Morning Draw Results:
Tennessee (TN)
Pick 3: 716
Pick 4: 4703
Texas (TX)
Pick 3: 173
Pick 4: 7907
Repeating Digit String:
11334677777900
From this string, the digits that appear more than once are:
1, 3, 4, 7, 9, 0 (all repeating)
6 appears only once
2, 5, 8 are missing
So we have:
Repeating Digits (Hot): 0, 1, 3, 4, 7, 9
Single Occurrence (Mild): 6
Missing Digits (Cold): 2, 5, 8
🔍 Analysis of Draws:
Digits in the draws:
Pick 3: 716, 173 → digits: 1, 3, 6, 7
Pick 4: 4703, 7907 → digits: 0, 3, 4, 7, 9
Total digits from all draws:
7, 1, 6, 4, 7, 0, 1, 7, 3, 9, 0, 7
=> Unique digits: 0, 1, 3, 4, 6, 7, 9
✅ Repeating (Hot) Digits Hit in Draws:
✅ 0 (in 4703, 7907)
✅ 1 (in 716, 173)
✅ 3 (in 4703, 173)
✅ 4 (in 4703)
✅ 7 (in all)
✅ 9 (in 7907)
🔥 All hot digits appeared!
❌ Missing (Cold) Digits:
❌ 2, 5, 8 are still missing
📌 Summary:
The hot/repeating digits (0,1,3,4,7,9) are strongly represented in the morning draws.
The cold digits (2,5,8) are completely absent in these results.
Digit 6 (mild/appears once) showed up in 716.
Let’s calculate the probability that all 3 missing digits (2, 5, 8)
will appear in Pick 3 and Pick 4 draws across all states.
🎯 Definitions
Pick 3: 3-digit numbers → each digit is drawn independently from 0–9.
Pick 4: 4-digit numbers → each digit is drawn independently from 0–9.
Goal: At least one of 2, 5, 8 appears in the result.
Assumption: We're calculating per draw probability and then estimating
for many draws across all states.
✅ Step-by-Step
1. Probability a single digit is not 2, 5, or 8:
There are 10 digits (0–9). The non-2/5/8 digits are: 0, 1, 3, 4, 6, 7, 9 → 7 digits.
So:
Probability a single digit is NOT 2/5/8 = 7/10 = 0.7
Probability a single digit IS 2/5/8 = 3/10 = 0.3
2. Probability in a single Pick 3 draw:
We want the chance that at least one of 2, 5, or 8 appears.
Complementary Probability:
P(None of the digits are 2/5/8) =
0.7
3
=
0.343
0.7
3
=0.343
So, P(At least one digit is 2/5/8) =
1
−
0.343
=
0.657
1−0.343=0.657 or 65.7%
3. Probability in a single Pick 4 draw:
P(None of the digits are 2/5/8) =
0.7
4
=
0.2401
0.7
4
=0.2401
So, P(At least one digit is 2/5/8) =
1
−
0.2401
=
0.7599
1−0.2401=0.7599 or 75.99%
4. For one state’s Pick 3 and Pick 4, combined:
Pick 3: ~65.7%
Pick 4: ~76%
Assume independence:
P(2/5/8 appear in either Pick 3 or Pick 4)
=
1
−
(
1
−
0.657
)
×
(
1
−
0.76
)
1−(1−0.657)×(1−0.76)
=
1
−
(
0.343
×
0.24
)
=
1
−
0.0823
=
0.9177
1−(0.343×0.24)=1−0.0823=0.9177
➡️ 91.77% chance that at least one of 2, 5, or 8 appears in either
Pick 3 or Pick 4 for a single state.
5. Across all 50 states (approximation):
Let’s estimate the probability that at least one of the missing digits
appears somewhere in all state results.
Let’s assume 40 states have Pick 3 and/or Pick 4 draws daily.
P(No 2, 5, or 8 in one state) = 1 - 0.9177 = 0.0823
P(No 2, 5, or 8 in any of 40 states) =
0.0823
40
≈
4.1
×
10
−
40
0.0823
40
≈4.1×10
−40
➡️ Virtually 100% probability that 2, 5, or 8 will appear somewhere
in the Pick 3 or Pick 4 draws across all states.
✅ Conclusion:
Level Probability that 2, 5, or 8 appears
Single Pick 3 65.7%
Single Pick 4 76%
One state's Pick 3 or 4 91.77%
All states Pick 3 & 4 ~100%
In all state draws, it is almost certain that at least one of
the missing digits 2, 5, or 8 will appear.
