Ask the Wizard #242
Suppose there are two football games that I feel have a player advantage. Let’s say each has a 55% chance of winning, and I have to lay 110. Which is more profitable, to bet the games straight up or as a single parlay?
Rob from Las Vegas
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.
In London, there is a royal match side bet in baccarat. It pays if the Banker or Player get a king and queen in the first two cards. Do you have any odds on it?
Assuming eight decks, the house edge is 4.5%. For more information, visit my page on baccarat side bets.
I was at the Four Queens casino, which offers both 10/7 double bonus and 9/6 jacks or better. I only knew 9/6 strategy, so I played that. Another video poker player later rebuked me, saying I would have been better off playing the 9/6 strategy on the 10/7 machine. I disagree. There is a $5 bet riding on it. Who is right?
James from Las Vegas
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.
9/6 Jacks or Better Return Table with Optimal 9/6 Strategy
|Three of a kind||3||1484003070324||0.074449||0.223346|
Using the probabilities above, but applying them to the 10/7 Double Bonus pay table, we get the following return table.
10/7 Double Bonus Return Table with 9/6 Strategy
|Three of a kind||3||1484003070324||0.074449||0.223346|
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%.
There is a promotion being advertised by a Las Vegas card room: Make a flush in all four suits and you get $400. You have to use both of your hole cards, and there is a five-hour time limit. Assuming 35 hands per hour, and that the clock starts with the first flush, what is the probability of achieving the other three flushes within five hours? Thanks.
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.
In casinos that offer 5X odds, if you bet a multiple of $15 they tend to allow you to bet $75 on points of 4 or 10, $100 on points of 5 or 9, and $125 on points of 6 or 8. That amounts to 5X on the 4 and 10, 6.67X on the 5 and 9, and 8.33X on the 6 and 8. I was wondering what the house edge is on this? Presumably it is slightly better than the 5X odds house edge you list on your page which assumes a straight 5X for all the numbers.
Zach from New York
Compared to 5X odds, that lowers the overall house edge from 0.326% to 0.269%. I don’t see why they would allow the extra odds bet on the 6 and 8, because a 5X bet at $75 would pay an even $90. However, as long as they did, I would take advantage of the extra odds, as long you are comfortable with the additional risk.