Last Updated: December 13, 2004

Do you have any advice for betting on the flip of a coin?

Anonymous

Yes! My advice is bet on the face up side at the start of the flip. According to Science News Online the probability that a coin will land on the same side it started on is 51%. The article says the reason is because a flipped coin does not spin perfectly around its axis and sometimes appears to be flipping when it actually isn’t. The hypothesis only applies if coin is caught in the palm of the hand, so that bouncing is not an issue. The article also says that a spinning penny will land on tails 80% of the time, due to the heavier head side gravitating towards falling down first. However I’m skeptical of this. I tried this 20 times and got 11 heads and 9 tails. The probability of getting 9 or fewer tails in 20 spins with a probability of success of 80% is 1 in 1775.

Dear Wizard - Mathematically, why doesn’t the cancellation system work? (The system goes by a lot of other names too. Just to be clear, I mean the system where you start with a series of number and bet the total of the outside numbers, canceling them out when you win, etc.) It seems that all you have to win is 1/3 plus two of you bets to come out ahead. In roulette you have about a 45% chance of winning. So you should win over the long haul, but you don’t. Why not?

Nathan

As with most betting systems the cancellation system usually does result in a session win, at the cost of occasional huge losses. When the cancellation system does lose the results can be your worst nightmare. During those times when you just seem to lose almost all the time the bet sizes start adding up geometrically, which can quickly deplete a bankroll if the cards don’t run your way.

In the game of Yahtzee if only the Yahtzee itself is left on the card what is the probability of making it?

Anonymous

The following table shows the probability of success on the last roll according to the number of additional dice you need to make a Yahtzee.

### Last Roll Yahtzee Probabilities

 Needed Probabilityof Success 0 1 1 0.166667 2 0.027778 3 0.00463 4 0.000772

The next table shows the probabilities of improvement. The left column shows how many dice you need before any given roll and the top column shows how many you need after the roll. The body shows the probability of the given degree of improvement.

### Probabilities of Improvement

 Need Before Roll 0 1 2 3 4 Total 0 1 0 0 0 0 1 1 0.166667 0.833333 0 0 0 1 2 0.027778 0.277778 0.694444 0 0 1 3 0.00463 0.069444 0.37037 0.555556 0 1 4 0.000772 0.01929 0.192901 0.694444 0.092593 1

The next table shows the probability on the initial roll of needing 0 to 4 more dice to make a Yahtzee.

### First Roll Yahtzee Probabilities

 Needed Probability 0 0.000772 1 0.019290 2 0.192901 3 0.694444 4 0.092593

The next table shows the probability of improvement and then eventual success according to the number needed after the first roll. For example, if the player needs 3 more dice to make a Yahtzee the probability of improving to needing 2 more after the second roll and making the Yahtzee on the third roll is 0.010288066.

### Probabilities of Yahtzee after first roll according to number needed before and after second roll

 Need Before Roll 0 1 2 3 4 Total 0 1 0 0 0 0 1 1 0.166667 0.138889 0 0 0 0.305556 2 0.027778 0.046296 0.01929 0 0 0.093364 3 0.00463 0.011574 0.010288 0.002572 0 0.029064 4 0.000772 0.003215 0.005358 0.003215 7.1e-05 0.012631

To get the final answer take the dot product of the number needed after the first roll two tables up and the probability of eventual success in the final column one table up. This is 0.092593*0.012631+ 0.694444*0.029064 + 0.192901*0.093364 + 0.019290*0.305556 + 0.000772*1 = 4.6028643%. To confirm this I did a 100,000,000 game simulation and the simulated probability was 4.60562%.

I understand random number generators, virtual reel stops, and physical reel stops. What I don’t understand, and can’t find any information on anywhere, is how the game determines what the pay-out will be for the selected symbols. For example, in an IGT Red, White, and Blue game number SS4335 the top jackpot which is the Red Seven, White Seven, Blue Seven correspond to virtual reel positions 044, 043, 044 and physical reel stops 08, 08, 08 respectively. Each of the three reels have seven symbols, the Red Seven, White Seven, Blue Seven, Red Bars, White Bars, Blue Bars, and Blanks. That equals 343 symbol combinations. I know that the SS chip does not contain a table with all possible combinations and pay-outs. It has to be indexed somehow. How does the machine know that reel stops 08, 08, 08 correspond to the Red, White, and Blue Sevens and how does the machine know how much to pay out? I hope you can answer this question. If you cannot could you recommend any articles or books that can.

Anonymous

There is a "look up" table that maps the various random numbers to stops on the reels. However I wasn’t sure how they go from there to actually determing what the player won. So I asked a former slot machine mathematician, who asked not to be identified, about this one. Here is what he said, "Your first idea is correct. The position on each reel strip is independently selected via the RNG. The code then examines the symbols along each bet-upon payline to determine winning outcomes. Scatter awards could also be determined this way. ALL of the major video-based slot manufacturers do it this way. You could view the algorithm as a big series if-then-else’s but actual implementation might be a bit more eligant." I hope that helps.

p.s. After this column appeared I received another e-mail regarding this question. It is rather long so I offer this link.

First of all thanks for the great site. Is there no point in card counting if it is a single deck that is reshuffled after every hand?

Anonymous

You’re welcome, thanks for the compliment. There is still some point, especially with a full table. However under typical single deck rules (dealer hits soft 17, no double after split) I don’t think it is enough to overcome the 0.19% house edge.

What is the probability of getting four aces in four-card stud?

Anonymous

1/combin(52,4) = 1 in 270725.

Dear Wizard, Can you please explain to me how the house advantage on craps place bets are calculated. For instance, how does the nine to five payout on a four/ten place bet work out to a 6.67% house advantage when the true odds are two to one? No matter how I do it, I can’t come up with that 6.67% figure. This is driving me nuts. I would greatly appreciate an explanation.

Amanda

I prefer to calculate the house edge as 1-(pr(win)*payout - pr(lose)). In this case it would be 1-((1/3)*1.8 - (2/3)) = 6.67%. However if you know the fair payout and the actual payout a convenient formula for the house edge is (f-a)/(f+1), where f=fair payout and a=actual payout. In this case (2-1.8)/(2+1) = 0.2/3 = 6.67%.

If you were playing \$50 per play, what would you personally chose between these two games of video poker (assuming both games have the same pay schedules and you bet the maximum of 5 coins for each hand): single-play at \$10 or ten-play at \$1 per hand? Thanks for your time and consideration.

Anonymous

Mathematically they of course have the same expected return. However I would play the 10-play because the volatility is less and I think it is more fun.

Thanks for the great site. You recently stated that the average craps shooter lasts approx. 8.5 throws. I normally bet the pass with full odds followed by come bets with full odds. Does it make more sense to quit making come bets after, say four throws given the long term probability that the thrower will 7 out in only three to four more throws?

Fred

You’re welcome, thank you. The dice do not have a memory so at four throws you do not get any closer to sevening out. You could roll 1000 non-sevens and still be no closer or further away from a seven than you were the first throw. There is no optimal number of come bets, just make as many as you find the most fun.

What are the odds of playing 15 spins on European roulette, covering at eight numbers and not getting any of them?

Anonymous

The probability of losing any one spin is 1-(8/37) = 78.38%. So the probability of losing 15 spins is .783815 = 2.59%.

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