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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/(1p). So, the expected number of points per shooter is 0.406061/(10.406061) = 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*(x1)
If x<2: 100/(x1)
In the example above, the lines in the American format are:
New York Giants: 100*(2.7501) = +175
Dallas Cowboys: 100/(1.5131) = 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.