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dark_magician_92

Help Needed – 2

Question

My friend Found this somewhere and i calculated a different answer than what he told was. Plz mention ur approach and answer of this problem!!

Rules of Game:-

"We each get one die, the highest die wins. If we tie, I win, but since you always lose when you roll a one, if you roll a one you can roll again. If you get a one the second time you have to keep it."

What is each person's probability of winning?

What are the probabilities of winning if you can keep rolling until you get something besides a one?

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11 answers to this question

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a)37/72 and 35/72

b)1/2 for both

a)62.5% and 37.5%

b)61% and 39%

i might be wrong bcoz the 2nd part of the question doesnt make much change in the probabilities of 1st part, even though the condition is quite effective.

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Chance of the person who wins ties winning

if he rolls a 6 _ 6/6 = 1

if he rolls a 5 _ 5/6-1/36 = 0.805555556

if he rolls a 4 _ 4/6-2/36 = 0.611111111

if he rolls a 3 _ 3/6-3/36 = 0.416666667

if he rolls a 2 _ 2/6-4/36 = 0.222222222

if he rolls a 1 _ 1/6-5/36 = 0.027777778

final probablility = 0.513888889 (same as the person who gave you the puzzle

the first fraction covers the probability if there was no rerolling of 1's by the opponent., but you have to subtract from that the chance of the oponent rolling a 1 then rolling higher then you.

I still haven't worked out the second question but suspect it's just a hair under 50%

*edit: corrected some typo's

Edited by spamwolf
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a)37/72 and 35/72

b)31/60 and 29/60 since the sum to infinity of 1/6^n converges on .2 so .5*(5/6)+.5*sum of 1/6^n = 31/60

Edited by tonyd
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Started with the simplest game. P1 wins if >= P2. P1 wins 21 out of 36 combinations. 58.3%

Now if P2 can re-roll if they get a 1 then P1 still wins 16 out of 36. (If P1 rolls a 6 nothing P2 will re-roll matters)

In the case that P2 is re-rolling instead of a win...

if P1 rolled a 1 then they now have a 1/6 chance to win that set

if P1 rolled a 2 then they now have a 2/6

if 3 then 3/6

if 4 then 4/6

if 5 then 5/6

That sums up to 15/6 or 2.5 making a total of 18.5 out of 36 (37/72) or 51.38%

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Getting 50% for #2

In the case of P2 re-rolling every 1.

P1 still wins 16 out of 36 combinations.

P1 will now always lose when they roll a 1 so we need to look at P1 rolling 2-5 when P2 gets to re-roll.

if P1 rolled a 2 then P1 now has a 1/6 chance of winning with a 1/6 chance of a re-roll with another 1/6 chance of winning and so on...

1/6 + (1/6 * 1/6) + (1/6 * 1/6 * 1/6) + ... approaches 1/5

if P2 rolled a 3 then there is a 2/6 chance of winning with a 1/6 chance of a re-roll with another 2/6 chance of winning and so on...

2/6 + (2/6 * 1/6) + (2/6 * 1/6 * 1/6) + ... approaches 2/5

if 4 then 3/6...approaches 3/5

if 5 then 4/6...approaches 4/5

Those sum up to 10/5 or 2 making it 18 out of 36 combinations or 50%.

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Getting 50% for #2

In the case of P2 re-rolling every 1.

P1 still wins 16 out of 36 combinations.

P1 will now always lose when they roll a 1 so we need to look at P1 rolling 2-5 when P2 gets to re-roll.

if P1 rolled a 2 then P1 now has a 1/6 chance of winning with a 1/6 chance of a re-roll with another 1/6 chance of winning and so on...

1/6 + (1/6 * 1/6) + (1/6 * 1/6 * 1/6) + ... approaches 1/5

if P2 rolled a 3 then there is a 2/6 chance of winning with a 1/6 chance of a re-roll with another 2/6 chance of winning and so on...

2/6 + (2/6 * 1/6) + (2/6 * 1/6 * 1/6) + ... approaches 2/5

if 4 then 3/6...approaches 3/5

if 5 then 4/6...approaches 4/5

Those sum up to 10/5 or 2 making it 18 out of 36 combinations or 50%.

I concur the answer you have for #2.

Since the second roller is allowed to reroll every time he rolls 1, this means his only options are 2,3,4,5. Thus, istead of 36 combinations, we have only 30 combinations possible. Of these, 15 gives the win to the first roller and 15 to the second roller. Ergo - probability of winning is exactly 50% for both.

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Your probability of winning given that I have a six is 1/6.

Your probability of winning given that I have a five is 2/6.

Your probability of winning given that I have a four is 3/6.

Your probability of winning given that I have a three is 4/6.

Your probability of winning given that I have a two is 5/6.

Your probability of winning given that I have a one is X.

Where X is one sixth the sum of the above without the final entry.

That is 15/36= 5/12

Your chances are one sixth the sum of the above with the calculated X (5/12).

That is 5/12 * (1 + 1/6) = 5/12 * 7/6 = 35/72

Well, I'll be darn. Never give a sucker an even break.

I haven't viewed others but this looks good.

Edited by Common Zenz
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Thank you to all of you i.e.Common Zenz , Current, thoughtfulfellow, tonyd and spamwolf. All your answers are right. And now the funny thing, after checking your answers and method (which was quite different from mine), i decided to post my solution and it was while typing when i realized That I MADE ONE OF THE MOST SILLIEST MISTAKES bcoz of which the answer which shud have been = 37/72 = 111/216 , turned out to be 135/216. Similar sort of error was in the 2nd part. However i am glad that you all participated. Will post some problems, if i come across some and hopefully not mess up the answers :D :D

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For the first question, I used a 6x6 grid with the 36 possible outcomes, and came up with odds of 17 1/2 out of 36 that I would win if I was allowed to reroll one time if I got a dice roll of 1. This matches what others have calculated with math, at a 48.61% chance to win.

For the second part, you can't use a simple grid but what others posted looks good.

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For the first question, I used a 6x6 grid with the 36 possible outcomes, and came up with odds of 17 1/2 out of 36 that I would win if I was allowed to reroll one time if I got a dice roll of 1. This matches what others have calculated with math, at a 48.61% chance to win.

For the second part, you can't use a simple grid but what others posted looks good.

Yes, a grid will work for the second situation. It just needs to be a 6X5 grid since the second roller must ultimately have something other than 1. The second rollers odds are the same as using a five sided die instead of the standard six sided die.

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I know that people have already solved this one, but I'm just going to do a little add-on-type-thing.

These are the PLAYER'S odds, NOT the dealer's.

First, we assume no rerolling allowed.

p(iwin | iroll6) = 5/6

p(iwin | iroll5) = 4/6

p(iwin | iroll4) = 3/6

p(iwin | iroll3) = 2/6

p(iwin | iroll2) = 1/6

p(iwin | iroll1) = 0 -> remember, no rerolls allowed for this one.

Total p = (5/6 * 1/6 + 4/6 * 1/6 + 3/6 * 1/6 + 2/6 * 1/6 + 1/6 * 1/6 + 0 * 1/6) = 15/36

Now, we allow one reroll. Yay! :)

p(iwin | iroll6) = 5/6

p(iwin | iroll5) = 4/6

p(iwin | iroll4) = 3/6

p(iwin | iroll3) = 2/6

p(iwin | iroll2) = 1/6

p(iwin | iroll1) = 15/36 (as above)

Remember, if we reroll after a one, we are now basically playing a game with no rerolls allowed.

Total p = (5/6 * 1/6 + 4/6 * 1/6 + 3/6 * 1/6 + 2/6 * 1/6 + 1/6 * 1/6 + 15/36 * 1/6) = 105/216 or ~48.6% (odds go to house. hmm)

I think someone already did this correctly, but I'll show it again, jic.

With unlimited rerolls allowed:

In this game, you basically have a FIVE-sided die, since all ones get rerolled. So the contingent probability is the same as above, because the dealer has a six-sided die, but p(event) is now 1/5 for each roll 2-6. Here we go! :)

Contingent probabilities:

p(iwin | iroll6) = 5/6

p(iwin | iroll5) = 4/6

p(iwin | iroll4) = 3/6

p(iwin | iroll3) = 2/6

p(iwin | iroll2) = 1/6

Total p = (5/6 * 1/5 + 4/6 * 1/5 + 3/6 * 1/5 + 2/6 * 1/5 + 1/6 * 1/5) = 15/30 = 1/2. Well well.

AAAWWWOOOOOOOOOOO! (don't worry, i didn't hit my thumb with a hammer. that was my wolf howl.)

Edited by AryaWolf
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