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On November 11 2009 01:07 meeple wrote:
In that case, he's right in that the odds are stacked in the favour of the guy who gets an extra roll, but to get an actual percentage is a huge case-by-case solution. It basically boils down to how many cases are there that the guy who gets to go first can win straight up. There are basically only 2 cases. He gets 456 in his first roll or his second roll, and if he doesn't score any of these combinations, well it turns into a perfect normal game, except he's down one chance to beat the other guy. The probability that he gets 456 on his 1st of 2nd roll is very low, its 2/216 = less than 1%. Therefore, we can essentially consider this a problem where they roll at the same time, but one guy has 1 less roll. This doesn't give us a 100% accurate solution but its very close, since the probability of the guy winning with 456 is so low.
If we then solve the simplified probability case, you find that the odds favouring the guy with the extra roll to be about 67%, if I did everything right.
Firstly, I reckon the odds of getting 4-5-6 on the first or second roll at ~2.75% (1 - (71/72)^2). Secondly, what did you do to estimate the odds with the 4-5-6 rule ignored?
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furthermore, a 123 is just as likely as a 456. the odds of either insta-winning or insta-losing when you go first will cancel each other out over time. i don't see how going first affects the probabilities whatsoever tbh
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Firstly, I reckon the odds of getting 4-5-6 on the first or second roll at ~2.75% (1 - (71/72)^2)
2(1/6^3)
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On November 11 2009 01:12 Hawk wrote:normally each person gets three rolls, someone was just angry they lost a lot of money in poker and stacked the odds to try to entice people to play him for a big pot in a dumb luck game 
Oh. yeah basically the guy with 3 dice will obviously have 1 extra chance to get a higher number?
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On November 11 2009 01:15 MakkurtE wrote:Show nested quote +Firstly, I reckon the odds of getting 4-5-6 on the first or second roll at ~2.75% (1 - (71/72)^2) 2(1/6^3)
But it's not 1/6^3--it's 3/6^3*. And it's not simply odds-of-getting 4-5-6 in one roll times 2--it's 1 - the odds of not getting a 4-5-6 in either roll (the first way treats 4-5-6 4-5-6 as a double win, which it isn't).
*edit: actually 6/6^3
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On November 11 2009 01:07 meeple wrote:
In that case, he's right in that the odds are stacked in the favour of the guy who gets an extra roll, but to get an actual percentage is a huge case-by-case solution. It basically boils down to how many cases are there that the guy who gets to go first can win straight up. There are basically only 2 cases. He gets 456 in his first roll or his second roll, and if he doesn't score any of these combinations, well it turns into a perfect normal game, except he's down one chance to beat the other guy. The probability that he gets 456 on his 1st of 2nd roll is very low, its 2/216 = less than 1%. Therefore, we can essentially consider this a problem where they roll at the same time, but one guy has 1 less roll. This doesn't give us a 100% accurate solution but its very close, since the probability of the guy winning with 456 is so low.
If we then solve the simplified probability case, you find that the odds favouring the guy with the extra roll to be about 67%, if I did everything right.
what? the chance of getting 456 in 1 roll is ~2,77%. You can go 6/5/4, 5/6/4, 4/6/5, etc. There is also the chance of getting 123 to lose the game instantly, which is also ~2,77%. That means there is a chance of ~5,55% (12/216) to finish the game instantly in the first roll.
The chance to finish the game instantly in 2 rolls is ~11% (the chance of not finishing the game is (204/216)^2 which is ~89%).
That means that in 89% the person with 3 rolls has an advantage. in the other 11% there is no advantage for anyone because person 1 could either win or lose instantly by the same chance. Which means the person with 3 rolls has an advantage overall.
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On November 11 2009 01:19 qrs wrote:Show nested quote +On November 11 2009 01:15 MakkurtE wrote:Firstly, I reckon the odds of getting 4-5-6 on the first or second roll at ~2.75% (1 - (71/72)^2) 2(1/6^3) But it's not 1/6^3--it's 3/6^3. And it's not simply odds-of-getting 4-5-6 in one roll times 2--it's 1 - the odds of not getting a 4-5-6 in either roll (the first way treats 4-5-6 4-5-6 as a double win, which it isn't).
You're right... I counted it twice. Shouldn't it be 1/216 + (1-1/216)(1/216) - 1/216?
Probability of getting it first try, then (prob[not getting it first])*prob[getting it second] - prob[getting autolose in the first roll]
I still get this as less than 1%
And I estimated in the second part wrongly too, I was just going by a trend in a couple cases. The actual problem would be very lengthy to do.
But thats not really the point, I think you could spend an hour on the problem and come up with a actual number but the fact is that you give up way more than you gain.
Edit: The guy above me it right too... we didn't account for order into the mix, but the argument still stands
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shit, you're right, i overlooked the 456 456 possibility. oops
although i'm actually thinking now it's neither 1/6^3 or 3/6^3 but 6/6^3 (3!/6^3)
edit: damn your ninja edit 
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as far as the odds of 4-5-6, the main thing you missed is that there are multiple ways of getting it, as Keniji pointed out. (I also miscounted them, I said 3, when there are really 6). So you're off by a factor of nearly 6. Anyway, Keniji's post is right, although he doesn't calculate the actual odds, of course.
I went as far as making a table of the odds per roll of each outcome:
Auto-loss 6/216
Whiff 168/216
Hit-1 5/216
Hit-2 5/216
Hit-3 5/216
Hit-4 5/216
Hit-5 5/216
Hit-6 5/216
Trip-1 1/216
Trip-2 1/216
Trip-3 1/216
Trip-4 1/216
Trip-5 1/216
Trip-6 1/216
Auto-win 6/216
Using this to calculate the exact odds that the guy with 3 rolls wins would be possible though lengthy, as meeple says, but I don't have the time. Someone good at programming could use these numbers to write a Monte Carlo simulation in a few minutes, though. The figure 67% seems very dubious to me, though (I think you withdrew it, meeple): seems unlikely that the guy with a 3:2 advantage in dice has a 2:1 advantage in odds of winning.
edit: crossposted with makkurte's last post and meeple's edit
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On November 11 2009 01:32 MakkurtE wrote:shit, you're right, i overlooked the 456 456 possibility. oops although i'm actually thinking now it's neither 1/6^3 or 3/6^3 but 6/6^3 (3!/6^3) edit: damn your ninja edit
its 3/6*2/6*1/6. with the first dice you can either hit 4,5,6 = 3. with the second you can hit 4,5,6 except the one you hit with the first dice. so it's 3-1 = 2...
edit: i see that's exactly what you said :D but with an explanation.
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its 3/6*2/6*1/6. with the first dice you can either hit 4,5,6 = 3. with the second you can hit 4,5,6 except the one you hit with the first dice. so it's 3-1 = 2...
which is 3*2*1 = 6 = 3! (factorial 3), which is what i said. no?
edit:lol at the multiple simultaenous postings
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Well, you guys are fast at posting these things. I type up my stuff and pretty much everything I wrote has already been said already.
+ Show Spoiler +The first player (that gets to roll twice) has about a 5.4% chance of rolling 456 or 123 during his two rolls, and a 69% chance to roll a hit of some kind. The odds for the second player are about 8.1% and 82% respectively. I didn't calculate the exact odds of either side actually winning since that would take a lot more calculations than I'm willing to do right now. Someone more versed in statistics can probably crank out the answer a lot faster than me.
Some calculation info:
There are 216 (6^3) outcomes from rolling 3 dice. There are 6 (3! - that's 3 factorial) ways to roll a "456" so rolling a "456" on one roll is 6/216, which comes to about 2.7778%. We'll call this probability "P". The odds of the first player rolling 456 in his rolls is the chance of rolling 456 on the first roll, and the chance of rolling anything but 456 and 123 on the first roll and then a 456 on the second roll
probability(456 on 2 rolls without failure) = P + (1-2*P)*P
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You guys are focusing on the wrong part of the simulation. We've established that the auto win/auto lose affects nothing. But the other part is more important:
The next guy would have to hit that or beat it.
That means that the person who goes first will have a chance to go for his gun while the other person is throwing dice. This ensures a win in the event of a unlucky roll.
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EPT![[image loading]](http://i36.tinypic.com/2rz9j7c.jpg)
