Easy math puzzles

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April 8th, 2025 at 7:23:42 AM permalink

Quote: Ace2
Playing 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 post

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.

charliepatrick

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April 8th, 2025 at 7:39:41 AM permalink

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.

ThatDonGuy

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April 8th, 2025 at 8:11:48 AM permalink

Quote: charliepatrick
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.
link to original post

The question said "no limit on doubles"; I assumed that meant that you can double on any number of cards.

aceside

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April 8th, 2025 at 8:26:17 AM permalink

Deleted.

Last edited by: aceside on Apr 8, 2025

Ace2

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April 8th, 2025 at 10:36:38 AM permalink

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

It�s all about making that GTA

aceside

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April 8th, 2025 at 11:07:36 AM permalink

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!

ThatDonGuy

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April 8th, 2025 at 11:34:44 AM permalink

Quote: Ace2
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

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

Ace2

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April 8th, 2025 at 12:04:27 PM permalink

Quote: Wizard
Quote: Ace2
Use any format you like. Flip a fair coin to determine who goes first
link to original post

I think it was implied other randomization methods are not allowed.
link to original post

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!

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

charliepatrick

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April 8th, 2025 at 1:59:57 PM permalink

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.

ThatDonGuy

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April 8th, 2025 at 2:25:42 PM permalink

Quote: charliepatrick
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.

link to original post

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.

charliepatrick

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April 8th, 2025 at 2:47:53 PM permalink

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.

ThatDonGuy

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April 8th, 2025 at 3:50:00 PM permalink

Quote: charliepatrick
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.

link to original post

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.

Wizard
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April 8th, 2025 at 3:55:39 PM permalink

Quote: Ace2
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.
link to original post

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)

Ace2

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April 8th, 2025 at 6:21:29 PM permalink

Quote: Wizard
Quote: Ace2
Playing 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 post

I get $100*(30/11) = $272.73. Math available upon request.

Note: edited 4/8/25 6:41 AM

link to original post

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 moment

Charliepatrick also gave this answer after Wizard

It�s all about making that GTA

Wizard
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April 10th, 2025 at 6:59:58 AM permalink

Here 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?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

unJon

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April 10th, 2025 at 9:51:29 AM permalink

Quote: Wizard
Here 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?
link to original post

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.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

ThatDonGuy

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April 10th, 2025 at 12:06:27 PM permalink

Quote: Wizard
Here 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?
link to original post

It's an oldie, all right - about 8 months old, from the looks of it (see #5)

Wizard
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April 10th, 2025 at 5:01:56 PM permalink

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.

link to original post

I agree!

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Ace2

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April 13th, 2025 at 6:27:48 PM permalink

Playing 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.

It�s all about making that GTA

Ace2

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April 14th, 2025 at 5:19:17 PM permalink

Anybody working on this problem ^^ or shall I post the answer?

It�s all about making that GTA

ThatDonGuy

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April 14th, 2025 at 5:43:17 PM permalink

Quote: Ace2
Anybody 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.

Wizard
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April 14th, 2025 at 5:55:26 PM permalink

Quote: Ace2
Playing 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.
link to original post

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.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Ace2

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April 14th, 2025 at 6:51:31 PM permalink

Hints:

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

Last edited by: Ace2 on Apr 14, 2025

It�s all about making that GTA

Wizard
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April 14th, 2025 at 7:48:45 PM permalink

Thank you. That absolutely helped.

Answer being corrected (8:01 PM PST)

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Ace2

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April 14th, 2025 at 8:05:02 PM permalink

Quote: Wizard
Thank you. That absolutely helped.

(12^2 - 1)/(12^10-1) - (12^2 - 1)/(12^11-1)=~ 1 in 472,351,157

link to original post

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 answer

Last edited by: Ace2 on Apr 14, 2025

It�s all about making that GTA

charliepatrick

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April 15th, 2025 at 1:14:34 AM permalink

Probem 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	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

Wizard
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April 15th, 2025 at 6:33:41 AM permalink

Quote: Ace2
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 answer
link to original post

I retracted my answer shortly after posting it. You must have seen it the short time it was up.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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April 15th, 2025 at 7:22:43 AM permalink

Exact estimate: 1 in 937,083,985,208,030.
Modified estimate: 1 in 5,921,355,908,379

Last edited by: Wizard on Apr 15, 2025

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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April 15th, 2025 at 3:48:38 PM permalink

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.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Ace2

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April 15th, 2025 at 5:20:14 PM permalink

Quote: Wizard

Exact estimate: 1 in 937,083,985,208,030.
Modified estimate: 1 in 5,921,355,908,379

link to original post

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)

It�s all about making that GTA

Ace2

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April 15th, 2025 at 5:27:16 PM permalink

Quote: charliepatrick
Probem 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
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

link to original post

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

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 minute

It�s all about making that GTA

Ace2

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April 15th, 2025 at 5:34:37 PM permalink

Allow 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

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.

Last edited by: Ace2 on Apr 15, 2025

It�s all about making that GTA

Ace2

Threads: 32
Posts: 2730

Joined: Oct 2, 2017

April 15th, 2025 at 6:31:43 PM permalink

Quote: 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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It appears you�ve calculated the probability starting with one 8. For this problem you must start with two 8s

It�s all about making that GTA

aceside

aceside

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Joined: May 14, 2021

April 16th, 2025 at 6:53:50 AM permalink

Quote: Ace2
Allow 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?

Ace2