#What is probability of rolling 11 with 3 dice

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white geyser
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ornate hamlet
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Do you mean adding the sum up? You cannot roll an 11 out of any 6-sided dice.

white geyser
earnest dagger
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Three 6-sided dice have 6^3 different possible outcomes.
146, 155, 236, 245, 335, 344 are the combinations that add up to 11. Each if those has 6 variations (for example: 146, 164, 416, 461, 614, 641), except for the ones with a repeating number, those have 3 variations (for example: 155, 515, 551).
So 6+6+6+3+3+3 of the 6^3 possibilities add up to 11.
That's 1 in 8 or 12,5%.

white geyser
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Ok how do you know 146, 155, 236, 245, 335, 344 are the only ones

ornate hamlet
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If the first is 1, then the 2nd and third must sum to 10. There are 3 possibilities for that.
If the first is 2, then the 2nd and third must sum to 9. There are 4 possibilities for that.
If the first is 3, then the 2nd and third must sum to 8. There are 5 possibilities for that.
...
If the first is 6, then the 2nd and third must sum to 5. There are 4 possibilities for that.
There are 3+4+5+6+5+4 possibilities, which is 27 possibilities out of 6^3, which is 1/8.

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This is one of the many ways to calculate 27 possibilities

earnest dagger
# white geyser Ok how do you know 146, 155, 236, 245, 335, 344 are the only ones

I just went down the list.
1: At least 4 is needed to make 11 archievable with the 3rd die. 146. Then for 5: 155. With 6 you get a combination we already had (146).
2: At least 3 needed to make 11 achievable, 236. 245. 2-6 was already done.
3: Same process. 335. 344.
4, 5 and 6 don't offer any new combinations once we reach the halfway point.

white geyser
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I figured out

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3/36+4/36+5/36+6/36+5/36+3/36

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ODS=one dice sum
TDS=two dice sum

P(ODS=1)P(TDS=10)+P(ODS=2)P(TDS=9)+P(ODS=3)P(TDS=8)+P(ODS=4)P(TDS=7)+P(ODS=5)P(TDS=6)+P(ODS=6)P(TDS=5)

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=(1/6)*(3/36)+(1/6)*(4/36)+(1/6)*(5/36)+(1/6)*(6/36)+(1/6)*(5/36)+(1/6)*(4/36)=27/216

white geyser
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I have another question

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I made a game

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Start with a score of 0. Then roll 3 dice, if you roll a sum of 11, you gain 10 points. If not, calculate the distance between your roll and 11 and subtract that number from your score. Then roll again and repeat

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Example game:
Starting score: 0
Roll 1: 2,3,4 (sum=9)
Score: 0-2
Roll 2: 5,5,3 (sum=13)
Score: -2-2
Roll 3: 1,2,3 (sum=6)
Score -4-5=-9
Roll 4: 3,4,4 (sum=11)
Final score: -9+10=1

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Suppose n = the expected number of rolls before rolling a 11

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And m = expected # of points gained per turn

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Then is the final expected score m*n?

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@earnest dagger

sonic thistle
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We figured that the ratio of games that add up to 11 to total games is 1/8

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Let’s call the number of games x

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Then it would be 11 (x/8) + n (7x/8)

earnest dagger
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I made a chart

IMG_1232.jpg
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On the left there’s each possible outcome and the resulting points (keep in mind that combinations with the structure xyz are actually 6 of the possibilities, xyy 3, and xxx 1)
On the right are the probabilities of each amount of points (the last column are percentages that are all rounded up or down to two decimals when needed)

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Then for your question, if you were to play 216 rounds and get the ‘expected’ result (which I interpreted as getting +10 27 times, -1 52 times, -2 46 times, etc.) then you would end up with -252 points.

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Which means that on average you lose -1,167 points per round (-252/216)

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The longer you play, the more predictable your final score will become.

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If you were to play only about 10 rounds then your total amount of points could easily climb into the positives with a few lucky rolls. But if you play lots, then your odds of getting lucky get smaller and smaller and your total will slowly drop lower and lower due to the average amount of points you get being negative.

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Seems like a good game for a casino to pick up huh

white geyser
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@earnest dagger you forgot an important detail. The game ends as soon as you roll an 11.

earnest dagger
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May I ask if you thought up this game to actually play it, or just made it for this math-question?

earnest dagger
white geyser
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Oh fuck

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How did I leave that out

white geyser
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It was actually pretty hard to come up with one where it wouldn't be extremely easy nor extremely hard to determine the best strategy/expected final score

earnest dagger
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m=-2,76...
((-1 times 52-2 times 46-3 times 36...)/(216-27), because we should leave out the +10 possibilities)
But I'm unsure on how to calculate n (the expexted amount of rolls to get 11)
I think it's just 8, but I'm not sure.

earnest dagger
white geyser
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Which works out to 7

white geyser
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Idk if that was wrong

earnest dagger
earnest dagger
white geyser
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True

earnest dagger
# white geyser It was actually pretty hard to come up with one where it wouldn't be extremely e...

Btw, a few comments:

  1. I'm not sure if I'm allowed to help you with assignments according to the rules of this server. The rules say 'cheating is prohibitied' but idk if asking for help on assignments like this are included in that. So if someone could inform me abt that that'd be nice.
  2. I'm not sure if this was a requirement in your assignments, but this game has no strategy at all, since the player has no controll over anything in this game at all. But if it's not a requirement it's fine because I think the math behind it is still interesting.
  3. Idk how hard you want to go on this assignment, but I think the thing I mentioned earlier about your odds evolving the longer you play was pretty interesting too (and how it can relate to the psychology of many other gambling games). So you could write a part about that too. Just a suggestion.
white geyser
white geyser
earnest dagger
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Ah cool