[bonanova]

Curious about this place called Morty's, mentioned more than

once here at the Den, you wandered in last night to meet the

boys and have a cool one.

You must not have blended too well, because Alex saw you

coming and was striking up a conversation before you could

say Sam Adams.

Ya look like a smart lad who likes a challenge, he began.

I've got a pair of dice here, and I'm sure you know the chances

of rolling a 7 are 1/6, but to roll a 12 they are only 1/36. Well,

then; even up, I'm willing to bet you'll roll a 12 before you roll

two 7's in a row. Keep in mind two dice are thrown each roll.

It sounded like an even bet; after all, 1/6 x 1/6 = 1/36.

Do you take the bet?

Edited February 8, 2009 by bonanova
Added text in red to remind that each roll is of two dice

[Guest]

This does indeed sound like an even bet. The variance and standard deviation probably are different, but I believe that on average, each outcome would take 12 rolls. If that is correct (which, knowing Bonanova, it probably isn't), then your odds of winning the bet are still 50-50.

When Alex said "two 7's in a row" what he meant was "a sum of 14", the odds of which of course are 0. Perhaps Alex is resorting to verbal trickery to win money.

Also, can we assume the dice aren't weighted?

[Guest]

In Wargames, the WOPR comes to the conclusion that "the only winning move is not to play.".

[Guest]

Given that you can't possibly win after just 1 roll, don't take the bet.

[Guest]

Ok, I'm an idiot (I've been saying that a lot lately, what a shame ). But my last post was very incorrect and I plan on fixing it now.

The average number of times it will take to roll a 12 is not 12 (as I previously said), it is 36. This is because (as everyone knows except, evidently, me) the probability of rolling a 12 is 1/36. There's a statistics formula that would prove that I am correct, it basically says in a geometric sequence (which this is), the average number of times a trial will need to happen before it is a success is 1/probability. Which in this case is 1/(1/36), or 36.

Now, the probability of rolling ONE 7 is 1/6. Therefore, it will take on average 6 rolls to reach a 7.

Now the probability of you rolling a SECOND 7 is still, unsurprisingly, 1/6. So it will take on average another 6 rolls to reach it.

Therefore, it will only take an average of 12 rolls to obtain two 7s, but an average of 36 rolls to obtain a 12.

So I don't take the bet. Even if Alex offered me 2:1 odds I wouldn't take it!

The reason why this is a confusing question (for me at least) is that the probabilities of both events are happening are the same; however, in the case of rolling two 7s, the events are not independent. That is, you have to roll the first 7 before you roll the second 7. Which improves the odds on the 7s.

So say the bet Alex gave you were "I'll roll this pair of dice twice; you roll this pair of dice once. If I obtain a 7 on both rolls, I win. If you obtain a 12, you win. If that does not happen, we will repeat until one of us wins." THEN, the odds would be equal, because the 7s would not be cumulative (as in, if Alex rolled a 7 once that would not count for his next roll, he would still have to roll two 7s next time to win).

But that's not the bet, is it????

[bonanova]

Also, can we assume the dice aren't weighted?

Yes.

You can assume, as the OP says, p[12] = 1/36 and p[7] = 1/6.

[bonanova]

Given that you can't possibly win after just 1 roll, don't take the bet.

Good start.

But consider that the probability of getting a 2nd roll is 35/36 - a pretty good chance.

[bonanova]

Ok, I'm an idiot (I've been saying that a lot lately, what a shame ). But my last post was very incorrect and I plan on fixing it now.

The average number of times it will take to roll a 12 is not 12 (as I previously said), it is 36. This is because (as everyone knows except, evidently, me) the probability of rolling a 12 is 1/36. There's a statistics formula that would prove that I am correct, it basically says in a geometric sequence (which this is), the average number of times a trial will need to happen before it is a success is 1/probability. Which in this case is 1/(1/36), or 36.

Now, the probability of rolling ONE 7 is 1/6. Therefore, it will take on average 6 rolls to reach a 7.

Now the probability of you rolling a SECOND 7 is still, unsurprisingly, 1/6. So it will take on average another 6 rolls to reach it.

Therefore, it will only take an average of 12 rolls to obtain two 7s, but an average of 36 rolls to obtain a 12.

So I don't take the bet. Even if Alex offered me 2:1 odds I wouldn't take it!

The reason why this is a confusing question (for me at least) is that the probabilities of both events are happening are the same; however, in the case of rolling two 7s, the events are not independent. That is, you have to roll the first 7 before you roll the second 7. Which improves the odds on the 7s.

So say the bet Alex gave you were "I'll roll this pair of dice twice; you roll this pair of dice once. If I obtain a 7 on both rolls, I win. If you obtain a 12, you win. If that does not happen, we will repeat until one of us wins." THEN, the odds would be equal, because the 7s would not be cumulative (as in, if Alex rolled a 7 once that would not count for his next roll, he would still have to roll two 7s next time to win).

But that's not the bet, is it????

getting 12 [on any single roll] before getting two 7's [on any two consecutive rolls].

You're right about p=1/36 => expecting 36 rolls to get 12 once.

But you got a little off in considering 6 rolls for the first 7, then 6 more rolls for the 2nd 7.

The 7's have to be consecutive.

That's why the OP suggested p[two consecutive 7's] = 1/6 x 1/6, suggesting 36 rolls to get that result as well.

Question is do you believe it's that straightforward?

[Guest]

getting 12 [on any single roll] before getting two 7's [on any two consecutive rolls].

You're right about p=1/36 => expecting 36 rolls to get 12 once.

But you got a little off in considering 6 rolls for the first 7, then 6 more rolls for the 2nd 7.

The 7's have to be consecutive.

That's why the OP suggested p[two consecutive 7's] = 1/6 x 1/6, suggesting 36 rolls to get that result as well.

Question is do you believe it's that straightforward?

I learned in Statistics and Probability class last semester that "consecutive" probabilities have a different equation to figure them out. However...I forgot what that is <_<

[Guest]

getting 12 [on any single roll] before getting two 7's [on any two consecutive rolls].

You're right about p=1/36 => expecting 36 rolls to get 12 once.

But you got a little off in considering 6 rolls for the first 7, then 6 more rolls for the 2nd 7.

The 7's have to be consecutive.

That's why the OP suggested p[two consecutive 7's] = 1/6 x 1/6, suggesting 36 rolls to get that result as well.

Question is do you believe it's that straightforward?

Oh I see I misunderstood (as you can tell by my post). I though it was just a 12 before two 7s period, didn't realize it had to be consecutive 7s.

I'm not sure.

I'll have to think on it some more and I'm pretty tired right now. But it does seem like it would be the same probability.

[Guest]

Your odds of getting a 12 before two consecutive 7s might be slightly greater, because you can get a 12 with your initial roll but the 7s take two rolls.

[Guest]

Consider two consecutive rolls. The probability of getting a seven on both rolls is (1/6)(1/6)= 1/36= 2.78%. The probability of getting at least one 12 in those two rolls is 1-(35/36)(35/36), which is 1-the probability of not getting a 12 on either roll, which equals 71/1296= 5.49%.

And, of course, if you don't get two 7s and you don't get a 12, then you keep rolling, and you can consider the next two rolls.

[Guest]

Good start.

But consider that the probability of getting a 2nd roll is 35/36 - a pretty good chance.

[Guest]

If we look at the probability of not rolling a consecutive 7 or a single 12 until the n th roll, then the only time the odds favour rolling consecutive 7's is with the first 2 rolls (i.e. 1/6 x 1/6 = 2.78% versus 35/36 x 1/36 = 2.70%)

For example, the probability of not obtaining a consecutive 7 until the 4th roll is 6/6 x 5/6 x 1/6 x 1/6 = 2.31% whereas the probability of not obtaining a 12 until the 4th roll is 35/36 x 35/36 x 35/36 x 1/36 = 2.55%.

My reasoning for the "6/6" probability on the first roll is that the outcome is irrelevant if the next roll is not a 7.

Good start.

But consider that the probability of getting a 2nd roll is 35/36 - a pretty good chance.

[Guest]

Your chances of getting a 12 on the first roll are 1/36, and the chances of getting two consecutive 7s is 0. On the second roll, you have had two chances to get a 1/36 roll, so 2x(1/36). You've had one chance to get a 7 (1/6) and then one chance to get another, so (1/6)x(1/6)=1/36. On the third, it's 3x(1/36), and 2x(1/6)x(1/6).

So the chances of having gotten a 12 by any given turn n are nx(1/36), while those of having gotten two consecutive 7s is (n-1)x(1/36).

Unless I'm thinking about this wrong.

[Guest]

you should take the bet since the chances of getting 12 at a single roll is far better than getting 2 7's in 2 consecutive rolls

[Guest]

Ok, so let's look at the probabilities of winning or losing on any given roll. For losing it is pretty straightforward:

p_lose(n)=(35/36)^(n-1)*(1/36)

For winning it's a bit more complicated:

p_win(1) = 0

p_win(2) = (1/6)*(1/6)

p_win(3) = (5/6)*(1/6)*(1/6)

p_win(4) = (6/6)*(5/6)*(1/6)*(1/6)

p_win(n) = (5/6)*(1/6)*(1/6)*(1-p_win(n-3))*(1-p_win(n-4))*(1-p_win(n-5))...

The result is that at any n, p_lose is always greater than p_win. Therefore, do not take the bet.

Edited February 9, 2009 by voltage

[Prof. Templeton]

that if you ran this experiment just once, it would be a pretty even bet. The more times you run it, however, the more the difference will become apparent. Let's say we roll the dice 36 times. Out of those rolls you would expect a roll of "twelve" (double sixes) to come up just once. Probability of 1/36. Out of the same 36 rolls you would expect "sevens" to come up six times or 1/6. But were looking for "sevens" in a row. One of the 36 rolls has been used up for the first pair of "sevens" so theres only 35 rolls available to get the next pair of "sevens". So I think our probability is reduced to ¹/₆ x ^.9722.../₆. Or .0277... for "twelve" versus .027 for two "sevens" in a row during the same number of rolls.

[HoustonHokie]

Here's how I think about it:

To obtain a 12 is pretty straight-forward. I think everyone agrees that if I roll a pair of dice 36 times, I should expect to see a 12 once.

To obtain two 7s is a little bit different. It requires two rolls of the dice. If I roll two dice twice in a row, I should expect to see two 7s once every 36 times. Seems like the same probability, except for the pair of rolls. I see it like this:

To obtain 12 requires 36 rolls of 2 dice, on average.

To obtain two 7s requires 36 pairs of rolls of 2 dice, on average.

Now pairs of rolls overlap. For example, the first pair is rolls 1 & 2. The second pair is rolls 2 & 3. The third pair is rolls 3 & 4. You get the picture. The 36th pair is rolls 36 & 37.

So, after 36 rolls of the dice, I'll expect to see a 12.

But I'll expect to see two 7s after 37 rolls of the dice.

I don't take the bet.

[Guest]

The question is "Do you take the bet". My answer is yes, eventually. Though I'm not much of a mathmatician I believe the odds of my taking the bet increase every time I suck down one of those cold ones mentioned in the puzzle. As to the odds of two sevens poping up one after another it seems like it would be 36:1 because the two numbers must apear sequencially. In any case, I think I could pull it off, and it would be worth trying just to see the look on that guys face peace

[Prof. Templeton]

The question is "Do you take the bet". My answer is yes, eventually. Though I'm not much of a mathmatician I believe the odds of my taking the bet increase every time I suck down one of those cold ones mentioned in the puzzle. As to the odds of two sevens poping up one after another it seems like it would be 36:1 because the two numbers must apear sequencially. In any case, I think I could pull it off, and it would be worth trying just to see the look on that guys face peace

LOL I want to sit on the stool next to yours.

[Guest]

No I would not take the bet. We all know the chances of rolling a 12 is 1/36=0.0277', and the chances of rolling 2 7's in a row is 1/36=0.0277' so therefore we have even chances right? NO

there is a 35/36 chance that we get a 2nd role which means that;

rolling a 12: 1/36=0.0277'

rolling 2 7s: 1/3=0.0277'*35/36(=0.9722') which means that the chances are not the same.

rolling a 12:0.0277'

rolling 2 7s:0.027

i.e. you have a slightly greater chance of losing that you do of winning

[Prof. Templeton]

Here's how I think about it:

To obtain a 12 is pretty straight-forward. I think everyone agrees that if I roll a pair of dice 36 times, I should expect to see a 12 once.

To obtain two 7s is a little bit different. It requires two rolls of the dice. If I roll two dice twice in a row, I should expect to see two 7s once every 36 times. Seems like the same probability, except for the pair of rolls. I see it like this:

To obtain 12 requires 36 rolls of 2 dice, on average.

To obtain two 7s requires 36 pairs of rolls of 2 dice, on average.

Now pairs of rolls overlap. For example, the first pair is rolls 1 & 2. The second pair is rolls 2 & 3. The third pair is rolls 3 & 4. You get the picture. The 36th pair is rolls 36 & 37.

So, after 36 rolls of the dice, I'll expect to see a 12.

But I'll expect to see two 7s after 37 rolls of the dice.

I don't take the bet.

That's an easier way to look at it.

[Guest]

I learned in Statistics and Probability class last semester that "consecutive" probabilities have a different equation to figure them out. However...I forgot what that is <_<

the way it is:

P(not E)=5/6

P(not E and not E)=5/6,^2=25/36

P(not not E) (which is P(E)= 1-P(not E)= 11/36

E being the event of rolling a 7

If Alex is saying that I win if I get 2 7's first, I would take the bet

[Guest]

I think I've got it:

The answer is not to take the bet. Since you are the one rolling the dice and not alex he is betting that you will get 12 before the double 7's. So the chance of you rolling double sevens is 1/36. But the trick is that you are rolling the same dice. Lets say you get the first 7 – now the chances of you getting a second 7 would be 1/6 your chances that would make you win. However the probability of alex not winning and you not losing is not 5/6. It is actually 5/6 – 1/36 (the 1/36 being the case that you roll a 12 and alex wins) = 29/36. Factoring that in, if you roll the die 36 times you are statistically guaranteed to get a 12 before the double 7’s because of the small chance that a 12 will appear right after your first 7.

If someone can put this in a better wording, please do, I realize its somewhat confusing.

Edited February 9, 2009 by Danny The Russian

[Guest]

the way it is:

P(not E)=5/6

P(not E and not E)=5/6,^2=25/36

P(not not E) (which is P(E)= 1-P(not E)= 11/36

E being the event of rolling a 7

If Alex is saying that I win if I get 2 7's first, I would take the bet

somehow, i think my own answer looks flawed