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This is not hard, but fun.

Suppose you have a die such that its 1,5,6 and 4 face are broken. Whenever 1 face lands, it gets rolled over to an adjacent face. If 5 face lands, it is equally likely that it may stick or get rolled over to the adjacent face. If 4 lands, it has 1/4 probability of rolling over to the adjacent face. While if 6 shows up it always sticks. (EDIT: Not to confuse 6 is no different from 2,3. They behave like regular die faces.That 6 statement is just to provide element to the puzzle.)

(By land I mean on top)

What is the probability of getting 4 on this die?

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    $\begingroup$ will the roll over probability for 5, 1 and 4 only occur during the first time the die lands, or does it always happen whenever that side moves to the top? i.e. if it lands on 5 which rolls over to 1, will it continue rolling over to another adjacent face? $\endgroup$ – votbear Dec 5 '17 at 3:43
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    $\begingroup$ What about 2 and 3? What makes them different from 6? $\endgroup$ – Deusovi Dec 5 '17 at 3:43
  • $\begingroup$ (it's actually kind of amusing to imagine a die repeatedly flopping back and forth between landing on 1 and 5) $\endgroup$ – votbear Dec 5 '17 at 3:45
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    $\begingroup$ Any face on a die has 4 adjacent faces. If 4 lands, I assume the 1/4 probability of it rolling over is to any of its 4 neighbours (with equal chance)? $\endgroup$ – Xenocacia Dec 5 '17 at 3:48
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    $\begingroup$ If it lands on say 5, and rolls over to 1, is that then the end? Or will it roll over again since it's on 1? $\endgroup$ – Lolgast Dec 5 '17 at 6:54
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A slightly differently explained answer from Jesse's, assuming the dice can only roll over once:

       base     Roll over from 1     Roll over from 4    Roll over from 5     Total
P(1) = 1/6*0    +0                  +1/4*1/4*1/6        +1/4*2/4*1/6        =( 0+0+1+2)/96
P(2) = 1/6      +1/4*1/6            +1/4*1/4*1/6        +0                  =(16+4+1+0)/96
P(3) = 1/6      +1/4*1/6            +0                  +1/4*2/4*1/6        =(16+4+0+2)/96
P(4) = 1/6*3/4  +1/4*1/6            +0                  +1/4*2/4*1/6        =(12+4+0+2)/96
P(5) = 1/6*1/2  +1/4*1/6            +1/4*1/4*1/6        +0                  =( 8+4+1+0)/96
P(6) = 1/6      +0                  +1/4*1/4*1/6        +1/4*2/4*1/6        =(16+0+1+2)/96

Giving total probabilities:

- 1: 3/96
- 2: 21/96
- 3: 22/96
- 4: 18/96
- 5: 13/96
- 6: 19/96

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  • $\begingroup$ I think you miss something, when 4 lands up (1/6) we got : 1/6 * 1/4 for the 4 stays on top, and 1/6 * 3/4 that another number roll, so if 4 land's up 1,2,5 and 6 can land's up after the last roll, so we got 1/6 * 3/4 *1/4 for the others numbers $\endgroup$ – Skyvask Dec 5 '17 at 9:11
  • $\begingroup$ @Skyvask From OP: "If 4 lands, it has 1/4 probability of rolling over to the adjacent face" $\endgroup$ – Lolgast Dec 5 '17 at 10:08
  • $\begingroup$ Ok i'm confused, just need to read more carrefuly next time $\endgroup$ – Skyvask Dec 5 '17 at 10:09
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I used a probability tree to figure this out:

First you roll the die and it lands with equal probability on 1 of 6 sides. For each side it lands on it has varying probabilities to end up on one of 5 numbers (1-6 excluding whichever is on the opposite side ie. not adjacent. note: Opposite = 7 - Landed) so you have the given stick probability for the same number again and then an equal probability for all the 4 adjacent sides (*1/4) the given amount.

Once I did this for every number, go through and add all the last branches for each number to get the total probability that the die will end up on each number, I got:

  • 1: 3/96
  • 2: 21/96
  • 3: 22/96
  • 4: 18/96
  • 5: 13/96 (edit: thanks for the find)
  • 6: 19/96

Edit: i added them all to verify and it equals 100%

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  • $\begingroup$ You miss +2 on the 1 probability, it's 5/96 ( since 3/96 when 4 lands up, and 2/96 when 5 lands up). $\endgroup$ – Skyvask Dec 5 '17 at 8:09
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    $\begingroup$ It's actually 1/96 from 4, so that one is correct. The probability for 5 is incorrect, it should be 13/96 $\endgroup$ – Lolgast Dec 5 '17 at 9:00

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