How to calculate the probabilities here? (3-person card game)

In summary: I have.In summary, the probabilities of winning in a 3-person card game are 9.38% for Mr. Black and 90.62% for Mr. Mismatch. These results were obtained through a simulation of 1,000,000 games, and were confirmed using mathematical calculations. It is not possible for Mr. Red to win in this game, as there are fewer red cards and more ways to get a mismatched pair. The probability of Mr. Mismatch winning is approximately 90.65%, as calculated using a formula that takes into account the number of red and black pairs in a game. It is impossible for a tie between Mr. Black and Mr. Mismatch to occur as long as
  • #1
murshid_islam
457
19
TL;DR Summary
Probabilities of winning in a 3-person card game
Suppose I have a shuffled pack of cards, out of which 4 red cards have been lost or removed. So now I have 22 red cards and 26 black cards in random order. I'm dealing cards in pairs. If a pair has 2 red cards, they are dealt to Mr. Red, if the pair has 2 black cards, they're dealt to Mr. Black, and if the pair contains 1 red and 1 black card, they are dealt to Mr. Mismatch.

After all the cards have been dealt in pairs, whoever has the most cards wins. Now, Mr. Mismatch will almost always win because there are more ways to get a mismatched pair (card-1 is red and card-2 is black or vice versa) than a pair of reds or blacks. Also, since there are 4 fewer red cards, Mr. Red will never win.

I want to know the probabilities of Mr. Black and Mr. Mismatch winning. I already ran a simulation (in Python) 1,000,000 (one million) times to see what percentage of games are won by the three players. Here's my code in case anyone wants to check it out

This is the result I got:

Mr. Red won = 0 times = 0.00%
Mr. Black won = 93799 times = 9.38%
Mr. Mismatch won = 906201 times = 90.62%

No matter how many times I run my code, I get similar results: Mr. Black winning little more than 9% and Mr. Mismatch winning a little more than 90%. However, Someone claimed that Mr. Mismatch should win about 95.6% of the time. I was wondering if it's my code that is wrong. So, I wanted to calculate the probabilities using math to check if my results are correct or not. Any ideas on how I can calculate the probabilities of their winning?
 
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  • #2
Equivalent formulation. An urn contains 22 red balls and 26 black balls. The balls are collected at random two at a time. The score starts at zero.
In the event of
  • two black balls, add -1 to score.
  • two red balls, add +0 to score.
  • mixed colours, add +1 to score.
Determine the probability that the score is positive after the urn is emptied. Note that if there occurs +1 odd number ##n## of times, then +1 will have occurred at least ##n+1## times.
 
Last edited:
  • #3
murshid_islam said:
Summary:: Probabilities of winning in a 3-person card game

Suppose I have a shuffled pack of cards, out of which 4 red cards have been lost or removed. So now I have 22 red cards and 26 black cards in random order. I'm dealing cards in pairs. If a pair has 2 red cards, they are dealt to Mr. Red, if the pair has 2 black cards, they're dealt to Mr. Black, and if the pair contains 1 red and 1 black card, they are dealt to Mr. Mismatch.

After all the cards have been dealt in pairs, whoever has the most cards wins. Now, Mr. Mismatch will almost always win because there are more ways to get a mismatched pair (card-1 is red and card-2 is black or vice versa) than a pair of reds or blacks. Also, since there are 4 fewer red cards, Mr. Red will never win.

I want to know the probabilities of Mr. Black and Mr. Mismatch winning. I already ran a simulation (in Python) 1,000,000 (one million) times to see what percentage of games are won by the three players. Here's my code in case anyone wants to check it out

This is the result I got:

Mr. Red won = 0 times = 0.00%
Mr. Black won = 93799 times = 9.38%
Mr. Mismatch won = 906201 times = 90.62%

No matter how many times I run my code, I get similar results: Mr. Black winning little more than 9% and Mr. Mismatch winning a little more than 90%. However, Someone claimed that Mr. Mismatch should win about 95.6% of the time. I was wondering if it's my code that is wrong. So, I wanted to calculate the probabilities using math to check if my results are correct or not. Any ideas on how I can calculate the probabilities of their winning?
It might be diffcult to calculate the probabilities in this case. I'll have a think about it, but a Python simulation would be the way to go.

Your code looks okay at first glance.
 
  • #4
Here's an idea. You only need to count the red pairs. If there are more than 6 red pairs, then black wins; otherwise, mismatch wins.

You would need to calculate the probability that there are 7-11 red pairs in your 48 random cards.

That might be not too hard, although I haven't thought about it yet.
 
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  • #5
I made a table for the cases and confirmed that RR has no chance to win.

1651739509234.png


BB=RR+2
BR=24-BB-RR=22-2*RR
 
Last edited:
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  • #6
PeroK said:
Here's an idea. You only need to count the red pairs. If there are more than 6 red pairs, then black wins; otherwise, mismatch wins.

You would need to calculate the probability that there are 7-11 red pairs in your 48 random cards.

That might be not too hard, although I haven't thought about it yet.
This is not too hard. Suppose there are precisely ##k## red pairs. That means there are precisely ##22-2k## mismatch pairs and ##2 + k## black pairs. The number of ways of achieving this is:
$$n(r = k) = \binom{24}{k}\binom{24-k}{2+k}2^{22-2k}$$Note that there are two ways of getting every mismatch pair.

The total number of random arrangements is ##N = \binom{48}{26}##. And the probability that mismatch wins is:
$$p(m) = \frac{1}{N}\sum_{k = 0}^{6}n(r = k)$$Checking this on a spreadsheet confirms that$$p(m) \approx 0.9065$$
 
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  • #7
N
2.73857E+13​
k(24, k)(24-k, 2+k)2^(22-2k)n(k)p(k)
0​
1​
276​
4194304​
1157627904​
4.22713E-05​
1​
24​
1771​
1048576​
4.4569E+10​
0.001627446​
2​
276​
7315​
262144​
5.2925E+11​
0.019325919​
3​
2024​
20349​
65536​
2.6992E+12​
0.098562189​
4​
10626​
38760​
16384​
6.748E+12​
0.246405473​
5​
42504​
50388​
4096​
8.7724E+12​
0.320327115​
6​
134596​
43758​
1024​
6.031E+12​
0.220224892​
0.906515306​
7​
346104​
24310​
256​
2.1539E+12​
0.078651747​
8​
735471​
8008​
64​
3.7694E+11​
0.013764056​
9​
1307504​
1365​
16​
2.8556E+10​
0.001042731​
10​
1961256​
91​
4​
713897184​
2.60683E-05​
11​
2496144​
1​
1​
2496144​
9.11479E-08​
0.093484694​
 
  • #8
anuttarasammyak said:
I made a table for the cases and confirmed that RR has no chance to win.

View attachment 301050

BB=RR+2
BR=24-BB-RR=22-2*RR
If there are more black cards, then you can never have more red pairs than black pairs. Or, even as many red pairs as black pairs.
 
  • #9
Note that as long at the number of black cards is not a multiple of ##3## there can never be a tie between black and mismatch in a game.

And, as long as the number of black cards is less than three times the number of red cards, then mismatch has a chance.
 

Related to How to calculate the probabilities here? (3-person card game)

1. What is the basic concept behind calculating probabilities in a 3-person card game?

The basic concept behind calculating probabilities in a 3-person card game is to determine the likelihood of a certain outcome occurring based on the number of possible outcomes and the cards that have already been played.

2. How do I calculate the probability of getting a specific card in a 3-person card game?

To calculate the probability of getting a specific card in a 3-person card game, you need to know the total number of cards in the deck, the number of cards already dealt, and the number of cards of the desired value remaining in the deck. The probability is then calculated by dividing the number of desired cards by the total number of remaining cards.

3. What is the difference between theoretical and experimental probabilities in a 3-person card game?

Theoretical probability is based on mathematical calculations and assumes that all outcomes are equally likely. Experimental probability is based on actual results from playing the game and may differ from theoretical probability due to chance or other factors.

4. How do I calculate the probability of winning in a 3-person card game?

The probability of winning in a 3-person card game depends on the specific rules and objectives of the game. In general, it can be calculated by determining the number of possible winning outcomes and dividing it by the total number of possible outcomes.

5. Can I use previous outcomes to predict future probabilities in a 3-person card game?

While previous outcomes can give an indication of the likelihood of certain cards being dealt, they do not guarantee future outcomes. Each round of the game is independent and the probabilities remain the same regardless of previous outcomes.

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