Jackpot Risk [Single]

[.Nu-13]

Jackpot Risk
Quick-Play Spell Card
Activate only when you activate a card's effect that lets you draw cards. Roll a six-sided die 3 times. If at least 2 results are the same, the effect of a card you activated becomes "Draw 4 cards". If all results are the same, the effect of a card you activated becomes "Draw 7 cards" If all results are different, discard all cards in your hand.

Because Gamble Turbo is abviously Tier 1 Deck [/sarcasm]

Risky, but if you can roll well (or cheat), it's +6 out of nowhere. And discarding effect can be also good in Fableds/Dark Worlds

[SP0RE]

[quote name='Laevatein' timestamp='1282760629' post='4564614']
Jackpot Risk
Quick-Play Spell Card
Activate only when you activate a card's effect that lets you draw cards. Roll a six-sided die 3 times. If at least 2 results are the same, the effect of a card you activated becomes "Draw 4 cards". If all results are the same, the effect of a card you activated becomes "Draw 7 cards" If all results are different, discard all cards in your hand.

Because Gamble Turbo is abviously Tier 1 Deck [/sarcasm]

Risky, but if you can roll well (or cheat), it's +6 out of nowhere. And discarding effect can be also good in Fableds/Dark Worlds
[/quote]
When your hand is Empty
Activate Jar of Greed, Activate this, Roll Dice, Profit!

It means you lose 1 card, a minimal risk!

I would use this, 3 times in my deck!

[CrabHelmet]

The probability of matching at least two of the three dice is higher than you might guess - it's 4/9, which is almost 50%. And if you're lucky, the massive advantage it gives you basically wins you the duel. Terrible design.

[SP0RE]

[quote name='Crab Helmet' timestamp='1282764290' post='4564868']
The probability of matching at least two of the three dice is higher than you might guess - it's 4/9, which is almost 50%. And if you're lucky, the massive advantage it gives you basically wins you the duel. Terrible design.
[/quote]

4/9?

I got 0.0277777778!

1/6*1/6

[CrabHelmet]

[quote name='Thunderpants' timestamp='1282765701' post='4564955']
4/9?

I got 0.0277777778!

1/6*1/6
[/quote]

That's the probability of all three matching. However, if you are fluent in the English language, you will see that I was referring to the probability of matching at least two of the three, not all three. Similarly, if you have a basic understanding of mathematics, you will see that two and three are not the same number.

[SP0RE]

[quote name='Crab Helmet' timestamp='1282766899' post='4565027']
That's the probability of all three matching. However, if you are fluent in the English language, you will see that I was referring to the probability of matching at least two of the three, not all three. Similarly, if you have a basic understanding of mathematics, you will see that two and three are not the same number.
[/quote]

I know what I mean in maths...

1/6 of a chance of getting 1 number.
1/6 of a chance of getting another number.

1/36 of a chance to get the same number twice...

or 1/216 chance of same number 3 times...

[CrabHelmet]

[quote name='Thunderpants' timestamp='1282767360' post='4565057']
I know what I mean in maths...

1/6 of a chance of getting 1 number.
1/6 of a chance of getting another number.

1/36 of a chance to get the same number twice...

or 1/216 chance of same number 3 times...
[/quote]

You lack the slightest comprehension of combinatorial mathematics, so I'm going to need to give you a refresher course.

Let's begin with the question of matching all three numbers, because it is the simplest.

You are correct that the chance of rolling a given number three times in a row is 1/216; however, we are not asked to roll a specific given number three times in a row but rather ANY number three times in a row. We have a 1/216 chance of triple 1's; a 1/216 chance of triple 2's; a 1/216 chance of triple 3's; a 1/216 chance of triple 4's; a 1/216 chance of triple 5's; and a 1/216 chance of triple 6's. These are obviously mutually exclusive - if you rolled three 1's, you certainly didn't roll three 2's - and they all satisfy the Draw 7 condition, so the probability is their sum: 1/36.

To put it another way, think of it like this: suppose that, instead of rolling the three dice simultaneously, we roll them one at a time, rolling Die A before Die B and Die B before Die C. Obviously, this won't have any impact on the final probability of matching numbers - when we roll the dice doesn't affect their numbers, after all - but it does make the situation easier to analyze.

Now, we start by rolling Die A, and we get a number. That number might be 1. It might be 2. It might be 3, 4, 5, or 6. It doesn't matter; we can roll any of the six numbers and this will work fine. Call the number we roll X. Since we define X to be whatever we rolled on Die A, the probability of rolling X on Die A is obviously 1, certainty. Next, we roll Die B. Since we desire three of a kind, we need to match Die A, so Die B needs to also roll X, which it does with a probability of 1/6. Finally, we roll Die C, which also needs to roll X, again with probability 1/6. Thus, the final probability of rolling X on all three dice is 1*1/6*1/6=1/36.

Remember, you're not trying to match the three dice to some previously-specified number; you're trying to match them to each other. In other words, you're really trying to match [b]two[/b] dice - B and C - to whatever value happened to show up on Die A.

That's why the probability of matching all three dice is 1/36. I could explain where the 4/9 number for matching only two dice comes from if you like, but it's a bit more complicated and I fear you'll have issues understanding even this simpler problem.

[SP0RE]

[quote name='Crab Helmet' timestamp='1282768605' post='4565117']
You lack the slightest comprehension of combinatorial mathematics, so I'm going to need to give you a refresher course.

Let's begin with the question of matching all three numbers, because it is the simplest.

You are correct that the chance of rolling a given number three times in a row is 1/216; however, we are not asked to roll a specific given number three times in a row but rather ANY number three times in a row. We have a 1/216 chance of triple 1's; a 1/216 chance of triple 2's; a 1/216 chance of triple 3's; a 1/216 chance of triple 4's; a 1/216 chance of triple 5's; and a 1/216 chance of triple 6's. These are obviously mutually exclusive - if you rolled three 1's, you certainly didn't roll three 2's - and they all satisfy the Draw 7 condition, so the probability is their sum: 1/36.

To put it another way, think of it like this: suppose that, instead of rolling the three dice simultaneously, we roll them one at a time, rolling Die A before Die B and Die B before Die C. Obviously, this won't have any impact on the final probability of matching numbers - when we roll the dice doesn't affect their numbers, after all - but it does make the situation easier to analyze.

Now, we start by rolling Die A, and we get a number. That number might be 1. It might be 2. It might be 3, 4, 5, or 6. It doesn't matter; we can roll any of the six numbers and this will work fine. Call the number we roll X. Since we define X to be whatever we rolled on Die A, the probability of rolling X on Die A is obviously 1, certainty. Next, we roll Die B. Since we desire three of a kind, we need to match Die A, so Die B needs to also roll X, which it does with a probability of 1/6. Finally, we roll Die C, which also needs to roll X, again with probability 1/6. Thus, the final probability of rolling X on all three dice is 1*1/6*1/6=1/36.

Remember, you're not trying to match the three dice to some previously-specified number; you're trying to match them to each other. In other words, you're really trying to match [b]two[/b] dice - B and C - to whatever value happened to show up on Die A.

That's why the probability of matching all three dice is 1/36. I could explain where the 4/9 number for matching only two dice comes from if you like, but it's a bit more complicated and I fear you'll have issues understanding even this simpler problem.
[/quote]


Thank you for the explanation. I see where my calculations where astray, and where yours where correct. I was wrong and I admit it. I was calculating getting triple X or triple U, V, W, X, Y or Z.

[.Nu-13]

[quote name='Crab Helmet' timestamp='1282768605' post='4565117']
You lack the slightest comprehension of combinatorial mathematics, so I'm going to need to give you a refresher course.

Let's begin with the question of matching all three numbers, because it is the simplest.

You are correct that the chance of rolling a given number three times in a row is 1/216; however, we are not asked to roll a specific given number three times in a row but rather ANY number three times in a row. We have a 1/216 chance of triple 1's; a 1/216 chance of triple 2's; a 1/216 chance of triple 3's; a 1/216 chance of triple 4's; a 1/216 chance of triple 5's; and a 1/216 chance of triple 6's. These are obviously mutually exclusive - if you rolled three 1's, you certainly didn't roll three 2's - and they all satisfy the Draw 7 condition, so the probability is their sum: 1/36.

To put it another way, think of it like this: suppose that, instead of rolling the three dice simultaneously, we roll them one at a time, rolling Die A before Die B and Die B before Die C. Obviously, this won't have any impact on the final probability of matching numbers - when we roll the dice doesn't affect their numbers, after all - but it does make the situation easier to analyze.

Now, we start by rolling Die A, and we get a number. That number might be 1. It might be 2. It might be 3, 4, 5, or 6. It doesn't matter; we can roll any of the six numbers and this will work fine. Call the number we roll X. Since we define X to be whatever we rolled on Die A, the probability of rolling X on Die A is obviously 1, certainty. Next, we roll Die B. Since we desire three of a kind, we need to match Die A, so Die B needs to also roll X, which it does with a probability of 1/6. Finally, we roll Die C, which also needs to roll X, again with probability 1/6. Thus, the final probability of rolling X on all three dice is 1*1/6*1/6=1/36.

Remember, you're not trying to match the three dice to some previously-specified number; you're trying to match them to each other. In other words, you're really trying to match [b]two[/b] dice - B and C - to whatever value happened to show up on Die A.

That's why the probability of matching all three dice is 1/36. I could explain where the 4/9 number for matching only two dice comes from if you like, but it's a bit more complicated and I fear you'll have issues understanding even this simpler problem.
[/quote]

Will sig