Okay. I’m debating something with a friend of mine.

If you’re playing a game with a single die where you need to roll 6 and are only give two rolls, he believes your chances of winning are 2-in-6.

I argue that based on that logic that if you were to roll six times you’d have a 100% chance of winning, which is just not true. You could just as easily roll a 1 six times.

When I said that he said the chances of rolling a 1 all six times is close to impossible.

Now I agree that multiple rolls should technically give you more chances at winning, but that it doesn’t give you better odds at winning (if that makes sense). I also agree that rolling the same number multiple times in a row is probably more difficult the more rolls that are involved. So maybe it wouldn’t be just as likely to roll a 1 all six times as rolling a 6 once, but I don’t really know.

I hope the teeming masses can clear things up for me.

Your chance of not rolling a 6 after one roll is 5 in 6. Your chances of not rolling a 6 after two rolls is (5 in 6) times (5 in 6) or 25 in 36. Thus your chance of rolling at least one 6 in two rolls is 11/36, which is slightly less than 12/36 = 1/3 = 1/6 + 1/6.

To your point, if you roll 6 times, your chance of not rolling a 6 are 5^6/6^6 or 33%, not zero as your friend would say.

Wow! Thanks for the quick response. I showed my friend via IM and he concedes defeat. But the debate continues.

Now he proposes that in this game, that there is still a way to give someone a 2 in 6 chance of rolling a 6. That is by not counting a roll if it was the same as your last.

Example if you roll a 1, and then roll a 1 again, you get a third roll. If you roll a 1 a third time you get a fourth. He says this effectively removes that number from the die meaning that the second time you roll instead of having a 1-in-6 chance you have a 1-in-5 chance.

Using your math. The chances of not rolling a six the first time are 5-in-6 and the chances of not rolling it the second time become 4-in-6, or 20 in 36. Which means there is a 4/9 or 44% chance of rolling a 6. Of course by second time I mean the roll after the first roll which is not the same as your first roll (I know it’s confusing).

So this method actually gives the player a higher chance of success. I guess this is where I am confused. Shouldn’t it be closer to a 33% chance?

Say then your first roll is a 1. Then after a few more attempts you keep rolling a 1, but eventually roll a 2. Then after a few more times of rolling 2s you get a 3. And so on until you have rolled everything, but a 6.

Your chances of not rolling 6 the first time are 5-in-6, then 4-in-6, 3-in-6, 2-in-6, 1-in-6, and finally 0-in-6.

Roll 1: 16.666…%
Roll 2: 44.444…%
Roll 3: 72.222…%
Roll 4: 90.740…%
Roll 5: 98.456790123…%
Roll 6: 100%

But this doesn’t add up for me. First if anything the second roll should be less than 33.333…% chance, but more than the 11/36 (30.555…%) chance previously.

Also isn’t there a very small chance that out of an infinite number of rolls you never roll a 6.

So isn’t a 100% chance a pipe dream? Can’t it only be a 99.999…% chance? I know in math repeating 9’s after a decimal point are equal to 1, but still there is that chance that you’ll never roll a 6. Or maybe I am wrong.

To expand on what JWT Kottekoe said, if you need to roll a 6 to win, and are given n rolls, your chances of losing are (5/6)[sup]n[/sup]. (That’s the chance that the first roll isn’t a 6 x the chance that the second roll isn’t a 6 x … x the chance that the last roll isn’t a 6.)

So the probability of winning is 1 - (5/6)[sup]n[/sup].

The greater n is, the closer this comes to 1, but it never reaches 1 (for any finite number of rolls).

Of course, there is in reality no such thing as an infinite number of rolls.

If I understand you correctly, there is a very small chance that, out of any particular finite number of rolls, you never roll a 6.

If you take the calculus approach, and let x be a number that approaches infinity, then the limit of (the probability that you never roll a 6 in x rolls) as x approaches infinity is 0.

If you take an approach based on infinite sets, if you consider the set of all infinite sequences of die rolls, there would be some such sequences that don’t include any sixes ({1, 1, 1, 1, 1, …} for example). But the subset of such sequences would, I think, have measure 0, meaning that (at least under some models of probability) the probability of randomly selecting such a sequence is 0.

If you eliminate already-rolled numbers from the non-six outcomes, they should be removed from the total number of possibilities as well. It’s like rolling a 5-sided die.
In other words, the odds you don’t roll a 6 the first time is 5/6, and the odds you don’t roll a 6 the “second” time is 4/5. This gives a result of 5/6 * 4/5 = 20/30, so your friend is correct.

Now, what about the repeated die rolls? In that case, you’d need to keep account of a series of outcomes and split. It becomes more complicated - the chance you roll a non-six, and then any different number + chance you roll a non-six, then repeat, then a different number + …
You’ll end up with an infinite series as one of the possible outcomes. Fortunately, however, it will sum to 0, and doesn’t affect the value of the probability. (Important to note about the way probability is defined : just because something has 0 probability doesn’t mean it can’t happen).

I think he’s right. I modeled this as a Markov chain with four states: first roll, reroll, win (i.e., roll a six) and loss. From first roll, you have a 1/6 chance of going to win, and a 5/6 chance of going to reroll. From reroll, you have a 1/6 chance of staying in reroll, a 1/6 chance of going to win, and a 2/3 chance of going to loss. From win, you’re guaranteed to stay in win, and likewise for loss. The initial distribution guarantees that you’ll start in first roll. This is an absorbing Markov chain, and so we can apply the standard theory to show that, given that you start in first roll, you have a 1/3 chance of ending up in win, and a 2/3 chance of ending up in loss.

You’re correct about the 6th die roll being 100%. I know you’re concerned about that possibility that you’ll roll repeats for eternity, but it doesn’t really matter. Saying that roll 6 is 100% chance a 6 (assuming 1-5 have been rolled previously) is saying that “If a sixth roll is made, it will be a 6 for sure.” The 100% part tells you the latter of this statement while you’re concerned with the former.

Rolling 1-5 for eternity is no different than your wife coming in and asking you to take out the garbage, or a train running you over, or the die spontaneously combusting. All those things prevent a sixth roll, yes, but that’s not what the 100% calculation is telling you. It presumes a sixth roll.

However, OP, probabilities asside, there’s an easier way to give a player a 1/3 chance of winning. They win on a 6 or a 5. No multiple rolls necessary.

What if you were only given one roll, but rolling a 1 automatically gives you a reroll?

Is that a 1/5 chance?

If so how is that different from rolling a 1 the first time and getting a second roll where now 1 is a reroll. Why does it go from a 1/5 to a 1/3 chance.