April 24th, 2020 at 2:40:31 PM
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It's been more then a decade since I last asked the wizard :-)
Corona times, me and the guys been using this app for our private Texas Hold'em table.
we do suspect the RNG for this app, but we still having great friendly fun using it combine with a digital wallet app and Zoom Meeting !
Yesterday we had 3 times SF as a winning hand, at the same table, all within a period of less the 150 hands.
1.Would you suspect the RNG for this app we are using ?
2.How can we ensure / test RNG reliability ??
3.can someone recommend a solution for our friendly online meetings ?
Thank you!
Ani Badama
Corona times, me and the guys been using this app for our private Texas Hold'em table.
we do suspect the RNG for this app, but we still having great friendly fun using it combine with a digital wallet app and Zoom Meeting !
Yesterday we had 3 times SF as a winning hand, at the same table, all within a period of less the 150 hands.
1.Would you suspect the RNG for this app we are using ?
2.How can we ensure / test RNG reliability ??
3.can someone recommend a solution for our friendly online meetings ?
Thank you!
Ani Badama
Last edited by: PaguaBadama on Apr 24, 2020
April 24th, 2020 at 3:52:01 PM
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Which app?
You would think that an app that shuffles a deck of cards would use the date and time as a seed for the random number generator and/or run through shuffles in the background between actual shuffles.
You would think that an app that shuffles a deck of cards would use the date and time as a seed for the random number generator and/or run through shuffles in the background between actual shuffles.
April 24th, 2020 at 5:08:09 PM
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Quote: ThatDonGuyWhich app?
You would think that an app that shuffles a deck of cards would use the date and time as a seed for the random number generator and/or run through shuffles in the background between actual shuffles.
In order to avoid irrelevant prejudice I didn't mentioned the app name.
nope,
for me it feels like as if deliberately some cards of the same suits and/or few following cards or 4 of the same value cards taken out of the deck prior to shuffling, resulting potential for monsters hands and making the game more exciting.
Last edited by: PaguaBadama on Apr 24, 2020
April 24th, 2020 at 10:01:00 PM
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Texas hold'em involves making a 5 card poker hand from 7 cards -your two hole cards and 5 common cards.
When using 7 cards, the frequency of straight flushes is 0.0311%.
However, each deal may involve 8 or 9 players, each with a different set of 2 hole cards + the 5 common cards. So, in the absence of rigorously modeling, lets estimate that the chance of a straight flush in a dealt round of Texas Hold'em is on the order of 0.2%.
1 in 500.
So, OP reports an observance of 3 straight flushes in about 150 hands. 1 in 50.
I am too much into my stores of hard liquor tonight to do a statistical analysis. But my besotted intuition is that this is plausible.
When using 7 cards, the frequency of straight flushes is 0.0311%.
However, each deal may involve 8 or 9 players, each with a different set of 2 hole cards + the 5 common cards. So, in the absence of rigorously modeling, lets estimate that the chance of a straight flush in a dealt round of Texas Hold'em is on the order of 0.2%.
1 in 500.
So, OP reports an observance of 3 straight flushes in about 150 hands. 1 in 50.
I am too much into my stores of hard liquor tonight to do a statistical analysis. But my besotted intuition is that this is plausible.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
April 24th, 2020 at 11:26:59 PM
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When using 7 cards, the frequency of straight flushes is 0.0311%(by gordonm888), assuming 8 players, so prob straight flushes in a round = 0.000311 * 0.999689^7 * 8 = 0.248%, so it is about 1 in 403.
METHOD 1
Let straight flushes are normally distributed, so to find probability MORE THAN 3 straight flushes in 150 round(8 players in a round),
mean = 0.00248 = 1/403, Var = 0.0099( based on 1 straight flush +402 no straight flush)
P(X>3) =P[ Z> (3-150/403)/(0.0099*150)^0.5] = P[Z >2.156], Prob = 0.0155 = 1.55% !
I am not 100% sure, please check and give your comments.
METHOD 2
Probability of success on a single round = 0.00248, So P(X>3) = 1 - P(X=0) - P(X=1) - P(X=2) - P(X=3)
P(X=0) = 0.99752^150 = 0.6890
P(X=1) = 0.99752^149 * 0.00248 * 150 = 0.2569
P(X=2) = 0.99752^148 * 0.00248^2 * 150 * 149/2 = 0.0476
P(X=3) = 0.99752^147 * 0.00248^3 * 150 * 149 * 148 /6 = 0.005837
P(X>3) = 1 - P(X=0) - P(X=1) - P(X=2) - P(X=3) = 1-0.6890-0.2569- 0.0476-0.005837 = 0.000663 = 0.0663%,
So prob EXACTLY 3 times SF as a winning hand, at the same table, all within a period of 150 hands = P(X=3) = 0.005837 =0.5837%
I think METHOD 2 is correct because the value of Variance(Var) used in METHOD 1 is NOT ACCURATE and straight flushes may not conform to normal distribution. Please give your comments
METHOD 1
Let straight flushes are normally distributed, so to find probability MORE THAN 3 straight flushes in 150 round(8 players in a round),
mean = 0.00248 = 1/403, Var = 0.0099( based on 1 straight flush +402 no straight flush)
P(X>3) =P[ Z> (3-150/403)/(0.0099*150)^0.5] = P[Z >2.156], Prob = 0.0155 = 1.55% !
I am not 100% sure, please check and give your comments.
METHOD 2
Probability of success on a single round = 0.00248, So P(X>3) = 1 - P(X=0) - P(X=1) - P(X=2) - P(X=3)
P(X=0) = 0.99752^150 = 0.6890
P(X=1) = 0.99752^149 * 0.00248 * 150 = 0.2569
P(X=2) = 0.99752^148 * 0.00248^2 * 150 * 149/2 = 0.0476
P(X=3) = 0.99752^147 * 0.00248^3 * 150 * 149 * 148 /6 = 0.005837
P(X>3) = 1 - P(X=0) - P(X=1) - P(X=2) - P(X=3) = 1-0.6890-0.2569- 0.0476-0.005837 = 0.000663 = 0.0663%,
So prob EXACTLY 3 times SF as a winning hand, at the same table, all within a period of 150 hands = P(X=3) = 0.005837 =0.5837%
I think METHOD 2 is correct because the value of Variance(Var) used in METHOD 1 is NOT ACCURATE and straight flushes may not conform to normal distribution. Please give your comments
Last edited by: ssho88 on Apr 25, 2020