Difference between revisions of "Talk:Surly gambler"
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Ohh, nice box :P | Ohh, nice box :P | ||
| − | Anyway, 44/183={{statrate|183|44}} => 0.24*0.24*0.76 (chance of 2, assuming independent odds) =~4.37% | observed chance of 2 => 6/183=~3.2%. | + | Anyway, 44/183={{statrate|183|44}} <del>=> 0.24*0.24*0.76 (chance of 2, assuming independent odds) =~4.37%</del> | observed chance of 2 => 6/183=~3.2%. |
Conclusion: Unless you can get that 6 to 60 there's too much chance (±6.3% lol) and little statistic.<br>Also, if the chance is indeed 24% statistically you ought to have already seen 2 occurrences of 3 cards together, which proves 200 is still a small sample. [[User:Patojonas|Patojonas]] 12:29, 23 July 2012 (PDT) | Conclusion: Unless you can get that 6 to 60 there's too much chance (±6.3% lol) and little statistic.<br>Also, if the chance is indeed 24% statistically you ought to have already seen 2 occurrences of 3 cards together, which proves 200 is still a small sample. [[User:Patojonas|Patojonas]] 12:29, 23 July 2012 (PDT) | ||
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:At, say, a 10% drop rate, there would be a {{#expr: (1-0.10)*(1-0.10)*(1-0.10)*100 round 1}}% chance of exactly 1 dropping, a {{#expr: 3*0.10*(1-0.10)*(1-0.10)*100 round 1}}% chance of exactly 1 dropping, and a {{#expr: 3*0.10*0.10*(1-0.10)*100 round 1}}% chance of exactly 2 dropping, and a {{#expr:0.10*0.10*0.10*100 round 1}}% chance of exactly 3 dropping. This is closer, but still not quite the same as the data. In any case, I agree with one part of your conclusion: 200 is too small a sample. | :At, say, a 10% drop rate, there would be a {{#expr: (1-0.10)*(1-0.10)*(1-0.10)*100 round 1}}% chance of exactly 1 dropping, a {{#expr: 3*0.10*(1-0.10)*(1-0.10)*100 round 1}}% chance of exactly 1 dropping, and a {{#expr: 3*0.10*0.10*(1-0.10)*100 round 1}}% chance of exactly 2 dropping, and a {{#expr:0.10*0.10*0.10*100 round 1}}% chance of exactly 3 dropping. This is closer, but still not quite the same as the data. In any case, I agree with one part of your conclusion: 200 is too small a sample. | ||
:Of course, all this is assuming the drop rate for each card is the same. If it isn't, that'll be a lot of "fun".... --[[User:Hoyifung04|hoyifung04]] 13:19, 23 July 2012 (PDT) | :Of course, all this is assuming the drop rate for each card is the same. If it isn't, that'll be a lot of "fun".... --[[User:Hoyifung04|hoyifung04]] 13:19, 23 July 2012 (PDT) | ||
| + | ::Hope you don't mind, but I fixed your formulas on the first line, there was a 3* in the 1 and 2 drops that was messing the values.<br>If you check now, you'll see that the theoretical values assuming 24 are within or very close to the experimental data you obtained (except for the no drops case, a small change in the numbers influences that one too much).<br>Also, no point in doing statistics for the 3 drops when you haven't observed any occurrences ;) [[User:Patojonas|Patojonas]] 14:03, 23 July 2012 (PDT) | ||
| + | :::Wouldn't it have a 3* in the formula? Because it's <code>(#1 drops)(#2 DOESN'T drop)(#3 DOESN'T drop) + (#1 DOESN'T drop)(#2 drops)(#3 DOESN'T drop) + (#1 DOESN'T drop)(#2 DOESN'T drop)(#3 drops)</code>, or, since we're assuming all three have the same rate, <code>3*(drops)(DOESN'T drop)(DOESN'T drop)</code>. In addition, without multiplying by three, there's only 63.5% covered... where did the other 36.5% go? With the multiplication by 3, there's 100% over the four outcomes. Also, if I'm reading correctly, then the chance of something happening in a {{wikipedia|binomial distribution}} is <code>C(n,k)p<sup>k</sup>(1-p)<sup>n-k</sup></code>, and C(3,1) = C(3,2) = 3. | ||
| + | :::Also, the stats for 3 drops is just for completeness. --[[User:Hoyifung04|hoyifung04]] 15:25, 23 July 2012 (PDT) | ||
| + | :::: I submit lol, University clearly dulled my brain as I feared o.O [[User:Patojonas|Patojonas]] 16:07, 23 July 2012 (PDT) | ||
| + | ---- | ||
| + | <pre> | ||
| + | surly gambler 563 | ||
| + | playing card (1) 128 | ||
| + | playing card (2) 23 | ||
| + | playing card (3) 0 | ||
| + | </pre> | ||
| + | |||
| + | {| cellpadding="3" cellspacing="0" border="1" align="center" style="text-align:center" | ||
| + | |- | ||
| + | ! !! No drops !! One drop !! Two drops !! Three drops | ||
| + | |- | ||
| + | | Observed || {{statrate|563|412|+}}% || {{statrate|563|128|+}}% || {{statrate|563|23|+}}% || {{statrate|563|0|+}}% | ||
| + | |- | ||
| + | | Predicted (10%) | ||
| + | | {{#expr: (1-0.10)*(1-0.10)*(1-0.10)*100 round 1}}% | ||
| + | | {{#expr: 3*0.10*(1-0.10)*(1-0.10)*100 round 1}}% | ||
| + | | {{#expr: 3*0.10*0.10*(1-0.10)*100 round 1}}% | ||
| + | | {{#expr:0.10*0.10*0.10*100 round 1}}% | ||
| + | |} | ||
| + | --[[User:Hoyifung04|hoyifung04]] 18:08, 15 August 2012 (PDT) | ||
Latest revision as of 17:08, 15 August 2012
"Your opponent attacks ... "Draw," she says. You go for your weapon, but a professional gambler is always better at the draw. She hits you for 3 damage."
I don't know how to tell the difference between all the types of messages, but this one wasn't on the page. Usi 02:39, 30 March 2008 (MST)
Playing card drop rate
I don't know how to find the probabilities for each playing card, so I'm just going to put my numbers here. Note that each line is exclusively that number (i.e. the "playing card (1)" line means that EXACTLY one playing card dropped).
surly gambler 183 playing card (1) 38 playing card (2) 6 playing card (3) 0
If someone is able to extract the drop rate of each playing card, I'd be grateful. In the meantime, I'm going to collect more data. --hoyifung04 09:29, 23 July 2012 (PDT)
Ohh, nice box :P
Anyway, 44/183=(24% ± 6.3%) => 0.24*0.24*0.76 (chance of 2, assuming independent odds) =~4.37% | observed chance of 2 => 6/183=~3.2%.
Conclusion: Unless you can get that 6 to 60 there's too much chance (±6.3% lol) and little statistic.
Also, if the chance is indeed 24% statistically you ought to have already seen 2 occurrences of 3 cards together, which proves 200 is still a small sample. Patojonas 12:29, 23 July 2012 (PDT)
- Pretty sure you can't actually get 24% that way, but using that value, there would be a 43.9% chance of no cards dropping, a 41.6% chance of exactly 1 dropping, and a 13.1% chance of exactly 2 dropping, and a 1.4% chance of exactly 3 dropping.
- Current rates are 76 ± 6.3% for no cards dropping, 20.8 ± 6% for exactly 1 dropping, 3.3 ± 2.6% for exactly 2 dropping, and 0 ± 0.1% for exactly 3 dropping.
- At, say, a 10% drop rate, there would be a 72.9% chance of exactly 1 dropping, a 24.3% chance of exactly 1 dropping, and a 2.7% chance of exactly 2 dropping, and a 0.1% chance of exactly 3 dropping. This is closer, but still not quite the same as the data. In any case, I agree with one part of your conclusion: 200 is too small a sample.
- Of course, all this is assuming the drop rate for each card is the same. If it isn't, that'll be a lot of "fun".... --hoyifung04 13:19, 23 July 2012 (PDT)
- Hope you don't mind, but I fixed your formulas on the first line, there was a 3* in the 1 and 2 drops that was messing the values.
If you check now, you'll see that the theoretical values assuming 24 are within or very close to the experimental data you obtained (except for the no drops case, a small change in the numbers influences that one too much).
Also, no point in doing statistics for the 3 drops when you haven't observed any occurrences ;) Patojonas 14:03, 23 July 2012 (PDT)- Wouldn't it have a 3* in the formula? Because it's
(#1 drops)(#2 DOESN'T drop)(#3 DOESN'T drop) + (#1 DOESN'T drop)(#2 drops)(#3 DOESN'T drop) + (#1 DOESN'T drop)(#2 DOESN'T drop)(#3 drops), or, since we're assuming all three have the same rate,3*(drops)(DOESN'T drop)(DOESN'T drop). In addition, without multiplying by three, there's only 63.5% covered... where did the other 36.5% go? With the multiplication by 3, there's 100% over the four outcomes. Also, if I'm reading correctly, then the chance of something happening in a binomial distribution isC(n,k)pk(1-p)n-k, and C(3,1) = C(3,2) = 3. - Also, the stats for 3 drops is just for completeness. --hoyifung04 15:25, 23 July 2012 (PDT)
- I submit lol, University clearly dulled my brain as I feared o.O Patojonas 16:07, 23 July 2012 (PDT)
- Wouldn't it have a 3* in the formula? Because it's
- Hope you don't mind, but I fixed your formulas on the first line, there was a 3* in the 1 and 2 drops that was messing the values.
surly gambler 563 playing card (1) 128 playing card (2) 23 playing card (3) 0
| No drops | One drop | Two drops | Three drops | |
|---|---|---|---|---|
| Observed | 73.2 ± 3.7% | 22.7 ± 3.5% | 4.1 ± 1.7% | 0 ± 0.1% |
| Predicted (10%) | 72.9% | 24.3% | 2.7% | 0.1% |
--hoyifung04 18:08, 15 August 2012 (PDT)
