deciding the flip
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Swordfish1989
New Member
It's still 50/50. The only reason you think it's not is because you see one result before the other. So yes, you may think that if you roll first and get a 6, your opponent only has a 1/6 chance of tying with you. And that may be in that specific roll, just like if you only flip one coin and it comes heads, it came up heads 100% of the time. If you and your friend both do a high roll 100 times, you'll have won 50 times and so will he.
vaporeon
Moderator
I just don't get how it's still 50/50. If I roll 11 using 2 dice, there's not going to be a 50/50 chance of my opponent rolling something higher or tying it. The only way a 50/50 can happen is if I roll 3 with both dice, then and only then does my opponent have a 50 percent chance of rolling higher or lower.
I'm talking about the number rolled using 2 dice and they chances based on the number rolled. not heads or tails, even or odd.
Swordfish1989
New Member
Because you're only looking at one individual roll, like I said. Also, you see one result before the other - like I already said. So you have the illusion that it's not 50/50 when really, it is.
Do it 100 times, 1000 times, etc, and you will only have won 50% of the time. And that's what matters in statistics, large sample size.
Do it 100 times, 1000 times, etc, and you will only have won 50% of the time. And that's what matters in statistics, large sample size.
PokeMedic
New Member
Why don't you rock-paper-scissors to decided who gets to flip. Or why don't you roll a die who decides to get a flip. Why not play a pick-up match to decide who gets to flip? Why don't you have a staring contest to decide who gets to flip?
Just flip it.
50/50 chance this isn't good enough.
Just flip it.
50/50 chance this isn't good enough.
...
are you 10 or a super senior, because i'm pretty sure that your required to take either 'algebra 1-2' or 'google.com' freshman year.
vaporeon
Moderator
No, I'm 23 in masters. I understand algebra so don't try insults here. What I don't understand is how a dice roll of 12 still has a 50 percent chance of anything.
This isn't Math Class. Either accept what's given as fact or prove it wrong. Everyone here has tried to explain it to you, and unfortunately you don't understand. That is not our fault, as we have each provided a variety of resources and time in addition to the fact that you are already on the internet and have full access to Google, Yahoo, MSN, and Wikipedia.
vaporeon
Moderator
Then explain to me how 12 equals 50/50 when the best outcome with 2 dice is twelve.
PokeMedic
New Member
Each 6-sided die has 6 numbers, 1 through six. 3 even numbers, 2 4 6, and 3 odd numbers, 1 3 5. 3 and 3 is equal to 50/50. There is a 50/50 chance of getting an even or odd number, even and odd being used to represent heads or tails instead of flipping coins.
There -I explained it. If you still don't get it, I'm not sure if troll or just ignorant.
NoPoke
Active Member
because you don't always roll a 12, just as often you will roll a 2. So the number of times you have an easy win after rolling a 12 is exactly balanced by the number of times you lose hard after rolling a 2.
vaporeon
Moderator
See PokeMedic, I understand it. I understand that is a 50 percent chance at flipping heads or tails, even or odd but that is not how we decide who goes first. If we did even or odd then it would be no different then heads or tails.
We decide it on high roll and doing that, the numbers are different. There's no 50/50 unless you roll 3 with both of the dice.
---------- Post added 11/14/2011 at 05:24 AM ----------
Yes, I know this. I've rolled snake eyes before. Thats just how the numbers work sometimes.
We decide it on high roll and doing that, the numbers are different. There's no 50/50 unless you roll 3 with both of the dice.
---------- Post added 11/14/2011 at 05:24 AM ----------
Yes, I know this. I've rolled snake eyes before. Thats just how the numbers work sometimes.
NoPoke
Active Member
try writing out the wins and losses for each possible scenario.
I'll start you off with a single six sided dice used for high roll.
I roll 6 , I win 5/6 times
I roll 5, I win 4/6 times and lose 1/6 times
I roll 4, I win 3/6 times and lose 2/6 times
I roll 3, I win 2/6 times and lose 3/6 times
I roll 2, I win 1/6 times and lose 4/6 times
I roll 1, I lose 5/6 times.
Now add up all the wins and losses. What is the ratio of wins to losses?
To convince yourself, as I suspect you wont be convinced, do the same table for high roll with a pair of dice.
I'll start you off with a single six sided dice used for high roll.
I roll 6 , I win 5/6 times
I roll 5, I win 4/6 times and lose 1/6 times
I roll 4, I win 3/6 times and lose 2/6 times
I roll 3, I win 2/6 times and lose 3/6 times
I roll 2, I win 1/6 times and lose 4/6 times
I roll 1, I lose 5/6 times.
Now add up all the wins and losses. What is the ratio of wins to losses?
To convince yourself, as I suspect you wont be convinced, do the same table for high roll with a pair of dice.
NoPoke
Active Member
. In which case I'm going to say that you are misapplying Bayesian inference. The odds for who goes first in our game are completely described by the frequentist approach that has been described by several posters in this thread.---
In the example I gave have you carried out the calculation I suggested and computed the ratio of wins:losses?
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psychup2034
New Member
NoPoke said:
I already did it 2 pages ago... :nonono: *sigh*
Ignorant panda is ignorant.
psychup2034 said:
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