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Ask The Wizard #186

When the house awards me free play on a slot machine, ex. $100, how much does it actually cost them?

David from W. Palm Beach

It depends. If the slot play may be used in any machine in the casino, including video poker, then it can be worth 99 cents on the dollar or more, depending on the video poker offerings. For example, the MGM Mirage casinos award $1 in free slot play for every one point earned. It can be used in any machine in any casino connected to the MGM Mirage player card. Most MGM Mirage properties offer 9/6 Jacks or better, so the value of $1 in free slot play is worth 99.54 cents, with correct strategy.

However, sometimes free slot play must be done on particular promotional machines that don’t accept money. The value of this kind of free play is hard to estimate closely, but generally very little. For example, the Las Vegas World used to sell “$1000” vacation packages for $400. Of the alleged $1000 value $600 was in promotional slot machine play. In his book “Million Dollar Video Poker,” Bob Dancer writes that he did this deal numerous times over, and estimates the value of the free slot play to be about ten cents on the dollar.

We’ve been given a challenge at work -- just for fun, and none of us can work it out. A farmer has 5 trailers full of sheep. Four of the trailers contain sheep weighing 39kg and the 5th trailer contains sheep weighing 40kg. All of the sheep are identical. He goes to the market. He wants to find out which of the trailers contains the sheep weighing 40kg, and he can only use the large weighing scales once!!! How does he do it? Please help, it is driving us all mad at my work place -- it’s a vet’s!!

Becca

The answer is at the end of the column.
I am a regular blackjack player at a 50-table casino. There are some dealers I have identified that, when I am playing 6th or 7th position, expose their facedown card perhaps 25% of the time. I am wondering what the proper play would be without them being moved right away. I am afraid to take full advantage of seeing the card. If I know the dealer has 20 should I still hit with a 19 or would that draw too much heat? Or should I wait on it and take advantage in more marginal situations?

Earl from Canada

That is what my blackjack appendix 16 is for. To answer your question, I wouldn’t make any plays that would set off a red flag, like hitting a hard 19. My advice is stick to more marginal, believable, errors.
I was recently in Las Vegas and played a new game for me -- World Poker Tour: All in Poker. I did not bet on the bonuses and used your advice on how to play the game. I bet $15 on the blind, so on average my overall bets were approximately $100. At one casino I was rated as a $15 player and at another a $45 player. I don’t understand why two casinos, both in the same corporate family, would give me two different ratings for the same bets in the same game, and I don’t understand why I wasn’t rated as a 90-100 player.

Mark from Merrick

The first casino was correctly basing your bet size on the ante only. The second was counting the raise bets. If the second casino does include raises in the average bet then it should be using a lower house edge for purposes of rating. In my opinion many casinos do not comp players accurately. Each casino has its own policies, regardless of which corporate family it is in. What is important in your case is which casino gives you back the most for your play. There are lots of factors that go into that decision besides the average bet size.
I have a friend who starts complaining when his first card is a six, without waiting to see what his second card and the dealer’s up card are. I think he should wait because he could get a two, three, four, five, etc. (i.e., a decent second card) OR the dealer could show a two through six (a good card for the table). What do you think? How much worse are his odds of winning with a first card of six without knowing this second card or the dealer’s up card? Or is my friend just a whiner? Thank you for your time.

Scott from Long Beach

My blackjack appendix 14 shows that if your first card is a six your expected value is already about −21%. For example, if he bet $100, a fair price to sell the hand and bet would be about $79. Maybe you can take advantage of his complaining by offering to buy his hand for less than the fair 79 cents on the dollar. I’d suggest 75 cents on the dollar, to give you an edge, without taking too much advantage.
Great site! A must see for Vegas newbies and those thinking of striking it rich through gambling. Is there an optimal ratio to wager between the ante and the bonus bet in the Texas Hold’em Bonus table game? Is it simply 10% bonus to ante since you calculated 90.4% of bonus bets are losers?

Thomas from Austin

Thanks for the kind words. I get asked this question a lot regarding every game where there are two or more bets to choose from. You should bet 100% on the better bet. In the case of Texas Hold ’em bonus the element of risk on the ante is 0.53% and on the bonus bet it is 8.54%, assuming Vegas rules both ways. For comparing one bet to another I believe the element of risk should be used. So in this case, the ratio of ante bet to bonus bet should be infinity, because the bonus bet should be zero. Same thing with Three Card Poker, which is usually the venue this question is asked about. In that game you should bet 100% on the ante and 0% on the Pairplus.
I read somewhere that half a counter’s edge is in taking insurance. In the UK insurance is only offered on a blackjack. Hence in the UK a counter’s edge must be about half of what it is under rules allowing insurance on any ace. Have I got this right?

Grame from London, UK

According to the famed “Illustrious 18” table in Don Schlesinger’s “Blackjack Attack” insurance is the single most valuable hand to the counter, in terms of value gained by making count dependent strategy changes. The value of taking insurance, when the true count is +3 or more, adds 0.117% to the player’s overall advantage, based on the hi-low count, spreading 1 to 8. However this does not come close to half the total advantage from strategy changes. Looking only at the top 18 count dependent plays (the Illustrious 18), plus the top 4 count dependent surrenders (the Fab Four) the total value to the counter is 0.469%. So correct insurance accounts for only 25% of the value of strategy changes. A good counter will gain about another 1% by betting more in good counts. So taking away the insurance option reduces the total value of counting by only about 8%. In the UK it would be slightly less because you can still insure with a blackjack. The advantage figures are rough and depend on lots of things. The value of all surrender deviations is about the same of insurance deviations.
I’m a long-time subscriber of your newsletter and still loving your web site. I came across a casino web site that offers roulette in which the wheel doesn’t contain any zeros. It just has number 1-36, and all the standard roulette rules apply. Do you see any way to take advantage of this? I know you don’t like betting systems but in this case there is no house edge. There must be a money management system that could work profitably with these table limits. Any advice is appreciated.

Mark from Gatineau, Quebec

Thanks for the kind words. I think I’ve answered this one before, but no, even with zero house edge there is still no betting system that can win, over the long run.
What is the probability of seeing a "broken board" in Texas Hold’Em? That is, five cards on the board where no pair exists, no flush is possible and no straight is possible.

Tim from Arcata

The number of combinations of five different ranks on the board is combin (13,5)*45 = 1287 × 1024 = 1,317,888.

The probability that these five ranks will represent three suits, two of two, and one of one, is combin(4,2)*2*combin(5,2)*combin(3,2)=360. Combin(4,2) is the number of ways to choose two suits out of four for the suits represented twice. 2 for the two ways to choose the suit represented once. Combin(5,2) for the number of ways to choose two ranks out of five for the first suit of two cards. 45 for the number of ways to choose two ranks out of the three left for the other suit of two.

The probability these five ranks will represent four suits, one of two, and three of one, is 4*combin(5,2)*3*2=240. 4 is the number of ways to choose one suit out of four for the suits represented twice. Combin(5,2) is the number of ways to choose two ranks out of five for that suit of two cards. 3 is the number of ways to choose one rank out of the three left for the first suit of one. 2 is the number of ways to choose one rank out of two for the second suit of one.

There are 45=1024 ways to arrange four suits on five different ranks.

So the probability that no more than two of one suit will be present is (360+240)/1024 = 600/1024 = 58.59%.

There are combin(13,5)=1287 ways to arrange 5 ranks out of 13. The number of these combinations in which no three ranks are within a span of 5 is 79. There is no easy formula for this one. I had to cycle through every combination. So the probability the ranks will be sufficiently spaced apart is 79/1287 = 6.14%.

So, the probability of a broken board is (1317888/2596960)*(600/1024)*(79/1287) = 1.825211%.

I have been challenged on my number of broken straights. Here is a list of all 79 possible.

2378Q
2378K
2379Q
2379K
237TQ
237TK
237JQ
237JK
237QK
2389K
238TK
238JK
238QK
2479Q
2479K
247TQ
247TK
247JQ
247JK
247QK
 2489K
248TK
248JK
248QK
257TQ
257TK
257JQ
257JK
257QK
258TK
258JK
258QK
267JQ
267JK
267JA
267QK
267QA
267KA
268JK
268JA
 268QK
268QA
268KA
269JA
269QA
269KA
278QK
278QA
278KA
279QA
279KA
289KA
3489K
348TK
348JK
348QK
358TK
358JK
358QK
368JK
 368JA
368QK
368QA
368KA
369JA
369QA
369KA
378QK
378QA
378KA
379QA
379KA
389KA
469JA
469QA
469KA
479QA
479KA
489KA
Answer to sheep question

anonymous

Take one sheep from trailer 1, two from trailer 2, three from trailer 3, four from trailer 4, and zero from trailer 5. If all the sheep weighed 39 kg then the total weight would be 39 * 10 = 390 kg. However 0 to 4 sheep are one kg heavier. If the total weight is 391, then there is one heavy sheep on the scale; thus it must have come from trailer 1. Likewise, if the total weight is 392, then there are two heavy sheep on the scale, which must have come from trailer 2. In the same manner a weight of 393 means the heavy sheep are in trailer 3, a weight of 394 means the heavy sheep are in trailer 4, and a weight of 390 means the heavy sheep are in trailer 5.