ChatterBank1 min ago
Just For Fun - Triple Die Game
7 Answers
In a game called "Triple Die" , three standard dice are used. Punters bet on a die face,
if none of that face are rolled they lose their stake. Otherwise they keep their stake
and win their stake times the number of times that face shows. In the Solvay version,
all rolls of three-of-a-kind are void with no money changing hands. Which version of
the game is better for the punter?
if none of that face are rolled they lose their stake. Otherwise they keep their stake
and win their stake times the number of times that face shows. In the Solvay version,
all rolls of three-of-a-kind are void with no money changing hands. Which version of
the game is better for the punter?
Answers
1 to 7 of 7
Best Answer
No best answer has yet been selected by kuiperbelt. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.
Perhaps surprisingly, I make it the Solvay version. Let your stake be one, then in the standard version your expected score is:
125/216*(-1) + 75/216*(0) + 15/216*(1) + 1/216*(2) =-108/216 (=-0/5)
While in the Solvay version your expected score would be, I think:
120/216*(-1) + 75/216*0 + 15/216*1 + 6/216*0 = -105/216 (=-0.4861...)
which is very slightly better. I think my initial (tired) intuition was that of course you'd do better with a chance of getting three times your original stake, but conversely the Solvay version also removes five ways you could lose, which ends up being in your favour.
Still, in both versions you end up losing in the long run, so perhaps better to get it out of the way by playing the standard rules.
I rushed over the working here -- which could also mean I've made a mistake, but I think it's all correct -- but can go back to the beginning if you need more info on what the hell I've done and where everything above comes from.
125/216*(-1) + 75/216*(0) + 15/216*(1) + 1/216*(2) =-108/216 (=-0/5)
While in the Solvay version your expected score would be, I think:
120/216*(-1) + 75/216*0 + 15/216*1 + 6/216*0 = -105/216 (=-0.4861...)
which is very slightly better. I think my initial (tired) intuition was that of course you'd do better with a chance of getting three times your original stake, but conversely the Solvay version also removes five ways you could lose, which ends up being in your favour.
Still, in both versions you end up losing in the long run, so perhaps better to get it out of the way by playing the standard rules.
I rushed over the working here -- which could also mean I've made a mistake, but I think it's all correct -- but can go back to the beginning if you need more info on what the hell I've done and where everything above comes from.
1st version=
prob 0 correct = (5/6)^3 = -125/216
potential loss is therefore (125/216) * S (where S = Stake)
prob 1 face = 3C1 * 1/6 * (5/6)^2 = (75/216) *S
prob 2 faces = 3C2 * (1/6)^2 * 5/6 = (15/216)
but the payout is twice the stake = (30/216)* S
prob 3 faces = 3C3 * (1/6)^3 = 1/216
But the payout is three times the stake = (3/216)* S
Therefore the net result is (-125 +75 + 30 + 3)/216
= -17/216 of your stake
In the solvay version there are 5 three of a kinds where the stake is now not lost. Therefore the potential loss = (125-5=120)/ 216 =-120/216
The other 3 of a kind was the winner 1/216 * 3= 3/216 which is now void
Therefore the Solvay expected result = (-120 + 75 + 30)/216 = -15/216
Therefore the Solvay version is better by (17 -15)/216 or 2/216
prob 0 correct = (5/6)^3 = -125/216
potential loss is therefore (125/216) * S (where S = Stake)
prob 1 face = 3C1 * 1/6 * (5/6)^2 = (75/216) *S
prob 2 faces = 3C2 * (1/6)^2 * 5/6 = (15/216)
but the payout is twice the stake = (30/216)* S
prob 3 faces = 3C3 * (1/6)^3 = 1/216
But the payout is three times the stake = (3/216)* S
Therefore the net result is (-125 +75 + 30 + 3)/216
= -17/216 of your stake
In the solvay version there are 5 three of a kinds where the stake is now not lost. Therefore the potential loss = (125-5=120)/ 216 =-120/216
The other 3 of a kind was the winner 1/216 * 3= 3/216 which is now void
Therefore the Solvay expected result = (-120 + 75 + 30)/216 = -15/216
Therefore the Solvay version is better by (17 -15)/216 or 2/216
JJ and I have reached the same conclusion but with slightly different answers. I think the error is mine, but it depends on the "keep their stake and win their stake times the number of [scoring dice]". For whatever reason, I interpreted this to mean that one die matching your bet is break-even, but I think the "and" means that this was wrong. If so, mea culpa, it was late and I was in a mood, but also for whatever reason the version where you break even with one scoring die makes more sense to me, ie basically you get back either 0,1,2, or 3 times your stake (rather than JJ's 0,2,3,4).
A different approach...
Putting aside for a moment the 6 unique '3 of a kind' combos, the other 210 outcomes offer an identical chance of winning/losing in both versions of the game.
With the stipulation that a punter holding onto their stake = +1, focus is now drawn to the other 6 unique combos.
Under the ** Solvay rules, five of them get to keep their stakes, whereas those stakes would have been lost under the standard version, therefore +5. The other punter is worse off since they would have gained 3 times their stake, loss = -3.
Conclusion; punters are better off by +2 stake monies everytime
'3 of a kind' are rolled.
Hence Solvay version of the game is more advantageous to the punters.
Alternative description...
Consider playing the game to standard rules. 6 stakes are layed.
The 'banker' takes 2 of the stake monies everytime '3 of a kind' rolls up. Whereas under the Solvay version no monies change hands.
** This reminded me of the 1927 Solvay convention held in Belgium - A meeting of great quantum minds :-)
Putting aside for a moment the 6 unique '3 of a kind' combos, the other 210 outcomes offer an identical chance of winning/losing in both versions of the game.
With the stipulation that a punter holding onto their stake = +1, focus is now drawn to the other 6 unique combos.
Under the ** Solvay rules, five of them get to keep their stakes, whereas those stakes would have been lost under the standard version, therefore +5. The other punter is worse off since they would have gained 3 times their stake, loss = -3.
Conclusion; punters are better off by +2 stake monies everytime
'3 of a kind' are rolled.
Hence Solvay version of the game is more advantageous to the punters.
Alternative description...
Consider playing the game to standard rules. 6 stakes are layed.
The 'banker' takes 2 of the stake monies everytime '3 of a kind' rolls up. Whereas under the Solvay version no monies change hands.
** This reminded me of the 1927 Solvay convention held in Belgium - A meeting of great quantum minds :-)
Goodness - who do I award BA to LOL??
I shall print verbatim the answer as it was presented to me (I admit not to fully understanding the answer LOL) See below:
"Imagine 6 punters all betting on different numbers. The overall outcomes are identical 210/216 times. For the other six, the outcome is worse (by 4) for one punter, and better (by 1)
for the other five. So the overall outcome in the Solvay version is better for the punters by 1".
I shall print verbatim the answer as it was presented to me (I admit not to fully understanding the answer LOL) See below:
"Imagine 6 punters all betting on different numbers. The overall outcomes are identical 210/216 times. For the other six, the outcome is worse (by 4) for one punter, and better (by 1)
for the other five. So the overall outcome in the Solvay version is better for the punters by 1".
Unfortunately Kuiperbelt all of us think that answer is wrong. We are consistent that the difference is 2!
The gain by 5 punters = 1
but the loss for one punter is 3 (not 4)
Three times the stake is paid for 3 correct, not 4
The stake is irrelevant as that is returned in both methods, it is only the winnings (3) that is lost
The gain by 5 punters = 1
but the loss for one punter is 3 (not 4)
Three times the stake is paid for 3 correct, not 4
The stake is irrelevant as that is returned in both methods, it is only the winnings (3) that is lost
1 to 7 of 7
Related Questions
Sorry, we can't find any related questions. Try using the search bar at the top of the page to search for some keywords, or choose a topic and submit your own question.