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Are Don't Pass superior to Pass bets at craps, and by how much?20 September 2010
Is Don't Pass indeed a better option? If so, why and by how much?
Only the "flat" portions of the alternatives need be examined to answer these questions. Although Odds taken or laid after a "point" is established may seem to differ for the two cases, they're genuinely opposite but equal and therefore a wash. At least, until you evaluate skewness, which hardly anybody does.
Flat bets are the wagers made prior to a shooter's come-out. They win even money, 1-to-1, during both the come-out and point phases of a roll. For purposes of comparison, consider the results of 1980 statistically-correct decisions.
On the come-out, Pass wins if the shooter throws a seven or 11 and loses on a two, three, or 12. Any other number on the come-out is established as the point and wins or loses later. The probability of a seven is 6/36 and that of an 11 is 2/36. The chance of Pass winning on the come-out is consequently 8/36. For the hypothetical 1980 instances, this is (8/36) x 1980 or 440 wins. The chance of Pass losing on the come-out is 1/36 for the two, plus 2/36 for the three, plus 1/36 for the 12 -- 4/36 in all. In the 1980 coups, this is (4/36) x 1980 or 220 losses.
Finding expected performance on points requires two steps. The first involves the prospect of a number becoming the point, the second the chances of its subsequently winning or losing.
The likelihood of having six or eight as a point is the total of ways either can form, out of the 36 dice combinations. That's 5/36 for each, 10/36 for either. The chance the number will then win is the five ways it can show out of the sum of its five plus the six ways a seven can appear -- that is, 5/11. Conversely, the chance the number will lose as a point is 6/11. Overall, the 1980 postulated events will have (10/36) x (5/11) x 1980 or 250 wins and (10/36) x (6/11) x 1980 or 300 losses on sixes or eights.
Analogous logic holds for the other numbers. The chance of either five or nine as the point is 8/36; that of then winning is 4/10 and losing 6/10. So ultimately, the idealized 1980 rolls will have (8/36) x (4/10) x 1980 or 176 wins and (8/36) x (6/10) x 1980 or 264 losses on five or nine. The chance of either four or 10 as the point is 6/36, that of then winning is 3/9 and losing 6/9. The 1980 model rolls will have (6/36) x (3/9) x 1980 or 110 wins and (6/36) x (6/9) x 1980 or 220 losses on four or 10.
All told, on Pass, the 1980 prototype cycles yield 976 wins and 1004 losses. The difference is the root of house advantage. Per dollar bet, the inherent net loss is $1 x (1004 - 976) or $28. That's $28/1980 or 1.41 cents per roll per dollar. Look familiar? It should, since the house's edge on this bet is 1.41 percent.
What about Don't Pass? During the come-out, Don't Pass bets win on two or three -- probability is 3/36, lose on seven or 11 -- probability is 8/36, and push on 12 -- probability is 1/36. For the 1980 rolls, that's 165 wins, 440 losses, and 55 pushes. On points, Don't Pass wins on sevens and loses on the numbers. The amounts are accordingly the flip side of those for Pass: 300 wins with 250 losses on sixes and eights, 264 wins with 176 losses on fives and nines, and 220 wins with 110 losses on fours and 10s.
Tallies for the come-out and the points add up to 949 wins, 976 losses, and 55 pushes for Don't Pass. On the 1980 rolls, loss per dollar bet due to the house edge will be $1 x (976 - 949) or $27. Divide by 1980 or (1980 - 55) to get 1.36 or 1.40 cents per roll per dollar including or ignoring those with pushes, respectively.
Strictly in terms of edge, advocates of Don't Pass are correct. Depending on how pushes are accounted for, edge is 0.05 or 0.01 percent less. But, the house still siphons away $1.39 or $1.40 for every $100. As the rhymer, Sumner A Ingmark, reflected:
For they believe it is a sin to end a day with no net win.
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