# Ask the Wizard #125

"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.

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.

"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 | Probability of 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 | 0.000071 | 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%.

"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.

"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.

"Anonymous" .

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

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%.

"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.

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.

"Anonymous" .

The probability of losing any one spin is 1-(8/37) = 78.38%. So the probability of losing 15 spins is .7838^{15} = 2.59%.