## Wizard Recommends

# Ask the Wizard #262

Ibeatyouraces

For those not familiar with the rules, they are explained at the Price Is Right web site. Please take a moment to go there if you’re not familiar with the game, because I’m going to assume you know the rules. There are several YouTube videos of the game as well. Here is an old one, which shows a second chance, but the maximum prize at the time was $10,000 only. It is now $25,000.

First, let’s calculate the expected value of a prize that is not paired with a second chance. The following table shows that average is $1371.74.

### Punch a Bunch Prize Distribution with no Second Chance

Prize | Number | Probability | Expected Win |

25000 | 1 | 0.021739 | 543.478261 |

10000 | 1 | 0.021739 | 217.391304 |

5000 | 3 | 0.065217 | 326.086957 |

1000 | 5 | 0.108696 | 108.695652 |

500 | 9 | 0.195652 | 97.826087 |

250 | 9 | 0.195652 | 48.913043 |

100 | 9 | 0.195652 | 19.565217 |

50 | 9 | 0.195652 | 9.782609 |

Total | 46 | 1.000000 | 1371.739130 |

Second, calculate the average prize that does have a second chance. The following table shows that average is $225.

### Punch a Bunch Prize Distribution with Second Chance

Prize | Number | Probability | Expected Win |

500 | 1 | 0.250000 | 125.000000 |

250 | 1 | 0.250000 | 62.500000 |

100 | 1 | 0.250000 | 25.000000 |

50 | 1 | 0.250000 | 12.500000 |

Total | 4 | 1.000000 | 225.000000 |

Third, create an expected value table based on the number of second chances the player finds. This can be found using simple math. For example, the probability of 2 second chances is (4/50)×(3/49)×(46/48). The expected win given s second chances is $1371.74 + s×$225. The following table shows the probability and average win for 0 to 4 second chances.

### Punch a Bunch Prize Return Table

Second Chances | Probability | Average Win | Expected Win |

4 | 0.000004 | 2271.739130 | 0.009864 |

3 | 0.000200 | 2046.739130 | 0.408815 |

2 | 0.004694 | 1821.739130 | 8.551020 |

1 | 0.075102 | 1596.739130 | 119.918367 |

0 | 0.920000 | 1371.739130 | 1262.000000 |

Total | 1.000000 | 1390.888067 |

So the average win per punch (including additional money from second chances) is $1390.89.

The following table shows my strategy of the minimum win to accept, according to the number of punches remaining. Note the player can get to $1,400 with prizes of $1,000 + $250, + $100 + $50 via three second chances.

### Punch a Bunch Strategy

Punches Remaining | Minimum to Stand |

3 | $5,000 |

2 | $5,000 |

1 | $1,400 |

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

JimmyMac

Given that a point is established, the probability that the shooter makes the point is pr(point is 4 or 10) ×pr(making 4 or 10) + pr(point is 5 or 9) × pr(making 5 or 9) + pr(point is 6 or 8) ×pr(making 6 or 8) = (6/24) × (3/9) + (8/24) × (4/10) + (10/24) × (5/11) = 201/495 = 0.406061.

If the probability of an event is p, then the expected number of times it will happen before failure is p/(1-p). So, the expected number of points per shooter is 0.406061/(1-0.406061) = 0.683673.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Anon E. Mouse

There are several ways that are about equally as good. However, the one I find the easiest to understand is called the merge sort. Here is how it works:

- Divide the list in two. Keep dividing each subset in two, until every subset is size 1 or 2.
- Sort each subset of 2 by putting the smaller member first.
- Merge pairs of subsets together. Keep repeating until there is just one sorted list.

The way to merge two lists is to compare the first member of each list, and put the smaller one in a new list. Then repeat, and put the smaller one after the smaller member from the previous comparison. Keep repeating until the two groups have been merged into one sorted group. If one of the original two lists is empty, then you can append the other list to the end of the merged list.

The following table shows the maximum number of comparisons necessary according to the number of elements in the list.

### Merge Sort

Elements | Maximum Comparisons |

1 | 0 |

2 | 1 |

4 | 5 |

8 | 17 |

16 | 49 |

32 | 129 |

64 | 321 |

128 | 769 |

256 | 1,793 |

512 | 4,097 |

1,024 | 9,217 |

2,048 | 20,481 |

4,096 | 45,057 |

8,192 | 98,305 |

16,384 | 212,993 |

32,768 | 458,753 |

65,536 | 983,041 |

131,072 | 2,097,153 |

262,144 | 4,456,449 |

524,288 | 9,437,185 |

1,048,576 | 19,922,945 |

2,097,152 | 41,943,041 |

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Anon E. Mouse

I show after a 21 point loss or more the team will cover the spread 51.66% of the time. However, that is well within the margin of error. The following table shows the outcomes against the spread of the last game, according to the win or loss by the same team the previous game. The results never stray far from 50% and are always within a standard deviation of it. Basically, I find no statistical correlation between the win/loss against the spread and how many points the team won or lost by the previous game.

### Win, Loss, or Tie Against Spread According to Margin of Victory or Defeat the Previous Game

Previous Game Outcome | Win Against Spread | Loss Against Spread | Tie Against Spread | Win Ratio | Standard Deviation |

Win by 21 or more | 233 | 247 | 17 | 48.54% | 2.28% |

Win by 14 to 20 | 235 | 219 | 11 | 51.76% | 2.35% |

Win by 10 to 13 | 188 | 180 | 8 | 51.09% | 2.61% |

Win by 7 to 9 | 198 | 181 | 12 | 52.24% | 2.57% |

Win by 4 to 6 | 164 | 170 | 12 | 49.10% | 2.74% |

Win by 3 | 202 | 212 | 14 | 48.79% | 2.46% |

Loss by 2 to win by 2 | 184 | 188 | 14 | 49.46% | 2.59% |

Loss by 3 | 209 | 207 | 12 | 50.24% | 2.45% |

Loss by 4 to 6 | 174 | 163 | 9 | 51.63% | 2.72% |

Loss by 7 to 9 | 187 | 195 | 9 | 48.95% | 2.56% |

Loss by 10 to 13 | 173 | 189 | 14 | 47.79% | 2.63% |

Loss by 14 to 20 | 220 | 232 | 15 | 48.67% | 2.35% |

Loss by 21 or more | 249 | 233 | 15 | 51.66% | 2.28% |

Table based on every NFL game from week 1 of the 2000 season to week 4 of the 2010 season.

Anon E. Mouse

Let’s look at the October 25, 2010 Monday Night Football game as an example. The European odds are posted as:

New York Giants 2.750

Dallas Cowboys 1.513

Both figures represent what you will get back for one unit wagered if you win, including your original wager. When the decimal odds are greater than or equal to 2, then the translation is easy: just subtract one, and then multiply by 100. If the odds are less than 2, then (1) subtract 1, (2) take the inverse, and (3) multiply by -100.

For those of you who prefer a formula, if the decimal odds pay x, here is the calculation for the equivalent American odds:

If x>=2: 100*(x-1)

If x<2: -100/(x-1)

In the example above, the lines in the American format are:

New York Giants: 100*(2.750-1) = +175

Dallas Cowboys: -100/(1.513-1) = -195

You can also automatically convert all the lines by selecting "American Odds" in the pulldown menu in the upper left of Pinnacle’s web site, above the logo.