Ask The Wizard #201
Thanks for the great website! My father and I are having an argument about hedging bets and could really use your help! The particular situation involves a bet on the Super Bowl. Prior to the start of the season (my father can’t remember the year) my uncle placed a bet that New England would win the Super Bowl. The bet paid 60 to 1. Just before the Super Bowl (in which New England was playing) my uncle hedged his bet (father can’t remember the details) giving up the potential $6,000 pay day but guaranteeing a $3,000 one instead. I’m convinced that this was a sucker bet but my father won’t listen to me. I’m arguing that by hedging in this spot he was giving up expected value and the smart bettor never does this. My father is arguing that it’s fine to give up the expected value because of the money involved and the fact that this bet doesn’t come up often, exactly the same as insuring your house. Of course, I argue that home ownership is inherently different than sports betting in that one can be avoided while the other really can’t. What are your thoughts? Please help us settle this!
The Seventh of my Ten Commandments of Gambling is, "Thou shalt not hedge thy bets." However, in my remarks I add, "Exceptions can be made for insuring life-changing amounts of money." So if the surrender value of $3,000 is a life-changing amount of money to him, and if the probability of winning were not much more than 50%, then I wouldn’t begrudge his decision. However, unless this was in 2002, the probability of New England winning was much higher than 50%. The other two years they played in the Super Bowl, 2004 and 2005, they were 7-point favorites. I would estimate the probability of winning either year was about 71%. A fair surrender value would have been 0.71 × $6100 (including his original bet back) = $4,331. The house edge on the offer, which was equivalent to an even money bet on the other team, was 29%-71% = 42%. So, if I’m right about the year, he made a very bad decision. He could have gotten much better odds on the open market by betting the other team on the money line. Whoever offered only $3,000 was either very ignorant of the game or took unfair advantage. Interestingly, New England won all three of their recent Super Bowls by three points.
Recently my room-mate got into my online casino account and lost a great deal of MY money. I would like to "charge back" all the transactions he made through the site. Besides being put on a global list are there any other negatives to disputing charges of this sort?
I think being put on the chargeback database will be the only drawback. So that will pretty much end your online gambling. However, I don’t think it is fair to the casino to charge back. It was not their fault that your room-mate used your credit card and lost your money. The right thing would be for your room-mate to pay you back for what he lost. I feel strongly about this, having been stiffed many times myself. It is not a coincidence that "Thou shalt honor thy gambling debts" is the first of my Ten Commandments of Gambling. If your room-mate refuses, and you go ahead with the chargeback, be honest with any investigation. It will be easy to see the charges came from the same I.P. address, and you may be asked about it. Give him a chance to pay first, and if he refuses don’t protect him over this.
Suppose that five cards are dealt from a 52-card deck and the first one is a king. What is the probability of at least one more king? I saw an Ace problem you were doing like this one but I couldn't really follow it. Thank you for any help.
The way I prefer to answer probability questions is to use the combinatorial function. Doing it that way, there are combin(48,4) = 194,580 ways to choose four cards that are not kings out of the 48 non-kings in the deck. There are combin(51,4)=249,900 ways to choose any four cards out of the remaining 51 cards in the deck. So, the probability of getting no kings in the next four cards are 194,580/249,900 = 77.86%. Thus, the probability of getting at least one kind is 100% - 77.86% = 22.14%.
Several people have said that the combinatorial function is probably over the heads of the type of people asking these kinds of simple probability questions. I don’t disagree with that, but a major reason for this site is to try to teach my readers something about math. The combinatorial function is extremely useful in probability, and saves a lot of time. However, the question at hand can be easily answered without it.
The probability the second card is not a king is 48/51. That is because there are 48 non-kings left in the deck, and 51 total cards left. If the second card is not a king, then the probability the third card is also not a king is 47/50 (47 non-kings divided by 50 cards left). Following this to the end, the probability that none of the other four cards are kings is (48/51)×(47/50)×(46/49)×(45/48) = 77.86%. The probability that this is not the case, in other words at least one king, is 100% - 77.86% = 22.14%.
Do you think any individual player can beat the odds over the long run in California games that every seated player has the same opportunity to bank, once every round, assuming players have to pay $1 to the house every hand?
Yes and no. Those games usually have a banker advantage, so if you took it at every opportunity, you would have a long-term advantage. However, there are agreements among the casinos and banking organizations not to let ordinary players do this excessively, as if it were a business, as opposed to recreational gambling.
First, choose 5 cards from a single 52-card deck. Second, add their blackjack values (T,J,Q,K = 10, A = 1). What are the odds the sum is even/odd? I would think that with the over-abundance of even cards, the sum would be much more likely to be even.
Surprise, an odd total is more likely at 50.03%, despite 30 of the 52 cards being even. The following table shows the probability of each even/odd split.
Odd/Even Question
| Evens | Odds | Combinations | Probability | Sum |
| 0 | 5 | 15504 | 0.005965 | Odd |
| 1 | 4 | 155040 | 0.059655 | Even |
| 2 | 3 | 565440 | 0.217564 | Odd |
| 3 | 2 | 942400 | 0.362607 | Even |
| 4 | 1 | 719200 | 0.276726 | Odd |
| 5 | 0 | 201376 | 0.077483 | Even |
| Total | 2598960 | 1 |
The bonus package for Mohegan Sun includes two ten dollar bets in coupon form. These are not match play. A ten dollar bet on an even money proposition e.g. the one at the Big Six wheel will return ten dollars. The house keeps coupons wagered win or lose. The player does not have to add any money of his own. The only games that can be played are the Big Six wheel or Sic Bo . Where are the best places to use these coupons. I have bet high and low in sic bo only losing if a triple shows. I have also bet the one and two on the wheel (same spin).
Usually these free bet coupons are limited to even money bets, so this is an interesting case. My advice is to use the free bet on a long-shot, to minimize the effect of the rule that you lose the free bet, even if you win. The biggest long-shot in Big Six is the joker/logo, with a probability of winning of 1/54. I’m not sure whether the Mohegan Sun pays 40 or 45 on the joker, but assuming 45 the value of the free bet is (1/54)×45 = 83.33% of face value. In Sic Bo the biggest long-shots are on the six triples. I’m also not sure what they pay for a specific triple, but I would guess 180. In that case the value of any one of the six triple bets would be (1/216)×180 = 83.33% of expected value. So, we have a tie in terms of expected value. In that case I would go for the bet with the greater probability of winning, the joker/logo in Big Six, but that is up to you.
In San Diego casinos Super Fun 21 has a $1 side bet that on the first hand of single deck a diamond suited blackjack pays $300. What are the correct odds for getting this with 6 players and you are sitting at 1st base?
There is one way to get the ace and four ways to get the 10-point card, for a total of 1*4=4 winning combinations. There are combin(52,2)=1,326 ways to choose 2 cards out of 52. So the probability of winning is 4/1326 = 0.30%. Fair odds would be 330.5 to one. The expected return is 0.0030*300 + 0.9970*-1 = -0.0920. So the house edge is 9.2%.
The reason they limit this bet to the first hand after the shuffle is a card counter could take advantage it otherwise. Without tracking the cards, you can assume the house edge is 9.2% all the time.
Thank you for all the great information on your webpage. I am currently on active duty in the Air Force and will be giving a seminar on responsible gambling.
My history professor at NMSU told our class that the only way to win at Blackjack was to bet a little at a time and walk away will small profits..$25. This logic doesn’t seem to work in my books...I know it’s false. My question is...let’s say I have $1,000,000 to gamble with in my lifetime. Do I have "better odds" by betting the whole million at one hand of Blackjack vs. small hands or are the odds always the same regardless? You have a great website and keep up the good work. Thanks a lot for your help!
You’re welcome. Your history professor is wrong. This "small win" strategy is nothing new. Usually it does result in a small win, but the occasional big losses more them wipe them out. To answer your question, it depends on what you mean by “better odds.” If you mean which way results in greatest average balance, it doesn’t make any difference. The expected loss is the same with one bet of $1,000,000 or one million bets of $1, assuming basic strategy, and you have reserve money to double or split. However, if you mean which has the greater probability of a net win, your chances are much better off with a single bet. If you make one million bets of $1 the expected loss is $2,850, with a standard deviation of $1,142. The probability of showing a profit is 0.6%. Betting one hand of $1,000,000, the probability of a win is 42.4%, with a push at 8.5%, and a net loss at 49.1%.