My wife and I go back and forth on this one. Assume that you already have your small straight. Lets say your 5 dice are 1,2,3,3,4. To get a large straight, should you re-roll the 3, or re-roll the 1 AND 3? My wife says just the 3, because that way you only need 1 die to go your way instead of 2. I say the 1 AND 3 because it doubles your chances of getting a 5 which you need, and you still have a 1/3 chance that the second dice is a 1 or 6.

Looking for some probability experts to help me through this one.

In: Mathematics

You can get a total of 36 combinations out of 2 die. You need 5;1 or 5;6 or 1;5 or 6;5 to win.

So 4 out of 36 combos are a win. That’s only 1 in 9.

Just rolling the 3 will result in a 1/6 chance of a large straight (obviously)

Rolling the 1 and 3 together will give you your straight if you roll a (1,5),(5.6),(6,5), or (5.1).

Out of 36 possible outcomes only those 4 work. That is a 1/9

Your wife is correct. What you aren’t considering is that if one of your dice land on the right spot you are now in the same exact position as your wife started in: needing a single 1/6 throw to get the one leftover number. However you had to throw the other die to get to that point with some probability of failing, while your wife started there with no luck needed

Assuming that you’ll take the 1 or 6 on your first roll if you don’t get the five, then your odds of getting a large straight are as follows:

1/9 to get LS on first roll

7/36 to get a 5, but no 1 or 6

[1/3 to get LS on subsequent roll]

4/9 to get 1 or 6, but no 5

[1/6 to get LS on subsequent roll]

1/4 to get no 1,5 or 6

[1/9 to get LS on subsequent roll]

So your odds are (1/9) + (7/36 x 1/3) + (4/9 x 1/6) + (1/4 x 1/9) = 10/36

Your wife needs to get a 5 on either of her rolls, which would be 1-(5/6)^2 =11/36

So your wife has the better strategy, beating yours by 1/36.

Let’s do the math !

First launch :

1,2,3,3,4

Your wife plays. She launches the 3. Her strategy is : “I launch this die until I get a five !”

1. She gets a 5 (1/6).

2. She does not (5/6). She launches again, and gets it (1/6).

So she has 1/6+5/6*1/6=1/6+5/36=0.3055…

Now, you. The strategy is : “Launch the 1 and the 3 until you get a 5 (you store it) or a 1-6 (you store it)”.

1,2,3,3,4

You take the 1 and 3 and launch them. You need a 5, and either a 1 or a 6.

So, what can happen if you plan on winning this :

1. The first dice is a 5 (1/6), and

1. the second dice is a 1 or a 6 (1/3)

2. the second dice is neither a 1 or a 6 (2/3), but you throw it again and it’s one (1/3)

2. The first dice is a 1 or a 6 (1/3), and

1. the second dice is a 5 (1/6)

2. it’s not (5/6), but you launch it again and you get it (1/6)

3. The first dice is neither a 5, a 1 or a 6 (1/2), and

1. the second is a 5 (1/6). You launch the first one again, and it’s a a 1 or a 6 (1/3)

2. the second is a 1 or a 6 (1/3). You launch the first one again, and it’s a 5 (1/6)

3. the second is neither a 5, a 1 or a 6 (1/2). You launch the two of them again and

1. the first one is a 5 (1/6), the second one a 1 or a 6 (2/3)

2. the exact opposite

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So, we get :

1/6*(1/3+2/3*1/3)+1/3*(1/6+5/6*1/6)+1/2*(1/6*1/3+1/3*1/6+1/2*(1/6*2/3*2))=0.305…

Well. It’s the same thing, exactly. (And TIL, too, I was instictively going for your strategy).

My old math teacher used to tell me that when I get twice the same result, I should try to understand the why, because there should be a shortcut in the reasoning. Here I just bruteforced it. But it’s beer-time with my roomate, I’ll edit later if I find the courage to dig into it further.

One roll left or two rolls left?