Swlong1981
Swlong1981
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January 12th, 2012 at 6:12:16 AM permalink
First time posting, I have 2 questions.

In 15 rolls of a a 6 sided die, what is the probability that any 1 number does not appear?

In 3 rolls of 5 six sided dice, what is the probability that any 1 number does not appear?

I understand that the probability of 1 number not appearing is 15.38/1. What I am more interested in is if you didn't have to choose which number will not appear, what is the overall chance of any specific number not showing up.

Thanks in advanced for your answers.
Ayecarumba
Ayecarumba
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January 12th, 2012 at 9:30:44 AM permalink
Homework?

The probability any one number not appearing in a single roll of a six sided die is 5/6. The odds of it not appearing in 15 rolls is (5/6)^15. This is the same whether a single die is rolled 15 times in sucession, or five dice are thrown three times, as each roll is independent.

If it doesn't matter which number, I think you multiply the above by 6 to get the probability of any of the six numbers not appearing in 15 rolls. But that is just a guess on my part.
Simplicity is the ultimate sophistication - Leonardo da Vinci
guido111
guido111
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January 12th, 2012 at 9:49:46 AM permalink
Quote: Swlong1981

First time posting, I have 2 questions.

In 15 rolls of a a 6 sided die, what is the probability that any 1 number does not appear?

What I am more interested in is if you didn't have to choose which number will not appear, what is the overall chance of any specific number not showing up.

Thanks in advanced for your answers.

I will answer the first question then look at the second one when I have more time.

For "any" number instead of a "specific" number this can be solved with inclusion-exclusion principle.
In' rel='nofollow' target='_blank'>https://wizardofodds.com/ask-the-wizard/roulette/]In Roulette (0,00) board, what is the probability that any number will not have hit by the 200th spin?

5th question from the bottom shows how to do the math.

He also covers the principle here

What is the expected number of rolls of two dice for every total from 2 to 12 to occur at least once?
My answer using his formula is: For at least 1 number not appearing.

0.389432829 - 0.034255919 + 0.000610352 = ~0.3557872614049
simulation results below
64.41% showed all 6 numbers.
p = 1/6
n = 15
stat: number of different values

group middle freq freq/100
---------------------------------------------
1.50 <= x < 2.50 2.00 11 0.00%
2.50 <= x < 3.50 3.00 6146 0.06%
3.50 <= x < 4.50 4.00 324968 3.25%
4.50 <= x < 5.50 5.00 3227537 32.28%
5.50 <= x < 6.50 6.00 6441338 64.41%

---------------------------------------------
2.00 11
3.00 6146
4.00 324968
5.00 3227537
6.00 6441338
---------------------------------------------
cumulative
---------------------------------------------
1.50 <= x < 2.50 2.00 11 0.00%
2.50 <= x < 3.50 3.00 6157 0.06%
3.50 <= x < 4.50 4.00 331125 3.31%
4.50 <= x < 5.50 5.00 3558662 35.59%
5.50 <= x < 6.50 6.00 10000000 100.00%

---------------------------------------------
2.00 11
3.00 6157
4.00 331125
5.00 3558662
6.00 10000000


added
And a table for rolls 6 to 40.
1 - 6 sided die
at least 1 number not showing in n rolls


rollsProb of "at least 1 number" 
1 dieNot Rolling1 in
1  
2  
3  
4  
5  
60.9845679011.015673981
70.9459876541.057096248
80.8859739371.128701374
90.8109567901.233111323
100.7281878721.37327198
110.6437935811.553292901
120.5621843191.778776045
130.4861418062.05701297
140.4171546512.397192497
150.3557872612.810668364
160.3019956023.311306497
170.2553675483.915924352
180.2152928844.644835369
190.1810769235.52251487
200.1520124596.57840815
210.1274225157.847906613
220.1066834669.373523764
230.08923544211.20630974
240.07458483513.4075512
250.06230214016.05081298
260.05201725719.2243894
270.04341361823.03424689
280.03622196827.60755583
290.03021428333.09692937
300.02519811439.68551043
310.02101146047.59307647
320.01751823857.08336542
330.01460432568.47286706
340.01217413382.14137276
350.01014768698.54463535
360.008458122118.229561
370.007049579141.8524396
380.005875412170.2008212
390.004896686204.2197681
400.004080910245.0433565
guido111
guido111
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January 12th, 2012 at 9:51:00 AM permalink
Quote: Ayecarumba

Homework?

The probability any one number not appearing in a single roll of a six sided die is 5/6. The odds of it not appearing in 15 rolls is (5/6)^15. This is the same whether a single die is rolled 15 times in sucession, or five dice are thrown three times, as each roll is independent.

If it doesn't matter which number, I think you multiply the above by 6 to get the probability of any of the six numbers not appearing in 15 rolls. But that is just a guess on my part.

For any "specific" number the above formula is correct.
That is to say before any trials begin, a number is chosen.
Swlong1981
Swlong1981
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January 12th, 2012 at 10:19:39 AM permalink
Thanks for the quick response.

I assure you it's not homework, just a question that came up around the office. We have been playing shake a day in the office, which no one has one yet, and have been trying to come up with other "games" with the 5 dice.

So my understanding is that ~35% of the time at least one number will not show up when you roll 5 dice 3 times?

We are trying to figure out if the "player" chooses a specific set of numbers (i.e 1, 3, 5 or 2,3,4,6) or all of the numbers 1-6. What would be the likely hood of those events happening

Again, thanks for the help!
guido111
guido111
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January 12th, 2012 at 10:33:06 AM permalink
Quote: Swlong1981

Thanks for the quick response.

I assure you it's not homework, just a question that came up around the office. We have been playing shake a day in the office, which no one has one yet, and have been trying to come up with other "games" with the 5 dice.

So my understanding is that ~35% of the time at least one number will not show up when you roll 5 dice 3 times?

No.
I only answered your first question above.
"In 15 rolls of a a 6 sided die, what is the probability that any 1 number does not appear?" answer at least 1 number ~35%
exactly 1 number ~32%
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