Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
The game of Yahtzee is played with five dice. On the first turn, a player rolls all five dice, and then may decide to keep any, all, or none of the dice aside before rolling again. Each player has a maximum of three rolls to try to get a favorable combination of dice "kept" on the side.
If a player rolls two 2s and two 4s on his/her first roll, and keeps all four of these dice aside, what is the probability of getting a full house (three of one value and two of another) in one of his/her next two rolls? (ie what is the probability of getting either a 2 or a 4 in one of the next two rolls?)
HintThink of the probability of NOT getting a full house.
The answer is NOT 2/3 because you cannot add probabilities. On each roll, the probability of getting a 2 or a 4 is 1/3, so therefore, the probability of not getting a 2 or a 4 is 2/3. Since the die is being rolled twice, square 2/3 to get a 4/9 probability of NOT getting a full house in two rolls. The probability of getting a full house is therefore 1 - 4/9, or 5/9.
May 19, 2006
|You can add probabilities together, but you have to be looking at the right probabilities.|
The odds of getting a full house on the first roll is 1/3
the odds of not getting on the first roll is 2/3, and the odds of getting it on the second roll (not taking into consideration the first roll) are 1/3
So the odds of getting it on the second roll are
2/3 * 1/3 = 2*1 / 3*3 = 2/9
The odds of getting it either on the first or second roll are
1/3 (odds on roll one) + 2/9 (odds on it going to roll two and getting it then)
1/3 + 2/9 = 3/9 + 2/9 = (3+2)/9 = 5/9
Of course, the odds of not getting it on either roll are 2/3 * 2/3 = 4/9
The reason why this becomes relevant is if one of the roll doesn't end up as a full house, his opponent doesn't notice the roll has been made and he makes another role, it's easier to work out the odds of getting this on the third role after not getting it the first two, and add this to the odds this time around.
May 19, 2006
|Blagh kind-of already said this, but the teaser doesn't specify that the answer must be the probability of getting a 2 or 4 on the second roll, but rather the probability on either roll. If the first roll does come up a 2 or a 4, you wouldn't bother rolling again, taking the second roll completely out of the equation. Thus, a single probability of 5/9 does not work for both rolls, as the first roll is specifically a 1/3 probability. If you want to ignore the fact that you are playing Yahtzee and throw the dice twice no matter what numbers you got, the probability of getting a 2 or 4 on 'both' rolls is 1/9. |
Anyway, my point is that while your math is technically correct, in my opinion your answer should reflect both roll probabilities as your question clearly allows for both.
Nice teaser, I liked it!
May 22, 2006
|It seems that AwwwSweet makes a valid point. I'm not a math person but there is a logical factor here for a second roll if the desired point is missed.|
May 22, 2006
|But getting a 2 or a 4 on the first roll is a subset of getting it on "either" roll. I thought it was difficult to calculate the probabilities of each scenario so I thought my answer was simplest -- find the probability of not getting a 2 or 4 and subtracting that from one. That avoids all of the "what if you get it on the first roll" problems, I think.|
Thanks for all of your comments by the way -- this is my first teaser.
May 22, 2006
|Agreed Buenos, IF you had qualified finding a 2 or 4 only on the second roll. However, you said either roll, and in Yahtzee if you did get the 2 or 4 on the first roll you simply wouldn't roll again.|
Again, it was a terrific teaser and I am not trying to detract from a great probability problem; I just think it could be qualified more clearly.
May 22, 2006
|OK, I see what you mean AwwwSweet. However, I meant that after your first roll you have two 2s and two 4s, and ::at that point:: you want to assess the probabilities. You want to get a full house in one of your next two turns -- and I think my answer reflects the probability right after the initial roll, when there are two rolls remaining. Therefore, it doesn't really matter which turn you get the full house -- you decide after the first roll whether you want to go for the full house or maybe for more fours.|
May 29, 2006
|Buenos, you are quite right in saying it has to be assessed from where you are in the game (which is how it states it in your teaser). I'm not sure about the hint, though. I just calculated it like Blagh. I suppose it avoids the complication about the 3rd roll only being necessary if the 2nd wasn't successful - Nice to see there's more than one way to look at the problem.|
Jul 05, 2006
|The fact that you wouldn't roll again if you got a 2 or 4 on the first roll decreases the number of outcomes and alters the probability. The way I see it is this:|
If you roll a:
1, 3, 5, or 6 - you roll again, and get the full house 2/6 of the time. This totals to 8 full houses in 24 attempts.
2 or 4 - you don't roll again, and this counts as only 1 outcome, for a total of 2 full houses in 2 attempts.
We can add these totals since they are based on the 26 possible outcomes. The result is a full house 10/26 of the time.
Does this make sense to anyone else?
Aug 09, 2006
|Plman yes there are 26 diffrent possible outcomes however they don't all have the same probability. For instance the probability of roling a 2 the first time is 1/6 but the probability of roling a 3 then a 2 is only 1/36 so you cannot use the method of counting outcomes because they are no longer equally likely.|
May 19, 2007
|I do not get it|
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