In the dice game Yahtzee, a player has

three tries at rolling some or all of a set of five dice. Each player is trying

to achieve results such as three of a kind, two pairs, full house, so on. A

yahtzee occurs when a player rolls five of a kind. If Cheryl rolls a pair of

2s on the first toss, and then rolls only the non-2s showing on the subse-

quent two tosses, find the probability that she gets a yahtzee.

Update:

Unit is on binomial distribution

Relevance
• 2 years ago

I think the intended answer is to just have Cheryl continually rolling for twos, since she already has 2 twos.

I'll answer that way first, but then add a slight modification based upon what a smart player would actually do.

She's already rolled 2 twos on her first roll.

Binomial probability formula:

P(X=k) = C(n,k) * p^k * q^(n-k)

p = 1/6

q = 5/6

n = number of dice being rolled

k = number of desired outcomes

To get a Yahtzee with all twos, there are 4 cases.

CASE 1:

She could roll 3 twos on the second roll and then stop -- she has a Yahtzee.

P(X=3) = C(3,3) * (1/6)^3 * (5/6)^0

= 1 * 1/216 * 1

= 1/216

CASE 2:

She could roll 2 twos and then on the next roll get 1 two

Next roll (3 dice, 2 twos)

P(X=2) = C(3,2) * (1/6)^2 * (5/6)^1

= 3 * 5/216

= 15/216

Last roll (1 die, 1 two)

P(X=1) = C(1,1) * (1/6)^1 * (5/6)^0

= 1 * 1/6 * 1

= 1/6

15/216 * 1/6

= 15/1296

CASE 3:

She could roll 1 two and then on the next roll get 2 twos

Next roll (3 dice, 1 two)

P(X=1) = C(3,1) * (1/6)^1 * (5/6)^2

= 3 * 25/216

= 75/216

Last roll (2 dice, 2 twos)

P(X=2) = C(2,2) * (1/6)^2 * (5/6)^0

= 1/36

75/216 * 1/36

= 75/7776

CASE 4:

She could roll 0 twos and then on the next roll get 3 twos

Next roll: (3 dice, 0 twos)

P(X=0) = C(3,0) * (1/6)^0 * (5/6)^3

= 1 * 1 * 125/216

= 125/216

Next roll (3 dice, 3 twos)

P(X=3) = C(3,3) * (1/6)^3 * (5/6)^0

= 1 * 1/216 * 1

= 1/216

125/216 * 1/216

= 125/46656

ALL 4 CASES:

1/216 + 15/1296 + 75/7776 + 125/46656

Get a common denominator of 46656:

216/46656 + 540/46656 + 450/46656 + 125/46656

= 1331/46656 (~2.853%) <--

Okay, that's probably the intended answer, but I'm going to have you consider one additional case. Let's assume that on the second roll, she gets 3 numbers that are the same, but not twos (say three 3s, or three 4s). In that case, she would switch and try to get a Yahtzee in the new number.

We need to split the case of getting 0 twos on the second roll.

5 cases are she gets 3 of the same number that aren't twos --> shift strategy to the new number

120 cases are she gets 3 non-twos that aren't all the same --> keep the old strategy and try to get 3 twos on the last roll.

If she gets 3 of the same, then she will be rolling *2* dice to get a match. Otherwise it is the remaining 3 dice as before.

Case 4a - 3 of the same number, then 2 more that match:

P(3 of the same, then 2 more that match)

= 5/216 * 1/36

= 5/7776

Case 4b - no matches, then 3 more twos.

= 120/216 * 1/216

= 120/46656

The revised sum is:

1/216 + 15/1296 + 75/7776 + 5/7776 + 120/46656

=

36/7776 + 90/7776 + 75/7776 + 5/7776 + 20/7776

= 226/7776

= 113/3338(~2.906%) <--

1331/46656 (~2.853%) if she blindly keeps rolling for twos no matter what.

113/3338 (~2.906%) if she switches strategy when getting 3 of another number.

• 2 years ago

You'd have to break this down into several pieces.

We know that we are only rolling 3 dice on the 2nd roll, so there are 4 possible outcomes:

rolling 3 2's

rolling 2 2's

rolling 1 2

rolling 0 2's

We have to determine the probability of each of these events happening. If we do it right, the sum of the four should equal 1.

So first, the easy 2. 3 2's and 0 2's:

3 2's : 1/6 chance for 3 dice, or 1/6³ or 1/216

0 2's : 5/6 chance for 3 dice, or (5/6)³ or 125/216

Now we have 1 2. Let's first see what happens if we roll a 2 first, then two non-2's:

1/6 * 5/6 * 5/6 = 25/216

Then multiply this by 3 since it could be x2x or xx2, so:

25/216 * 3 = 75/216

and similar logic for rolling two 2's:

22x = 1/6 * 1/6 * 5/6 = 5/216

Then it could be 2x2 or x22, so multiply this by 3:

5/216 * 3 = 15/216

Now let's double check to make sure we did all this math right. The sum should be 1:

1/216 + 125/216 + 75/216 + 15/216 = 216/216 = 1

So now we're good with roll 2.

If we roll the 1/216 chance of three 2's, we're done so we don't have to roll a third time. So we're done here.

For the other three cases, we have to roll what's left and find the probability of getting all 2's in the final roll. So starting with 0 2's, which was 125/216. In order to get three 2's now is the same 1/216 as it was before. The probability of both events happening (rolling 0 2's followed by 3 2's) is the product of the two probabilities, so:

125/216 * 1/216 = 125/46656

Now for 1 2's, which was 75/216. If this happens we have to roll the last 2 dice as 2's, so 1/36. Multiply this:

75/216 + 1/36 = 75/7776

And finally 2 2's, which was 15/216, we need the one last 2, so 1/6:

15/216 * 1/6 = 15/1296

Now for these four cases, add up the probabilities to get the event that any one of them could happen:

1/216 + 125/46656 + 75/7776 + 15/1296

Getting a common denominator of 46656:

216/46656 + 125/46656 + 450/46656 + 540/46656