A strip of 41 squares is numbered 0,1,2,…,40 from left
to right and a token is placed on the square marked 0. Pinar
rolls a pair of standard six-sided dice and moves the token right a number of squares equal to the total of the dice roll. If Pinar rolls doubles, then she rolls the dice a second time and moves the token in the same manner. If Pinar gets doubles again, she rolls the dice a third time and moves the token in the same manner. If Pinar rolls doubles a third time she simply moves the token to the square marked 36.
The expected value of the square that the token ends on can be expressed as a/b where a and b are coprime positive
integers. What is the value of a+b.
Solve using a probability tree.
First calculate the outcomes with distinct rolls for a sum of P(x)=30/36.
For simplicity, the probability is multiplied by 36 to use integers.
sum=x 36*P(x) x*36*P(x)
3 2 6
4 2 8
5 4 20
6 4 24
7 6 42
8 4 32
9 4 36
10 2 20
11 2 22
Sums of 2 and 12 are always doubles so they do not appear in the table above.
Doubles have a probability of 6/36=1/6, so the above table should be multiplied by (1+1/6) to account for the first double then non-doubles.
Getting doubles twice has a probability of (6/36)², so we multiply the above probabilities by (1+1/6+1/36)=43/36 for all cases except 3 doubles.
3 doubles have a probability of (1/6)³, so the complete table becomes:
sum=x P(x) x*P(x)
3 43/648 43/216
4 43/648 43/162
5 43/324 215/324
6 43/324 43/54
7 43/216 301/216
8 43/324 86/81
9 43/324 43/36
10 43/648 215/324
11 43/648 473/648
36 1/216 1/6
for a sum of probabilities of 1.
Expected value of outcome
=Σ x*P(x)
=1541/216
^Wrong a+b is < 1000
The way I interpreted the question is that she does not move at all if she rolls doubles.
After rereading, it looks like that when she rolls doubles, she moves the token before rolling again, and on the third double, she advances it to 36.
If this latter interpretation is correct, it will be necessary to adjust the calculations for the new situation.
Try the same principle as above for the new situation. If you don't get the right answer, post what you've got and we'll take it from there.
even with the latter interpretation i still cannot get a result of a+b<1000. could you try working on it?
122
To find the expected value of the square that the token ends on, we need to calculate the probabilities of each possible outcome and multiply them by the value of the square.
Let's consider the possible outcomes:
1. Pinar rolls non-doubles on all three rolls.
2. Pinar rolls doubles on the first roll, but not on the second and third rolls.
3. Pinar rolls doubles on the first and second rolls, but not on the third roll.
4. Pinar rolls doubles on all three rolls.
For each outcome, we need to calculate the probability and the expected value.
1. Pinar rolls non-doubles on all three rolls:
The probability of this outcome is (5/6)^3 since there are five possible non-doubles on each roll out of six possible outcomes. The expected value for this outcome is 0 since the token does not move from square 0.
2. Pinar rolls doubles on the first roll, but not on the second and third rolls:
The probability of this outcome is (1/6) * (5/6) * (5/6) since there is a 1/6 chance of rolling doubles on the first roll and a 5/6 chance of rolling non-doubles on the second and third rolls. The expected value for this outcome is the sum of the values of the squares she can end up on, which is 1 * (1 + E) where E is the expected value.
3. Pinar rolls doubles on the first and second rolls, but not on the third roll:
The probability of this outcome is (1/6) * (1/6) * (5/6) since there is a 1/6 chance of rolling doubles on the first and second rolls and a 5/6 chance of rolling non-doubles on the third roll. The expected value for this outcome is the sum of the values of the squares she can end up on, which is 2 * (1 + E) where E is the expected value.
4. Pinar rolls doubles on all three rolls:
The probability of this outcome is (1/6) * (1/6) * (1/6) since there is a 1/6 chance of rolling doubles on each roll. The expected value for this outcome is 36 since the token moves directly to square 36.
Now we can set up an equation to solve for the expected value:
E = (1/6) * (5/6) * (5/6) * (1 + E) + (1/6) * (1/6) * (5/6) * (2 + E) + (1/6) * (1/6) * (1/6) * 36
Simplifying the equation, we get:
E = (25/216) * (1 + E) + (10/216) * (2 + E) + (1/216) * 36
E = (25/216) + (25/216)E + (10/216) + (10/216)E + (36/216)
191/216 E = 71/216
E = 71/191
Hence, the value of a is 71 and the value of b is 191. Therefore, a + b = 71 + 191 = 262.