Frank Scoblete Presents: When less is more, by Skinny

Frank Scoblete presents an article by Skinny.

A reader on the private pages of the GTC website (www.goldentouchcraps.com) read my recent article, "Probability or Odds: Same or Different?" and posted a question about it. I thought the question is one which is common among many craps players. Hence, I chose to use both for a follow-up article to my prior one. I hope you find it helpful if you have had similar ideas.

The reader wrote:

I was glad to see Skinny's article posted here. I'll admit I'm having trouble grasping the "probability versus odds" concept. I'm wondering if the likelihood of a bet winning ("probability") and the varying "payoffs" once a bet wins ("odds") are two separate and unrelated factors of the game that are sometimes confusingly mixed with each other in our discussions or recommended bets, etc. I'm not questioning the dollar amount of the bets or their payoffs here. I'm simply questioning the likelihood that a given bet will win.

Can somebody clear this up a bit for me (and perhaps for others as well)? Anyway, here are my confused thoughts on this at the moment:

As there are 5 ways to roll a 6 or 8, the "probability" of these numbers being rolled is second only to the 7, of which there are six ways to roll. As there are 4 ways to roll a 5 or 9, the "probability" of these numbers being rolled are less than only the 6, 8 (5 ways to roll) and 7 (6 ways to roll). In turn and obviously, there are three ways to roll a 4 or 10.

Following the logic of this and discussing the point (or box) numbers, it would seem that with five ways to roll the 6 or 8, placing either or both of these numbers, the player would have the highest "probability" of winning the bet. With four ways to roll the 5 or 9, it would seem that this bet has the second highest "probability" of winning, inferior only to the 6 or 8.

When making a come bet, we don't know what number the bet will end up on until it is resolved. Given this and strictly in terms of the "probability" of winning the resolved come bet, how can a come bet ever be superior to a place bet if the come bet resolves to a number other than 6 or 8? Wouldn't it make sense to place the 6 and/or 8, since the five ways to roll these numbers have the highest "probability" of winning, then to spread out by placing the 5 and/or 9, etc?

I've read that buying the 4 or 9 (vig is paid on wins only) is a superior bet because of the lower "house advantage" one can attain. This may be true in terms of the payout and getting the bet closer to "true odds." But in terms of the "probability" of the bet winning, I'm not yet grasping how this bet could ever be superior to placing or having a come bet resolve to the 6, 8, 5 or 9. For example, when buying the 4 or 10 for $39 and vig paid on wins only, the house advantage is reduced to 0.85%. When placing the 5 or 9, the house advantage is 4%. Even though the 4 and 10 have the lower house edge, there are still only three ways to roll a 4 or 10 compared to four ways to roll a 5 or 9. To me, it seems more "probable" that the 5 or 9 will win even though they have a higher house advantage. At this point I'm thinking the 5 or 9 seem to be the smarter bet.

Again, I'm not questioning the bet payoffs but just the "probability" of a given bet winning. Hope I haven't expressed this too clumsily.

Answers…? - Class…? Class…? Anyone…? Bueller…? Hehe!

Thanks!

Dear Reader,

Your question goes to the heart of why I believe many gamblers make poor decisions with their wagers at the craps table. The math behind the frequency of winning/losing bets and their corresponding payouts can indeed be very confusing. So much so, it becomes difficult for many to figure out what is a good bet. In trying to come up with an answer it is easy to misapply basic principles, which can lead to erroneous conclusions.

On the surface you seem to be making a valid point. You are asking if one makes bets that have the highest probability of winning, won't you win more money by making those bets because you will get paid more often. It seems logical if you make bets that win more frequently than other bets you should win more money with them. If other bets have a smaller probability of winning, it stands to reason you will win those bets less frequently. How can someone who wins fewer bets make more money than someone who wins a higher percentage of the bets made?

The answer to this question goes right back to your first paragraph. There you said you wondered if the likelihood of a bet winning ("probability") and the varying "payoffs" once a bet wins ("odds") are two separate and unrelated factors of the game that are sometimes confusingly mixed with each other.

The chance of your winning a bet and the payoff on that win are most definitely not two separate and unrelated factors of the game. Tying the two together is critical and essential to determining whether or not you should make that bet.

Let me give you an example to illustrate that point. I am going to set up two games where you bet on the outcome of the roll of two dice. Each game has a player and a house. You can decide whether you want to be the player or the house. In each game, the player throws the dice and wins his bet on any number he throws other than a 12. The house only wins the bet if a 12 is thrown. We know there are 36 ways to throw a number with two dice and only one way to throw a 12. Thus the player will win the bet 35 out of 36 rolls on average and the house will only win the bet one out of 36 rolls.

Clearly the player is going to win this bet far more often than the house. But in order to determine if this is a good bet or not for the player, we MUST take into consideration the payout for the bet. In Game A, the house pays the player $1 each time he wins and collects $1 each time a 12 is thrown. In Game B, the house pays the player $1 each time he wins and collects $100 when a 12 is thrown. Even though the player will win 35 times out of 36 on average in the long run in both games, I think it is clear you would want to be the player in Game A and the house in Game B.

So why isn't it best to always be the player? The player has a 35/36 probability of winning in both games. This is a very high likelihood of winning the bet. If you have the highest probability of winning the bet as the player, would not you always be better off being the player? I think by now you can see the answer is obvious. It is possible to win more money by making a bet that wins less frequently. In Game B it is better to be the house even though the house will only win 1 bet out of every 36. Of course the answer is apparent in this artificial game that I have made up. In Game A the player can expect to win $35 and lose $1 on average every 36 rolls for a net win of $34. In Game B the player can expect to win $35 and lose $100 on average every 36 rolls for a net loss of $65. Even though the player wins more times far more often than the house in Game B it is better to be the house. It is better to make the bet that wins less frequently, because the payout is better in relation to the number of times the bet will be won than lost.

This is why I said you can not separate the probability of winning a bet from its corresponding payout to determine which bets are best to make. I hope it is clear from my artificial example that you want to make bets that provide the best payout with respect to the frequency of winning the bet irrespective of the regularity with which the bet can be won. And that is exactly what the House Advantage (HA) does for you. It tells you how much you can expect to lose in relation to the initial amount wagered on bets that are resolved. The HA essentially looks at the payout in relation to the probability of winning the bet and does the calculation for you. If you make bets that have the lowest HA, you can expect to lose less money on those bets in the long run.

When the HA is lower on the 4 and 10 than it is on the 6 and 8 or 5 and 9, the 4 and 10 become better bets even though you will win them less often. It is the same as being the house in Game B. You win less frequently but you lose less money because the HA is lower.

When the casino collects the vig on wins only on the 4 and 10, they typically collect $1 vig on a wager from $20 to $25 and $2 vig on wagers from $26 to $50. Paying $1 vig on $25 or $2 vig on $50 on a win only for the 4 and 10 has a HA of 1.33%. This is less than the HA on the 6 and 8, which is 1.52%. Even though there are only three ways to roll a 4 or 10 and five ways to roll a 6 or 8, the payout on the 4 and 10 is better in relation to the frequency with which you will win the bet compared to the 6 and 8. Now if you can actually buy the 4 or 10 for $39 and only pay $1 vig on a win only, that is a fantastic bet. It has a very low HA of 0.85%. I don't know of any casinos that do that, but if you do then go for it!

The same reasoning holds true for making pass/come bets instead of place bets on the 6 and 8. The pass/come bet has a HA of 1.41% and that can be reduced even lower by taking odds. 1x odds reduces the HA to 0.85%, 2x odds has a 0.61% HA, 3x-4x-5x odds carries a 0.37% HA and 5x odds has a 0.33% HA. You are correct when you said you don't know what number your come bet will end up on. The 1.41% HA on the base bet takes into consideration all the possible ways your base bet can be resolved. It looks at all the ways you can win or lose that bet on the come out roll if it does not go to a box number. It also looks at the number of times you can expect it to go to each and every one of the box numbers and how many times you can expect to win or lose each of those as well. The end result of ALL those possibilities is you can expect to lose 1.41% of your initial wager on the pass/come bet once they are resolved. Since the odds bet is paid at true odds, it does not matter which box number your bet ends up on. You will be paid at true odds, which is the same as the frequency with which you can expect to win/lose that bet in the long term. You will still lose the 1.41% of your initial wager on the base bet but you will lose less in relation to your total wager by taking odds. That is why the HA gets lowered when you take odds. You can still expect to lose the same amount of money but it will be less in relation to the total you bet when you take the odds because you are now increasing the total amount you are wagering.

Since the HA on pass/come bets with or without odds are all lower than the 1.52% HA on the 6 and 8, pass/come bets are better bets to make than the place bet on the 6 and 8. It does not matter that some times your pass/come bet goes on the 4 and 10 or 5 and 9, which win less often than the 6 and 8. That is all taken into consideration in the calculation of the HA when comparing the payout to the win frequency as I previously stated. In the long run the lower HA is the better bet just like the house is the better bet in Game B.

I am not saying place bets are a bad bet. Place bets on the 6 and 8 or 4 and 10 (when the vig is paid on a win only) are still good bets to make because they have a reasonably low HA. But you can get a lower HA with pass/come bets.