# Help with the Probabilty of Rolling Two Ten-Sided Dice Multiple Times Until 100 is Reached

I need some help figuring out the probability of reaching or exceeding 100 based on a number of rolls of two, ten-sided dice. Here's the scenario.

I am starting from zero. I am rolling two (fair) ten-sided dice, to generate a result between 2 and 20. After the roll, I'm recording the number rolled as the 'total' and then rolling again. I'm taking the new result and adding it the total, then rolling again and so on. I'm trying to have the total reach or exceed 100.

Example: On my first roll, I get 12. I record 12 and roll again. On my second roll, I get 7. I add 7 to the current total of 12 to have a new total of 19. Then I roll again.

• How many rolls must I make to have a 25% chance of reaching or exceeding 100?
• How many rolls must I make to have a 50% chance of reaching or exceeding 100?
• How many rolls must I make to have a 75% chance of reaching or exceeding 100?
• How many rolls must I make to have a 90% chance of reaching or exceeding 100?
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$$16 \mbox{ rolls: } \approx 16\%;$$ $$17 \mbox{ rolls: } \approx 30.8\%;$$ $$18 \mbox{ rolls: } \approx 48.4\%;$$ $$19 \mbox{ rolls: } \approx 65.3\%;$$ $$20 \mbox{ rolls: } \approx 79.2\%;$$ $$21 \mbox{ rolls: } \approx 88.7\%;$$ $$22 \mbox{ rolls: } \approx 94.5\%.$$ –  Oleg567 Mar 31 '14 at 14:56

(Scenario with 1 dice):

Denote $P(r,s)$ probability to get sum $s$ by $r$ rolls.

Denote $Q(r,s) = 1-\sum\limits_{q=1}^{s-1}P(r,q)$ $~$ probability to get sum $\ge s$ by $r$ rolls.

Obviously: $$P(1,1)=P(1,2)=\ldots=P(1,10)=0.1;$$

then $$P(r,s) = \sum\limits_{q=s-10}^{s-1} P(r-1,q)\cdot 0.1$$

(i.e. sum $s$ we can get of "previous" sums $s-10$, or $s-9$, ..., or $s-1$).

Step-by-step, we can fill table of probabilities $P(r,s)$.

A few conclusions: $Q(16,100) = 1-\sum\limits_{q=1}^{99}P(16,q) \approx 0.159924$;
$Q(17,100) \approx 0.307628$;
$Q(18,100) \approx 0.483771$;
$Q(19,100) \approx 0.654096$;
$Q(20,100) \approx 0.791809$;
$Q(21,100) \approx 0.887108$;
$Q(22,100) \approx 0.944590$;
$Q(23,100) \approx 0.975252$.

Now you can choose closest values:

$25 \%$: ~ 17 rolls;
$50 \%$: ~ 18 rolls;
$75 \%$: ~ 20 rolls;
$90 \%$: ~ 21 rolls.

(Scenario with 2 dices):

As dices are independent, then all that we need is to consider even number of single rolls, and divide them by $2$.

$Q_2(8,100) =Q(16,100) \approx 0.159924 (\approx 15.99 \%);$
$Q_2(9,100) =Q(18,100) \approx 0.483771 (\approx 48.38 \%);$
$Q_2(10,100)=Q(20,100) \approx 0.791809 (\approx 79.18 \%);$
$Q_2(11,100)=Q(22,100) \approx 0.944590 (\approx 94.46 \%).$

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