# Ask the Wizard #81

I do not understand why you should lay the odds on the don’t pass or don’t come bets. It seems that you have already dodged the 7 and ll bullet, so the bet is now in your favor. Why would you dilute a bet that is already heavily in your favor with a large (relative speaking)bet at true odds? It seems that you are working in the houses favor by reducing the house edge on the entire bet.

I understand that taking the odds on the pass side reduces the overall house edge, however I don’t understand how laying the odds can reduce the house edge on the don’t side. I’m very curious? By the way, I discussed this with several casino bosses and dealers yesterday and they all had opinions, but not reasons for these opinions. Thanks for your time.

Mike

Let’s say you have a \$10 don’t pass bet and the point is a 4. You have a 2/3 chance of winning the bet, so the expected value is (2/3)*\$10 + (1/3)*-\$10 =\$ 10/3 = \$3.33. Now consider adding a \$40 odds bet on top of it. Now you have a 2/3 chance of winning \$30 and a 1/3 chance of losing \$50. The expected value of both bets combined is (2/3)*\$30 + (1/3)*-\$50 = \$10/3 = \$3.33. So either way your expected gain is 3 dollars and 33 cents. With the don’t pass alone the player edge is \$3.33/\$10 = 33.33%. With the don’t pass and odds the player edge is \$3.33/\$50 = 6.67%. So, yes, the player edge as a percentage drops by making the odds bet. However that player edge is effective over more money. The way I think gamblers should view the house edge is as the price to pay for entertainment. If you want to pay as little as possible then taking or laying the odds is getting entertainment for free.

After performing my own infinite deck analysis for Blackjack with the same rules as yours (dealer stands all 17s, re-splitting allowed to 4 hands except Aces, which can only be split once, doubling after splitting, draw only one card to split Aces), I came across your site. In comparing expected values, I obtained the same numbers as you in all cases, except for pair splitting, which were slightly different. So I’m wondering how you went about your calculation of expected values for splitting?

Anonymous

It took me years to get the splitting pairs correct myself. Cindy of Gambling Tools was very helpful. Peter Griffin also addresses this topic in chapter 11 of the The Theory of Blackjack Let’s say I want to determine the expected value of splitting eights against a dealer 2. Resplitting up to four hands is allowed. Here is how I did it.

1. Take a 2 and two 8’s out of the shoe.
2. Determine the probability that the player will not get a third eight on either hand.
3. Go through all ranks, except 8, subtract that card from the deck, play out a hand with that card and an 8, determine the expected value, and multiply by 2. For each rank determine the probability of that rank, given that the probability of another 8 is zero. Take the dot product of the probability and expected value over each rank.
4. Multiply this dot product by the probability from step 2.
5. Determine the probability that the player will resplit to 3 hands.
6. Take another 8 out of the deck.
7. Repeat step 3 but multiply by 3 instead of 2.
8. Multiply dot product from step 7 by probability in step 5.
9. Determine the probability that the player will resplit to 4 hands.
10. Take two more 8’s out of the shoe.
11. Repeat step 3 but multiply by 4 instead of 2, and this time consider getting an 8 as a third card, corresponding to the situation where the player is forced to stop resplitting.
12. Multiply dot product from step 11 by probability in step 9.
13. Add values from steps 4, 8, and 12.

The hardest part of all this is step 3. I have a very ugly subroutine full of long formulas I determine using probability trees. It gets especially ugly when the dealer has a 10 or ace up.

Eight golfers went to a new course. The caddy master put 8 bags on four carts at random. The golfers put 8 marked golf balls in a hat. The balls were thrown in the air. The 2 closes balls to each other were partners. In every case the partners’ golf bags were already on the same cart. What is the probability of that the golf bags were paired up correctly before the throw?

Anonymous

The formulaic answer for the number of combinations would be combin(8,2)*combin(6,2)*combin(4,2)/fact(4) = 25*15*6/24 = 105. Another way to solve the number of combinations would be to take one golfer at random. There are 7 possible people to pair him with. Then pick another golfer at random from the six left. There are 5 possible people to pair him with. Then pick another golfer at random from the four left. There are 3 possible people to pair him with. So the number of combination is 7*5*3 = 105. Thus the answer is 1 in 105.