[SVWB] 12%中虹 每10包保底1虹 每包虹卡期望值=?
Let’s break it down step-by-step!
**Base Rate:**
Each pack has a 12% chance for a legendary, unless 9 consecutive packs have
no legendary, then the 10th is guaranteed legendary.
**We need the expected (average) number of legendaries per pack given this “
pity timer” rule.**
---
### **Step 1: Model the Sequence**
- Packs 1–9: Each has 12% chance for legendary, 88% chance for none.
- If packs 1–9 all fail (probability \(0.88^9\)), pack #10 **guaranteed**
legendary.
### **Step 2: Stationary Process**
It’s a periodic process with resets after each legendary. EV per pack will
be average legendaries per cycle divided by cycle length.
#### **The 'cycle':**
A cycle is from the last legendary to the next. There are two ways a cycle
ends:
1) You get a legendary before hitting the 10-pity.
2) You get nothing for 9 packs, then a guaranteed legendary on the 10th.
Let’s compute **expected legendaries per cycle** (which is always 1), and
**expected cycle length**.
---
### **Step 3: Expected Cycle Length**
Let’s define $X$ = number of packs until a legendary is obtained (with pity
timer).
#### **Probability you get legendary in pack $k$** (for $k$ = 1 to 9):
For pack $k$:
- No legendary in packs 1 to $k$ – 1 ($0.88^{k-1}$),
- Legendary in pack $k$ ($0.12$):
$P(\text{legendary in pack }k) = 0.88^{k-1} \cdot 0.12$
For packs 1–9, sum over $k=1$ to $9$.
For pack 10 (pity):
- No legendary in all packs 1–9 ($0.88^{9}$),
- Guaranteed legendary in pack 10.
So, $P(\text{legendary in pack 10}) = 0.88^9$
**Expected cycle length:**
$E = \sum_{k=1}^9 k \cdot (0.88^{k-1} \cdot 0.12) + 10 \cdot (0.88^9)$
---
### **Step 4: Compute the Value**
Let’s compute numbers step by step:
#### Calculate $S = \sum_{k=1}^{9} k \cdot 0.88^{k-1} \cdot 0.12$
You can factor $0.12$ out:
$S = 0.12 \times \sum_{k=1}^9 k \cdot 0.88^{k-1}$
Let's make a table:
| $k$ | $0.88^{k-1}$ | $k \cdot 0.88^{k-1}$ |
|---|----------|------------------------|
| 1 | 1 | 1 |
| 2 | 0.88 | 1.76 |
| 3 | 0.7744 | 2.3232 |
| 4 | 0.681472 | 2.725888 |
| 5 | 0.599698 | 2.99849 |
| 6 | 0.527734 | 3.166404 |
| 7 | 0.464410 | 3.25087 |
| 8 | 0.408679 | 3.269434 |
| 9 | 0.359641 | 3.236769 |
Sum the last column:
$T = 1 + 1.76 + 2.3232 + 2.725888 + 2.99849 + 3.166404 + 3.25087 + 3.269434
+ 3.236769 =$
Let's add step-wise:
- $1 + 1.76 = 2.76$
- $2.76 + 2.3232 = 5.0832$
- $5.0832 + 2.725888 = 7.809088$
- $7.809088 + 2.99849 = 10.807578$
- $10.807578 + 3.166404 = 13.973982$
- $13.973982 + 3.25087 = 17.224852$
- $17.224852 + 3.269434 = 20.494286$
- $20.494286 + 3.236769 = 23.731055$
So,
$S_1 = 0.12 \times 23.731055 = 2.8477266$
Now add pity timer contribution:
$0.88^9 = 0.359641$
$S_2 = 10 \times 0.359641 = 3.59641$
Sum for expected cycle length:
$E = S_1 + S_2 = 2.8477266 + 3.59641 = 6.4441366$
---
### **Step 5: Compute EV legendary per pack**
Each cycle yields **exactly 1 legendary**.
$EV_{\text{legendary/pk}} = \frac{1}{E} = \frac{1}{6.4441} \approx 0.1551$
---
## **Final Answer**
**The expected number of legendaries per pack, given 12% rate and a pity
timer at 10 packs, is about:0.155 egendary per pack.
Or about **15.5%** chance on average per pack over the long run.
答案是15.5% 也就是平均每6.45包會拿到一虹
你答對了嗎
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