Found a way to cut the house edge in half

I was trying to hand calculate the house edge in craps yesterday. Why? To see if I can still do basic math (questionable), to kill some time (successful) and to build gambling-math self-confidence (failure). I came up with .7% house edge on passline bets with no odds and cannot find my mistake. It might just be calculation error, but I thought I'd check in here and see if my reasoning is flawed from the outset.

I approached it figuring I can figure the house edge by multiplying the amount won by the chance of winning, and subtracting the amount lost times the chance of losing. First question: is that correct? So in general if some game pays out 3 to 1 for rolling say a certain number on a die, I'd figure: (1/6)*3 - (5/6)*1 = -2/6 = -1/3 = house edge of 33%.

Doing the same with craps, figuring the pass line first with its wins and losses, then the box numbers, and totalling everything. I tried to keep everything weighted correctly, meaning that for a four, for example, my calculation looked like this. (3/36)*[ (3/9)*($5) - (6/9)($5)]. My reasoning being that you have 3 chances of initially rolling a 4. Then you have three chances to roll a 4 and win, six chances to roll a 7 and lose, and all other numbers are meaningless.

It seems telling that my answer was off by one-half, though maybe that is just a coincidence. I read the WOO section on calculating the odds, though only briefly, and did not follow the summation from 1-infinity of all the possibilities step. So I may have major theory problems here, or I may have just not handled my fractions right.

Is my thinking relatively sound? And is there some easy-to-spot theory reason why I'd be off by one-half? (I didn't take the short cut of doing only half of each complementary pair of box numbers or anything like that, that I know of, but maybe something else?)

Thanks, A

Quote: AndyGB

So in general if some game pays out 3 to 1 for rolling say a certain number on a die, I'd figure: (1/6)*3 - (5/6)*1 = -2/6 = -1/3 = house edge of 33%.

What you are computing is the expected value of the wager (EV) 3/6 + -5/6 = -2/6
we then divide that by the $Bet, in your example that is 1
HE = EV/$Bet

Quote: AndyGB

Doing the same with craps,
figuring the pass line first with its wins and losses, then the box numbers, and totalling everything.
I tried to keep everything weighted correctly, meaning that for a four, for example,
my calculation looked like this. (3/36)*[ (3/9)*($5) - (6/9)($5)].

Your method is computing the EV.
Since you were using $5 in your calculations , your value , the expected value should be 35/495 = 0.070707...

HE = EV/$Bet

0.070707 / 5= you got it

So close but yet so far!! I totally got to .07070. (And figured that looked like about half of 1.4%, oops.) And I did not realize at all that very basic difference between what I wanted to do and what I was actually doing. Probably because I've always used $1 bets as examples when trying to calculate stuff in the past, so it has been a wash (Divide by 1 at the end...).

Awesome, thank you very much for your help. Soon I will be smarter. And nice to know my fraction arithmetic still works.

A

I think you're forgetting to pick up your original bet.

There are a few methods.
One is to look at all the outcomes count how many often you lose (in this case) one Unit and then how often you win one Unit (and in other games how many times you StandOff (with Don't this is needed).
So 1 in 36 (roll 2) you lose, similarly for 3, 7, 11, 12. Remember 1/36*-1, 2/36*-1, 6/36*+1, 2/36*+1, 1/36*-1.
So (3 in 36)*(3 in 9) you win on a point 4. Remember 3/36*3/8*+1
and (3 in 36)*(6 in 9) you lose on a point of 4. Remember 3/36*6/9*-1
and continue....then total.

Another method (say roulette). After all 37 possible spins (UK single 0) you'd win 1-time at "35 to 1" and lose 36-times "-1", so lose 1 unit every 37-spins; or you could think 1-time I get back "36 for 1" and nothing else pays out so get back 36 for every 37 bet.

Another example.
Card games: (Uses "to 1" method) There are N possible hands you can be dealt. N1 of these (say Straight Flush) pay X1 to 1, N2 (say Quads) pay X2 etc. N99 (say No Pair) lose 1 Unit. Just total {N1*X1, N2*X2, ....N99*-1} and compare the overall loss to the number of possible hands N.
Fruit Machines/Lotteries: (Uses "for 1" method) These always take your bet and then sometimes pay you out, so you total all the possible payouts*combinations and see what you get back for covering every combination (3 bells - 2 chances - pays 1000; etc.).

Quote: AndyGB

my calculation looked like this. (3/36)*[ (3/9)*($5) - (6/9)($5)].

Dude, you calculated HE in dollars ($.07) for a $5 bet. You didn't calculate HE in percent. So... 7 cents per $5 is what percent?

(6/36 + 2/36 - 4/36 - (6/36)*(6/9) + (6/36)*(3/9) - (8/36)*(6/10) + (8/36)*(4/10) - (10/36)*(6/11) + (10/36)*(5/11) = -0.0141414 = -1.414141%

Translates to:

7-winner + 11-winner - crap-loser -four/ten_loser +four/ten_winner -5/9_loser +5/9_winner -6/8_loser + 6/8_winner = HOUSE EDGE (or actually, player's edge [which is negative]).

Quote: AndyGB

I approached it figuring I can figure the house edge by multiplying the amount won by the chance of winning, and subtracting the amount lost times the chance of losing. First question: is that correct? So in general if some game pays out 3 to 1 for rolling say a certain number on a die, I'd figure: (1/6)*3 - (5/6)*1 = -2/6 = -1/3 = house edge of 33%.

Sounds right to me.

Quote: AndyGB

Doing the same with craps, figuring the pass line first with its wins and losses, then the box numbers, and totalling everything. I tried to keep everything weighted correctly, meaning that for a four, for example, my calculation looked like this. (3/36)*[ (3/9)*($5) - (6/9)($5)]. My reasoning being that you have 3 chances of initially rolling a 4. Then you have three chances to roll a 4 and win, six chances to roll a 7 and lose, and all other numbers are meaningless.

It seems telling that my answer was off by one-half, though maybe that is just a coincidence.

Can you "show your work"? Here's what I get, for each of the come-out roll numbers:
2: 1/36 x (-1) = 1/36 x (-1)
3: 2/36 x (-1) = 1/36 x (-2)
4: 3/36 x ( (3/9 x 1) + (6/9 x (-1) ) = 3/36 x (-3/9) = 1/36 x (-1)
5: 4/36 x ( (4/10 x 1) + (6/10 x (-1) ) = 4/36 x (-2/10) = 1/36 x (-4/5) = 1/36 x (-44/55)
6: 5/36 x ( (5/11 x 1) + (6/11 x (-1) ) = 5/36 x (-1/11) = 1/36 x (-5/11) = 1/36 x (-25/55)
7: 6/36 x 1 = 1/36 x 6
8: 5/36 x ( (5/11 x 1) + (6/11 x (-1) ) = 1/36 x (-25/55) (same as for 6)
9: 4/36 x ( (4/10 x 1) + (6/10 x (-1) ) = 1/36 x (-44/55) (same as for 5)
10: 3/36 x ( (3/9 x 1) + (6/9 x (-1) ) = 1/36 x (-1) (same as for 4)
11: 2/36 x 1 = 1/36 x 2
12: 1/36 x (-1)
The sum = 1/36 x ( (-1) + (-2) + (-1) + (-44/55) + (-25/55) + 6 + (-25/55) + (-44/55) + (-1) + 2 + (-1) )
= 1/36 x (-28/55) = -28 / (36/55) = 0.014141414, which is correct.

Your method seems sound; apparently, it is just being implemented wrong for some reason.

Quote: ThatDonGuy

Your method seems sound; apparently, it is just being implemented wrong for some reason.

Read the whole thread, Don.
He didn't calculate percent, he calculated dollars on a $5 bet.

Quote: indignant99

Read the whole thread, Don.

But...but...that requires effort, and work, and waiting 16 minutes for my burger and hot dog when I want them in 9.