Two questions related to probability theory and pedagogy

A housemate of mine and I disagree on the following question:

Let's say that we play a game of yahtzee. Of the five dice you throw, two dice obtain the value 1, two other dice obtain the value 2, and one die shows you six dots on the top side. Given the fact that you haven't thrown a "full house" yet, you start throwing the die which value is 6 again and again, until you throw a one or a two. You get to throw the die of six one or two times. If you throw a one or a two the first time, you stop, because now you have the "full house" already. If you haven't thrown a one or a two with the die of six, you throw it again, hoping for a one or a two this time.

Now, what is the probability that you throw a one or a two with the fifth die after two turns? (Given the way a rational person operates in this situation.)

My take on this question was the following: the probability that you throw a one or a two the first time with the fifth dice is $1/3$, and the probability that you don't throw a dice of which the value is one or two the first time, but you do throw a one or a two the second time is $2/3 \cdot 1/3 = 2/9$. Adding these values gives you the probability: $1/3 + 2/9 = 3/9 + 2/9 = 5/9$.

My housemate, however, argues that the chance to throw a one or a two the first time is $1/3$, and believes that throwing the fifth dice again, gives you a probability of throwing a one or a two of $1/3$ again. Adding these values gives the expected probability of throwing a full house of $1/3 + 1/3 = 2/3$.

Who is right, my housemate or me?

I strongly believe I am right, but even if you tell me I'm right, I might not be able to convince my housemate of the truth. He argues that my way of reasoning implies that the probability of throwing a one or a two with the fifth dice the second time is smaller than throwing it the first time. Could you please also provide me with a pedagogically sound way to explain him why the probability is $5/9$?

• What if the housemate decided to throw the dice four times? Are the odds of full house 1/3 + 1/3 + 1/3 + 1/3 = 4/3? – Jack Schmidt Nov 23 '11 at 21:41
• If people had a better understanding of probabilities, casinos would go broke, and lotteries would be laughed out of existence. – André Nicolas Nov 23 '11 at 21:47

You are correct. And you can tell your housemate that you are using conditional probability for the second part of your calculation. Given that you did not complete a full house on roll 1, the probability of rolling it on the second try is $\frac{1}{3}$, just as he says. However, that scenario is only one of two that might occur. As you have written, the scenario where you miss on the first try will happen with probability $\frac{2}{3}$. That reduces the weight that we would apply to that $\frac{1}{3}$ on the second roll.

Here's an alternative approach. Make a $6\times6$ table where the columns represent rolls on try 1 and the rows represent rolls on try 2. (Pretend that if you succeed during try number 1 that you throw again just for fun.) If you count, $20$ of these entries represent situations where you make your full house. These $36$ situations are all equally likely to occur, so the probability of eventual success is $\frac{20}{36}$, or $\frac{5}{9}$.

Lastly, as one more response to your house mate, bring up this idea of tossing the second die just for fun even when you have already made the full house the first time. His probability of $\frac{1}{3}$ on roll 2 is counting the times that you would roll a 1 or a 2 on the second try even though you already succeeded on roll number 1. His probability over counts because of this.

Actually, that might be the most convincing argument for him. He is fine to add $\frac13+\frac13$, but then he has double counted those situations where you would roll a 1 or 2, and then again roll a 1 or 2. So to account for this double counting, he can correct by subtracting the chance that you succeed on both rolls: $\frac13+\frac13-\frac19$

You are right. The easiest way to see this is to recognize the probability that you roll a $1$ or a $2$ as

$$1 - Pr(\text{not rolling a }1 \text{ or } 2) = 1 - (4/6)^2 = 5/9.$$

Imagine throwing that die twice. Repeat this 999 times.

Out of the 999 trials, how many of them would have the first throw showing 1 or 2? Well, around 1/3 of them. 333 of the 1000 trials have the first throw resulting in 1 or 2.

Now, how many of the trials will have the second throw resulting in 1 or 2, but not the first? Well, around 666 of the trials do not have the first throw showing 1 or 2 (see above). And around 1/3 of these 666 will have the second throw showing 1 or 2. So one third of 666 is 222.

So the probability of the first throw showing 1 or 2 or the second showing 1 or 2, but not the first is $(333+222)/999=5/9$.

The last sentence bears repeating: you want the probability that either

1) the first throw results in 1 or 2

or

2) the first throw does not result in 1 or 2 and the second throw results in 1 or 2.

You add the probabilities of 1) and 2).

• And if your housemate is really obstinate, you could actually perform this experiment many times and establish the result empirically. (I'll run off and hide now, saying that with all these mathematicians around...) – David Mitra Nov 23 '11 at 21:52

We can write the process out as an extensive-form game tree:

• First throw:

• Probability $\frac 13$: rolled 1 or 2, stop.
• Probability $\frac 23$: rolled 3, 4, 5 or 6, roll again:

• Probability $\frac 13$ (total probability $\frac 23 \cdot \frac 13 = \frac 29$): rolled 1 or 2, stop.
• Probability $\frac 23$ (total probability $\frac 23 \cdot \frac 23 = \frac 49$): rolled 3, 4, 5 or 6, no more throws left.

The overall probability of getting 1 or 2 is then $\frac 13 + \frac 23 \cdot \frac 13 = \frac 59$, while the probability of getting 3, 4, 5 or 6 is simply $\frac 23 \cdot \frac 23 = \frac 49$.