Craps on a $100 bankroll

22 June 2019

By Jerry Stickman

From Brian,

Assuming I have $100 bankroll at a $10 minimum table. The point roll is not a six or an eight. Would it not be better to cover six and eight with $10 bet including the pass for a total of $30 than a pass bet of $10 plus $20 on odds on more difficult numbers to hit like 4/10 or even 5/9?

Hi Brian,

This is a question with which many craps players wrestle. First, a slight clarification. Place bets on the 6 or 8 are paid at a rate of 7 to 6. Therefore, in order to get the full amount possible, the player has to bet a multiple of $6 to get a pay of a multiple of $7. On a $10 table, the minimum place bet on a 6 or 8 needs to be $12. A winning $10 bet on a placed 6 or 8 would be only $11 ($7 for $6 and even money on the remaining $4). A winning $12 bet would pay $14. The player gets an extra $3 for the additional $2 bet.

In craps, the dice do not have a memory. With each and every roll each of the possible outcomes has the same odds as the previous roll. Because of this we can use math to calculate the house edge.

First let’s look at the place bets. The first row below shows the results for placing a 6 and 8 for $10 each. The second row shows the results for $12 each.

• Amount bet: $20, True odds: 6/5, Paid odds: 7/6, True odds win: $24, Actual win: $22, Difference: -$2, House edge: 4.55%

• Amount bet: $24, True odds: 6/5, Paid odds: 7/6, True odds win: $28.80. Actual win: $28, Difference: -$0.80, House edge: 1.52%

Now let’s look at the $10 pass line or come bets with $20 odds.

• Number: 6 or 8, Odds: 6/5, Base win: $10 + Odds win: $24 = $34, True odds win: $36, Difference -$2, House edge: 3.03%

• Number: 5 or 9, Odds: 3/2, Base win: $10 + Odds win: $30 = $40, True odds win: $45, Difference -$5, House edge: 6.67%

• Number: 4 or 10, Odds: 2/1, Base win: $10 + Odds win: $40 = $50, True odds win: $60, Difference: -$10, House edge: 11.11%

It looks like your theory about using the extra $20 for place bets on the 6 and 8 when the point is not a 6 or 8 is correct.

Or is it?

While it is true that the house edge for placing the 6 and 8 (4.55% or 1.52% depending on whether $10 or $12 is placed) is less than the house edge on a $10 pass line bet with $20 in odds when the point is a 4, 5, 9, or 10 (6.67 percent or 11.11%), the above is comparing apples with oranges.

Calculating the house edge after the point is established disregards the 2-to-1 player advantage (six ways to throw a 7 and two ways to throw an 11 for a win vs. one way to throw a 2 or 12 and two ways to throw a 3 for a loss) that the base portion of the bet has on come out rolls. The odds shown above are actually for what is called a “put” bet. The put bet is made after a point is established.

Taking both the come out and point cycle into consideration, the house edge for a pass line or come bet with no odds is 1.41 percent. This is even less than placing the 6 or 8. The house edge for pass line or come bets with double odds is a mere 0.61% – far below any of the house edges shown above.

I know the explanation may not seem logical, but look at it this way. What happens with the $20 after a point is established? If you use it to place a 6 or 8, there is house edge on it – either 4.55% or 1.52%. If, instead you choose to put the $20 behind the line or come bet as odds, the $20 has no house edge – zero. Certainly, zero is better than a negative, don’t you think?

In craps, the math rules.

All supposedly winning craps betting systems have some sort of flaw in the logic. In the long run, the math will win out, meaning that you lose. The casino wins.

May all your wins be swift and large and all your losses slow and small.

Jerry “Stickman”

Jerry “Stickman” is an expert in craps, blackjack, video poker and advantage slot machine play. He authored the video poker section of "Everything Casino Poker: Get the Edge at Video Poker, Texas Hold'em, Omaha Hi-Lo, and Pai Gow Poker!" You can contact Jerry “Stickman” at stickmanjerry@aol.com.