House edge on placing the 6 and 8 in craps

I recently received the following.

Jerry:

Let me say up front I love the GTC site, etc. Just wish I had known about this stuff instead of counting cards in the late seventies.

I have never seen a publication with the correct data re: the following 2 points.

By the way I expect full RFB and a hooker in return for this info … LOL

1) Placing the 6 and 8.

Consider a $6 place bet on the 6. There are only 11 happenings out of 36 in this universe that produce a decision: 5 wins x $7 and 6 losses of $6 for a net loss of 1 unit. Having bet 11 X $6 the loss is 1/66 or -1.51515%.

However, consider a $6 place bet on BOTH the 6 and 8 simultaneously. Now we have a 16 possible game. If we look at the P & L as each of the possible outcomes occurs, we see that we get 10 X $7 = +$70 and 6 X -$12 = -$72 for a loss of $2. However our total money at risk was $12 X 16 = $192. A loss of $2 in $192 is a losing percent of only 1.0416666667 (or 1 and 1/24 %).

2) < the second point will be addressed in a future article >

Tony

Dear Tony,

You have an interesting way of looking at these bets. On the surface it appears quite logical. However, you must do a separate calculation for each bet -- placing the 6 and placing the 8. You cannot lump them together, since they are two distinct bets. Each bet would have about a -1.52% edge.

Since you will probably not take the statement above as proof, let's look at this in another way.

You are using the correct formula for calculating the edge for each bet: Sum of all profit (or loss) divided by the sum of all bets made. Consider the following table.

In order to determine the edge of any bet, we need to determine the profit from each roll. To do that we need to multiply the bet by the roll frequency to come up with the "Pro-rated Profit." Then we must determine the "Pro-rated Bets/Roll" by multiplying the new bets after roll by the roll frequency. You calculate the edge by dividing the "Pro-rated Profit" by the "Pro-rated Bets/Roll."

In the above table I used larger bets to get more accuracy in the final calculations. Using the numbers above the edge would be -.56 divided by 36.67 or -1.527 percent -- not -1.04 percent.

Tony, you still may not take the above as proof, possibly thinking I am playing with numbers, but will you take the results of a simulation?

Using the Smart Craps simulator, I set up a simulation that assumed a random thrower betting a $30 6 and a $30 8. The simulation ran for 10,000,000 rolls of the dice (actually 10,000,003 rolls in order to finish the last hand). The result: an expected edge of -1.521498 percent.

Three strikes and you are out. All the other books are right. The house edge on placing the 6 and 8 is 1.52 percent, not 1.04 percent.

Sorry, Tony, no RFB and no hooker on this one.