If a wagering proposition was such that you could expect to win 191 times and lose 14 times, what odds would you have to offer the other player to provide yourself with a 2% profit margin. I have figured out the answer to this problem but I think I went about it the hard way. I am curious to see if there is an easier way.
Stephen Adams,
The true odds would be to offer (191+14)/14 = 14.64... for 1.
Then, if the player played 205 times (at one unit per try, so a total of 205 units bet) he would receive (14.64.../win)*(14 wins) = 205 units. This would be a 0% EV bet.
If instead you want a 2% advantage, then you would offer odds of 14.64.../1.02 = 14.36... for 1. Now, if the player played 205 times (at one unit per try, so a total of 205 units bet) he would receive (14.36.../win)*(14 wins) = 200.98... units. This would be an -2% EV bet (from the player's perspective, of course).
Note that, in BJ, the odds are given as x TO y (e.g., BJ pays 3 TO 2), rather than z FOR y. The relationship is that x = z - 1, so the 0% EV offer would be 14.64... for 1 = 13.64... to 1, while the -2% EV offer would be 14.36... for 1 = 13.36... to 1.
Hope this helps!
Dog Hand
P_win = probability you will win = 191/(191+14)
P_lose = probability you will lose = 1-P_win = 14/(191+14)
e_w = return for win
e_l = return for lose
profit = (2%)*e_l (since e_l is what you will invest)
P_w e_w - P_l e_l = profit = (2%)e_l
P_w e_w/e_l - P_l = (2%)
e_w/e_l = ((2%)+P_l)/P_w
e_w/e_l = ((0.02)(191+14)+14)/(191) = 0.0948
or the other way
e_l/e_w = 10.5525
if profit = (2%)*e_w (from another perspective)
e_l/e_w = ((0.02)(191+14)+191)/(14) = 13.9357 (close to previous post)
See my response over at ap.com.
Don
Thanks for the clearification Don.
I had the player with a 2% profit at odds 13.9357 to 1.
The house with a 2% profit needs odds 13.35 to 1.
The equation form is still correct. I forgot to change the sign for -2% profit.
The proper equation is
P_w = 14/205 P_l = 191/205
P_w e_w - P_l e_l = profit = (-2%)e_l
e_w/e_l = ((-2%) + P_l)/P_w = 13.35
