# Die wager

wolfgang59
Posers and Puzzles 30 Aug '13 03:33
1. wolfgang59
Mr. Wolf
30 Aug '13 03:33
Both players can wager \$1, \$2, 3\$, \$4, \$5 or \$6.

The catch: You can only score up to your wager.
eg If you wager \$3 and roll a 4,5 or 6 you score ZERO.

If its a tie then the pot is shared.
Otherwise winner takes it all.

2. AThousandYoung
30 Aug '13 06:18
I'm not sure I understand the rules. Isn't \$6 the obvious choice?
3. forkedknight
Defend the Universe
30 Aug '13 20:121 edit
The results are kind of interesting (I ran a simulation, didn't take the time to do the math)

The worst bet is 2 (loses to everything except another 2)

Betting 1 works surprisingly well, it handily defeats 2 and 3, barely edges out 4, and loses to 5 and 6. 1 is the only bet that wins out over any higher bet.

http://pastebin.com/CHSFJqqN
4. talzamir
Art, not a Toil
30 Aug '13 21:29
So..

Bet 1: win 1 on 1, zero on 2..6
Bet 2: win 2 on 1 or 2, zero on 3..6
Bet 3: win 3 on 1..3, zero on 4..6
Bet 4: win 4 on 1..4, zero on 5..6
Bet 5: win 5 on 1..5, zero on 6
Bet 6: win 6 on 1..6

Highest bet gives best odds for the best score, so of course, bet 6.

If you meant "you need to roll at least what you bet", things get more interesting.

Bet \$1: 6/6 chance of 1
Bet \$2: 5/6 chance of 2
Bet \$3: 4/6 chance of 3
Bet \$4: 3/6 chance of 4
Bet \$5: 2/6 chance of 5
Bet \$6: 1/6 chance of 6

There are three variables here. what you bet, what the opponent bets, and the roll of the die. If the opponent bets randomly, on average if

I bet \$1: I win \$1.52
I bet \$2: I win \$1.06
I bet \$3: I win \$0.63
I bet \$4: I win \$0.04
I bet \$5: I lose \$0.98
I bet \$6: I lose \$2.36

Simply because I only lose \$1 at most. Split pot is very nice when most of the money was the opponent's.

But if the opponent picks a specific value..

1 loses to 2, beats 3..6.
2 loses to 3, beats 4..6
3 loses to 4, beats 5..6
4 loses to 5, beats 6
5 beats 6

So I wouldn't put \$6 in the pot as a. I'd risk a lot of money and b. usually I'd end up with zero points and either lose or split, losing money in both cases. Other than that the optimal strategy seems to be "bet low, or \$1 higher than your opponent." So I would mix \$1..\$3 bids.
5. wolfgang59
Mr. Wolf
31 Aug '13 02:39
Originally posted by AThousandYoung
I'm not sure I understand the rules. Isn't \$6 the obvious choice?
Didn't seem obvious when I posted but perhaps it is!

2 players
Lets say A wagers \$6 and B wagers \$5

Out of 36 possibilities
A wins when he rolls a 6 6 times
A wins when he rolls a 5 5 times and one tie
A wins when he rolls a 4 4 times and one tie and one loss
A wins when he rolls a 3 3 times and one tie and 2 losses
A wins when he rolls a 2 2 times and one tie and 3 losses
A wins when he rolls a 1 1 time and one tie and 4 losses

So in 36 possibilities
A wins 21 (winning \$104)
A ties 5 (losing \$2.50)
A loses 10 (losing \$60)

A net gain of \$41.50 making it a good bet.

Looks like \$6 is the best wager.
(Against \$1 wager it wins 35/36 and draws once giving a gain of \$32.50)

How should the game be structured so that every wager is equally valid
ie how much should it cost to have 1, how much 1 & 2, how much 1,2 & 3 etc.
6. DeepThought
31 Aug '13 19:39
Originally posted by wolfgang59
Didn't seem obvious when I posted but perhaps it is!

2 players
Lets say A wagers \$6 and B wagers \$5

Out of 36 possibilities
A wins when he rolls a 6 6 times
A wins when he rolls a 5 5 times and one tie
A wins when he rolls a 4 4 times and one tie and one loss
A wins when he rolls a 3 3 times and one tie and 2 losses
A wins when he rolls a 2 2 ...[text shortened]... equally valid

ie how much should it cost to have 1, how much 1 & 2, how much 1,2 & 3 etc.[/b]
The rules for this are a little unclear. You have 2 players, each of whom bets into a pot. I assume the winner takes the pot, with a tie splitting the pot evenly, or do you just get back what you bet? If the latter then since only way one can lose is by rolling more than one has bet then betting \$6 is a no-brainer. So I assume that the pot is split evenly between the two players if neither goes bust, as otherwise it's trivial.

So if A bets 6 and B bets 1, then A has a 5/6 chance of winning 7 and a 1/6 chance of winning \$3.50, for an average profit of \$0.42 and B wins \$3.50 every 6th hand but puts \$1 into the pot on each roll for a loss of \$0.42.

If A bets 6 and B bets N then A wins with a probability of (1 - N/6) and ties with a probability N/6. B therefore wins (N + 6)/2 with a probability of N/6 and pays out N every round, B's average profit is then N*(N + 6)/12 - 6N = N*(12N - 30)/12 < 0 if N < 6.

Assuming I've understood the rules correctly then the only bet that makes any sense is \$6. So I've been trying to think of a rule change that would make it more of a game. The first thing I thought was if you make it so a player wins if they roll their exact bet then in the case where one player bets \$6 and the other \$1 the player who has bet \$1 wins \$6 net on one hand and loses \$1 on the other 5. But if the other player bets \$2 then the player who has bet \$1 wins \$0.66 (=(2 + 0 + 4*1/2)/6) as they tie on 4 of the hands. So all I've done is changed the best strategy.

To make the game non-trivial over repeated trials you need a rule that made your correct bet depends on what the other player has done, in such a way that the bets are complimentary, for each bet by A there is a best bet unique to A's bet that B can make. We can get this by making the player who is cyclically closest the winner (on a roll of 1 this would have bets of \$6 beating bets of \$5). With the player who bet more winning in the event of a tie. In other words if A bets \$6 and B \$4, then A wins on 1, 2, 5, and 6. B wins on 3 and 4. I haven't worked out if there's still an automatic strategy though. Ideally one would be indifferent to which bet one made against a random better. The ideal strategy should depend of one's opponents so that cautious individuals might prefer low bets, so that when faced with one of them one would choose a bet that scored well against low bets.
7. forkedknight
Defend the Universe
01 Sep '13 05:13
Originally posted by DeepThought
The rules for this are a little unclear. You have 2 players, each of whom bets into a pot. I assume the winner takes the pot, with a tie splitting the pot evenly, or do you just get back what you bet? If the latter then since only way one can lose is by rolling more than one has bet then betting \$6 is a no-brainer. So I assume that the pot is split e ...[text shortened]... t when faced with one of them one would choose a bet that scored well against low bets.
One item where I think we read the rules differently is regarding who rolls / how many dice are rolled.

I read the rules as meaning that each player rolls their own dice, whereas from reading your post, I get the impression there is only a single dice roll.

With one roll per player, I think it would be possible to change the wager for each value to make it more fair, and potentially into a reasonable game
8. DeepThought
01 Sep '13 12:51
Originally posted by forkedknight
One item where I think we read the rules differently is regarding who rolls / how many dice are rolled.

I read the rules as meaning that each player rolls their own dice, whereas from reading your post, I get the impression there is only a single dice roll.

With one roll per player, I think it would be possible to change the wager for each value to make it more fair, and potentially into a reasonable game
I wasn't sure, the thread title said "die wager" and not "dice wager" implying only one die and the analysis in the other posts seemed to indicate one die. It makes more sense with two dice rolls though - whoever rolls highest wins but not if they've gone bust. If player A bets \$6 and B bets \$N then there are N ties, (36 - N^2) cases where A rolls greater than N (and wins as B is lower or bust) or B rolls greater than N (and goes bust), N(N - 1)/2 cases where A wins and the same number of cases where B wins. It's easier to keep track of B's winnings:

A bets \$6 and rolls Ra and B bets \$N and rolls Rb; B's winnings per hand = Wb, the profit = Pb

(Ra > N) | (Rb > N) Wb = 0; Pb = -N
(Ra = Rb) & (Rb <= N) Wb = 3 + N/2; Pb = 3 - N/2
(Ra > Rb) & (Rb <= N) Wb = 0; Pb = -N
(Ra < Rb) & (Rb >= N) Wb = 6 + N; Pb = 6

B pays out 36N. <x> = expectation of x

36<Wb> = (3 + N/2)N + (6 + N)N(N - 1)/2
36<Pb>/N = (6 + N)N/2 - 36
<Pb> = N[N(N + 6)/2 - 36]/36

which is negative definite for N < 6 (<Pb>(5) = -42.5/36 ~ -\$1.18). In other words the best strategy is just to bet to the maximum. So adding a second die doesn't change anything. Having a rule where you always win if you roll what you bet (if your opponent misses) doesn't do enough as if B bets \$5 he gains \$5.50 when both roll a 5 but still loses to A's 6 as A has hit as well for an average gain of \$0.15, not enough to overturn the deficit. I agree one could try tinkering with the bet sizes to correct this, but it needs to force the players to play adaptively.

I like the idea of only one die roll. Each player has two unknowns, the other player's bet and the die roll. Really rolling two dice is the same as rolling one bigger one, it just makes the range of possibilities bigger without changing anything fundamental. One would then have to work out a win table so over the 90 possible non-trivial cases (where A and B do not bet the same, or a rule where A bets first and if B bets the same B can bet again with knowledge of A's bet) so that any bet would always have negative expectation against at least one other bet - for example betting one less than the opponent could be made to be the best move. With betting 2 less one gets break-even but betting \$1 against \$6 is a disaster (6 counting as 1 less than 1 for this). That way if A tries to bet \$6 every time then B can just bet \$5. That way a strategy needs to be responsive and the game becomes interesting.
9. talzamir
Art, not a Toil
01 Sep '13 18:01
A simpler variant of the game would be..

Player 1: pick a number between 1 and 20, in secret.
Player 2: pick a number between 1 and 20, in secret.

Reveal numbers. If they are the same, the round is a draw.

Roll a 20-sided die.

Round win criteria:
* if neither player chose a number in excess of the die roll: high number wins.
* if one chose in excess of the die roll: low number wins.
* if both chose in excess of the die roll: draw.

Game win criteria:
* single round: maximize your chance to win.
* single round: minimize your chance to lose.
* set number of rounds, win gives a point. Outperform the opponent.
* play to a set score, win gives a point. Reach the target score first.
* try to reach as many points as possible during a set number of rounds. Opponent points don't matter.

While in all cases it's good to win each round, those are all different games, so the optimal strategy could vary.

* you pick a number, out loud. Opponent has to choose a) a different number or b) a different number that is at least 2 greater or lower than yours or c) at least 3 greater or lower than yours. Who has the edge in this game?
10. DeepThought
01 Sep '13 23:181 edit
Originally posted by talzamir
A simpler variant of the game would be..

Player 1: pick a number between 1 and 20, in secret.
Player 2: pick a number between 1 and 20, in secret.

Reveal numbers. If they are the same, the round is a draw.

Roll a 20-sided die.

Round win criteria:
* if neither player chose a number in excess of the die roll: high number wins.
* if one chose in lower than yours or c) at least 3 greater or lower than yours. Who has the edge in this game?
Both players losing on a tie is a good idea, but I don't think twenty sided dice really adds anything. I'll assume a 6 sided dice in what follows. With the original betting the problem is that as soon as player A has a lead in money (assuming each player starts with \$100) they can just choose 6 each time to force player B (who'll run out of money quicker) to choose some other number. This leaves player A a favourite. This is partially cured by the 20 sided dice, but only really by dilution, if the betting has stabilized so player B chooses 19 and the other 20 then player B is not at a great disadvantage. A better cure is to make an exact roll win except against a higher exact roll. Also I don't think it should just be 1 point for a win. The advantage with Wolfgang59's betting regime is that there is a penalty for betting \$6 when one loses. Also choosing the right strategy when short of money is an interesting problem.

I'll suggest these rules:
On entering the game each player has committed all the money they've brought to the table, if a player drops out they forfeit their remaining money equally between the other player and the house. Each player makes a secret bet, costing an increasing price for each die face, for now I'll stick with 1 costs \$1, 2 costs \$2 etc.. The bets are made into a pot which can be won by either player or the house. Once both players have bet the bets are revealed and if identical the house wins. Each player rolls a die. A player is bust if they roll over the level of their bet, if both players are bust then the house wins, if only one player is bust the other wins. If a player hits their bet exactly then they win provided their opponent does not also hit their bet, in which case the higher roll wins. If both players roll the same then unless one of them is bust the house wins. Otherwise the player rolling the highest wins the pot.

The rule about hitting exactly makes the ending more interesting. Say player A has \$50 and player B \$10. Player B choosing 1 has a 5/36 chance of winning \$6 each go, whereas player A has a 31/36 chance of winning \$1 each go. So it's actually quite a close contest.

I think the 20 sided dice is too big, the numbers below 15 are irrelevant to strategies. I quite like the idea of having a fixed target, rather than the pot based system of the original game. The problem is the correct strategy is a no-brainer, one alternately bets 5 and 6 in the hope the other does so as well in order to maximize one's combined score, if the other player regards themselves as an opponent just bet 5.