Ask the Wizard! (No. 242)

Good question. Straight up, the advantage per wager is 0.55×(10/11) - 0.45 = 0.05. As a parlay, the advantage is (0.55)²×((21/11)²-1)-(1-(0.55)²) = 10.25%. So, it would seem the parlay is the way to go to maximize advantage.

However, the variance is greater as a parlay. If you are following the Kelly Criterion, then you will have to protect your bankroll for the parlay with a smaller wager. In this example, the optimal Kelly bet straight up is 5.48% of bankroll if the two games overlap, 5.50% if you first game ends before you bet on the second game, and 3.88% for the parlay. Multiplying the wager times the advantage, we get 0.00275 straight up (based on a 5.50% advantage) and 0.00397 for the parlay. Thus, the parlay results in the greater profit.

I considered the general case for this kind of question, looking also at 3-team and 4-team parlays and money line wagers. Assuming a small advantage for all bets, as a rule of thumb, if the probability of each event winning is less than 33%, then you should bet straight up. If each probability is between 33% and 52%, then you should do a 2-team parlay. If each probability is between 52% and 64%, then you should do a 3-team parlay. If each probability is greater than 64%, then you should do a 4-team parlay. If you are doing straight up wagers, then you are about equally well off doing 2-team or 3-team parlays, again assuming you have an advantage to begin with.

I should stress that if you are a recreational gambler going against a house edge (what sports bettor will admit to that?), then betting straight up minimizes the house advantage.

The other video poker player is right. Here is the return table for 9/6 Jacks or Better, breaking down the four of a kinds, assuming optimal strategy.

Using the probabilities above, but applying them to the 10/7 Double Bonus pay table, we get the following return table.

You can see the return is 99.63% playing 9/6 strategy on a 10/7 machine. You gain 0.63% from the better pay table but lose 0.54% from errors, for a net gain of 0.09%.

Let's say your first flush is in spades. At 35 hands per hour, in five hours 175 hands could be played. You then have 175 hands to make a flush in hearts, diamonds, and clubs. I'm going to assume the player never folds a hand that has a possibility of attaining a flush in one of the suits he needs.

The probability of a flush of a specific suit, let's say hearts, using both hole cards is combin(13,2)×[combin(11,3)×combin(39,2) + combin(11,4)×39 + combin(11,5)]/(combin(52,2)×combin(50,5)) = 10576566/2809475760=0.003764605.

In the next 175 hands the probability of missing a heart flush would be (1-0.003764605)¹⁷⁵=0.51682599.

It would be incorrect to say the probability of failing to make the other three suits would be pr(no heart flush)+pr(no dimaond suit) + pr(no club flush), because you would double counting the probability of faling to make two of them. So you should add back in pr(no heart or diamond flush) + pr(no heart or club flush) +pr(no club or diamond flush). However, that would incorrectly over-subtract the probability of not making all three flushes. So you should add back in pr(no club, diamond, or heart flush).

The probability of going 175 hands and never get either of two specific suits is (1-2×0.003764605)¹⁷⁵=0.266442448.

The probability of going 175 hands and never getting any of the three suits left is (1-3×0.003764605)¹⁷⁵=0.137015266.

So the answer is 1-3×0.51682599 + 3×0.266442448 - 0.137015266 = 0.111834108.

I would like to thank dwheatley for his help with this problem. It is discussed on my bulletin board at Wizard of Vegas.