Easy math puzzles
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46 members have voted
April 8th, 2025 at 3:37:01 AM
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There 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 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.
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.
Last edited by: charliepatrick on Apr 8, 2025
April 8th, 2025 at 7:23:42 AM
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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.
April 8th, 2025 at 7:39:41 AM
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I 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.
April 8th, 2025 at 8:11:48 AM
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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.
April 8th, 2025 at 8:26:17 AM
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Deleted.
Last edited by: aceside on Apr 8, 2025
April 8th, 2025 at 10:36:38 AM
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Clarification. 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
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
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
It’s all about making that GTA
April 8th, 2025 at 11:07:36 AM
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The pair splitting problem was derived by Chesterdog here three years ago. He came up with a formula for the EV of a non-aces pair splitting up to three times, but he hasn’t given a formula for this with an infinite times of splitting. I’d like to see a formula of this. Interesting!
April 8th, 2025 at 11:34:44 AM
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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
April 8th, 2025 at 12:04:27 PM
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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
Last edited by: Ace2 on Apr 8, 2025
It’s all about making that GTA
April 8th, 2025 at 1:59:57 PM
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There 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.
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.
April 8th, 2025 at 2:25:42 PM
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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.
April 8th, 2025 at 2:47:53 PM
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Assume 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.
(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.
April 8th, 2025 at 3:50:00 PM
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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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You're right - if you are dealt another 2, then you are dealt an additional card to each of those 2s, not just one of them.
April 8th, 2025 at 3:55:39 PM
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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.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 8th, 2025 at 6:21:29 PM
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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
It’s all about making that GTA
