A math question that I can't quite get my head around....

[Ojas]

That is where you are making your mistake. You are looking at the potential total number combinations which would be 6 squared (6 numbers, 2 dice) exceeding the potential combinations of 2, 3, 3 and 4. However, the initial hypothetical was based upon a single number between 3-6 and not the total of two dice.

Actually, the values in my grid do not represent a total. naaman defined win as "if you roll two dice and need at least one (both are not necessary) to come up three or better". The W in my grid represents this win criteria (at least once die rolled a 3,4,5,6).

Here's yet another way to calculate the probability...

You roll both dice and want to determine if you won. You look at D1 first:

1 -> need to look at D2
2 -> need to look at D2
3 -> win
4 -> win
5 -> win
6 -> win​

So we can say...

4 out of 6 times you win regardless of D2's outcome, but
2 out of 6 times you need to look at D2 (which we know has a 4/6 chance of winning)
​

... which is equivalent to ...

4/6 +
2/6 * 4/6​

... which is 8/9 or 88.89% (the same outcome as my grid).

 

[Ojas]

I think this is exactly what I was trying to explain a couple posts up.

Right and if you want to figure out your chances with three dice, visualize a 3D cube instead of the 2D grid. It's easier to count the losing (L) cells. In a cube, you'll have 8 possible scenarios where you'll lose. There are a total of 6*6*6 or 216 possible outcomes. So, you win with 216-8 or 208 outcomes, which means with three dice, your chances are 208/216 = 96.30%.

Or another way...

<table><tr><td><tr><td>rolling 1 die:</td><td>2/3</td><td>= 66.67%</td></tr><tr><td>rolling 2 dice:</td><td>2/3 + (1/3 * 2/3)</td><td>= 88.89%</td></tr><tr><td>rolling 3 dice:</td><td>2/3 + (1/3 * 2/3) + (1/3 * 1/3 * 2/3)</td><td>= 96.30%</td></tr></table>
Like I said, the more dice you add, the closer your chances are to 100% (but you'll never actually get to 100%, of course).

I'm still up for betting on this! How about a $25 donation to NSX Prime?

 

[naaman]

Ojas, I think you may be the only one who understands my question at this point, so I will attempt a re-wording of the problem. Let's look at it generically:

You are playing a game of chance. You are allowed fixed number of attempts to win the game. How do you calculate the likelihood that you will EVENTUALLY win given a set number of tries to overcome given odds? You do not lose until you run out of tries (and fail on all tries).

 

[Lotus_Esprit]

But, what if the likelihood doubled again? For example, you roll a die and you are looking for it to come up as a 3 or higher (66.67% chance). Now, if you roll two dice and need at least one (both are not necessary) to come up three or better, the likelihood is doubled, but what percentage is it that one or the other will come up three or better? It's obviously not 133%.

The only thing I could think of would be to cut the likelihood of failure in half. So for example, on a single die you have a 33.3% chance to fail, while the chances of both dice failing is only 16.67%. But then, that's not really "doubling" the chance for success, since by that method your chances for success are 83.3%, which is not twice as much as 66.67%.

Since the outcome of the first die has no impact on the outcome of the second die, the odds should be doubled, but I can't figure out how to represent that with a percentage.

So what is the answer? I'm stumped.

1 die: 4/6 (3,4,5,6) = 66.66%

if you were to row two dice and both of the dice must be higher then 3, we have to use the statistic formula of probability since the first die is dependent on the second die.

4/6 * 4/6 = 44.44% that the roll will produce both die to be over 3.

so 44.44% is your answer. Im a computer science major minor math major. :smile:

 

[Vega$ NSX]

Ojas, I think you may be the only one who understands my question at this point, so I will attempt a re-wording of the problem. Let's look at it generically:

You are playing a game of chance. You are allowed fixed number of attempts to win the game. How do you calculate the likelihood that you will EVENTUALLY win given a set number of tries to overcome given odds? You do not lose until you run out of tries (and fail on all tries).

Based on the wording of this equation, Ojas is correct. Just apply his formula to whatever the proper odds are. In the case of the rolling of the dice 3 and greater, the correct overall odds would be 8/9 or 88.89%

 

[Lotus_Esprit]

Right and if you want to figure out your chances with three dice, visualize a 3D cube instead of the 2D grid. It's easier to count the losing (L) cells. In a cube, you'll have 8 possible scenarios where you'll lose. There are a total of 6*6*6 or 216 possible outcomes. So, you win with 216-8 or 208 outcomes, which means with three dice, your chances are 208/216 = 96.30%.

Or another way...

<table><tr><td><tr><td>rolling 1 die:</td><td>2/3</td><td>= 66.67%</td></tr><tr><td>rolling 2 dice:</td><td>2/3 + (1/3 * 2/3)</td><td>= 88.89%</td></tr><tr><td>rolling 3 dice:</td><td>2/3 + (1/3 * 2/3) + (1/3 * 1/3 * 2/3)</td><td>= 96.30%</td></tr></table>
Like I said, the more dice you add, the closer your chances are to 100% (but you'll never actually get to 100%, of course).

I'm still up for betting on this! How about a $25 donation to NSX Prime?

i think this is wrong. the more die you add the more chances of it happening is better but not by that much.

1 die = 4/6 = 44.44%

so to have both die to get higher then 3 you have to multiply them
1 dice = 2/3 * 2/3
1 dice + 1 die = 2/3 * 2/3 *2/3

since the die roll is dependent of each other. however if the second roll is not dependent (doesnt have to be 3 or higher) and let say that the first roll needs to be 3 or higher. we can represent it some what liek your way.

2/3 + 1/6 = 5/6

 

[Vega$ NSX]

1 die: 4/6 (3,4,5,6) = 66.66%

if you were to row two dice and both of the dice must be higher then 3, we have to use the statistic formula of probability since the first die is dependent on the second die.

4/6 * 4/6 = 44.44% that the roll will produce both die to be over 3.

so 44.44% is your answer. Im a computer science major minor math major. :smile:

That is for both dice to be 3 or greater. But it's 88.87% for either one of the dice to be 3 or greater (essentially double the odds).

 

[Vega$ NSX]

1 die = 4/6 = 44.44%

Huh??? 4 divided by 6 is .6667 or 66.67%.

 

[Lotus_Esprit]

That is for both dice to be 3 or greater. But it's 88.87% for either one of the dice to be 3 or greater (essentially double the odds).

for BOTH of them to be greater then we have to mulitply. I do not know how you guys are getting 88.87% because you cannot just add 44.444 + 44.44 together because both of the probability of the dice rolling higher then a 3 is DEPENDENT of each other thus needing to mulitiply. The only time you can add both probability if they are indenpendent of each other.

then how do you do 3 dice

44.44 + 44.44 + 44.44 ?

 

[RSO 34]

But the real issue is what are the odds of the plane taking off from the conveyor belt?

 

[Vega$ NSX]

for BOTH of them to be greater then we have to mulitply. I do not know how you guys are getting 88.87% because you cannot just add 44.444 + 44.44 together because both of the probability of the dice rolling higher then a 3 is DEPENDENT of each other thus needing to mulitiply. The only time you can add both probability if they are indenpendent of each other.

then how do you do 3 dice

44.44 + 44.44 + 44.44 ?

No we aren't adding 44.44% + 44.44%. What we are saying is that for one die to roll 3 or greater the odds are 66.67%. For two dice to both roll 3 or greater it's 44.44%. But for either one of the die to roll 3 or greater then the odds are 88.89%. Which makes sense intuitively. It's harder to have two dice simutanously roll 3 or greater than one (44.44% is less than 66.67%) and it's easier to have only one of two dice roll 3 or greater than one (88.89% is greater than 66.67%).

 

[Ojas]

j/k I kid!!!

Im a computer science major minor math major. :smile:

As such, you should be capable of running the simple Monte Carlo simulation I posted...

... and then requesting a refund from the institution responsible for your education.

 

[Lotus_Esprit]

But the real issue is what are the odds of the plane taking off from the conveyor belt?

it flys, myth buster proved it.

 

[Daedalus]

How do you calculate the likelihood that you will EVENTUALLY win given a set number of tries to overcome given odds?

In this case the easiest way to calculate the answer is to calculate the odds of *losing*, then subtract from 1.

Odds of rolling only a 1 or 2 (one roll) is 1/3.
Odds of rolling only a 1 or 2 (two rolls) is 1/9 = 1/3 * 1/3
Odds of rolling only a 1 or 2 (three rolls) is 1/27 = 1/3 * 1/3 * 1/3.

Ojas is correct. For the original question, odds are 8/9 = 1 - 1/9.

 

[Ojas]

You are playing a game of chance. You are allowed fixed number of attempts to win the game. How do you calculate the likelihood that you will EVENTUALLY win given a set number of tries to overcome given odds? You do not lose until you run out of tries (and fail on all tries).

Knowing your chances of winning is not enough to answer this question: We also need to know the risk and reward: how much it costs to play the game and how much it pays if you win. This is what Vega$ NSX was speaking towards.

Sample Scenario:

Let's say you're playing a game at an arcade. It costs $0.75 to play, but if you win, it gives you $1.00 (a net gain of $0.25). You've figured you have a 2/3 chance of winning at this game.
​

Q:

Should you play?​

A:

One way to find out is to calculate how much, on average, you'll win or lose per play. You have a 2/3 chance of coming out ahead by $0.25, but a 1/3 chance of losing $0.75. So, (2/3*.25) + (1/3*-.75) = -0.08. On average, you'll lose about $0.08 per play, so you don't play this game.​

You can use a similar approach to determine whether or not play games of chance. You just need to know: cost of playing, potential win, and chances of winning the game.

 

[Vega$ NSX]

Knowing your chances of winning is not enough to answer this question. In order to answer this question, we also need to know the risk and reward: how much it costs to play the game and how much it pays if you win. This is what Vega$ NSX was speaking towards.

Sample Scenario:

Let's say you're playing a game at an arcade. It costs $0.75 to play, but if you win, it gives you $1.00 (a net gain of $0.25). You've figured you have a 2/3 chance of winning at this game.
​

Q:

Should you play?​

A:

In order to answer that, calculate how much, on average, you'll win or lose per play. You have a 2/3 chance of coming out ahead by $0.25, but a 1/3 chance of losing $0.75. So, (2/3*.25) + (1/3*-.75) = -0.08. On average, you'll lose about $0.08 per play, so you don't play this game.​

You can use a similar approach to determine whether or not play games of chance. You just need to know: cost of playing, potential win, and chances of winning.

Spoken like a true Vegas bookie. :biggrin: If you ever needed a new career change, I'm sure Sin City would be interested. :biggrin:

 

[docjohn]

the probability that a random event will occure is the same no matter how many times the "things" needed to make it happen act.All thing being equal.Kinda like the plane,,,,,aw forgetaboutit!!!

 

[queenlives]

But the real issue is what are the odds of the plane taking off from the conveyor belt?

+1

i was actually starting to drift to the zaino or zymol discussion.

:tongue:

 

[sahtt]

Re: Who wants to bet?

just count the W's

32/36 = 8/9 = 88.89%

This is correct based off the original question and those definitions of success and failure.

 

[NCC-1701D]

Ojas is completely right on his solution..

Another way to look at the problem is to say this

What are the odds of someone rolling a 1 or 2 on 2 consectutive rolls?

so the odds are now 1/3 x 1/3 = 1/9 so if we apply this to the OP question.

so a person has a 1/9 chance of losing which gives him a 8/9 chance of winning..



opps I just noticed that i wrote the same thing Daed had..

 

[lrsudog]

Are you absolutely sure? I mean... is there some kind of formula that applies to probabilites like this?

When it comes to statistical probabilities such as the ones listed, the odds are always even. Roulette always gives (assuming an America table woth both a 0 and 00) the same odds for a Black number with every new spin of the wheel. If it lands on Black five-hundred times in a row, the odds are still the same for the next spin.

 

[DocL]

Too much math going on here. Just flip a coin. It's a 50:50 chance each time.