ChrisBanks
Verified Member
I haven't read through this entire thread yet, but my calculation of the price of a free scratch when playing 8 - 8 is the following:
1 multiplied by the estimated probability that the player will scratch in the game.............
If the player scratches approximately once every 5 games, then we multiply 1 by .20, and the value of the free scratch is .20.
This makes logical sense. Clearly if you are playing Scott Frost, he is unlikely to scratch into the pocket in a given game. Therefore the value of the free scratch is less than one.
To further support this argument, ask Scott if he would prefer to receive a one ball spot, or a free scratch. If he is indifferent between the two choices, then this supports that the free scratch is worth 1 ball.
My hypothesis is that Scott will prefer to receive the one ball spot every time. This supports the claim that the free scratch is worth less than one ball.
I am very confident that this is the proper way to assign a ball value to a free scratch. Assigning it a value of one in an 8-8 game does not take into consideration the probability of a scratch occuring.
A method used in philosophy is to subject theories to their limitation. Suppose we gave Scott Frost 50 free scratches. Is the spot now 50-8? Clearly this does not make intuitive sense.
Instead, in this scenario we take the estimated probability of Scott scratching 1 time in a game, 2 times in a game, 3 times in a game.... up to 50.
So for example, the probability of Scott scratching in a game:
1 time: 10%
2 times: 5%
3 times 1%
4 times: 0%
5 times: 0%...........
6 times: 0%
Sum all of the totals we estimate that the percentage of Scott scratching any number of times in a game is 16%.
Therefore allowing him 50 free scratches is worth .16 balls.
I don't believe there is a better method of quantifying the value of a free scratch.
1 multiplied by the estimated probability that the player will scratch in the game.............
If the player scratches approximately once every 5 games, then we multiply 1 by .20, and the value of the free scratch is .20.
This makes logical sense. Clearly if you are playing Scott Frost, he is unlikely to scratch into the pocket in a given game. Therefore the value of the free scratch is less than one.
To further support this argument, ask Scott if he would prefer to receive a one ball spot, or a free scratch. If he is indifferent between the two choices, then this supports that the free scratch is worth 1 ball.
My hypothesis is that Scott will prefer to receive the one ball spot every time. This supports the claim that the free scratch is worth less than one ball.
I am very confident that this is the proper way to assign a ball value to a free scratch. Assigning it a value of one in an 8-8 game does not take into consideration the probability of a scratch occuring.
A method used in philosophy is to subject theories to their limitation. Suppose we gave Scott Frost 50 free scratches. Is the spot now 50-8? Clearly this does not make intuitive sense.
Instead, in this scenario we take the estimated probability of Scott scratching 1 time in a game, 2 times in a game, 3 times in a game.... up to 50.
So for example, the probability of Scott scratching in a game:
1 time: 10%
2 times: 5%
3 times 1%
4 times: 0%
5 times: 0%...........
6 times: 0%
Sum all of the totals we estimate that the percentage of Scott scratching any number of times in a game is 16%.
Therefore allowing him 50 free scratches is worth .16 balls.
I don't believe there is a better method of quantifying the value of a free scratch.