raydog wrote: ↑Sun Jan 09, 2022 12:32 pm
Irish, this is a straightforward math problem (not simple, but straightforward). You just need to define your probabilities very precisely.
I have worked on this sort of problem, and have a program which will calculate all the possible outcomes - of this and future games - and sum the probabilities of every outcome to determine the odds of either team winning the game to 10 pts.
As inputs I need to have the probability of each possible point outcome of each game (there are 6 of them: +4, +2 or +1 pt, for each of two teams). I already have a good estimate of this distribution for the future, random hands, and have simplified them to 4 situations:
1) both teams need at least 3 points to win [.067,.141,.459,.187,.121,.025]
2) dealing partnership is at 8 or 9 pts, and so won't bid alone [0, .223, .448, .187, .117, .025]
3) non-dealing partnership is at 8 or 9 pts, and so won't bid alone [.067, .139, .459, .184, .151, 0]
4) both teams have 8+ points and so won't bid alone [0, .147, .182, .449, .222, 0]
What do these 6 numbers mean? The first 3 are the odds of the dealing team winning +4, +2 or +1 pts, respectively. The last 3 are the odds of the other team scoring +1, +2 or +4 pts, respectively. I calculated these probabilities based on several million random hands played by my program, so they may actually vary depending on the aggressiveness of the bidding, how well the hands are played, if other score considerations are taken into account in addition to "don't bid alone with 8+ pts, etc. But it's a good start, I think.
The final input is the point probability distribution for the initial hand, which I tried to create from the information you provided. I came up with the following:
[0, .15, .3, 0, .55, 0] if dealer bids; EV = -0.5
This means that dealer has a 55% chance of getting euchred, a 30% chance of scoring 1 pt., and a 15% chance of scoring 2 pts., for an overall EV of -0.5. I basically fixed the euchre rate at 55%, then played with odds of 1 or 2 pts to get an EV of -0.5. If you propose a different point probability distribution, let me know, but it will likely change the EV.
For the case where dealer passes, it was a bit more difficult to decipher the odds you wanted, but I came up with the following (trying to get an EV of -0.7):
[.02, .02, .26, .44, .2, .06]
So the dealer's team scores points 30% of the time (2% chance of 4 pts, 2% chance of 2 pts, 26% chance of 1pt), and the opposing team scores points 70% of the time (6% chance of scoring 4 pts - 70% * 8%, 20% chance of scoring 2 pts, 44% chance of scoring 1pt). Again, if you want to propose a different point probability distribution, just let me know.
Using all the data above, here are the odds of winning the game to 10 pts:
bid: 40.2%
pass: 37.4%
It's just math, IF the input probabilities are correct. Let me know if you want to tweak those probabilities.