A (simple ?) probability question.

I’m not good at figuring probability beyond the most basic. Here’s the problem:

  • A pair of standard six-sided dice.

  • You get one throw.

  • If either of the die show a 6 you win.

  • If either of the die show a 4 you get another throw.

  • On the second throw if either of the die show a 6 you win.

  • No further throws no matter what comes up.

What are the odds you will win?

There’s an 11/36 chance of winning outright on the first throw (1-6, 2-6, 3-6, 4-6, 5-6, 6-6, 6-5, 6-4, 6-3, 6-2, 6-1).

There’s a further 9/36 chance of getting a second roll (1-4, 2-4, 3-4. 4-4, 5-4, 4-5, 4-3, 4-2, 4-1).

If you get a second roll, there’s again a 11/36 chance of winning on that throw.

So, that’s a total of:

 11/36 + (9/36)*(11/36)

= 495 / 1296
= 55/144
= 38.2%

You are not looking for combinations only sixes. I would say a 1 in 6 of winning on the first throw and also a 1 in 6 if getting another turn with the same 1 in 6 if you get the other turn so my guess wuld be closer to 7/36

I tried to edit but I was too late, I would say 1 in 3 for winning on first roll, 4 in 15 for loosing on first roll, 2 in 5 for getting another roll if you don’t win. and the same 1 in 3 on the third roll. Not sure what the total would be but it seems better than 50/50. Combinations don’t count here as they do in craps.

You could say that, but you’d be wrong. The calculations done by Great Antibob are correct.

A combination is not needed to accomplish winning. You could throw the dice one at a time and not even throw the second die if the first roll gave you a six. So you would have to add a 0-6 and a 6-0 to his possibilities.

the OP clearly states that you only get one throw, two on the exception that you get a ‘4’. even if we allow for your changes to the OP …

what is this? you either throw the dice or you don’t. you’re conflating the possiblities of two exclusive events.

So HoneyBadgerDC, what are the chances of winning that game if you start with 6 dice instead of 2?

The combination of 6-6 would have to be removed. These dice do not have to be thrown at the same time if one lands before the other and is a 6 the game is all over. Possible combinations are not part of this. The odds are 1/6 for each dye thrown and if you throw them 6 times the odds are 1/1 it will hit, 3 dies thrown would be 50/50 chance. You guys are looking at combinations as you would in craps.

The question is really about throwing 1 die twice, throwing pairs of dice to accomplish this is only a time saver.

Are you saying that if you roll six dice (or one die six times) you’re certain to roll a six?

So I would have a 1/1 chance to win the game, with 6 dice?

Final answer?

I never said that, the odds suggest you would be but odds don’t mean anything will actually happen. After 600 rolls you should have approx 100 of each number pop up.

This is true, but this perspective tends to make a mess out of the math, which is why Great Antibob simply listed all the combinations that fit the required criteria. That is, there is no way to roll and win that is not in the list. But, if you want a combinationless analysis, here you go.
Here is the analysis of the probability of winning on the first roll:

What is the probability that the first die rolled is a 6? 1/6.

What is the probability that the second die rolled is a 6 if the first is not? (You need this caveat because if there are two 6s, you don’t win twice. You win on the first one, and then you’re done.) 1/6 × 5/6, which is equal to 5/36.

Thus, the total probability of winning on the first roll is 1/6 + 5/36, which is 11/36, as Great Antibob stated.
Now, here is the probability of getting to roll again:

What is the probability that the first die rolled is a 4 if you don’t win on this roll? 1/6 × 5/6 = 5/36.

What is the probability that the second die rolled is a 4 if the first is not a 4 and if you don’t win on this roll? 1/6 × 4/6 = 4/36.

Thus the probability of rolling again is 5/36 + 4/36 which is 9/36.
Then, calculating the total probability is as Great Antibob explained
11/36 + 9/36 × 11/36 = 495/1,296 = 55/144.

(I’m pretty sure this is right, anyway.)

The possible combos are not as Great listed them, correctly they would be
There are no combinations involved in this.

A pair of dice have 36 possible combinations, it would take 6 dice to give you the same 36 possibles when only looking for results from 1 die.

If you don’t believe that RadicalPi’s answer is correct, please point out the mistake.

It doesn’t matter whether you roll the dice at the same time, or one and then wait to see if you should roll the other. The calculations to get there are different, but the result is the same. If you roll both at the same time – Great Antibob’s set method is correct, if you roll them one after the other, RadicalPi is correct.

The reason for this is that the roll of the second die is dependent on the first roll. Specifically, you have a 5/6 chance that you will make the second roll. This is basically what RadicalPi was showing.

How in the hell do you come up with that?

Throw one die. There’s a 1 in 6 chance that you win, and a 5 in 6 chance you don’t.

If you don’t win, throw the second die. Again, there’s a 1 in 6 chance you win.

So your odds of winning with both dice are:

1/6 + 5/6 * 1/6 = 11/36 or 11 in 36

Not 1 in 3. :rolleyes:

It’s an easy mistake. Since “+” is effectively the “or” operator in probability, “die one OR die two” looks a lot like 1/6+1/6=2/6=1/3. It’s the dependence relationship that gets you, so it ends up as 1/6+(5/6*1/6)=11/36 (1/6 OR 1/6 given you lose the first).