What is the average prize per punch and optimal strategy for the Punch a Bunch game on The Price is Right?

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.

What is the average number of points hit by a craps shooter before he sevens out?

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.

Comparing two elements at a time, what is the fastest way to sort a list, minimizing the maximum number of comparisons?

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:

1. Divide the list in two. Keep dividing each subset in two, until every subset is size 1 or 2.
2. Sort each subset of 2 by putting the smaller member first.
3. 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.

If an NFL team was soundly beaten its last game, is it better to bet on them, or against them, their next game? Same question for a huge victory. I keep hearing that after a big loss a team will "have something to prove," while a team with a big win may be overconfident and lazy. Any truth to that?

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.

Pinnacle sportsbook just went to posting their odds in decimal format. How do I convert sports book odds from the decimal format to the American format?

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.