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Each 2 could receive more 2s. For each starting 2, 12/13 times that's it, 1/13 you start again with two 2s.
x = 12/13 + 1/13*2x
13x = 12+2x
x = 12/11. (I thought this would be 13/12, which would agree with where wizard's 64/24 comes from, but the above calculation suggests it's 12/11!)
As you started with two 2s, in total you expect 24/11 hands.
Each hand 10/13 no double, 3/13 double - average number of bets = 16/13.
So expected number of bets = 24/11 * 16/13 = 384/143 = $268.53.
Quote: Ace2Playing blackjack, you are dealt a pair of 2s against a dealer 3. You play basic strategy which means you split 2s and double down with 9-11 against a dealer 3. This is an infinite deck game with no limit on resplitting or doubles
Assuming you're a $100 flat bettor, what's your total expected wager on this hand?
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I get $281.26.
First, calculate the expected number of hands.
For each of the 2s you are dealt, there is a 1/13 chance of being dealt another 2; for each additional 2, there is a 1/13 chance of being dealt a further 2.
The expected number of hands = 2 (12/23 x 1 + 1/13 x 12/13 x 2 + (1/13)^2 x 12/13 x 3 + ...)
= 2 x 12/13 x (1 + 1/13 x 2 + (1/13)^2 x 3 + ...)
= 24/13 x (1 + 1/13 + (1/13)^2 + ...)^2
= 24/13 x (13/12)^2
= 13/6
For each hand, assuming you are not dealt any Aces, the possible ways of getting to a hard 9, 10, or 11 starting with a 2 and your next card is not a 2 are:
2 3 2 2
2 3 2 3
2 3 2 4
2 3 3 2
2 3 3 3
2 3 4
2 3 5
2 3 6
2 4 3
2 4 4
2 4 5
2 5 2
2 5 3
2 5 4
2 6 2
2 6 3
2 7
2 8
2 9
Excluding the initial 2, there are 3 hands where you are dealt 1 card, 11 where you are dealt 2, and 5 where you are dealt 3
The probability of getting a doubling hand is 3 x 1/13 + 11 x 1/169 + 5 x 1/2197 = 655/2197, so the expected bet on a particular hand = 100 + 655/2197 x 100.
The expected total bet is 13/6 x 100 x 2852/2197 = 281.2623274.
Quote: charliepatrickI had assumed you could only double on 2-card totals of 9, 10 or 11 since it did reference "basic strategy" which assumes it's (fairly near) a regular Blackjack game.
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The question said "no limit on doubles"; I assumed that meant that you can double on any number of cards.
Incidentally, I’m in LV now and have been playing some blackjack. I’m always surprised at how often a split hand can turn into three or four total bets on the table with resplit/double, so I decided to do some math. Though I never really thought about it before, for most split situations it’s actually probable to catch a resplit/double. For the 22 vs 3 scenario, you have a 1 - (9/13)^2 =~ 52% chance of at least one double/resplit. From that I know the total expected bet on the hand is > $252.07
Quote: Ace2Clarification. You can only double on two-card totals. The only differences between this game and a standard game are infinite deck and unlimited resplits. I meant there is also no limit on the number of double-down bets you can make, given unlimited resplits
First, determine how many hands you can have.
For each of the initial 2s, there is:
12/13 chance of not getting another 2, so there is 1 hand
1/13 x 12/13 chance of getting one 2, so there are 2 hands
(1/13)^2 x 12/13 chance of getting two 2s, so there are 3 hands
and so on
The sum is 12/13 x (1 + 2 x 1/13 + 3 x (1/13)^2 + ...)
= 12/13 x (1 + 1/13 + (1/13)^2 + ...)^2
= 12/13 x (13/12)^2 = 13/12
Since you start with two 2s, the expected number of hands is 2 x 13/12 = 13/6.
The only way you will get a hand of 9, 10, or 11 with two cards if the first card is a 2 is if the next card dealt to the hand is a 7, 8, or 9; there is a 3/13 chance of that in an infinite deck, so the expected amount bet per hand = 3/13 x 200 + 10/13 x 100 = 1600/13
The expected total bet = 13/6 x 1600/13 = 1600/6 = $266.67
It’s been demonstrated time and time again on this thread that nothing is implied and there is zero room for ambiguity when posting a problem. IMO math guys don’t tend to be communication guys which is part of the issue. When I originally posted the Russian roulette problem, someone said a handgun shot to the head is only 90% fatal, so I had to amend the rules to assume it’s 100% fatal and the game ends when someone shoots himself. I was quite surprised I had to clarify that!Quote: WizardQuote: Ace2Use any format you like. Flip a fair coin to determine who goes first
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I think it was implied other randomization methods are not allowed.
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I posted my answer because I couldn't see any real difference between the RR problem as originally stated and your version. Are you implying that a "turn" can only consist of one pull? That could change one solution I gave which requires Alan to pull twice on his first turn. I'm guessing my rule 4 does not apply and in that case the most equitable* solution would be for Alan and Bob to make one pull with one bullet in cylinder on each turn, never respinning between turns. If the game continues to the sixth pull, that means Bob is committing suicide on that pull.
*By most equitable, I mean the players have similar chances of being alive at each round of the game. Until the sixth pull is made, there is never more than a 16.7% absolute difference of survival. Both players have a 50% chance of winning the game
Each 2 could receive more 2s. For each starting 2, 12/13 times that's it, 1/13 you start again with two 2s.
x = 12/13 + 1/13*2x
13x = 12+2x
x = 12/11. (I thought this would be 13/12, which would agree with where wizard's 64/24 comes from, but the above calculation suggests it's 12/11!)
As you started with two 2s, in total you expect 24/11 hands.
Each hand 9/13 no double, 3/13 double, 1/13 split - average number of bets per unsplit hand = 15/12.
So expected number of bets = 24/11 * 15/12 = 30/11 = $272.72.
The error I had made in earlier post was assuming there were 13 possible hands, but one of these is where you get another 2 so resplit. Therefore the expected number of doubles from hands that do not resplit is 1/4, meaning on average for a non-resplit hand you make 5/4 wagers.
Quote: charliepatrickThere is the starting position of two hands, each with a 2.
Each 2 could receive more 2s. For each starting 2, 12/13 times that's it, 1/13 you start again with two 2s.
x = 12/13 + 1/13*2x
13x = 12+2x
x = 12/11. (I thought this would be 13/12, which would agree with where wizard's 64/24 comes from, but the above calculation suggests it's 12/11!)
As you started with two 2s, in total you expect 24/11 hands.
Each hand 9/13 no double, 3/13 double, 1/13 split - average number of bets per unsplit hand = 15/12.
So expected number of bets = 24/11 * 15/12 = 30/11 = $272.72.
The error I had made in earlier post was assuming there were 13 possible hands, but one of these is where you get another 2 so resplit. Therefore the expected number of doubles from hands that do not resplit is 1/4, meaning on average for a non-resplit hand you make 5/4 wagers.
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You don't double the number of hands with each 2; you increase it by 1
x = 12/13 + 1/13 (x + 1) = 1 + 1/13 x
12/13 x = 1
x = 13/12
I missed that "it's 1/3 to get a hand that you double, not 3/13" as well.
(i) There is a 12/13 chance you won't get another 2, in that case (ignoring doubling opportunities) there's only one hand. This caters for the 12/13.
(ii) There is a 1/13 chance you will get another 2. In that case you split and hence land up with two hands that, until you take another card, are each a "single 2". This means there are 2x (not x+1).
Hence x (your original single 2) can turn out to be 12/13 (one hand) or 1/13 times (two single 2s) 2x.
This gives x = 12/13 + 1/13*2x.
Multiplying by 13 gives 13x = 12 + 2x,
Subtracting 2x gives 11x=12.
Hence x =12/11.
Since one starts with a pair of twos, prior to considering doubling, this gives 24/11.
Quote: charliepatrickAssume x is how many hands you expect to make from a "single 2" before the dealer gives you a second card.
(i) There is a 12/13 chance you won't get another 2, in that case (ignoring doubling opportunities) there's only one hand. This caters for the 12/13.
(ii) There is a 1/13 chance you will get another 2. In that case you split and hence land up with two hands that, until you take another card, are each a "single 2". This means there are 2x (not x+1).
Hence x (your original single 2) can turn out to be 12/13 (one hand) or 1/13 times (two single 2s) 2x.
This gives x = 12/13 + 1/13*2x.
Multiplying by 13 gives 13x = 12 + 2x,
Subtracting 2x gives 11x=12.
Hence x =12/11.
Since one starts with a pair of twos, prior to considering doubling, this gives 24/11.
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Quote: Ace2It’s been demonstrated time and time again on this thread that nothing is implied and there is zero room for ambiguity when posting a problem.
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You may subscribe to that, but I don't. I believe in sticking to the rules and items specified in the puzzle. If I ask a river crossing puzzle, I feel I don't need to state that you can't call a helicopter on Uber to take everyone across in one ride. So, I make no apologies for not stating you can't introduce a coin. However, I should have said the gun passes back and forth after every pull, not every turn.
Also, it wasn't me who mentioned the 90% success rate in suicides by handgun. I assumed a 100% accuracy was implied.
I believe you’re correct (🍺) , though I’d like to simulate or Markov it to be 100% sure. I don’t have a computer at the momentQuote: WizardQuote: Ace2Playing blackjack, you are dealt a pair of 2s against a dealer 3. You play basic strategy which means you split 2s and double down with 9-11 against a dealer 3. This is an infinite deck game with no limit on resplitting or doubles
Assuming you're a $100 flat bettor, what's your total expected wager on this hand?
link to original postI get $100*(30/11) = $272.73. Math available upon request.
Note: edited 4/8/25 6:41 AM
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Charliepatrick also gave this answer after Wizard
A blind man is placed at a table with 30 coins. he is told that 10 are face up heads and 20 are face up tails. Gloves are put on his hands, so he can't feel which side is face on any coin. He is tasked with separating the coins into two piles such that the number of tails will be the same in each pile. The piles may be of any size and may be unequal to each other. How can he do it?
Quote: WizardHere is an oldie:
A blind man is placed at a table with 30 coins. he is told that 10 are face up heads and 20 are face up tails. Gloves are put on his hands, so he can't feel which side is face on any coin. He is tasked with separating the coins into two piles such that the number of tails will be the same in each pile. The piles may be of any size and may be unequal to each other. How can he do it?
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Quote: WizardHere is an oldie:
A blind man is placed at a table with 30 coins. he is told that 10 are face up heads and 20 are face up tails. Gloves are put on his hands, so he can't feel which side is face on any coin. He is tasked with separating the coins into two piles such that the number of tails will be the same in each pile. The piles may be of any size and may be unequal to each other. How can he do it?
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It's an oldie, all right - about 8 months old, from the looks of it (see #5)
Quote: unJon
I’ve always liked this one. Separate the coins into a pile of 10 and pile of 20. Then flip over every coin in the pile of 20.
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I agree!
What's the probability you re-split to exactly ten total hands (not more)?
Closed-form solutions only.
Quote: Ace2Anybody working on this problem ^^ or shall I post the answer?
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I started on it, but then I realized I was making the same mistake I did with the first one. I don't mind if you post the answer.
Quote: Ace2Playing infinite-deck blackjack, you are dealt a pair of 8s against dealer 6. You always split 8s and the casino allows unlimited re-splitting.
What's the probability you re-split to exactly ten total hands (not more)?
Closed-form solutions only.
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I have been working on this for at least three hours and have only many sheets of paper to show for my work, but no formulas of even answers without a closed form. It would be fairly easy to do with with a recursive program, but I know that wouldn't satisfy your closed-form condition.
Calculating the total number of 8s dealt can be framed as a risk-of-ruin problem. Think of an 8 as a win and a non-8 as a loss. Starting with a "bankroll" of two 8s, you have a 1/13th chance of receiving another 8 on the next card dealt as long as total 8s outnumber total non-8s. With this you can easily Markov the answer as: what's the chance of busting out the bankroll on the eighteenth additional card dealt? Considering two-card hands only, twenty total cards dealt is ten total hands. A Markov chain is fine for confirming the answer but does not count as closed-form solution
Answer being corrected (8:01 PM PST)
Disagree. That’s off by more than a factor of 1,000 from my answer. Maybe I’m wrong or maybe I’m correctly answering another problem, but my formulaic answer agrees to my Markov chain answerQuote: WizardThank you. That absolutely helped.
(12^2 - 1)/(12^10-1) - (12^2 - 1)/(12^11-1)=~ 1 in 472,351,157
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You start with two 8s, but land up with ten 8s and ten others
Chances are 1/13 ^ 8 * 12/13 ^ 10 * number of ways to get there (ensuring there are always more 8s than not).
So 88…88NN...NN works but NNN…. doesn't
43758 Total combinations of putting 8 items in 18 boxes
No of 8s | …. | …. | …. | …. | …. | …. | …. | …. | …. | |
No of Ns | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | |
1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |
2 | 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 | ||
3 | 5 | 14 | 28 | 48 | 75 | 110 | 154 | |||
4 | 14 | 42 | 90 | 165 | 275 | 429 | ||||
5 | 42 | 132 | 297 | 572 | 1001 | |||||
6 | 132 | 429 | 1001 | 2002 | ||||||
7 | 429 | 1430 | 3432 | |||||||
8 | 1430 | 4862 | ||||||||
9 | 4862 | |||||||||
10 | 4862 |
This happens to be 1/9th of COMBIIN (18 8)
It is not a coincidence that 1430 is COMBIN ( 16 7 ) / 8
etc.
So answer = COMBIN ( 18 8 ) / 9 * POWER (1/13,8) * POWER (12/13,10)
0.000 002 676 992
Quote: Ace2Disagree. That’s off by more than a factor of 1,000 from my answer. Maybe I’m wrong or maybe I’m correctly answering another problem, but my formulaic answer agrees to my Markov chain answer
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I retracted my answer shortly after posting it. You must have seen it the short time it was up.
Exact estimate: 1 in 937,083,985,208,030.
Modified estimate: 1 in 5,921,355,908,379
Hands | trees | Prob | Inverse |
---|---|---|---|
1 | 1 | 0.9230769230769 | 1.08 |
2 | 1 | 0.0655439235321 | 15.26 |
3 | 2 | 0.0093080128093 | 107.43 |
4 | 5 | 0.0016523099661 | 605.21 |
5 | 14 | 0.0003285065968 | 3,044 |
6 | 42 | 0.0000699777366 | 14,290 |
7 | 132 | 0.0000156163334 | 64,036 |
8 | 429 | 0.0000036037693 | 277,487 |
9 | 1430 | 0.0000008529631 | 1,172,384 |
10 | 4862 | 0.0000002059225 | 4,856,197 |
11 | 16796 | 0.0000000505114 | 19,797,522 |
12 | 58786 | 0.0000000125531 | 79,661,456 |
13 | 208012 | 0.0000000031540 | 317,058,368 |
14 | 742900 | 0.0000000007998 | 1,250,266,830 |
15 | 2674440 | 0.0000000002045 | 4,891,090,144 |
16 | 9694845 | 0.0000000000526 | 19,002,166,307 |
Total | 0.9999999999816 |
Here is the formula for n hands:
probability n hands = combin(2n-2,n-1)/n * (1/13)^(n-1) * (12/13)^n
I admit I had to do some outside research on full binary trees and Catalan numbers. I'll take whatever credit I can get, assuming I'm even right. I spent about four hours on this, so hopefully at least get an A for effort.
I’ve used Poisson to solve many problems, but I don’t believe it could be applied here since the order of events matters (8s must always outnumber non-8s)Quote: Wizard
Exact estimate: 1 in 937,083,985,208,030.
Modified estimate: 1 in 5,921,355,908,379
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I agree with your answer and you were starting to see certain patterns. You were essentially building a Pascal’s Triangle with those combinations and I believe that’s fundamental to the solution. I’ll post my full answer in a minuteQuote: charliepatrickProbem is to split 8s ensuring there land up only being 10 hands
You start with two 8s, but land up with ten 8s and ten others
Chances are 1/13 ^ 8 * 12/13 ^ 10 * number of ways to get there (ensuring there are always more 8s than not).
So 88…88NN...NN works but NNN…. doesn't
43758 Total combinations of putting 8 items in 18 boxes
No of 8s …. …. …. …. …. …. …. …. …. No of Ns 2 3 4 5 6 7 8 9 100 1 1 1 1 1 1 1 1 11 1 2 3 4 5 6 7 8 92 2 5 9 14 20 27 35 443 5 14 28 48 75 110 1544 14 42 90 165 275 4295 42 132 297 572 10016 132 429 1001 20027 429 1430 34328 1430 48629 486210 4862
This happens to be 1/9th of COMBIIN (18 8)
It is not a coincidence that 1430 is COMBIN ( 16 7 ) / 8
etc.
So answer = COMBIN ( 18 8 ) / 9 * POWER (1/13,8) * POWER (12/13,10)
0.000 002 676 992
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For a fair, even-money game starting with a bankroll of one unit, the probability of surviving n bets and winning at least k of them is:
combin(n,n-k)/ 2^n
So, for instance, the probability of surviving ten bets and winning >= six bets is:
combin(10,4) / 2^10 = 210 / 1,024
Note this is equal to the probability of winning exactly six bets without considering bankroll
This pattern also works for initial bankrolls of two units...you just increase the numerator of the combination formula by one.
So, for instance, the probability of surviving ten bets and winning >= six bets (starting with bankroll of two units) is:
combin(11,4) / 2^10 = 330 / 1,024
Back to the problem at hand. Think of free 8s (8s that haven’t been dealt a card yet) as your bankroll. Additional 8s increase your bankroll but once an 8 gets dealt a non-8 it decreases your bankroll.
Starting with a pair of 8s, the only way to get exactly ten total hands is: 1) exactly eight (not >= eight) of the next sixteen cards must be 8s; 2) total free 8s must be >0 until the final card is dealt 3) the final two cards must be non-8s. Therefore:
12^8 / 13^16 *
(combin(17,8) - combin(17,7)) *
(12/13)^2
=~ 2.68 x 10^-6
It is very easy to get this answer with a Markov chain. Starting with a "bankroll" of two free 8s, you have a 1/13 chance of going up one free 8 and a 12/13 of going down one free 8. Busting the bankroll ends a trial. The chance of busting on the eighteenth additional card confirms the answer. Additionally, the expected number of total hands of the Markov chain is 24/11 which agrees to last week's blackjack split puzzle
Thanks to all participants.
It appears you’ve calculated the probability starting with one 8. For this problem you must start with two 8sQuote: Wizard
Hands trees Prob Inverse 1 1 0.9230769230769 1.08 2 1 0.0655439235321 15.26 3 2 0.0093080128093 107.43 4 5 0.0016523099661 605.21 5 14 0.0003285065968 3,044 6 42 0.0000699777366 14,290 7 132 0.0000156163334 64,036 8 429 0.0000036037693 277,487 9 1430 0.0000008529631 1,172,384 10 4862 0.0000002059225 4,856,197 11 16796 0.0000000505114 19,797,522 12 58786 0.0000000125531 79,661,456 13 208012 0.0000000031540 317,058,368 14 742900 0.0000000007998 1,250,266,830 15 2674440 0.0000000002045 4,891,090,144 16 9694845 0.0000000000526 19,002,166,307 Total 0.9999999999816
Here is the formula for n hands:
probability n hands = combin(2n-2,n-1)/n * (1/13)^(n-1) * (12/13)^n
I admit I had to do some outside research on full binary trees and Catalan numbers. I'll take whatever credit I can get, assuming I'm even right. I spent about four hours on this, so hopefully at least get an A for effort.
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Quote: Ace2Allow me to digress for a moment. A few days ago I "discovered" the following ways to calculate risk-of-ruin for certain scenarios.
For a fair, even-money game starting with a bankroll of one unit, the probability of surviving n bets and winning at least k of them is:
combin(n,n-k)/ 2^n
So, for instance, the probability of surviving ten bets and winning >= six bets is:
combin(10,4) / 2^10 = 210 / 1,024
Note this is equal to the probability of winning exactly six bets without considering bankroll
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I’m learning about the calculation of risk-of-ruin (RoR). In many situations, there is also a tie probability. For example, the win/tie/loss probabilities are 0.4/0.1/0.5, respectively. How to calculate the RoR in this even money example?
I know of only a handful of RoR formulas and they only apply to even-money payouts and unrealistic scenarios such as: no edge, player has edge, initial bankroll of 1-2 units or unlimited bankrollQuote: acesideQuote: Ace2Allow me to digress for a moment. A few days ago I "discovered" the following ways to calculate risk-of-ruin for certain scenarios.
For a fair, even-money game starting with a bankroll of one unit, the probability of surviving n bets and winning at least k of them is:
combin(n,n-k)/ 2^n
So, for instance, the probability of surviving ten bets and winning >= six bets is:
combin(10,4) / 2^10 = 210 / 1,024
Note this is equal to the probability of winning exactly six bets without considering bankroll
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I’m learning about the calculation of risk-of-ruin (RoR). In many situations, there is also a tie probability. For example, the win/tie/loss probabilities are 0.4/0.1/0.5, respectively. How to calculate the RoR in this even money example?
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The only practical RoR estimate I know of is my conjecture: RoR for a given session/bankroll is double the chance of finishing the session with a loss >= bankroll amount. It's highly accurate for low-edge games, sessions with >50 bets and bankrolls >= 1 deviation
Btw the house edge of the your .4/.1./.5 even money game is 10%, so without doing any calculations I can say your RoR will be very high unless you have an enormous bankroll and/or play for a very short time