# Ask The Wizard #147

Your system for your NFL picks seems to be weighted too heavily towards the underdog. Only 8 out of 49 games have you picking the favorite. In one you pick a team on an even line. Is this a flaw in your system or are people more apt to pick favorites and your system is trying to take advantage of their tendency? I didn’t do an analysis but it tends to be from my observation that few bets are won in the spread range without the team actually winning the game.

Ian from Boulder

Historically speaking underdogs are a better bet. Here are the results every game played from the start of the 1983 season through week 10 of the 2005 season.

Favorite wins against spread: 2554 games
Underdog wins against spread: 2724 games
Game ends exactly on spread: 150 games

So on resolved bets the underdog have won 51.61% of the time. It is also well known that square bettors prefer to bet favorites, creating value on the underdogs.

In Hold Em poker, does not burning cards affect the odds? Does burning cards affect the odds and game at all?

Jimmy from Canberra, Australia

No. The probabilities do not change in any card gave by burning cards.

I understand you have already answered the probability of getting the "dead man’s hand", a two pair of aces and eights, is 0.0609% on April 3, 2005, but I believe the dead man’s hand is "two black Aces, two black eights and the Queen of clubs" what is the probability of drawing that exact hand from a single standard deck?

Sett from Gold Coast

There is only one way to get that exact hand. So the probability would be 1 in combin(52,5) or 1 in 2,598,960.

This year our "Football Bet Taker" upped the juice on over/under bets from 10% to 20%, and eliminated the 10% juice on parlays. So this year instead of making 2 separate o/u bets, I have been parlaying the 2 bets at a payoff of 2.5 to 1. Is this a good strategy to use?

Rob from St. Louis, Missouri

The expected return on the straight bets would be (0.5*1 + 0.5*(-1.2))/1.2 = -8.33%. The expected return on the parlay would be 0.25*2.5 + 0.75*-1 = -12.5%. However if I were to only bet two games and want to win or go bust trying then I would go with the parlay. More importantly I would boycott this bookie out of principle, because I’ve never heard of having to lay -120 on straight bets before.

Dear wiz, I wish I had your brain. In any event, my struggle with stats continues. I am trying to figure out a formula to figure out the probability of getting a flush on the flop, by the turn and by the river (texas holdem) crossed by whether my hole cards are suited or not. I tried C(50,2) / C(47,5) but that didn’t work out for a suited pair by the river...I should have paid more attention in school ! Thanks ! Your biggest fan

Eric from Toronto

Thanks for the kind words but I'm not that smart. A couple years ago I took the Mensa entrance exam, and didn't make the requisite top 2%. I'm still upset that they refused to tell me how well I did do. On January 13 Jeopardy tryouts are coming to Vegas, for which I have an appointment, and am sure I'll blow that too. Anyway, to answer your question here you go:

With suited hole cards:

Flush after flop: combin(11,3)/combin(50,3) = 165/19600 = 0.842%.
Flush after turn: (combin(11,2)*39/combin(50,3))*(9/47) = 2.096%.
Flush after river: (combin(11,2)*combin(39,2)/combin(50,4))*(9/46) = 3.462%.

With unsuited hole cards:

Flush after flop: 0%
Flush after turn: 2*combin(12,4)/combin(50,4) = 0.430%.
Flush after river: (2*combin(12,3)*39/combin(50,4))*(9/46) = 1.458%.

Here are the commulative probabilities.

With suited hole cards:

Flush by flop: 0.842%.
Flush by turn: 2.937%.
Flush by river: 6.400%.

With unsuited hole cards:

Flush by flop: 0.000%
Flush by turn: 0.430%.
Flush by river: 1.888%.

My boyfriend and I have been dating on and off for over a year. He has cheated on me before but he was really drunk and didn’t remember it. I am starting to feel he is cheating again, and in turn I have definitly become a snoop. I check his call log on the internet and today I noticed that he called this girl, Anne, five minutes after he called me and was wasted and yelled at me. I also heard on his voicemail the girl leaving a message at 2am saying to call him. He has been very short with me lately, and when I called him today, he said "did you need something?". I just don’t know what to do. If I break up with him,I will totally regret it because he is always able to make me feel like he WAS trying very hard, and it’s my fault that I wanted way too much. I forgot one thing. My birthday was over one month ago, and I have yet to get a present. Please help

Sally from New York City

Birthdays are such relationship killers. If he isn’t blatantly cheating he is at least hedging his bets by heating up things with those on the waiting list. However I can’t say that I blame him because you seem paranoid and possessive. My advice is to lower the temperature on this. Do as he does and heat up some friendships with other guys as a back up plan. Either he will get jealous and make a stronger effort or it will hasten the eventual ending, which are both better than continuing to go sideways.

I have been playing for "fun" at online casinos and am considering playing for real. However, some casinos state that when playing for fun the Windows RNG is used and when playing for real the UNIX RNG is used. Will the difference in RNG’s affect my win probability? Thank you!

Vicki from Mechanicsburg

That shouldn’t change the odds at all. The Windows RNG is probably not very good, but good enough for free play. However when real money is on the line a smart operation would use a proven good RNG on their own end.

The weekly salaries of teachers in one state are normally distributed with a mean of \$490 and a standard deviation of \$45. What is the probability that a randomly selected teacher earns more than \$525 a week? I can’t remember how to calculate a probability from just the mean and SD without the population.

Sue from Queen Creek

That would be \$35 above average, or 7/9 standard deviations. The probability of being more than 7/9 standard deviations above expectations would be 1-Z(7/9) = 1- 0.78165 = 0.21835.

What do you think of the Bible Code?

Vince from Manila

I would put those behind it on the same level as those selling get rich quick gambling schemes. The mathematically ignorant taking advantage of the mathematically ignorant.

I’ve been dating my ex-boyfriend’s good friend for about 5 months now. My ex boyfriend and I have completely moved on, however we are now friends. He recently came up to me and told me that my new boyfriend was cheating on me with at least three different girls, has unprotected sex with them and has an STD. He also said that the only reason he is still with me is because he "doesn’t want to break my heart." My question to you is, is he really cheating? He is with me, at work or school about 75% of the time and the other 25% I’m not sure where he’s at and I don’t question it. I never had reason to believe that he was cheating on me until now...what do you think the deal is? Thanks.

Amy from Jacksonville

I would ask your ex what his evidence is. Accusations should always be backed up with evidence. Maybe there is suddenly some bad blood between the two of them and this is a way of your ex seeking retribution. It seems unlikely he would make up this story out of thin air so there may be some basis in truth to it. However he is professing more knowledge than I think it totally believable. So probe him for more details.

Two people are playing rock paper scissors. It is presumed the game doesn’t involve strategy. If you are playing ’best of 3’, and player A wins the first round, what are the odds that player B will win the game?

Andrew from Pewaukee,WI

Player B would need to win the next two (not counting ties) so the probability is (1/2)*(1/2) = 1/4.

What are the odds a deck could be shuffled back into starting order, with either a totally random shuffling method or with a perfect rifiling shuffle, and how many times would it take?

Andrew from Pewaukee,WI

The probability of a random shuffle resulting in starting order is 1 in 52!, or 1 in 8.06582*1067. If you did a perfect shuffle, in which last card was the first to come down, thus remaining last, it would only take 8 shuffles to be back to the starting order. If the 26th card was the first two come down then it would take 72 shuffles to back to the starting order.