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Old 01-24-2017, 12:57 AM   #1
Sparky79
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Interesting probability game played

The game has 8 rounds played with many players. The goal of the game is to score as many points as possible.

Each round, two dices are tossed.

If the result is a pair (example: 22, 33, etc.) the remaining players still in the round get a score of 0 for this round (they lose all their accumulated points for this round).
If the result of the dice toss is not a pair, the players add up the dice values and add it to their score for the round. (Example: a 2 is rolled and a 4 is rolled. The player would get 6 points)
If the pair is exactly 1,1 (a one and another one) the players lose all their points from ALL rounds.

Players can roll the dice as much as they want per round (unless they get a pair at which point they get a score of 0 for this round - or lose ALL their points if the pair is 1,1, and move to the next round).
Players can also decide to stop rolling the dice, and move to the next round.

What is the optimal strategy for this game?




Our Solution (might be wrong):

1st round: Sit down (and stop rolling) if you have 35 or more.

2nd till 8th round: Sit down if you have more points than this EV equation for this round:
EV = 35 + (-1/36)(whatever you have) <35

Example: You scored 105 points after rounds 1, 2 and 3. At round 3, you should stop rolling after 31 points. --> 35 - 1/36(136) = 31.22 --> so if you have 31 points you should stop rolling. Thus, you would have 136 points in total.

If ever you have more than 210 points --> immediately sit out on all remaining rounds. You fear a 1,1 roll.
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Old 01-24-2017, 01:26 AM   #2
whosnext
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Re: Interesting probability game played

I just skimmed this.

Are you saying that there is no single winner to the game? You are just trying to get as many points as possible? (Like $100 payout per point.)

Each person has total control of their own rolls and gets points only from their own rolls, presumable. Meaning that the other players' rolls and decisions do not matter.

So this is essentially a one-player game (versus the house, say). Right?

Edit: that doesn't seem right now that I re-read the rules regarding doubles.
Please be more explicit with the rules.
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Old 01-24-2017, 01:40 AM   #3
Sparky79
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Re: Interesting probability game played

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Originally Posted by whosnext View Post
I just skimmed this.

Are you saying that there is no single winner to the game? You are just trying to get as many points as possible? (Like $100 payout per point.)

Each person has total control of their own rolls and gets points only from their own rolls, presumable. Meaning that the other players' rolls and decisions do not matter.

So this is essentially a one-player game (versus the house, say). Right?

Edit: that doesn't seem right now that I re-read the rules regarding doubles.
Please be more explicit with the rules.
This is a single winner game with the winner winning a wager after the game has been played (the winner is the one with the most points). The house is the one rolling for everyone; therefore, each round people (still in the round) get the same dice rolls. People who withdraw wait until all other players have finished taking dice rolls or until the house rolls a pair. Then the next round starts.
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Old 01-24-2017, 01:51 AM   #4
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Re: Interesting probability game played

Okay, thanks.

Can a person stop in one round and "re-enter" in a subsequent round? (The reason why this would be of interest is obvious.)
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Old 01-24-2017, 02:02 AM   #5
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Re: Interesting probability game played

Yes, for example, on round 1, the dice can be rolled 2,1 then 5,6 then 5,6 then 6,4. Player A withdraws. Player A would have 35 points entering round 2. Player B, C.... could continue taking dice rolls until they hit a pair (giving them 0 for this round) or until they withdraw. Then once every player has withdrawn or a pair has been rolled, round 2 starts with all players. And the dice are rolled...
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Old 01-24-2017, 02:11 AM   #6
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Re: Interesting probability game played

Gotcha.

It seems to me that the winning strategy will depend upon the total number of players playing, the number of players still "alive" in the round, what round it is, your score and the scores of all the other players. Right?

Sounds very complex. Maybe the optimal final round strategy can be determined. And then a backwards recursion can solve for the optimal strategy for the previous rounds?

I would start small. Try to solve the two-person two-round game, or something "simple" like that. Then expand to more players and/or more rounds.

If I have time tomorrow, I might noodle it.

Edit: Typically solution methods would assume that everyone uses the same strategy. A kind of Nash equilibrium. But this would mean that everyone would have the exact same score at all times. That doesn't sound interesting.
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Old 01-24-2017, 02:27 AM   #7
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Re: Interesting probability game played

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Originally Posted by whosnext View Post
Gotcha.

It seems to me that the winning strategy will depend upon the total number of players playing, the number of players still "alive" in the round, what round it is, your score and the scores of all the other players. Right?



Edit: Typically solution methods would assume that everyone uses the same strategy. A kind of Nash equilibrium. But this would mean that everyone would have the exact same score at all times. That doesn't sound interesting.
Yes, if everyone played GTO, everyone would have the same score making the game not very interesting.

I don't think so. Only on the last round we could expect behind/currently losing players to take an infinite number of dice rolls until they pass the leader, get zero for the round or bust (with a pair 1,1,).
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Old 01-24-2017, 03:26 AM   #8
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Re: Interesting probability game played

What is the process whereby players sit out (I presume they can do this "mid-round"). Of course, it is important to know if the players have to announce they are sitting out in some pre-determined order (like at a poker table), or from high score to low score, or all have to decide simultaneously, etc.

The more I think about this game, the more interesting it seems. There appear to be elements here of Chicken and Final Jeopardy wagering. Neat.

Edit: one other thing a strategy will depend upon is the scores of all players before the round began in addition to the "current" scores as the round progresses.
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Old 01-24-2017, 04:10 AM   #9
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Re: Interesting probability game played

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Originally Posted by whosnext View Post
What is the process whereby players sit out (I presume they can do this "mid-round"). Of course, it is important to know if the players have to announce they are sitting out in some pre-determined order (like at a poker table), or from high score to low score, or all have to decide simultaneously, etc.

The more I think about this game, the more interesting it seems. There appear to be elements here of Chicken and Final Jeopardy wagering. Neat.

Edit: one other thing a strategy will depend upon is the scores of all players before the round began in addition to the "current" scores as the round progresses.
Sitting out is done simultaneously. And, yes players do sit out mid-round. If they never sit out during the round, they will eventually hit a double/pair (22,33, etc.) and lose their points. Yes, I think the amount of points you have in the current round (and the ones you accumulated before the round started is important in deriving the strategy -like my solution considered), but I don't think our opponents' points matter until the final round starts.
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Old 01-24-2017, 07:41 AM   #10
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Re: Interesting probability game played

Let me rephrase the rules of this game, just to know if I got it.

1) N players, 8 rounds, one winner (the one with most points);
2) in each round the house rolls two dice (possibly) multiple times;
3) before each roll, each player still in play must simultaneously decide whether sitting out or keep rolling. If a player decides to sit out, he/she is done for the round (guess a player can sit out also before the first roll of a round. Am I correct?)
4) players know the players still in the round (can't see this mentioned, just assumed. Is that true?);
5) house rolls two dice. 3 things can happen:
- dice are 1-1. Every player in the round loses ALL the points accumulated (this round + previous rounds). The round is over.
- Dice are a pair (different from 1-1). Every player in the round loses the points made in the current round (they keep points from past rounds however). The round is over.
- Dice are unpaired. Every player in the round receives as many points as the sum of the two dice. Round continues unless every player sit out.
6) Points for each player are just the sum of the points made in each round.

If the above is correct, I guess that one should keep rolling pretty much always, unless they are the last player in play or their advantage is so big that they are very unlikely to be cought up.

Say that N=2 (A and B) and there are just two rounds. Of course, for the first roll both players should play. Now, if A sits out and B keeps rolling, B is a clear favourite (5 times out of six B will hold on the next roll and can stop with a higher score). Note that this argument is valid not just for the second roll, but for any roll.

The second round starts. If A and B have the same score, it's pretty much guarantee that the game will end in a tie. Both will keep rolling (for the same reasons as above) and the round will finish in the same time for both of them.

Say B>A. B tells A: I will keep rolling unless you stop. A must keep rolling (by stopping A wins just 1/36) and hope in 1-1 which will give him/her a tie. This will happen one time out of six (do you see why?). So B will win 5/6 and tie 1/6. Can B do better? It depends
on the point gap between them and the probability to realize a G run or better. If that's very low, B could stop at the beginning of the round hoping A to not fill the gap.

The example above should show that in the two-player game, the first stopping is in serious trouble. So, guess that optimal is always keep rolling. Can't see how the number of rounds can change the strategy (and also the number of players: if the goal is to arrive first, one must seek to get an advantage over the others and this means that is hard to see a situation in which one should stop). However, I might be seriously wrong of course.

Very different if players don't know who is in the game (and so if they are the last one). Need to think about that.
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Old 01-24-2017, 02:28 PM   #11
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Re: Interesting probability game played

This game is actually nuts even with just 1 round and 3 players (1 round/2 players is stupid as above).
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Old 01-24-2017, 02:39 PM   #12
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Re: Interesting probability game played

Nick, I think you have it correct. Also I think everybody knows who is still "alive" in each round (and everybody's pre-round and current scores, etc.).

Can I suggest a thought experiment? Suppose the game is exactly one round. Suppose there are 100 players, you and 99 others. The 100 players are labeled 1-100, and Player N must use a pre-announced strategy to sit out after N rolls. Player 1 sits out after 1 roll, Player 2 sits out after 2 rolls, etc.

In that game, what number should I pick to win the most games? Is it obvious what the optimal strategy is in this simple game?

Coke to Tom who posted something similar while I was typing.
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Old 01-24-2017, 03:28 PM   #13
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Re: Interesting probability game played

Maybe I'm on crack, but I don't think there's a NE for 1 round/3 players, and I don't think there's a sub-NE for B and C if A declares first he's rolling as long as anybody else is still in the game and B and C have to (non-cooperatively) pick strategies afterwards.
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Old 01-24-2017, 04:03 PM   #14
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Re: Interesting probability game played

Here are some simple, and maybe obvious, results of my simplified game of exactly one round with T players where player N (in 1 to T) pre-announces they will sit out after N rolls. (Prizes are split on ties.)

- For T less than 11, Player T wins the most prizes.
- For T at least 11, Player 1 wins the most prizes.

These are direct consequences of the geometric distribution.

Admittedly, these are not very realistic or interesting.
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Old 01-24-2017, 04:23 PM   #15
Sparky79
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Re: Interesting probability game played

Quote:
Originally Posted by nickthegeek View Post
Let me rephrase the rules of this game, just to know if I got it.

1) N players, 8 rounds, one winner (the one with most points);
2) in each round the house rolls two dice (possibly) multiple times;
3) before each roll, each player still in play must simultaneously decide whether sitting out or keep rolling. If a player decides to sit out, he/she is done for the round (guess a player can sit out also before the first roll of a round. Am I correct?)
4) players know the players still in the round (can't see this mentioned, just assumed. Is that true?);
5) house rolls two dice. 3 things can happen:
- dice are 1-1. Every player in the round loses ALL the points accumulated (this round + previous rounds). The round is over.
- Dice are a pair (different from 1-1). Every player in the round loses the points made in the current round (they keep points from past rounds however). The round is over.
- Dice are unpaired. Every player in the round receives as many points as the sum of the two dice. Round continues unless every player sit out.
6) Points for each player are just the sum of the points made in each round.




Say B>A. B tells A: I will keep rolling unless you stop. A must keep rolling (by stopping A wins just 1/36) and hope in 1-1 which will give him/her a tie. This will happen one time out of six (do you see why?). So B will win 5/6 and tie 1/6. Can B do better? It depends
on the point gap between them and the probability to realize a G run or better. If that's very low, B could stop at the beginning of the round hoping A to not fill the gap.
Yes, these rules for the game are correct.

If at a given time during a round player A chooses to sit out and Player B chooses to roll (he doesn't know A will sit out because they decide simultaneously), then 1/36 times player B will lose everything (Making A>B), 5/36 times player B will lose all his points for this round (making A>B assuming the sum of his points from this round and all previous rounds is greater than B's), and 30/36 times player B will score points (making him obviously greater than A given he was already in the lead).
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Old 01-24-2017, 06:04 PM   #16
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Re: Interesting probability game played

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Originally Posted by TomCowley View Post
Maybe I'm on crack, but I don't think there's a NE for 1 round/3 players, and I don't think there's a sub-NE for B and C if A declares first he's rolling as long as anybody else is still in the game and B and C have to (non-cooperatively) pick strategies afterwards.
This is weird- if you take the sub-case where one player has quit with a positive score and both remaining players have higher scores, (never roll always roll) and (always roll, never roll) are both NEs, but I have no idea how the play is supposed to resemble one.

There's no reason that I can see for the third player to ever drop out for 1/6 equity when he can just suicide pact to 1/3 equity waiting for doubles. And 2 players dropping out on the same roll is even worse. (1/12 each). So it looks like a full suicide pact is the only result as long as quitting would leave the situation above.

Now, what's neat here is that only one player max can declare to be the bully in the 2-player game. If 2 do, then the 3rd player can drop out early and watch them kill each other. So it looks like ((always roll, *****), (always roll, *****), (always roll, *****)) and the 3 permutations of ((always roll, bully), (always roll, *****), (always roll, *****)) are all NEs.
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Old 01-24-2017, 06:53 PM   #17
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Re: Interesting probability game played

Maybe this is related to what TomC just posted ...

Suppose this game is played with 3 people (A,B,C). Entering the final round C leads B by 1 point and B leads A by 1 point. Call the scores [10,11,12].

Player A pre-announces that she is stopping after her score exceeds 12. Players B & C both Roll for the first roll (along with A) of the final round. Suppose the sum of the dice is 7. New "current" scores are [17,18,19]. What should B and C do now?

If I did this correctly, I find that this sub-game Nash Equilibrium is that Player B should Roll with prob 1/5 and Stop with prob 4/5 and that Player C should Roll with prob 5/6 and Stop with prob 1/6. Overall expected payoffs are [1/36, 5/36, 30/36].
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