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Ask the Wizard #186
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.
The answer is at the end of the column.
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.
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.
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.
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.
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.
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.
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 |
| 2489K |
| 268QK |
| 368JA |
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.