.05555 = 216 rolls. At 3 rolls per minute (Is that the norm?), on average, you'll get to stand there an hour and 12 minutes.
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Anyone care to check my math? I have trouble believing that you can stand there for over an hour with only a $23 bankroll. Then again, with a lot of rolls that have no resolution, maybe the math is right.
I do know one thing: Betting just those two bets would be extremely boring. (So maybe that confirms the math.)
It's just an illustration of how difficult it is to calculate length of play using just expected value. You really need to use your risk of ruin probabilities. To use your example with such a small bankroll your risk of ruin is very high.
Quote: konceptumBut my question is more of what would be the expected loss if you played the betting system up until it lost.
From your example, it's clear to me you are wondering what the expected loss is if you play until you lose once (roll a 7).
You can safely ignore anything but 6, 7 and 8, so the probability of rolling 6 or 8 before a 7 is 10 / 16 = 5/8, while the probability of 7-out is 3/8. So, like you start to explain, the expectation is:
3/8 * -12 + 5/8 * 3/8 * -5 + 5/8 * 5/8 * 3/8 * 2 + 5/8 * 5/8 * 5/8 * 3/8 * 9 + ...
this is the sum from n = 0 to infinity of [(5/8)^n * 3/8 * (-12 + 7n)]
Although I have a math background, I don't like evaluating infinite series that much, so I'm going to stop here.
Quote: dwheatleythis is the sum from n = 0 to infinity of [(5/8)^n * 3/8 * (-12 + 7n)]
This is what I was looking for. Very simple now that I see it. Unfortunately, I still don't know how to evaluate an infinite series.
However, just by seeing this, I did realize that what I *could* do is use a spreadsheet. And that's something I know how to do.
At n=24, the sum was at -0.33 (rounded). At n=1000, the sum was still at -0.33. Therefore, I assume that my expected mathematical loss on such a system is -0.33 (a loss of 33 cents).
My idea on this was not that this is a system that I play, but just one of curiosity on what something like this is expected to lose. Now that I have some semblance of how to figure this out, I can use it to calculate other "systems" that people ask me about when they play craps. Yes, the idea is that you put down the $12 in bets, and play until you hit a seven. That could be one roll and lose it all, or several rolls and win some money or lose a slightly smaller amount. But if you did this a million times, on the average, you should see that you lost 33 cents each time. Note, that's not a million rolls. That's a million times of walking up to the table and putting down the $12 in bets and playing until a 7 is rolled.
There is no system that will guarantee winnings. But sometimes people have to see the numbers and math before they will believe it. (As a side note, I do have a system that guarantees winnings, but the casinos will not let me implement it. Very stingy of them, I say.)
Quote: konceptumI apologize if this has been answered. I tried searching, and didn't come up with anything.
I want to know if it is possible to calculate what the expected loss would be on a betting strategy in craps. I'd like to know the math on how to do it, so that I could then use that math for any betting strategy.
I think I understand the concept behind a single roll of the dice. If, for example, the betting strategy is $6 on Place 6, and $6 on Place 8, then there are 10 ways to win $7 and 6 ways to lose $12. So, (10/16)*$7 - (6/16)*$12 = -$0.125. Each roll of the dice, you should expect to lose 12 1/2 cents. Right?
Wrong! You have ignored the other 25 ways! On a per-roll basis, you have:
6/36 * -6 = -1.00
5/36 * +7 = + .9722
25/36 * 0 = 0
------
-.0278
Your ev per roll on a $6 place 6 or 8 is to lose 2.78 cents.
However, I don't consider this a particularly useful number. To figure the ev on the bet, regardless of how many rolls it takes to resolve it, is similar, but you ignore all the other rolls, reducing the possible rolls to 11, so:
6/11 * -6 = -3.2727
5/11 * +7 = +3.1818
------
-.0909
If you divide -.0909 by 6, you get -.01515, which is the usually-quoted house advantage (HA) on placing the 6 or 8.
Quote: konceptum
But my question is more of what would be the expected loss if you played the betting system up until it lost. In other words, if you walked up to a craps table, placed the bets, and rolled a 7, you would have lost $12. If you rolled one 6 or 8, and then a 7, you would have lost $5, and if you rolled two 6's or 8's, you would have won $2. (Rolls of other numbers are being ignored since they would have no impact on this betting strategy. You could roll 100 4's, and then a 7, and the strategy would still have resulted in a loss of $12.) (Also, I'm obviously assuming that bets are working on the come out all the time, etc, etc.)
So, would I take the probability of rolling a 7 and losing $12 multiplied by the probability of rolling one hit and then a 7 and losing $6 multiplied by the probability of rolling two hits and then a 7 and winning $2 multiplied by the probability of rolling three hits and then a 7 and winning $9, etc, etc? And, how do I calculate this since there are obviously an infinite number of possibilities to multiply together.
I remember some things from calculus in college, and I'm sure this is going to involve sum of infinite possibilities with limits approaching zero and maybe other stuff in that realm. If someone can try and explain it to me, I can try to understand it, but I may ask for several points of clarification. :) Thanks!
p(no hits) = 6/16 = .375
p(one hit) = 10/16 * 6/16 = .234375 (.609375)
p(two hits)= 10/16 * 10.16 * 6/16 = .14648 (.75586)
(Each time, the probability goes down by a factor of 10/16 and the result goes up $7.)
I put this into an Excel spreadsheet and took it out to 20 hits, at which point the probability is .00003101 and the win $128. The next column is the weighted amount, i.e. the amount of the net times the probability. I summed that column up and came up with -.34092.
So, the expected loss would be close to 34 cents.
Yeah, I have forgotten how to do integrals, too.
Cheers,
Alan Shank
Quote: goatcabinOn a per-roll basis
Right, but I wasn't thinking in terms of a per-roll basis, but more on the basis of just rolls that actually concern the bet at hand.
Quote: goatcabinIf you divide -.0909 by 6, you get -.01515, which is the usually-quoted house advantage (HA) on placing the 6 or 8.
So, would the way that I did it at the top of the post be the expected value of the bet I was talking about? Placing the 6 and 8 for $6 each?
Quote: goatcabinSo, the expected loss would be close to 34 cents.
And since I got 33 cents loss, I'm guessing I was pretty close to getting it right. :)
Quote: goatcabinYeah, I have forgotten how to do integrals, too.
I never would have guessed that I would have a reason to actually remember all that math stuff I learned in college. :)
