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OT: Home Sick for the Holidays, or Probability and Yahtzee
As my family has all gone out shopping, I am left alone to ponder the mysteries of life. My family are Yahtzee fanatics, and I find myself struggling to calculate probabilities for this game. For basic probability, I know the formula is
Chosen Outcome ----------------------------------- # of Possible Outcomes For example, on a standard d6, there is a 1 out of 6 chance that any number will come up. If I have a 2,3,4,5 and am rolling for a large straight, I should have a 2/6 or 1/3 chance to get it in one roll (since either a 1 or a 6 will work). IIRC, the odds for multiple dice with separate outcomes for each are calculated by multiplying the individual odds; i.e., the odds for rolling a Yahtzee (all 5 d6 on same number) in a single roll would be 1/6^5, or 1/7776. However, it's been quite some time since I've studied probability, so I don't remember if that's correct. My dilemma comes when you factor in multiple dice and multiple rolls for a single outcome. Rolling 6 d6 does not guarantee a roll of any particular number. Likewise, rolling 1 d6 six times does not guarantee a roll of any particular number. Obviously, simple experimentation shows there is something more to this than a simple formula such as: Chosen Outcomes * Rolls -------------------------------------- Possible Outcomes What, then is the formula? Using the large straight example, given a 2,3,4,5 and 2 rolls of a single die, what is the probability that I will roll a large straight? I impatiently await your elucid answers to these vital and perplexing questions about life's meaning (at least while my in-laws are here, so I can finally beat them http://forum.shrapnelgames.com/images/icons/icon10.gif )! |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
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Rolling 6 dice at once or 1 die at a time six times makes 0 difference. There will never be a guarantee of getting that 1 or 6. |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
I forget if you want to use permutations or combinations here... it will be one over one of the following formulae:
n is the total number of items, r is the number of those items you want. Combination (order does not matter): n! / [ ( n - r )! * r! ] Permutation (order matters): n! / (n - r)! So for probability, you have either [ ( n - r )! * r! ] / n! or (n - r)! / n! Depending on which is the one you want. http://forum.shrapnelgames.com/images/icons/icon12.gif |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
the meaning of life?
42 isint it? |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
Yahtzee has absolutely nothing to do with 'odds' or probabilities.. It's pure luck. I know. When I play and my friend Claude is also in the group playing he regularly gets 4-5 Yahtzee's per game. I don't remember a game where he did not get at least 1 Yahtzee. (And he rolls an unusually high number of Six's)
Luck, that's the only thing that matters in Yahtzee. So, each time you roll the dice, just say a simple prayer... "Thy will be done." and see just how well that works. (oh, ans you don't have to say this prayer out loud, so you can keep the secret.) Cheers! http://forum.shrapnelgames.com/images/icons/icon10.gif [ December 24, 2003, 19:56: Message edited by: David E. Gervais ] |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
There was a dice game I used to play with my D&D friends when I was in Korea. Forgot now all of the rules, but IIRC we threw 6 dice also. But that's all the similarities to Yahtzee there was. Dang, I wish I could remember how that game went now. I got 6 brass six-sided dice I got from Korea I call my "Dice of Death". They made quite a racket when I rolled them (they're quite heavy compared to regular dice).
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Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
there is no such thing as luck...
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Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
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</font><blockquote><font size="1" face="sans-serif, arial, verdana">code:</font><hr /><pre style="font-size:x-small; font-family: monospace;">[ ( 6 - 1 )! * 1! ] / 6! [ 5! * 1 ] / 720 [ 120 * 1 ] / 720 120 / 720 1/6, or 16 1/6%</pre><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">but, </font><blockquote><font size="1" face="sans-serif, arial, verdana">code:</font><hr /><pre style="font-size:x-small; font-family: monospace;">[ ( 6 - 5 }! * 5! ] / 6! [ 1! * 120 ] / 720 120 / 720 1/6, or 16 1/6% (should be 83 1/3%)</pre><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">http://forum.shrapnelgames.com/image...s/confused.gif Obviously, the odds that I will roll a 6 and the odds that I will roll less than 6 are not both 1/6 (at least on a standard d6 http://forum.shrapnelgames.com/images/icons/icon12.gif ). It's worse with the permutation formula: </font><blockquote><font size="1" face="sans-serif, arial, verdana">code:</font><hr /><pre style="font-size:x-small; font-family: monospace;">( 6 - 1 )! / 6! 5! / 720 120 / 720 1/6, or 16 1/6%</pre><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">but, </font><blockquote><font size="1" face="sans-serif, arial, verdana">code:</font><hr /><pre style="font-size:x-small; font-family: monospace;">( 6 - 5 )! / 6! 1! / 720 1/720, or ~.1389%</pre><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">http://forum.shrapnelgames.com/image...s/confused.gif Things are even more exciting when you factor in more dice, such as the odds to roll at least one six with five d6: </font><blockquote><font size="1" face="sans-serif, arial, verdana">code:</font><hr /><pre style="font-size:x-small; font-family: monospace;">[ ( 30 - 5 }! * 5! ] / 30! [ 25! * 120 ] / 2.6525285981219105863630848e+32 [ 15,511,210,043,330,985,984,000,000 * 120 ] / 2.6525285981219105863630848e+32 1,861,345,205,199,718,318,080,000,000 / 2.6525285981219105863630848e+32 7.0172483965587413863275932241449e-6 .0000070172483965587413863275932241449, or ~.0007%</pre><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">http://forum.shrapnelgames.com/image...s/confused.gif I'd have thought my chances of rolling a six would have been slightly better than that! That's pretty close to 1/6^4. http://forum.shrapnelgames.com/images/icons/shock.gif Offhand, I'd guess there's a problem with the denominator in this formula--could it be n!/r, which would make the formula: </font><blockquote><font size="1" face="sans-serif, arial, verdana">code:</font><hr /><pre style="font-size:x-small; font-family: monospace;">r [ ( n - r )! * r! ] / n! 5 [ ( 6 - 5 }! * 5! ] / 6! 5 [ 1! * 120 ] / 720 5 [ 120 ] / 720 600 / 720 5/6, or 83 1/3%</pre><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">but, then again: </font><blockquote><font size="1" face="sans-serif, arial, verdana">code:</font><hr /><pre style="font-size:x-small; font-family: monospace;">3 [ ( 6 - 3 )! * 3! ] / 6! 3 [ 3! * 3! ] / 720 3 [ 6 * 6 ] / 720 3 * 36 / 720 108 / 720 3/20, or 15% (should be 50%)</pre><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">http://forum.shrapnelgames.com/image...s/confused.gif Any insight here? [ December 25, 2003, 16:48: Message edited by: Krsqk ] |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
Those formulae are correct for combinations and permuations. In truth, I was hoping someone would come along and fill in the blanks... http://forum.shrapnelgames.com/images/icons/icon12.gif Try google...
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Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
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If you want, you could always write a program to run through all the permutations and check whether they match any of the scoring combos... I've written two Yahtzee programs myself so I could always borrow something from one of those and just run all the permutations through a loop instead of generating them randomly... http://forum.shrapnelgames.com/images/icons/tongue.gif |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
Who said there is no such thing as luck?
Yesterday, I had my friend Claude over and told him about this discussion. I then asked if he would like to join me in a test, he said sure, it'd be fun. This is the test we preformed. We each rolled 5 dice 20 times, we took note of how many 6's turned up. We then repeated this test 10 times to see what kind of averages we came up with. </font><blockquote><font size="1" face="sans-serif, arial, verdana">code:</font><hr /><pre style="font-size:x-small; font-family: monospace;">David / Claude 14 / 34 12 / 32 17 / 35 20 / 37 17 / 28 16 / 31 21 / 36 32 / 27 (wow I beat him! :eek: ) 18 / 33 15 / 33 --------------- 18.2 / 32.6</pre><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">Now if those numbers don't show some degree of luck I don't know how else to explain the results. Would someone here like to roll this test and see how their results compare? (It would be nice to see how an outside 'neutral' person did with this simple test.) Oh, now you can understand why I hate playing RISK against him, he always wins. (his best battle win against me in RISK was when I attacked him with 37 armies and he only had 5. He won. (and still had 3 of his 5 armies left after the battle.) Cheers! http://forum.shrapnelgames.com/images/icons/icon10.gif |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
It's not luck David, but probability you have demonstrated there. Your sample is too small to be meaningfull. If you and your friend rolled 100 times, or a thousand, or ten thousand, the scores would get much closer together. Eventually within percentage points of one another. On the other hand if you and your friend sat down and rolled 10 times each again he would be just as likely to have a better score then you the second time around.
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Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
I'd start checking your friend's dice, David. http://forum.shrapnelgames.com/images/icons/icon12.gif http://forum.shrapnelgames.com/images/icons/tongue.gif
Actually, Geo is absolutely right, with one clarification. The difference in percentage points will likely decrease, eventually dropping near the break even point; the numerical difference will likely increase. There is no guarantee that actual results will even out over time (the usual misapplication of the law of averages). Put another way, the odds of rolling a six with a single die are 1/6 every time, whether the "score" is 75 to 5 or 40 to 40. Previous results do not affect future results. ========= In a separate thought, I found one page (granted, one which wasn't a strictly mathematical approach) which said probabilities which are mutually independent (such as rolling two separate dice) should be added; i.e., the odds of rolling at least one six with two dice is 1/6+1/6, or 1/3. If that's true, though, the odds of rolling at least one six with twelve dice would be 12/6, or 200%. That obviously isn't correct. As I'm having much trouble finding any helpful information on the web, such as an intro to probability site or the like, I'd appreciate any links y'all can pass along. |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
geo, as I said, we rolled 5 dice 20 times (100 die rolls) and did this test 10 times. (10x100 = 1000 die rolls)
It took us about a half hour to make all the rolls. If you mean that we should have rolled this full test 10 more times to get 10,000 die rolls, that is a bit much. It wasn't my aim to prove the law of averages, but to see how in one quick session Claude always seems to out-six me. Besides rolling 10,000 or 100,000 dice is unrealistic and does not reflect real life applications. Life does not work like that. Averages and probabilities are at best a global "best guess" and no matter what the 'odds' are or how many million tests are used to achieve those 'averages' it does not guarantee your imediate results. Think of it this way, according to 'averages' if you roll a dice you have a 1/6 chance to hit a 6, but if you miss, your odds on the second throw don't improve, they stay exactly the same. Cheers! http://forum.shrapnelgames.com/images/icons/icon10.gif P.S... so, anyone want to try my test and share the results? |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
Wow, Claude should go to Las Vegas and get rich gambling! http://forum.shrapnelgames.com/images/icons/icon12.gif
Out of 100 dice rolls, the average number of 6's that should appear is 100/6 = 16.667 times. David's results are close to this. Claude's getting almost twice as much. I want to calculate what the odds are of getting Claude's results, but all my statistics and probability textbooks are at work so I can't refer to them... |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
This math problem is a cumulative binomial distribution.
Let's look at binomial distribution. Each die toss is an event with two mutually exclusive possible outcomes: either you roll a 6 (success), or you roll 1 -5 (failure). If we have N events, then the binomial distribution can be used to determine the probability of getting exactly r successes in N outcomes. The formula is: P(r) = (N! / (r!(N - r)!)) ( p^r) (1 - p)^(N-r) where N = number of trials, P(r) = probability of getting exactly r successes in the N outcomes p = probability of success on any one trial. So, in our situation, N = 1000 dice rolls, p = 1/6. David got 182 successes out of 1000 trials and Claude got 326 out of 1000. So, the probability of David getting exactly 182 out of 1000, by using the above formula, is P(182) = (1000! / (182!(1000-182)!))((1/6)^182)(1-1/6)^(1000-182) And the probability of Claude getting 326 out of 100 is: P(326) = (1000! / (326!(1000-326)!))((1/6)^326)(1-1/6)^(1000-326) But so far we are talking about the probability of getting exactly the number we got. I think we're more interested in the cumulative case, where we ask, what is the probability of getting at least 182 successes out of 1000, or at least 326 out of 1000? Well, then we have to add them up, i.e. P(182 or more) = P(182) + P(183) + P(184) + .... + P(1000) and P(326 or more) = P(326) + P(327) + P(328) + ... + P(1000) I thought, no problem, I'll do this on Excel, but I immediately found that Excel can't handle 1000! because it's too big a number. http://forum.shrapnelgames.com/images/icons/icon9.gif |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
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Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
OK, I did some more searching around and discovered that Excel can, in fact, do this calculation.
There is a handy pre-built function called BINOMDIST(number_s, trials, probability, cumulative). Now this is easy. The probability of getting 182 or more 6's in 1000 rolls is: P = 1 - BINOMDIST(181, 1000, 1/6, TRUE) P = 0.105008 I needed to subtract from 1 because the cumulative function adds from 0 to 181. To get the portion for 182 and above, we subtract from the whole (1) the part we don't want, the stuff at 181 and below. OK, so there's about a 10.5% chance of getting 182 or more. The probability of getting 326 or more 6's in 1000 rolls is: P = 1 - BINOMDIST(325, 1000, 1/6, TRUE) P = -0.5.08482E-14 Obviously there's something wrong. The answer can't be negative. I think Excel still can't handle this kind of big / small number. Does anybody have a better program and/or calculation method that can do this? Or am I doing something wrong? |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
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When people say there is no such thing as luck, I don't believe them. I have faced luck and lost too many times playing RISK. LOL Cheers! http://forum.shrapnelgames.com/images/icons/icon10.gif |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
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4.02387260077093773543702433923e+2567 http://forum.shrapnelgames.com/images/icons/shock.gif Anyone want to put a name to that number? http://forum.shrapnelgames.com/images/icons/icon10.gif The BINOMDIST function (with the TRUE parameter) gave me values steadily approaching 1 all the way up to 167, at which point Excel decided they were close enough to round to 1 (when I subtracted them from 1, the values were things like .1E-77 and the like). The odds of having no more than 167 sixes when rolling 1000 dice shouldn't be ~100%, should it? The problem apparently occurs at 260 successes. The function calculates that the probability of having no more than 260 successes is slightly more than 1. This may be due to the rounding limitations of this function. [ December 27, 2003, 21:46: Message edited by: Krsqk ] |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
Since Excel can't handle 1000 events in the calculations, I have settled for something less ambitious.
I calculated the probability for each of the results for the 100 dice rolls for David and Claude. The numbers are cumulative, so for example the first entry means that there's a 80% chance of getting 14 or more 6's out of 100 dice rolls. </font><blockquote><font size="1" face="sans-serif, arial, verdana">code:</font><hr /><pre style="font-size:x-small; font-family: monospace;">David Probability Odds Claude Probability Odds 14 0.7999947521 1 in 1.25 34 0.0000190716 1 in 52434 12 0.9222807788 1 in 1.1 32 0.0001238040 1 in 8077 17 0.5058410243 1 in 2.0 35 0.0000070251 1 in 142347 20 0.2197498431 1 in 4.6 37 0.0000008412 1 in 1188738 17 0.5058410243 1 in 2.0 28 0.0031013887 1 in 322 16 0.6123424483 1 in 1.6 31 0.0002957849 1 in 3381 21 0.1518878479 1 in 6.6 36 0.0000024818 1 in 402926 32 0.0001238040 1 in 8077 27 0.0062087189 1 in 161 18 0.4005925583 1 in 2.5 33 0.0000496373 1 in 20146 15 0.7125790826 1 in 1.4 33 0.0000496373 1 in 20146</pre><hr /></blockquote><font size="2" face="sans-serif, arial, verdana">From this, it appears that what Claude did is very, very, improbable! |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
I had a friend who played miniatures who said:
"I rather be lucky than good any day" |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
Claude should play SE4. He'll beat your fleet of 50 dreadnaughts using 3 escorts armed with DUC's. http://forum.shrapnelgames.com/images/icons/icon12.gif
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Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
I just think my friend Claude's karma is dice based while mine is pixel based. This would explain why I'm good at computers and he is good at dice.
with this anaolgy, what kind of karma would you all say you have? (you must all be good at something. http://forum.shrapnelgames.com/images/icons/icon12.gif ) Cheers! http://forum.shrapnelgames.com/images/icons/icon10.gif |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
I was incredibly lucky at Backgammon and Yahtzee until they made me use the cups.
I'm not sure just how my hands could make such a difference.... |
Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
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Re: OT: Home Sick for the Holidays, or Probability and Yahtzee
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Sorry, I couldn't resist http://forum.shrapnelgames.com/images/icons/tongue.gif Cheers! http://forum.shrapnelgames.com/images/icons/icon10.gif |
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